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# Differentiation and
## Integration of vectors 15.7
Introduction
The area known as vector calculus is used to model mathematically a vast range of engineering
phenomena including electrostatics, electromagnetic elds, air ow around aircraft and heat ow
in nuclear reactors. In this Block we introduce briey the dierential and integral calculus of
vectors.
have a knowledge of vectors, in cartesian
form
Prerequisites be able to calculate the scalar and vector
products of two vectors
Before starting this Block you should . . .
be able to dierentiate and integrate scalar
functions.
## Learning Outcomes Learning Style
After completing this Block you should be able To achieve what is expected of you . . .
to . . .
dierentiate and integrate vectors
allocate sucient study time
## attempt every guided exercise and most
of the other exercises
1. Differentiation of Vectors
Consider the following gure.
y P
r
C
If r represents the position vector of an object which is moving along a curve C, then the position
vector will be dependent upon the time, t. We write r = r(t) to show the dependence upon
time. Suppose that the object is at the point P with position vector r at time t and at the point
Q with position vector r(t + t) at the later time t + t as shown in the next gure.
P
PQ
r(t)
Q
r(t + t)
x
Then P Q represents the displacement vector of the object during the interval of time t. The
length of the displacement vector represents the distance travelled while its direction gives the
direction of motion. The average velocity during the time from t to t + t is dened as the
displacement vector divided by the time interval t, that is,
PQ r(t + t) r(t)
average velocity = =
t t
If we now take the limit as the interval of time t tends to zero then the expression on the right
hand side is the derivative of r with respect to t. Not surprisingly we refer to this derivative as
the instantaneous velocity, v. By its very construction we see that the velocity vector is always
tangential to the curve as the object moves along it. We have:
r(t + t) r(t) dr
v = lim =
t0 t dt
Now, since the x and y coordinates of the object depend upon the time, we can write the position
vector r in cartesian coordinates as:
## Engineering Mathematics: Open Learning Unit Level 1 2
15.7: Applications of Integration
Therefore,
r(t + t) = x(t + t)i + y(t + t)j
so that,
## x(t + t)i + y(t + t)j x(t)i y(t)j
v(t) = lim
t0
t
x(t + t) x(t) y(t + t) y(t)
= lim i+ j
t0 t t
dx dy
= i+ j
dt dt
This is often abbreviated to v = r = xi
+ yj
using notation for derivatives with respect to time.
So the velocity vector is the derivative of the position vector with respect to time.
This result generalizes in an obvious way to three dimensions. If,
## then the velocity vector is
v = r (t) = x(t)i
+ y(t)j
+ z(t)k
The magnitude of the velocity vector gives the speed of the object.
We can dene the acceleration vector in a similar way, as the rate of change (i.e. the derivative)
of the velocity with respect to the time:
dv d2 r
a= = 2 = r = xi + yj + zk
dt dt
## Example If w = 3t2 i + cos 2tj, nd
dw dw d2 w
(a) (b) (c)
dt dt dt2
Solution
dw
1. (a) If w = 3t2 i + cos 2tj, then dierentiation with respect to t yields: = 6ti 2 sin 2tj
dt
dw
(b) = (6t)2 + (2 sin 2t)2 = 36t2 + 4 sin2 2t
dt
d2 w
(c) = 6i 4 cos 2tj
dt2
## It is possible to dierentiate more complicated expressions involving vectors provided certain
rules are adhered to. If w and z are vectors and c is a scalar, all these being functions of time
t, then:
## 3 Engineering Mathematics: Open Learning Unit Level 1
15.7: Applications of Integration
Key Point
d dw dz d dw dc
(w + z) = + (cw) = c + w
dt dt dt dt dt dt
d dz dw d dz dw
(w z) = w + z (w z) = w + z
dt dt dt dt dt dt
## Example If w = 3ti t2 j and z = 2t2 i + 3j, verify the results
d dz dw d dz dw
(a) (w z) = w + z (b) (w z) = w + z
dt dt dt dt dt dt
Solution
(a) w z = (3ti t2 j) (2t2 i + 3j) = 6t3 3t2 . Then
d
(w z) = 18t2 6t
dt
Also
dw dz
= 3i 2tj = 4ti
dt dt
so
dz dw
w +z = (3ti t2 j) (4ti) + (2t2 i + 3j) (3i 2tj)
dt dt
= 12t2 + 6t2 6t = 18t2 6t
d dz dw
We have veried (w z) = w + z
dt dt
dt
i j k
d
(b) w z = 3t t2 0 = (9t + 2t4 )k
implying (w z) = (9 + 8t3 )k
2t2 3 0 dt
Also,
i j k
dz
w = 3t t 0
2
dt 4t 0 0
= 4t3 k
## Engineering Mathematics: Open Learning Unit Level 1 4
15.7: Applications of Integration
Solution (continued)
i j k
dw
z = 3 2t 0
dt
2t2 3 0
= (9 + 4t3 )k
and so,
dz dw d
w + z = 4t3 k + (9 + 4t3 )k = (9 + 8t3 )k = (w z)
dt dt dt
as required.
## More exercises for you to try
1. If r = 3ti + 2t2 j + t3 k, nd
dr d2 r
(a) dt
(b) dt2
dB d2 B
(a) dt
(b) dt2
dt
when t = 1.
## (a) nd wz, (b) nd dw
dt
, (c) nd dz
dt
, (d) show that d
dt
(wz) = w dz
dt
+ dw
dt
z
## (a) r , (b) r, (c) |r|
Show also that the position vector and velocity vector are perpendicular.
2. Integration of Vectors
If a vector depends upon time t, it is often necessary to integrate it with respect to time. Recall
that i, j and k are constant vectors and must be treated thus in any integration. Hence the
integral,
I = (f (t)i + g(t)j + h(t)k) dt
## is simply evaluated as three scalar integrals i.e.
I= f (t)dt i + g(t)dt j + h(t)dt k
## 5 Engineering Mathematics: Open Learning Unit Level 1
15.7: Applications of Integration
1
Example If r = 3ti + t j + (1 + 2t)k, evaluate
2
rdt.
0
Solution
1 1 1 1
2
rdt = 3t dt i + t dt j + 1 + 2t dt k
0 0 0 0
3
1
3t2 1
t 1 3 1
= i+ j + t + t2 0 k = i + j + 2k
2 0 3 0 2 3
## More exercises for you to try
1. Given r = 3 sin t i cos t j + (2 t)k, evaluate r dt.
0
2. Given v = i 3j + k, evaluate:
1 2
(a) v dt, (b) v dt
0 0
## 3. The vector, a, is dened by a = t2 i + et j + tk. Evaluate
1 3 4
(a) a dt, (b) a dt, (c) a dt
0 2 1
## 4. Let a and b be two three-dimensional vectors. Is the following result true?
t2 t2 t2
a dt b dt = a b dt
t1 t1 t1
## Engineering Mathematics: Open Learning Unit Level 1 6
15.7: Applications of Integration
End of Block 15.7
## 7 Engineering Mathematics: Open Learning Unit Level 1
15.7: Applications of Integration
1. (a) 3i + 4tj + 3t2 k (b) 4j + 6tk
## 2. (a) (tet + et )i sin tj (b) et (t 2)i cos tj
3. 4i + 2j 7k, 8i + 2j
## Engineering Mathematics: Open Learning Unit Level 1 8
15.7: Applications of Integration
1. 6i + 1.348k
2. (a) i 3j + k (b) 2i 6j + 2k
3. (a) 0.333i + 0.632j + 0.5k (b) 6.333i + 0.0855j + 2.5k (c) 21i + 0.3496j + 7.5k
4. no.
## 9 Engineering Mathematics: Open Learning Unit Level 1
15.7: Applications of Integration | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://de.scribd.com/document/339080967/integral-differensial-vektor-pdf", "fetch_time": 1590763984000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-24/segments/1590347404857.23/warc/CC-MAIN-20200529121120-20200529151120-00594.warc.gz", "warc_record_offset": 315261134, "warc_record_length": 86305, "token_count": 2482, "char_count": 6689, "score": 4.65625, "int_score": 5, "crawl": "CC-MAIN-2020-24", "snapshot_type": "latest", "language": "en", "language_score": 0.9038054347038269, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00025-of-00064.parquet"} |
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# Asset allocators create portfolios, often in the form of
Author Message
SVP
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Asset allocators create portfolios, often in the form of [#permalink]
### Show Tags
22 Nov 2005, 22:06
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Asset allocators create portfolios, often in the form of mutual funds,
with the intention to turn in good results in both “bullâ€
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5045
Location: Singapore
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### Show Tags
22 Nov 2005, 22:50
(A) 'intention of' should be the correct idiom
(B) the intention of which is
(C) intended
(D) inappropriate use of 'and'
(E) awkward
Bstween B and C, I take B. It's concise, and to the point.
Current Student
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### Show Tags
23 Nov 2005, 01:04
ywilfred wrote:
(A) 'intention of' should be the correct idiom
(B) the intention of which is
(C) intended
(D) inappropriate use of 'and'
(E) awkward
Bstween B and C, I take B. It's concise, and to the point.
Did you mean to say C? This ubiquitous "intended" seems to be an ETS favorite, so let`s keep an eye out for it.
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### Show Tags
23 Nov 2005, 02:06
I vote for (A).
To be successful is not the intention of mutual funds, but of asset allocators themselves. Thus, (B) is not appropriate.
Moreover, what is "which" referring to? mutual funds or portpolio? It's vague.
_________________
Auge um Auge, Zahn um Zahn !
Director
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### Show Tags
23 Nov 2005, 05:06
C is correct.
Where are you guyz getting these Qs from???
Man they are terse!
Senior Manager
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23 Nov 2005, 05:14
I do not find any fault with the original statement, and would choose A.
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23 Nov 2005, 10:20
[quote="vivek123"]Asset allocators create portfolios, often in the form of mutual funds,
with the intention to turn in good results in both “bullâ€
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### Show Tags
23 Nov 2005, 12:04
The idiom to check is 'intension of'
The OA is 'C'.
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### Show Tags
23 Nov 2005, 15:42
vivek123 wrote:
The idiom to check is 'intension of'
The OA is 'C'.
Please clarify here. Where in C do you see a of ???
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### Show Tags
23 Nov 2005, 21:07
sandalphon wrote:
vivek123 wrote:
The idiom to check is 'intension of'. The OA is 'C'.
Please clarify here. Where in C do you see a of ???
the correct idiom is "intent to" not "intent of" is also correct. however the problem here is correct participle phrese.
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### Show Tags
23 Nov 2005, 21:10
sandalphon wrote:
vivek123 wrote:
The idiom to check is 'intension of'. The OA is 'C'.
Please clarify here. Where in C do you see a of ???
the correct idiom is "intent to", not "intent of". however the problem here is use of correct participle phrase.
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14 Dec 2005, 22:40
[quote="vivek123"]Asset allocators create portfolios, often in the form of mutual funds,
with the intention to turn in good results in both “bullâ€
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15 Dec 2005, 15:35
To....forlorn
the portfolio is being modified by " in the form of mutual funds " and also "intended to turn in good results..."
In fact, both of phrase modifier portfolio but it's different form.
in the form of mutual funds ---prepositon phrase, apposition
intended to turn in good results----relative cluase
-----------------
Re: SC: Portfolios [#permalink] 15 Dec 2005, 15:35
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dripcima24
2022-09-07
How do I find the range of the function $y=-{2}^{x}+2$?
Houston Ellis
Here's my attempt:
y<2 is the range.
$y=-{2}^{x}+2$ Can also be written as
$2-y={2}^{x}$
and also,
$\mathrm{ln}\left(2-y\right)=x\mathrm{ln}2$
$⇒2-y>0⇒y<2$
tun1ju2k1ki
The range of $y=-{2}^{x}+2$ is $\left(-\mathrm{\infty };2\right)$
I'd start from a known fact that ${a}^{x}>0$ for all a>0 and $x\in \mathbb{R}$
So:
${2}^{x}>0$
$-{2}^{x}<0$
$-{2}^{x}+2<2$
y<2
$y\in \left(-\mathrm{\infty };2\right)$
Do you have a similar question? | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 21, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://plainmath.org/descriptive-statistics/93309-how-do-i-find-the-range-of-the-function", "fetch_time": 1718842450000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-26/segments/1718198861853.72/warc/CC-MAIN-20240619220908-20240620010908-00340.warc.gz", "warc_record_offset": 405922373, "warc_record_length": 19914, "token_count": 247, "char_count": 531, "score": 3.734375, "int_score": 4, "crawl": "CC-MAIN-2024-26", "snapshot_type": "latest", "language": "en", "language_score": 0.6388799548149109, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00061-of-00064.parquet"} |
# 52500 in Words
52500 in words is Fifty-two thousand Five hundred. If you paid an annual tax of Rs. 52500, you may express it as ” I paid Fifty-two thousand five hundred rupees as annual tax”. In this article, you will understand how to write the cardinal number 52500 in word form easily.
52500 in words Fifty-two thousand Five hundred Fifty-two thousand Five hundred in Numbers 52500
## 52500 in English words
We generally write numbers in words using the English alphabet. So, we can spell 52500 in English as “Fifty-two thousand Five hundred”.
## How to Write 52500 in Words?
The number 52500 has five digits, so we need to create a place value chart with five columns to obtain the place value for all the digits, as shown in the table below.
Ten thousands Thousands Hundreds Tens Ones 5 2 5 0 0
Here, ones = 0, tens = 0, hundreds = 5, thousands = 2, ten thousands = 5.
Now, expand these numbers, along with the place value.
5 × Ten thousand + 2 × Thousand + 5 × Hundred + 0 × Ten + 0 × One
= 5 × 10000 + 2 × 1000 + 5 × 100 + 0 × 10 + 0 × 1
= 50000 + 2000 + 500
= Fifty thousand + Two thousand + Five hundred
= Fifty-two thousand Five hundred
Thus, 52500 in words = Fifty-two thousand Five hundred.
### Facts About the Number 52500
52500 is a natural number that is the successor of 52499 and the predecessor of 52501.
52500 in words – Fifty-two thousand Five hundred
Is 52500 an odd number? – No
Is 52500 an even number? – Yes
Is 52500 a perfect square number? – No
Is 52500 a perfect cube number? – No
Is 52500 a prime number? – No
Is 52500 a composite number? – Yes
## Frequently Asked Questions on 52500 in Words
Q1
### How do you write 52500 in English words?
In English words, the number 52500 can be written as Fifty-two thousand Five hundred.
Q2
### How to write Rs. 52500 in words on a cheque?
On a cheque, we generally write Rs. 52500 in words as Fifty-two thousand Five hundred rupees only.
Q3
### Express the value in words for 50000 + 2500.
50000 + 2500 = 52500 The value of 50000 + 2500 is 52500, i.e. Fifty-two thousand five hundred. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://byjus.com/maths/52500-in-words/", "fetch_time": 1695814115000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-40/segments/1695233510297.25/warc/CC-MAIN-20230927103312-20230927133312-00654.warc.gz", "warc_record_offset": 171073356, "warc_record_length": 109426, "token_count": 587, "char_count": 2088, "score": 4.21875, "int_score": 4, "crawl": "CC-MAIN-2023-40", "snapshot_type": "latest", "language": "en", "language_score": 0.8394946455955505, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00026-of-00064.parquet"} |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Proportions Using Cross Products
## Cross-multiply to solve proportions with one variable
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A Golden Smile
Credit: eperales
Source: http://www.flickr.com/photos/96799823@N00/99860428/
Do you know where ratios can be found at the dentist's office? You might not be thinking of math as you have your teeth cleaned, but in order to determine a beautiful smile, dentists can use ratios.
#### News You Can Use
The Greeks believed that the key to all beauty could be found in mathematics. They began by taking geometric measurements in search of this code and discovered that there was indeed a particular ratio found in beautiful things. That ratio, called the golden ratio (also referred to by the Greek letter phi φ\begin{align*}\varphi\end{align*}), is approximately 1.618:1\begin{align*}1.618:1\end{align*} and can be found anywhere from nature, to art and architecture, to the human body. In 1509, Luca Pacioli, an Italian mathematician, published a book exploring the golden ratio and showed how it could be applied to the human face, as depicted in the illustration below.
Credit: Luca Pacioli, edited by Laura Guerin
Source: http://commons.wikimedia.org/wiki/File:Divina_proportione.png
More recently, esthetic dentists (dentists interested in developing beautiful smiles) studied this ratio and found that it describes an ideal relationship between certain teeth. When the golden ratio is present, the teeth come together in a beautiful smile.
Credit: Laura Guerin
Source: CK-12 Foundation
How does this work? Well, an esthetic dentist can measure the combined width across both an upper central incisor and its neighboring lateral incisor, as well as the width of the central incisor alone. If the relationship between these two widths forms the golden ratio, then the teeth are in proportion and so is the smile. If not, then the esthetic dentist can use the measurements to adjust the person's teeth until the correct proportion can be found. Now that is a lot to smile about!
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Excel
STATISTICAL
Master
Correlation and Covariance
for the Graduate Student and Business Manager
Clear and Complete - WITH LOTS OF SOLVED PROBLEMS
# Correlation and Covariance
Lots of Worked-Out, Easy-To-Understand, Graduate-Level Problems --->
( Scroll Down and Take a Look ! )
Introduction
Correlation and Covariance describe linear relationships between different variables. Both Correlation and Covariance describe whether variables move together in the same direction, move in opposite directions, or don't move in any
related way at all. Correlation and Covariance also describe how strong these linear relationships are between the variables.
## Basic Explanation of Correlation and Covariance
Correlation and Covariance are very similar ways of describing the direction and strength of linear relationships between two variables. Correlation is a more well-known concept and more widely used. It will therefore be covered in the first half of this course module. Covariance will be covered in the second half.
Correlation Analysis
Positive Correlation vs. Negative Correlation
Positive Correlation
If two variables are "positively correlated," they move in the same direction. When one goes up, the other goes up as well. Two variables that are positively correlated have a correlation coefficient that is between 0 and +1. The closer the correlation coefficient is to +1, the more exactly the two variables move together. A correlation coefficient between two variables of exactly +1.00 means that both variables move in lock-step with each other. A correlation coefficient between two variable of 0 indicates that there is no relationship between the movement of one variable and movement of the other variable.
Negative Correlation
If two variables are "negatively correlated," they move in opposite directions. When one goes up, the other goes down. When one variable goes down, the other goes up. Two variables that are "negatively correlated" have a correlation coefficient that is between -1 and 0. The closer the correlation coefficient is to -1, the more exactly the two variables move in opposite directions. A correlation coefficient between two variables of exactly -1.00 means that both variables move lock-step with each other in opposite directions. A correlation coefficient between two variable of 0 indicates that there is no relationship between the movement of one variable and movement of the other variable.
A general way to interpret the calculated r value is as follows:
0.0 to 0.2 - Very weak to negligible correlation
0.2 to 0.4 - Weak correlation
0.4 to 0.7 - Moderate correlation
0.7 to 0.9 - Strong correlation
0.9 to 1.0 - Very strong correlation
Calculation of Correlation Coefficient
The correlation also describes how linear a relationship is between two variables. The Correlation Coefficient can have values between -1 and +1. Below is the formula for calculating the Correlation Coefficient. Excel does such a great job in calculating correlation and covariance that it is not necessary to memorize the formulas of covariance and correlation, if you have access to Excel and know how to use the correlation functions. Here are the correlation formulas below:
***********************************************************
Population Correlation Coefficient
Correlation of variables x and y from a known population = ρ ("rho")
ρ = Population Correlation Coefficient
ρ
= ( Covariance of x and y) / ( Standard Deviation of x * Standard Deviation of y )
ρ = σxy / ( σx * σy)
(Covariance will be explained later in this module)
***********************************************************
Sample Correlation Coefficient
Correlation of variables x and y randomly sampled from an unknown population = r (This is the normal situation)
r = Sample Correlation Coefficient
The Sample Correlation Coefficient, r, is also known as the Product Moment Coefficient or Pearson's Correlation.
r = ( Sample Covariance of x and y) / ( Sample Standard Deviation of x * Sample Standard Deviation of y )
r = Sample Correlation Coefficient = Sxy / (Sx * Sy)
rxy = [ nΣxiyi - ΣxiΣyi ] / [ SQRT( nΣxi2 - (Σxi)2 ) * SQRT(nΣyi2 - (Σyi)2)]
Correlation does not mean causation. Correlation indicates that there might be causation between two variables but this may not be the case at all. There might be underlying causes there are not known and the correlation is merely incidental. Variables that provide causation but are not included in the correlation are often called "confounding variables."
The Pearson correlation, r, indicates the strength of a linear relationship between variables. There might also be non-linear relationships between variables. Visual examination of a graphing of the data points can sometimes show a non-linear relationship between the variables that would not be evident from the correlation analysis. It is always important to look at the data points on a graph in addition to any numerical analysis performed on the data.
Problem 1
Problem 1: Calculating Correlation Between Two Variables
Problem: Calculate the correlation between variables x and y based on the 6 pairs of x-y data given below:
x y 1 2 3 6 6 7 8 9 5 6 4 5
The Correlation of variables x and y is rxy:
Using the above formula:
r = Sample Correlation Coefficient = Sxy / (Sx * Sy)
rxy = [ nΣxiyi - ΣxiΣyi ] / [ SQRT( nΣxi2 - (Σxi)2 ) * SQRT(nΣyi2 - (Σyi)2)]
The data needs to be arranged in following way to facilitate correlation analysis:
X Y XY X2 Y2 1 2 2 1 4 3 6 18 9 36 6 7 42 36 49 8 9 72 64 81 5 6 30 25 36 4 5 20 16 25 Totals 27 35 184 151 231 ΣX ΣY ΣXY ΣX2 ΣY2
Number of samples = n = 6
Xavg = ΣX / n = 27 / 6 = 4.5
Yavg = ΣY / n = 35 / 6 = 5.83
rxy = Correlation Coefficient between X and Y
rxy = [ n* ΣXY – ΣX*ΣY ] / [ SQRT(n*ΣX2 – (ΣX)2) * SQRT ( n*Σy2 – (Σy)2 ) ]
rxy = [ 6*184 – 27*35 ] / [ SQRT(6*151 – (27)2) * SQRT ( 6* 231 – (35)2 ) ]
rxy = 0.94
This same problem above is solved in the Excel Statistical Master with only 1 Excel formula. If you found your statistics book confusing, You'll really like the Excel Statistical Master. Everything is explained in simple, step-by-step frameworks.
Problem 2
Problem 2: Calculating Correlation Between Multiple Variables
Problem: Determine the correlation between all of the variables below:
x y z a 1 2 10 24 3 6 9 45 6 7 8 56 8 9 7 46 5 6 6 67 4 5 5 23
We assume this data is sample data so the Correlation Coefficient is r, not σ which is the Correlation Coefficient for data from a known population.
Using the above formula:
rxy = [ nΣxiyi - ΣxiΣyi ] / [ SQRT( nΣxi2 - (Σxi)2 ) * SQRT(nΣyi2 - (Σyi)2)]
The correlation between x and y = rxy = 0.94
The correlation between x and z = rxz = -0.51
The correlation between x and a = rxa = 0.55
The correlation between y and z = ryz = - 0.39
The correlation between y and a = rya = 0.59
The correlation between z and a = rza = - 0.16
This same problem aboveis solved in the Excel Statistical Master with only 1 Excel formula. The Excel Statistical Master is the fastest way for you to climb the business statistics learning curve.
The closer to +1 the correlation between 2 variables is, the more they move together in the same direction.
The closer to -1 the correlation between 2 variables is, the more they move in opposite directions.
The closer to 0 the correlation between 2 variables is, the less related and more random is their movement.
r2 - Square of Correlation Coefficient
r2 - The square of the correlation coefficient, known as the Sample Coefficient of Determination, calculates what percentage of total variance of the output (dependent variable) of a regression is explained by the variance of the inputs (independent variables). In other words, r Square represents the proportion of the total variation that is explained by the regression equation.
Total Variance = Explained Variance + Unexplained Variance
Σ (Y – Yavg)2 = Σ (Yest – Yavg)2 + Σ (Y – Yest)2
r Square = Sample Coefficient of Determination
r Square = Explained Variance / Total Variance
Since Total Variance = Explained Variance + Unexplained Variance
Σ (Y – Yavg)2 = Σ (Yest – Yavg)2 + Σ (Y – Yest)2
r Square = Explained Variance / Total Variance
r Square = 1 - [ Unexplained Variance / Total Variance ]
r Square = 1 - [ Σ (Y – Yest)2 / Σ(Y – Yavg)2 ]
## Covariance Analysis
The covariance also describes how linear a relationship is between two variables.
The main difference between covariance and correlation is the range of values that each can assume. The Correlation between two variables can assume values only between -1 and +1. The Covariance between two variables can assume a value outside of this range. The more positive a covariance is, the more closely the variables move in the same direction. Conversely, the more negative a covariance is, the more the variables move in opposite directions.
Two independent variables will have a zero Covariance. A Covariance of zero does not that two variables are independent though. The two variables may have a nonlinear relationship. This may not be picked up at all by the Covariance calculation.
The Covariance calculation between two variables is very dependent upon the scale that the two variables are measured by. This is the main disadvantage of using Covariance instead of Correlation to compare two variables. The Correlation Coefficient is not dependent upon the scale used and provides the ability to compare different sets of data consistently.
Calculation of Covariance
Below is the formula for calculating the Covariance of variables. Excel does such a great job in calculating correlation and covariance that it is not necessary to memorize the formulas of covariance and correlation, but here they are, along with examples worked out in Excel:
Covariance of variables x and y from a known population = σxy
σxy = 1/n * Σ (xi - µx) * (yi - µy) as i goes from 1 to n
µx and µy represent population means and sxy represents a covariance from a population.
Covariance of variables x and y randomly sampled from an unknown population = sxy (This is the normal situation)
sxy = 1/ (n - 1) * Σ (xi - xavg) * (yi - yavg) as i goes from 1 to n
Problem 3
Problem 3: Calculating Covariance Between Two Variables
Problem: Calculate the covariance between variables x and y based upon the 6 pairs of x-y data given below:
x y 1 2 3 6 6 7 8 9 5 6 4 5
Using the above formula:
sxy = 1/ (n - 1) * Σ (xi - xavg) * (yi - yavg) as i goes from 1 to n
The Covariance of variables x and y is sxy:
sxy = 4.42
This same problem above is solved in the Excel Statistical Master with only 1 Excel formula. With the Excel Statistical Master you can do advanced business statistics without having to buy and learn expensive, complicated statistical software packages such as SyStat, MiniTab, SPSS, or SAS.
Problem 4
Problem 4: Calculating Covariance Between Multiple Variables
Problem: Determine the covariance between all of the variables below:
x y z a 1 2 10 24 3 6 9 45 6 7 8 56 8 9 7 46 5 6 6 67 4 5 5 23
We assume this data is sample data so the Covariance variable is sxy, not σxy which is the Covariance variable for data from a known population.
Using the above formula:
sxy = 1/ (n - 1) * Σ (xi - xavg) * (yi - yavg) as i goes from 1 to n
The covariance between x and y = sxy = 4.42
The covariance between x and z = sxz = -1.92
The covariance between x and a = sxa = 19.25
The covariance between y and z = syz = -1.42
The covariance between y and a = sya = 19.75
The covariance between z and a = sza = -4.25
This same problem above is solved in the Excel Statistical Master with only 1 Excel formula. The Excel Statistical Master is the fastest way for you to climb the business statistics learning curve.
The more positive the covariance between 2 variables is, the more they move together in the same direction.
The more negative the covariance between 2 variables is, the more they move in opposite directions.
Variance Relationships Derived From Covariance
Variance of a Sum = (σx+y)2 = σx2 + σy2 + 2Covxy
Variance of a Difference = (σx-y)2 = σx2 + σy2 - 2Covxy
Variance of a Variable + a Constant = (σx+C)2 = σx2
Variance of a Variable - a Constant = (σx-C)2 = σx2
Variance of a Variable times a Constant = (σCx)2 = σx2 * C2
Variance of a Variable divided by a Constant = (σx/C)2 = σx2 / C2
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### Transcript of dyna - beer
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Chapter 11 KINEMATICS OF PARTICLES
x
PO
x
The motion of a particle along astraight line is termed rectilinearmotion. To define the positionPof the particle on that line, we
choose a fixed origin O and apositive direction. The distancex from O toP, with the
appropriate sign, completely defines the position of the particle
on the line and is called the position coordinateof theparticle.
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x
PO
x
The velocity v of the particle is equal to the time derivative ofthe position coordinatex,
v =dx
dtand the accelerationa is obtained by differentiating v withrespect to t,
a =dv
dt
or a =d2x
dt2
we can also express a as
a = vdv
dx
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x
PO
x
v =dx
dta =
dv
dt
or a = d2
xdt2 a = v dvdxor
The velocity v and acceleration a are represented by algebraic
numbers which can be positive or negative. A positive value for
v indicates that the particle moves in the positive direction, anda negative value that it moves in the negative direction. A
positive value fora, however, may mean that the particle is truly
accelerated (i.e., moves faster) in the positive direction, or that
it is decelerated (i.e., moves more slowly) in the negativedirection. A negative value fora is subject to a similar
interpretation.
+-
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Two types of motion are frequently encountered: uniform
rectilinear motion, in which the velocity v of the particle is
constant and
x =xo + vt
and uniformly accelerated rectilinear motion, in which the
acceleration a of the particle is constant and
v = vo + at
x =xo
+ vot + at2
1
2
v2 = vo + 2a(x -xo )2
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x
O
xA
xB
xB/A
A B
When particlesA andB move along the same straight line, the
relative motion ofB with respect toA can be considered.
Denoting byxB/Athe relative position coordinate ofB with respect
toA , we have
xB =xA +xB/A
Differentiating twice with respect to t, we obtain
vB = vA + vB/A aB = aA + aB/A
where vB/A and aB/A represent, respectively, the relative velocityand the relative acceleration ofB with respect toA.
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A
B
C
xA
xB
xC
When several blocks are are connected by inextensible cords,
it is possible to write a linear relation between their position
coordinates. Similar relations can then be written between
their velocities and their accelerations and can be used toanalyze their motion.
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Sometimes it is convenient to use a graphical solution for
problems involving rectilinear motion of a particle. The graphical
solution most commonly involvesx - t, v - t, and a - tcurves.
a
tv
tx
t
t1 t2
v1
v2
t1 t2
v2 - v1 = a dtt1
t2
x1
x2
t1 t2
x2 -x1 = v dtt1
t2
At any given time t,
v = slope ofx - tcurve
a = slope ofv - tcurve
while over any given time interval
t1 to t2,
v2 - v1 = area undera - tcurve
x2 -x1 = area underv - tcurve
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x
y
r
P
Po
O
v
s
The curvilinear motion of a particle
involves particle motion along a
curved path. The positionPof the
particle at a given time is definedby theposition vectorr joining the
origin O of the coordinate system
with the pointP.
The velocity v of the particle is defined by the relation
v =dr
dtThe velocity vector is tangent to the path of the particle, and
has a magnitude v equal to the time derivative of the lengths ofthe arc described by the particle:
v =ds
dt
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x
y
r
P
Po
O
v
s
v =dr
dt
In general, the acceleration aof the particle is not tangent
to the path of the particle. It
is defined by the relation
v =ds
dt
a =dv
dtx
y
r P
Po
O
a
s
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x
y
zi
j
k
vx
vy
vz
xiyj
zk
P
x
y
z
i
j
k
r
ax
ay
az
P
Denoting byx,y, andzthe rectangular
coordinates of a particleP, the
rectangular components of velocity and
acceleration ofPare equal, respectively,to the first and second derivatives with
respect to tof the corresponding
coordinates:
vx =x vy =y vz=z. . .
ax =x ay =y az=z.. .. ..
r
The use of rectangular components is
particularly effective in the study of the
motion of projectiles.
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x
y
z
x
y
z
A
B
rA
rB rB/A
For two particles A andB moving
in space, we consider the
relative motion ofB with respect
toA , or more precisely, withrespect to a moving frame
attached toA and in translation
withA. Denoting by rB/Athe
relative position vector ofB with
respect toA , we have
rB = rA + rB/A
Denoting by vB/Aand aB/A, respectively, the relative velocityand
the relative acceleration ofB with respect toA, we also have
vB = vA + vB/A
aB = aA + aB/Aand
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x
y
C
P
an = en
O
v 2
r
at = etdv
dt
It is sometimes convenient to
resolve the velocity and acceleration
of a particlePinto components other
than the rectangularx,y, andz
components. For a particlePmoving
along a path confined to a plane, we
attach toP the unit vectors et
tangent to the path and en normal to
the path and directed toward thecenter of curvature of the path.
The velocity and acceleration are expressed in terms of tangential
and normal components. The velocity of the particle is
v = vet
The acceleration is
a = et + env2
r
dv
dt
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v = vet
In these equations, v is the speed of the particle and r is theradius of curvature of its path. The velocity vectorv is directed
along the tangent to the path. The acceleration vectora
consists of a component at directed along the tangent to thepath and a component an directed toward the center of
curvature of the path,
a = et + env2r
dvdt
x
y
C
P
an = en
O
v 2
r
at = etdv
dt
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x
P
O
eq
q
r = rer
erWhen the position of a particle moving
in a plane is defined by its polar
coordinates r and q, it is convenient to
use radial and transverse componentsdirected, respectively, along the
position vectorr of the particle and in
the direction obtained by rotating r
through 90
o
counterclockwise. Unitvectors er and eq are attached toPand are directed in the radial
and transverse directions. The velocity and acceleration of the
particle in terms of radial and transverse components is
v = rer+ rqeq. .
a = (r- rq2)er + (rq + 2rq)eq... .. . .
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x
P
O
eq
q
r = rer
erv = rer+ rqeq
. .
a = (r- rq2)er + (rq + 2rq)eq... .. . .
In these equations the dots represent differentiation withrespect to time. The scalar components of of the velocity
and acceleration in the radial and transverse directions are
therefore
vr= r vq= rq. .
ar= r- rq2 aq = rq + 2rq... .. . .
It is important to note that ar is not equal to the time derivative
ofvr, and that aq is not equal to the time derivative ofvq.
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Chapter 12 KINETICS OF PARTICLES:
NEWTONS SECOND LAW
Denoting by m the mass of a particle, by S F the sum, orresultant, of the forces acting on the particle, and by a the
acceleration of the particle relative to a newtonian frame of
reference, we write
S F = maIntroducing the linear momentum of a particle,L = mv,
Newtons second law can also be written as
S F = L.
which expresses that the resultant of the forces acting on a
particle is equal to the rate of change of the linear momentum
of the particle.
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To solve a problem involving the motion of a
particle, S F = ma should be replaced byequations containing scalar quantities. Using
rectangular components ofF and a, we have
SFx = max SFy= may SFz= maz
x
y
P
an
O
at
x
y
z
ax
ay
az
P
x
P
aq
O
ar
Using tangential and normal components,
SFt = mat= m dvdtv2
...
.. . .SFr = mar= m(r- rq2)
qr
SFn = man= m
SFq = maq= m(rq + 2rq)
• 7/28/2019 dyna - beer
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# How do I solve a large system of differential equations with indexed functions?
I strongly suspect that matrix methods are the right answer here, but I am having trouble visualizing how to set it up. Imagine a large number of coupled functions:
AB[m,n][t]
where m and n are whole number indices that can range up to hundreds or low thousands, and t is a continuous time variable. Think of AB[m,n] as the concentration of AB of type [m,n], and these concentrations of the different types can evolve with time. In addition, we have a single extra function:
B[t]
that is also a concentration that evolves with time, but for which there is only a single type. Initial conditions are:
AB[a,0][0] = A0
AB[anythingelse][0] = 0
B[0] = B0
Here, a is a constant whole number in the hundreds to low thousands.
AB[m,n] can transform into other types by two processes:
AB[m,n] + B --> AB[m-1,n+b]
AB[m,n] --> AB[m-1,n-1]
Here, b is a constant whole number significantly less than a. This means that the differential equations governing the evolution of the system are:
D[AB[m, n][t], t] == -k1[m, n] AB[m, n][t] B[t]
+k1[m + 1, n - b] AB[m + 1, n - b][t] B[t]
-k2[m, n] AB[m, n][t]
+k2[m + 1, n + 1] AB[m + 1, n + 1][t]
D[B[t], t] == -Sum[k1[m, n] AB[m, n][t],{m,0,a},{n,0,Infinity}] B[t]
Good guesses for the forms for k1 and k2, which we can use for testing, are:
k1[m_] = (m/a) k3
k2[m_, n_] = k4 n/(n + k5/m) + n k6 k1[m]
Here, k3, k4, k5, and k6 are positive real numbers.
How on Earth do I organize this system in a way to numerically solve this system using NDSolve (or ParametricNDSolve) or similar? The ultimate goal will be to have a measurable function that would be something like:
Sum[m AB[m,n],{m,0,a},{n,0,infinity}]
This function would then be fit to experimental data, varying k3-k6, and maybe a and b. Later generalizations may be to have a broader distribution of starting concentrations rather than all being identical, a range of different possible bs, etc.
• Are you really trying to solve a system of ODEs with an infinite number of variables (that is, does n really range from 0 to Infinity)? If not, have you tried creating a matrix equation to represent the ODEs, something like: AB'[t] == m1 . AB[t] . m2 B[t] + n1 . AB[t] . n2, where m1, m2, n1 and n2 encode the information about your k variables. Commented Nov 11, 2019 at 22:43
• Given the mechanism, the maximum value that n could take is a times b. I will look at your suggestion. Commented Nov 11, 2019 at 22:55
• Looks like I found a way to make your suggestion work! Thank you! I will post a completed answer shortly, once I debug one aspect. Commented Nov 12, 2019 at 0:58
a = 20;
b = 4;
Also, my equations above are slightly off. The first transformation should have been:
AB[m,n] + B --> AB[m-1,n+(b-1)]
which changes the differential equations somewhat. First, let's put in our definitions for the rate constants.
k1[m_, n_] = (m/a) k3
k2[m_, n_] = k4 n m/(n m + k5) + n k6 k1[m]
And some initial guess values:
k3 = k4 = k5 = k6 = 1;
AB0 = 10;
B0 = 50;
Now we'll make a vector for all of the species:
ABvector = Flatten@Table[AB[m, n][t], {m, 0, a}, {n, 0, a (b - 1)}];
We can index AB[m,n] via:
index[m_, n_] = m (a (b - 1) + 1) + n + 1;
AB[[index[m,n]]]
Now the k1 and k2 matrices:
k1matrix = SparseArray[
Flatten@Table[{{index[m, n], index[m, n]} -> -k1[m, n],
{index[m - 1, n + (b - 1)], index[m, n]} -> k1[m, n]},
{m, 1, a},
{n, 0, a (b - 1) - (b - 1)}],
{(a + 1) (1 + a (b - 1)), (a + 1) (1 + a (b - 1))}];
k2matrix = SparseArray[
Flatten@Table[{{index[m, n], index[m, n]} -> -k2[m, n],
{index[m - 1, n - 1], index[m, n]} -> k1[m, n]},
{m, 1, a},
{n, 1, a (b - 1)}],
{(a + 1) (1 + a (b - 1)), (a + 1) (1 + a (b - 1))}];
We can now define all of our rate equations (except for B) like this:
rateEq = Thread[D[ABvector, t] == k1matrix.ABvector B[t] + k2matrix.ABvector];
And our initial conditions are (except for B):
initCond = Thread[(ABvector /. t -> 0) == SparseArray[{index[a, 0] -> AB0},
{(a + 1) (1 + a (b - 1))}]];
We can now run NDSolve:
result = NDSolve[Join[rateEq, initCond,
{B[0] == B0, B'[t] == -Sum[k1[m, n] B[t] AB[m, n][t],
{m, 1, a},
{n, 0, a (b - 1) - (b - 1)}]}],
Append[ABvector, B[t]], {t, 0, 10}, WorkingPrecision -> 30];
And we can visualize the results like so:
fn[t_] = Table[AB[m, n][t], {m, 0, a}, {n, 0, a (b - 1)}] /. (result[[1]]);
frameTable = Table[ListPointPlot3D[fn[t], ImageSize -> Large, Filling -> 0,
FillingStyle -> Thickness[0.01], ColorFunction -> "Rainbow",
AxesLabel -> {"n", "m", "Concentration"},
ViewPoint -> {1.3, -2.4, 1.}], {t, 0, 3, 0.1}];
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## DEV Community
Abhishek Chaudhary
Posted on
# Smallest Subtree with all the Deepest Nodes
Given the `root` of a binary tree, the depth of each node is the shortest distance to the root.
Return the smallest subtree such that it contains all the deepest nodes in the original tree.
A node is called the deepest if it has the largest depth possible among any node in the entire tree.
The subtree of a node is a tree consisting of that node, plus the set of all descendants of that node.
Example 1:
Input: root = [3,5,1,6,2,0,8,null,null,7,4]
Output: [2,7,4]
Explanation: We return the node with value 2, colored in yellow in the diagram.
The nodes coloured in blue are the deepest nodes of the tree.
Notice that nodes 5, 3 and 2 contain the deepest nodes in the tree but node 2 is the smallest subtree among them, so we return it.
Example 2:
Input: root = [1]
Output: [1]
Explanation: The root is the deepest node in the tree.
Example 3:
Input: root = [0,1,3,null,2]
Output: [2]
Explanation: The deepest node in the tree is 2, the valid subtrees are the subtrees of nodes 2, 1 and 0 but the subtree of node 2 is the smallest.
Constraints:
• The number of nodes in the tree will be in the range `[1, 500]`.
• `0 <= Node.val <= 500`
• The values of the nodes in the tree are unique.
Note: This question is the same as 1123: https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/
SOLUTION:
``````# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def maxDepth(self, root):
if root:
a = 1 + self.maxDepth(root.left)
b = 1 + self.maxDepth(root.right)
return max(a, b)
return -1
def subtreeWithAllDeepest(self, root: TreeNode) -> TreeNode:
if root:
left = self.maxDepth(root.left)
right = self.maxDepth(root.right)
if left > right:
return self.subtreeWithAllDeepest(root.left)
elif right > left:
return self.subtreeWithAllDeepest(root.right)
else:
return root
`````` | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://dev.to/theabbie/smallest-subtree-with-all-the-deepest-nodes-3p7e", "fetch_time": 1719181079000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-26/segments/1718198864850.31/warc/CC-MAIN-20240623194302-20240623224302-00804.warc.gz", "warc_record_offset": 173445490, "warc_record_length": 76174, "token_count": 559, "char_count": 2052, "score": 3.765625, "int_score": 4, "crawl": "CC-MAIN-2024-26", "snapshot_type": "latest", "language": "en", "language_score": 0.7538585662841797, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00049-of-00064.parquet"} |
# The Unapologetic Mathematician
## Uniform Convergence
Today we’ll give the answer to the problem of pointwise convergence. It’s analogous to the notion of uniform continuity in a metric space. In that case we noted that things became nicer if we could choose our $\delta$ the same for every point, and something like that will happen here.
To reiterate: we say that a sequence $f_n$ converges pointwise to a function $f$ if for every $x$, and for every $\epsilon$, there is an $N$ so that $n>N$ implies that $|f_n(x)-f(x)|<\epsilon$. Just like we did for uniform continuity we’re going to move around the quantifiers so that $N$ can depend only on $\epsilon$, not on $x$.
We say that a sequence of functions converges uniformly to a function $f$ if for every $\epsilon$ there is an $N$ so that for every $x$, $n>N$ implies that $|f_n(x)-f(x)|<\epsilon$. In pointwise convergence, the value at each point does converge to the value of the limiting function, but the rates can vary widely enough to make it impossible to control convergence at two different parts of the domain simultaneously. But in uniform convergence we have “uniform” control of the convergence over the entire domain.
So let’s see how we can use this to show that the limiting function $f$ is continuous if each function $f_n$ in the sequence is. Uniform convergence tells us that for every $\epsilon$ there is an $N$ so that $n>N$ implies that $|f_n(x)-f(x)|<\frac{\epsilon}{3}$ for every $x$. But since $f_n$ is continuous at $x_0$ there is some $\delta$ so that $|x-x_0|<\delta$ implies that $|f_n(x)-f_n(x_0)|<\frac{\epsilon}{3}$.
And now we can use this $\delta$ to show the continuity of $f$. For if $|x-x_0|<\delta$, we find
\begin{aligned}|f(x)-f(x_0)|<|f(x)-f_n(x)|+|f_n(x)-f_n(x_0)|+|f_n(x_0)-f(x_0)|\\<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon\end{aligned}
The essential point here is that we were able to keep control of the convergence of the sequence both at the point of interest $x_0$, and at all points $x$ in the $\delta$-wide neighborhood.
Uniform convergence isn’t the only way to be assured of continuity in the limit, but it’s surely one of the most convenient. One thing that’s especially nice about uniform convergence is the way that we can control the separation of sequence terms from the limiting function by a single number $\epsilon$ instead of a whole function of them.
That is, instead of fixing an $\epsilon$, fix an $N$ and consider how far sequence terms can be from the limit. Take the maximum
$\max\limits_{n>N}|f_n(x)-f(x)|$
This depends on $x$, but if the convergence is uniform we can keep it down below some constant function. For pointwise convergence that isn’t uniform, no matter how big we pick the $N$ there will still be some differences that are “large” compared to others.
In this way, uniform convergence is more like convergence of numbers than pointwise convergence of functions. Uniform convergence just isn’t as floppy as pointwise convergence can be.
September 5, 2008 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 42, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://unapologetic.wordpress.com/2008/09/05/", "fetch_time": 1500771192000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-30/segments/1500549424200.64/warc/CC-MAIN-20170723002704-20170723022704-00642.warc.gz", "warc_record_offset": 723135896, "warc_record_length": 33262, "token_count": 783, "char_count": 3033, "score": 3.84375, "int_score": 4, "crawl": "CC-MAIN-2017-30", "snapshot_type": "longest", "language": "en", "language_score": 0.8577861189842224, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00030-of-00064.parquet"} |
0 votes
121 views
If the bandwidth of a medium is 100 Mbps and round-trip time is 50 microseconds then calculate the sequence bits in Go Back N ARQ flow control policy is applied and frame size 100 bits has to be transfer.
A ) 5
B) 6
C) 7
D) None of these
asked
edited | 121 views
0
what is d answer? 5 or 7
0
given answer is 6
0
Anwer will be 6. We have 50 microsec and we have bandwidth 100 mbps. So,we can send 5000 bits in that round trip time. So,each frame cntains 100 bits so,5000/100=50 frames. So, for 50 frames we must have log(50)=6 sequence bits.
+1
how r u processing this question.
ws+wr<=ASN(AVAILABLE SEQUENCE NUMBER)
we have to find ws(sender window size)
capacity of channel/frame bits will give the same
ceil of log(ws+1) will give the answer
0
We can send 50 frames. So,window size =50. So,sequence number =window size+1.So 51. Taking ceil(log(51)) will give 6 as the answer.
0
thanks rahul
i am doing a mistake now problem has been solved
i am taking RTT as propagation time which is wrong
0
L>=2*tp*B
=>RTT*B
=>50*100 bits=> 5000 bits
given frame size = 100 bits
no of seq required => 5000/100==> 50 +1=> 51
no of bits => log(51)==> 5.6==> 6 bits are required
Correct me if I'm wrong
0
yaa correct bro :)
## 1 Answer
+1 vote
Best answer
Round trip time = 2* Tp (propagation delay) = 50 * 10-6 sec
Transmission time ( Tt)= frame size(L) /Bandwidth (B) = 100 / (100 * 106 ) sec = 10-6 sec
Total window size = 1+ 2* (Tp / Tt )
= 1+ (50 * 10-6 / 10-6)
= 51
No of bits for sequence number = ceil of (log2 51) = 6
answered by (379 points)
selected
0
Here Will the bit sequence will be Ceil( log2(1+2a))
+15 votes
4 answers
1
0 votes
0 answers
2
0 votes
0 answers
3
0 votes
0 answers
4
+1 vote
2 answers
5
0 votes
0 answers
6
0 votes
0 answers
7 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://gateoverflow.in/262959/sudo-gate-test-series?show=263169", "fetch_time": 1548001019000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-04/segments/1547583722261.60/warc/CC-MAIN-20190120143527-20190120165527-00207.warc.gz", "warc_record_offset": 512574872, "warc_record_length": 20445, "token_count": 613, "char_count": 1808, "score": 3.59375, "int_score": 4, "crawl": "CC-MAIN-2019-04", "snapshot_type": "longest", "language": "en", "language_score": 0.8660494089126587, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00028-of-00064.parquet"} |
Tuesday, November 21, 2006
Mark I's Scribe Post
Today in class we corrected our homework. Our homework was to convert fraction's, decimal's, percent's and ratio's to fraction's, decimal's, percent's and ratio's. There are 12 in total, 3 for each topic.
Fraction's:
1. Fraction-Decimal: To convert a fraction to a decimal, you have to divide the numerator by the denominator.
2. Fraction-Percent: To convert a fraction to a percent, you have to divide the numerator by the denominator and multiply the answer by 100.
3. Fraction-Ratio: To convert a fraction to a ratio, you have to know that there are 2 parts to a ratio. The 1st part is the numerator and the 2nd part is the denominator-the numerator.
1. 4/5 4-:-5=0.80
2. 4/5 4-:-5=0.80 .80x100=80%
3. 4/5 5-4=1 4:1
Decimal's:
1. Decimal-Fraction: To convert a decimal to a fraction you put the decimal number over 100.
2. Decimal-Percent: To convert a decimal to a percent you have to multiply the decimal by 100.
3. Decimal-Ratio: To convert a decimal to a ratio, you 1st have to convert the decimal to a fraction. Then you put the numerator as the 1st part and the number you get when you - the denominator from the numerator as the 2nd part.
1. .3 30/100=15/50=3/10
2. .3 .3x100=30%
3. .3 3/10 3-10=7 3:7
Percent's:
1. Percent-Fraction: To convert a percent to a fraction, you have to just have to put the percent over 100 to get the fraction.
2. Percent-Decimal: To convert a percent to a decimal, you have to divide the the percent by 100.
3. Percent-Ratio: To convert a percent to a ratio, you have to convert the percent to a fraction by putting the percent over 100.Then you put the numerator as the 1st part and the 2nd part will be the answer of the denominator-the numerator.
1. 57% 57/100
2. 57% 57-:-100=.57
3. 57% 57/100 100-57=43 57:43
Ratio's:
1. Ratio-Fraction: To convert a ratio to a fraction, you put the 1st part of the ratio the numerator and add the numerator and the denominator and put is as the denominator.
2. Ratio-Decimal: To convert a ratio to a decimal, you have to 1st convert the ratio to a fraction by putting the 1st part of the ratio the numerator and the answer of the denominator-the numerator, put it as the denominator. Then you divide the numerator by the denominator.
3. Ratio-Percent: To convert a ratio to a percent, you first have to convert the ratio to a fraction by putting the 1st part as the numerator and putting the answer of the denominator-the numerator as the denominator. After you divide the numerator by the denominator and multiply the answer by 100.
1. 2:4 2/6
2:4 2/6 2-:-6=.33
3. 2:4 2/6 2-:-6=.33 .33x100=33.3
Homework:
Equivalents The Assignment
1. Make 4 different fractions using the digits below. You may only use each digit once. Convert these 4 fractions into decimals, percents and ratios.
1, 2, 3, 4, 5, 6, 7, 8, 9
2. Make 4 different decimal using the digits below. You may only use each digit once. Convert these 4 decimals into fractions, percents and ratios. (Do not use the decimals from the question above).
1, 2, 3, 4, 5, 6, 7, 8, 9
3. Make 4 different percents using the digits below. You may only use each digit once. Convert these 4 percents into fractions, decimals and ratios. (Do not use the fractions and decimals from the questions above).
1, 2, 3, 4, 5, 6, 7, 8, 9
4. Make 4 different ratios using the digits below. You may only use each digit once. Convert these 4 ratios into fractions, decimals and percents. (Do not use the fractions and decimals or percents from the questions above).
1, 2, 3, 4, 5, 6, 7, 8, 9
Remember only to use the digit only once and to make it into a fraction, decimal, percent and ratio also!
Convert 6:14 to a percent
Convert 36% to a decimal
Convert 78/100 to a ratio
Convert .23 to a fraction
THE NEXT SCRIBE WILL BE... MACKENZIE
julie_rose said...
Great scribe mark! ^_^
Mr. H said...
I love the questions. Come on people lets see you answer them. This is a great scribe and has a link and questions. Step up to the plate.
Harbeck
Mark I said...
Thanks! (:
Dan the Man said...
6/20 or 14/20
0.36
78:22
23/100
That was too easy Mark.
\ )
/ ) |
Mark I said...
Really well i'll make them harder next time just for you dan the man!
sorrel budden said...
wow all your blogs are fab we at nb have a lot to live up to hopefilly we will ;)
sorrel xxxxx
Mr. H said...
Sorrel they have been blogging for 2 years. Yes it takes a lot of work but I think you would agree it is worth it in the end. Thanks for commenting.
Mr. Harbeck
michelle 8-16 said...
Great link! It could really help people who are having trouble in class.
oranges said...
great scribe but it could use a little more colour but its really good! =)
Mark I said...
thanks Michelle! and to oranges I put a bit more color. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://816math.blogspot.com/2006/11/mark-is-scribe-post.html", "fetch_time": 1490377361000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-13/segments/1490218188550.58/warc/CC-MAIN-20170322212948-00428-ip-10-233-31-227.ec2.internal.warc.gz", "warc_record_offset": 4698252, "warc_record_length": 15782, "token_count": 1382, "char_count": 4785, "score": 4.40625, "int_score": 4, "crawl": "CC-MAIN-2017-13", "snapshot_type": "longest", "language": "en", "language_score": 0.8684899210929871, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00043-of-00064.parquet"} |
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I'm having trouble proving this one. I know its true. Any ideas? Here is what I have so far: If $a\mid b$, then there exists an integer $q_1$ such that $b = aq_1$. If $a\mid c$, then there exists an ...
172 views
### If $n$ is an odd integer, then there exist integers $a$ and $b$ such that $n=a^2-b^2$. [duplicate]
If $n$ is an odd integer, then there exist integers $a$ and $b$ such that $n=a^2-b^2$. Am I supposed to use induction or a direct proof?
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### Is this GCD proof valid?
I came across this theorem and wrote a proof, but I'm not sure if I made any incorrect assumptions. I also know that this isn't the easiest way to prove it - I just want to know if it works and ...
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### Modular arithmetic
How do I prove the following inequality with modular arithmetic? (No use of Fermat's last theorem is allowed.) $$3987^{12} + 4365^{12} \neq 4472^{12}$$
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### Proof of statement: If $a\mid b$ and $a\mid c$, then $a \mid b+c$
Statement: If $a$ divides both $b$ and $c$, then $a$ divides $b+c$ Proof: Assume that $a$ does not divide $b+c$. Then there is no integer $k$ such that $ak=b+c$. However, $a$ divides $b$, so $am=b$ ...
86 views
### How to write the proof for this?
Let $a,b,c \in \mathbb{Z}$, and $a \neq 0$. Use a proof by contradiction to show that if $(a \nmid (bc))$ then $(a \nmid b)$. The symbol $\nmid$ stands for "does not divide". I got the layout, but I ...
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### How can I prove by induction that $9^k - 5^k$ is divisible by 4?
Recently had this on a discrete math test, which sadly I think I failed. But the question asked: Prove that $9^k - 5^k$ is divisible by $4$. Using the only approach I learned in the class, I ...
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### Is the proof of the claim correct? Is the claim true?
We say that an integer a is divisible by the nonzero integer b, if a = bc for some integer c: When a is divisible by b, we write b | a and say b divides a. Claim: Let a and b be nonzero integers. If ...
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### Contradiction Proof regarding Well-Ordering Principle
Let $r_0$ be the smallest element of a set $S$ such that $S\subseteq\mathbb {N} \cup \{ 0 \}$. According to the Well-Ordering Principle, this implies that $r_0$ $\ge 0$ and $r_0 = a - q_0 b$ for some ...
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### Prove that there exist no positive integers $m$ and $n$ for which $m^2+m+1=n^2$
The problem: Prove that there exist no positive integers $m$ and $n$ for which $m^2+m+1=n^2$. This is part of an introductory course to proofs, so at this point, the mathematical machinery should not ...
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### Why is $n^2 - 2$ never a multiple of $3$?
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### Prove, for any positive integer $n$, that $n -3$ must be a multiple of $5$ if $n^3 -n -4$ is a multiple of $5$.
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228 views
### Show that $\gcd(a,b)=\operatorname{lcm}(a,b)$ if and only if $a=b$.
I know how to prove $a=b$ only if $\gcd(a,b)=\operatorname{lcm}(a,b)$, but I don't know how to prove the "if part". Can anyone help me?
### Proof Checking and input: Generators of $\mathbb{Z}_{pq}$
I'm self-studying abstract algebra (slowly but surely), and I have a question about my answer to the following prompt: Problem statement: Show that there are $(q-1)(p-1)$ generators of the group ... | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://math.stackexchange.com/questions/tagged/elementary-number-theory+proof-writing", "fetch_time": 1406146285000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2014-23/segments/1405997883466.67/warc/CC-MAIN-20140722025803-00075-ip-10-33-131-23.ec2.internal.warc.gz", "warc_record_offset": 237425217, "warc_record_length": 24099, "token_count": 2059, "char_count": 6866, "score": 3.53125, "int_score": 4, "crawl": "CC-MAIN-2014-23", "snapshot_type": "latest", "language": "en", "language_score": 0.9365872144699097, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00010-of-00064.parquet"} |
# math 18.pdf - Problem 1 Answer Part 1 T= 60 months R=...
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Problem 1: Answer. Part 1 T= 60 months; R= 1.5%p.m; principle=\$4000 F.V = P.V*e^rt =\$4000*2.71828^1.5%*60 =\$4000*2.71828^.9 =\$4000*2.45960311116 =\$9838.4 Part 2 T= 60 months; R= 1.5%p.m; principle=? P.V= F.V/e^r*t P.V= \$5000/2.71828^.9 =\$5000/2.4596031116 =\$2032.84. part 3 assume amount to be invested monthly be X F.V=p.v*(1+i)^n \$5000=X*(1+1.5%)^60+X*(1+1.5%)^59+ ................ X*(1+1.5%)^1)+X \$5000=X(1+1.5%)^60+(1+1.5%)^59+ ........... (1+1.5%)^1) \$5000=X(2.44+2.40+2.37++2.33+2.30+2.27+2.24+ ......... 1.015) X=\$5000/(2.44+2.40+2.37++2.33+2.30+2.27+2.24+ ......... 1.015 Problem 2: Solution Rate per month = 6.99%/12 = 0.5825% Total number of payments in 4 years = 12*4 = 48
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• Spring '18
• Saleem Malik
• Payment, Harshad number, payments, P.v, F.V
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# IMO- Mathematics Olympiad Class 8: Questions 731 - 737 of 1029
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Rs. 450.00 or
## Passage
Adjoing pie chart gives the expenditure (in percentage) on various items and saving of a family during a month.
Draw a pie chart below to show this data. Show your scale on the vertical axis and label the bars.
## Question number: 731 (3 of 9 Based on Passage) Show Passage
MCQ▾
### Question
If the monthly saving of the family is Rs. 4000, then the monthly expenditure on transport is________.
### Choices
Choice (4) Response
a.
1333.33 Rs
b.
1500.25 Rs
c.
1111.40 Rs
d.
2000.21 Rs
## Question number: 732 (4 of 9 Based on Passage) Show Passage
MCQ▾
### Question
The item on which expenditure is equal to the total savings of the family, is________.
### Choices
Choice (4) Response
a.
Education for children
b.
Clothes
c.
Food
d.
Question does not provide sufficient data or is vague
## Question number: 733 (5 of 9 Based on Passage) Show Passage
MCQ▾
### Question
The item on which expenditure is more 10 % to the total savings of the family, is________.
### Choices
Choice (4) Response
a.
Education for children
b.
Food
c.
Transport
d.
Question does not provide sufficient data or is vague
## Question number: 734 (6 of 9 Based on Passage) Show Passage
MCQ▾
### Question
Expenditure on clothes is________.
### Choices
Choice (4) Response
a.
12%
b.
20%
c.
10%
d.
15%
## Question number: 735 (7 of 9 Based on Passage) Show Passage
MCQ▾
### Question
If the monthly saving of the family is Rs. 3000, then the monthly expenditure on clothes is________.
### Choices
Choice (4) Response
a.
500 Rs
b.
2000 Rs
c.
1000 Rs
d.
1500 Rs
## Question number: 736 (8 of 9 Based on Passage) Show Passage
MCQ▾
### Question
The item on which expenditure is less 10 % to the total savings of the family, is________.
### Choices
Choice (4) Response
a.
Education for children
b.
Food
c.
Transport
d.
Question does not provide sufficient data or is vague
## Question number: 737 (9 of 9 Based on Passage) Show Passage
MCQ▾
### Question
If the monthly saving of the family is Rs. 4500, then the monthly expenditure on house rent is________.
### Choices
Choice (4) Response
a.
1500 Rs
b.
2000 Rs
c.
1000 Rs
d.
3000 Rs
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canonical model
Recall that a logic is a set of wff’s containing all tautologies and closed under modus ponens. Given a logic $\Lambda$, a set $\Delta$ of wff’s is $\Lambda$-consistent if $\perp$ can not be deduced from $\Delta$ given $\Lambda$. $\Lambda$ itself is said to be $\Lambda$-consistent if $\perp$ can not be deduced from the empty set. Let $\Lambda$ be a consistent normal modal logic. The canonical frame for $\Lambda$ is the Kripke frame $\mathcal{F}_{\Lambda}:=(W_{\Lambda},R_{\Lambda})$, where
1. 1.
$W_{\Lambda}$ is the set of all maximally consistent sets, and
2. 2.
$wR_{\Lambda}u$ iff $\square A\in w$ implies $A\in u$ for any wff $A$.
If we define $\Delta_{w}:=\{B\mid\square B\in w\}$, then the second condition above reads $wR_{\Lambda}u$ iff $\Delta_{w}\subseteq u$.
The canonical model of $\Lambda$ based on $\mathcal{F}_{\Lambda}$ is the pair $M_{\Lambda}:=(\mathcal{F}_{\Lambda},V_{\Lambda})$, where
• $V(p):=\{w\in W_{\Lambda}\mid p\in w\}$.
The main result regarding the canonical model of $\Lambda$ is:
Theorem 1.
$M_{\Lambda}\models A$ iff $\Lambda\vdash A$, where $A$ is any wff.
Since the logic $\Lambda$ is the intersection of all maximally consistent sets (see here (http://planetmath.org/LindenbaumsLemma)), the theorem is the result of the following:
Proposition 1.
$M_{\Lambda}\models_{w}A$ iff $A\in w$.
which is the result of the following:
Lemma 1.
For any world $w$ in $M_{\Lambda}$, $\square A\in w$ iff $A\in u$ for all worlds $u$ such that $wR_{\Lambda}u$.
Proof.
Suppose $\square A\in w$ and $wR_{\Lambda}u$. Then $A\in u$ by the definition of $R_{\Lambda}$. Conversely, suppose $A\in u$ for all $u$ such that $wR_{\Lambda}u$. In other words, $A\in u$ for all $u$ such that $\Delta_{w}\subseteq u$, or $A\in\bigcap\{u\mid\Delta_{w}\subseteq u\}$. But $\bigcap\{u\mid\Delta_{w}\subseteq u\}=\mbox{Ded}(\Delta_{w})$, the deductive closure of $\Delta_{w}$, so $\Delta_{w}\vdash A$, and therefore $\square\Delta_{w}\vdash\square A$ (see here (http://planetmath.org/SyntacticPropertiesOfANormalModalLogic)), or $w\vdash\square A$ (since $\Delta_{w}\subseteq w$), or $\square A\in w$ (since $w$ is maximally consistent). ∎
Proof of Proposition 1. We do induction on the number $n$ of logical connectives in $A$. If $n=0$, then $A$ is either a propositional variable or $\perp$. The former is just the definition of $V_{\Lambda}$. The later case is just the definition of $\Lambda$-consistency. Next, if $A$ is $B\to C$, then $M_{\Lambda}\models_{w}A$ iff either $M_{\Lambda}\not\models_{w}B$ or $M_{\Lambda}\models_{w}C$ iff $B\notin w$ or $C\in w$ iff $\neg B\in w$ or $C\in w$ iff $\neg B\lor C\in w$ iff $A\in w$. Finally, if $A$ is $\square B$, then $M_{\Lambda}\models_{w}\square B$ iff $\square B\in w$ iff $B\in u$ for all $u$ such that $wR_{\Lambda}u$ iff $M_{\Lambda}\models_{u}B$ for all $u$ such that $wR_{\Lambda}u$.
Recall that a logic is complete in a frame if it is complete in every model based on the frame. As a corollary to Theorem 1, we have
Corollary 1.
$\Lambda$ is complete in its canonical frame $\mathcal{F}_{\Lambda}$.
Proof.
Any wff valid in every model based on $\mathcal{F}_{\Lambda}$ is valid in $M_{\Lambda}$ in particular, and therefore a theorem of $\Lambda$ by Theorem 1. ∎
The converse is not true. There are in fact normal modal logics that are sound in no frames at all.
Canonical models are useful in proving the completeness theorems for many common normal modal logics. To prove that a logic is complete in a class of frames, by the corollary above, it is enough to show that the canonical frame is in the class. Here are two examples:
1. 1.
Let $\Lambda$ be the smallest normal logic containing the schema $\square A$. Then $\Lambda$ is complete in the class of null frames.
Proof.
By the discussion above, it is enough to show that $\mathcal{F}_{\Lambda}$ is a null frame: the assumption $\exists w\exists u(wR_{\Lambda}u)$ leads to a contradiction. Suppose $wR_{\Lambda}u$. Then $\Delta_{w}\subseteq u$. This means that if $\square A\in w$, then $A\in u$. But $\square A$ is a theorem (in $\Lambda$), $\square A\in w$ for any wff $A$. This means that $A\in u$ for any wff $A$, or that $u$ is inconsistent, a contradiction. ∎
2. 2.
Let $\Lambda$ be the smallest normal logic containing the schema $A\to\square A$. Then $\Lambda$ is complete in the class of weak identity frames (a binary relation $R$ is weak identity it is satisfies the condition $\forall x\forall y(xRy\to x=y)$).
Proof.
Again, we show that $R_{\Lambda}$ is weak identity. Suppose $wR_{\Lambda}u$. Then for any $A$, $\square A\in w$ implies that $A\in u$. Now, if $A\in w$, then applying modus ponens to $A\to\square A$, we get that $\square A\in w$ since $w$ is closed under modus ponens. But this means that $A\in u$. So $w\subseteq u$. But since both $w$ and $u$ are maximal, they are the same. ∎
Title canonical model CanonicalModel 2013-03-22 19:35:01 2013-03-22 19:35:01 CWoo (3771) CWoo (3771) 15 CWoo (3771) Definition msc 03B45 canonical frame | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 123, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://planetmath.org/CanonicalModel", "fetch_time": 1619025937000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-17/segments/1618039546945.85/warc/CC-MAIN-20210421161025-20210421191025-00382.warc.gz", "warc_record_offset": 553545827, "warc_record_length": 6712, "token_count": 1615, "char_count": 5039, "score": 3.609375, "int_score": 4, "crawl": "CC-MAIN-2021-17", "snapshot_type": "longest", "language": "en", "language_score": 0.7106069326400757, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00032-of-00064.parquet"} |
# Jason Grout
### Sidebar
courses:math_reasoning:spring_2014:home
# Math Reasoning, Spring 2014
• Full Book (through section 4.5, updated 05 May 2014)
• Section 4.5 ==Midterm== We had a midterm exam on Wed, 12 Mar, covering chapter 1.
# Class Log
## Mon, 05 May
We had a guest lecturer, Banu, introduce functions, one-to-one, onto, and bijections. We talked briefly about questions over the notes. In addition to the problems we have been preparing, please prepare problems 4.74, 4.80, 4.85, 4.86, 4.87, 4.91, 4.95, 4.96, 4.97, and 4.99. Corrections over 2.94 and 2.111 are due Wednesday.
We've talked a lot about proof techniques. Here is one proof technique that would probably not be accepted in a math journal:
## Wed, Apr 30
We worked a bit more on the one hole to fill with the proof 3.2, and then saw presentations of several more problems from section 3.1.
We didn't start presenting material from chapter 4, but we did very briefly talk about reflexive, transitive, and symmetric relations.
For Monday, type up and submit a proof of Theorem 3.9. Also, we will be calling for presentations for 4.31, 4.33, 4.34, 4.39, 4.41, 4.42, 4.47, and 4.56.
## Mon, Apr 28
Today we spent a while on some of the tricky proofs from section 3.1, like 3.2, 3.3, and 3.5. We saw some great work, and emphasized also some important points:
• make sure you don't use theorem A to prove theorem B, and then use theorem B to prove theorem A. Instead, do what Jay mentioned: make sure you prove one of the theorems independently.
• Be careful that your arguments don't allow false statements to be proven true. In David's example, he's right that you need to use more than just those lines about each prime being in the other list to show the two lists of primes are the same. I think David's idea about getting rid of each prime as you use it is a good direction to go to resolve this.
• Be careful when dealing with infinite processes. You can do it (and Courtney had a great process for how an infinite argument showed $\sqrt{2}$ was irrational), but infinity plays funny games with our intuition, so we have to be careful in how we deal with it. If you can do the same proof without appealing to an infinite process, it may be more understandable.
For next time, please continue working on problems from 3.1 or move on to section 4.1. We have a few problems from 3.1 that some of you will still put up. In section 4.1, we will call for presentations for problems 4.9, 4.10, 4.13, 4.21, 4.23, and 4.26.
For next Monday, please tex up your proof for Theorem 3.3. Feel free to come in and talk with me about it, check it with me, etc. I'm happy to give feedback on it before you turn it in.
## Mon, 14 Apr
We again discussed a lot of the proofs from 2.86-2.100. We'll spend one more day on this unit. For Wednesday, the problems we still have to do are: 2.86, 2.90, 2.94, 2.95, 2.96, 2.100, 2.101, 2.102. Some of these problems have a person who gets first shot at it, but everyone should do each of them.
I've also posted the next unit, section 2.5, on induction.
## Wed, 09 Apr
Today we talked about 2.79-2.88. We had a few more proofs up that we didn't have time to get to today. We talked for a while about proving most of the theorems we talked about. These problems are hard, probably the hardest we'll see this semester. It's tricky partly because there are so many quantifiers in each definition. Those of you that have been putting in the time have been making great progress. This is the kind of complicated thinking and reasoning that makes you as a mathematician extremely valuable.
For Monday, proofs for 2.79 and 2.87 are due at the beginning of class. These must be typed in LaTeX using the template at the bottom of this page (which is different than the template you've been using, so make sure to copy this new one). You are welcome to use the Sage Cloud; the icon in the upper right of the preview window will download the pdf file. Please note that I'm expecting printed copies turned in this time, not email or files on the Sage Cloud.
Problems that still have no claim are 2.86, 2.88, 2.90, 2.94, 2.96, 2.99, 2.100, 2.101, 2.102. If you haven't presented this week, you should especially make sure you are presenting one of these on Monday. If you've been missing some class periods, you should *especially* make sure you're coming and presenting.
The study group is meeting again Thursday around 7 or 8 at the library, around the cafe, again—email Chris Dorff for details.
On a final note, I think it's really valuable to see and learn how to handle failure. Ed Burger, a widely-respected math teacher, wrote about the importance of teaching failure. I highly recommend reading that article (and I'm open to adjusting the grade schema like he mentions if the class wants to).
## Mon, 07 Apr
Today we talked about 2.77-2.80, and then about the definition of limit points. We also talked about Cantor's diagonal argument for showing the real numbers are uncountable. It was some hard material we covered today, and you all did great. We'll be hitting it again next time, so do your best to prepare. Please go over your proofs starting with 2.78 to make sure they work well, and push ahead as far as you can, with a goal of going through 2.99.
Chris Dorff is organizing a study group, meeting at the library near the cafe at 8pm Tuesday night.
This math class may well be the most applicable math class you will ever have. Some lessons for life we learned today include:
- If you have an infinite number of things to do, don't try to do them all at once. You'll go crazy! Narrow it down to doing something specific to one thing. - Be careful about terms and definitions (in class, we emphasized that an open set and an open interval don't mean the same thing).
## Wed, 26 Mar
We finished section 2.2 and also reviewed a corrected proof or two from 2.1. We also talked about how big the rational numbers are (they are countably infinite), and discovered how to list them out.
There are several things for next time:
• We are making the last half of the midterm a take-home exam. I've graded problems 1-4. Problems 5-9 are due on Monday at the start of class. You may not use any resources in working on the problems on the exam—just as if you were taking the exam in class. Just you and your plain pencil/pen. You may not work together or discuss the problems. As a hint, you might try approaching the problem involving $a^2+b^2=c^2$ using even vs. odd arguments.
• I posted the next two sections. Section 2.3 starts with a lot of explanation and asks you to do some straightforward exercises, followed by two proofs similar to what we have done. Please prepare the problems in section 2.3 for discussion on Monday (i.e., problems 2.50-2.60).
• Please continue thinking about listing out all of the elements in $P(\mathbb{N})$. Is it possible? Note that from our discussion in class, your reasoning for why it is not possible should not apply to the rational numbers, since it is possible to list them out.
We talked a bit about the history of set theory. Here is a good article about its development and the discovery of the paradoxes that we've talked about.
## Mon, 24 Feb
We discussed 1.65-1.75, and had a great discussion about when it is equivalent to switch the order of various quantifiers.
For Wednesday, do 1.76-1.88.
We also talked about the Banach-Tarski paradox. It seems like it's like creating chocolate out of nothing, but the paradox is more fundamental than that. The paradox deals with the fundamental question, “what is the size of a set?” You might enjoy reading a non-technical explanation of the paradox.
## Wed, 19 Feb
We discussed 1.56-1.64, and had great proofs to show for it. I loved how we collectively came up with some really nice proofs by contradiction—good job everyone! We spent some time talking about quantifiers a bit more too.
We didn't turn in the problems today, but make sure to typeset your proof of Theorem 1.58 in the Sage Cloud system before class on Monday. Please prepare problems 1.68–1.75 to discuss and turn in on Monday.
We also talked a bit about Paul Erdős and the breaking news of the 13 gigabyte claimed proof of his discrepancy conjecture. For fun and math culture, we talked a bit about the Erdős number (see also the Erdős number project). Here is also a collection of stories and a collection of problems from him.
Much of Erdős' legacy comes from his amazing ability to come up with good problems. As was attributed to him in Some of my favorite problems and results, “Problems have always been an essential part of my mathematical life. A well chosen problem can isolate an essential difficulty in a particular area, serving as a benchmark against which progress in this area can be measured. An innocent looking problem often gives no hint as to its true nature. It might be like a 'marshmallow,' serving as a tasty tidbit supplying a few moments of fleeting enjoyment. Or it might be like an 'acorn,' requiring deep and subtle new insights from which a mighty oak can develop [….] In this note I would like to describe a variety of my problems which I would classify as my favorites. Of course, I can't guarantee that they are all 'acorns' but because many have thwarted the efforts of best mathematicians for many decades (and have often acquired a cash reward for their solutions), it may indicate that new ideas will be needed, which can in turn, lead to more general results, and naturally, to further new problems. In this way, the cycle of life in mathematics continues forever.”
## Mon, 17 Feb
Happy President's day! We talked about negation and had great presentations of 1.46–1.53. I thought Exercise 1.53, where we took the complicated tautology and rewrote it using implications in various ways, was a great exercise. We also covered a few common Latex issues I noticed while grading problems. We talked about proof by contradiction, and how it often comes up when you want to prove that a situation exists, even if we can't show how to construct it (like the Brouwer fixed-point theorem). We finished by introducing quantifiers.
For Wednesday, do 1.56–1.64 (get the updated pdf above, which includes the next section, section 1.4).
We were talking about computer proofs the other day, as well as SAT solvers. Just today, someone announced a proof that relied on a computer using a SAT solver to generate an example. The data is about 13 gigabytes, which is bigger than all of wikipedia. See arxiv or this writeup.
## Wed, 12 Feb
The Brouwer fixed-point theorem implies the swishing water-bottle fact. Francis Su has an interesting writeup of how the Borsuk-Ulam theorem implies the fixed point theorem.
The 3-SAT problem is the NP-complete problem we talked about.
## Latex
\documentclass[11pt]{article}
\usepackage{amssymb,amsmath,amsthm}
\usepackage{hyperref}
\usepackage[lmargin=1in,rmargin=2.5in]{geometry}
\usepackage{setspace}
\doublespacing
\theoremstyle{definition}
\newtheorem*{theorem}{Theorem}
\newtheorem*{corollary}{Corollary}
\newtheorem*{lemma}{Lemma}
\newtheorem*{conjecture}{Conjecture}
\newtheorem*{definition}{Definition}
\newtheorem*{example}{Example}
\title{Math 101, Spring 2014}
\date{ASSIGNMENT DUE DATE} % change to the date
\begin{document}
\maketitle
\begin{theorem}[THEOREM NUMBER]
THEOREM STATEMENT
\end{theorem}
\begin{proof}
\end{document} | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://jasongrout.org/courses/math_reasoning/spring_2014/home", "fetch_time": 1619140627000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-17/segments/1618039626288.96/warc/CC-MAIN-20210423011010-20210423041010-00587.warc.gz", "warc_record_offset": 419098656, "warc_record_length": 10107, "token_count": 2870, "char_count": 11411, "score": 3.65625, "int_score": 4, "crawl": "CC-MAIN-2021-17", "snapshot_type": "latest", "language": "en", "language_score": 0.9497054219245911, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00030-of-00064.parquet"} |
# Factorial number
In mathematics symbolized by placing the "!" (known as the exclamation mark or bang) after a number, it represents multiplying a number by all whole numbers smaller than it.
## Definition
A factorial is defined by the product
$n! = 1 \cdot 2 \cdot 3 \cdots (n{-}2) \cdot (n{-}1) \cdot n$
for $n ≥ 1$.
The same written as mathmatical product
$n! = \prod_{i = 1}^n i.$
and as recurrence relation
$n! = n \cdot (n-1)!$
## Examples
5! = 5 * 4 * 3 * 2 * 1 = 120
10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 3628800 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.rieselprime.de/z/index.php?title=Factorial_number&printable=yes", "fetch_time": 1563678680000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-30/segments/1563195526818.17/warc/CC-MAIN-20190721020230-20190721042230-00503.warc.gz", "warc_record_offset": 813450778, "warc_record_length": 6212, "token_count": 198, "char_count": 541, "score": 3.765625, "int_score": 4, "crawl": "CC-MAIN-2019-30", "snapshot_type": "latest", "language": "en", "language_score": 0.795872688293457, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00044-of-00064.parquet"} |
# JEE Conic Sections: Hyperbola
Have you ever wondered “What is the shape of the shadow that a household lamp casts on the walls?” From the cooling towers of thermal power plants to the McDonalds logo, you can find hyperbolic shapes in a lot of places around you if only you look for it.
The mathematical concept of Hyperbola is essential for engineering and architectural studies. It’s quite obvious that Hyperbola is very important for JEE. All its properties must be at your fingertips in order to solve the questions asked in JEE.
## Definition
You can think of a hyperbola to be two curves that resemble infinite bows.
A hyperbola is the locus of all points in a plane, the difference of whose distances from two fixed points (focus) is constant.
By Ag2gaeh – Own work, CC BY-SA 4.0, Link
Now let me introduce you to the various terms related to hyperbola:
By Ag2gaeh – Own work, CC BY-SA 4.0, Link
1) Focus: These are the two fixed points that we discussed in the definition.
2) Center: It is the mid-point of the line segment joining both the foci. It is also the intersection of the transverse and the conjugate axis.
3) Transverse Axis: It is the line joining both the foci. It is also the Major Axis.
4) Conjugate Axis: It is the line perpendicular the Transverse Axis and passes through the center of the hyperbola.
5) Vertex: The point at which the hyperbola intersects the transverse axis. Since the hyperbola intersects at two distinct points, there are two vertices in a hyperbola.
## General Form of Equation of Hyperbola
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
The expression B2 — 4AC is called the discriminant, which is used to determine the type of conic section.
For a hyperbola, B2 — 4AC > 0.
## Standard Form of Equation of Hyperbola
A hyperbola with Center as Origin (0, 0) and both the foci on the x-axis is given by the equation:
By Klaas van Aarsen – Own work, CC BY-SA 3.0, Link
The properties of the above hyperbola are:
1) The center is at O(0, 0).
2) The foci are S(ae, 0) and S1(–ae, 0)
3) The vertices are A(a, 0) and A1(–a, 0)
4) The directrix is given by the equation: x = a/e and x = –a/e
5) The length of the latus rectum is (2b2/a)
6) The length of the transverse axis = 2a
7) The length of the conjugate axis = 2b
8) The distance between 2 focii = 2ae
Now you might be wondering about ‘e’ used in various formulae. ‘e’ is known as the Eccentricity of a conic section. It is the ratio between the focus and the directrix which remains constant for a given conic section. For a hyperbola, e > 1.
By Deipnosopher – I created this work entirely by myself., CC BY-SA 3.0, Link
## Focal Distance of a Point
As the name suggests, it is the distance of any arbitrary point on the hyperbola from the focus.
Let P be any point present on the hyperbola,
|S1P — S2P| = 2a
Clearly, the focal distance of any point on the hyperbola is constant and is equal to the length of the Transverse Axis.
## Equation of Hyperbola with (h, k) as Center
The directions of the axes should be parallel to the co-ordinate axis.
## Parametric Equations
The parametric equations of the general hyperbola being discussed are:
x = a sec θ , y = b tan θ
Or
x = a cosh θ , y = b sinh θ
## Equations of Chord
1) The equation of a chord joining two points P( a sec θ₁ , b tan θ₁) and Q( a sec θ₂ , b tan θ₂) on the hyperbola
2) The equation of the chord of the hyperbola bisected at point (x₁, y₁) is given by,
T = S₁
3) Equations of the chord of contact of tangents drawn from a point (x₁, y₁) to the hyperbola is given by,
## Equation of Tangent to the Hyperbola
1) In point form the equation of the tangent to the hyperbola is,
2) In parametric form the equation of the tangent to the hyperbola is,
3) In Slope form the equations of the tangent to the hyperbola is,
y = mx ± √(a²m² — b²)
Here, m is the slope of the normal the tangent to the hyperbola.
4) The line y = mx + c touches the hyperbola if c² = a²m² — b².
## Equation of Normal to the Hyperbola
1) Point Form: In point form the equation of normal to the hyperbola is,
2) Parametric Form: The equation of normal to the hyperbola at a point P(a secθ, b tanθ) is,
ax cosθ + by cotθ = a² + b²
3) Slope Form: The equation of normal to the hyperbola is,
Here, m is the slope of the normal to the hyperbola being discussed.
4) A maximum of 4 normals can be drawn to a hyperbola from a point P(x₁, y₁)
## Diameter and Conjugate Diameter
1) Diameter: Diameter is the locus of mid-points of all the parallel chords of the hyperbola. The equation of the diameter, bisecting a system of parallel chords with each of slope m to the hyperbola is given by,
2) Conjugate Diameter: The diameters of a hyperbola are said to be conjugate if each diameter bisects the chords parallel to the other.
The diameters y = m₁x and y = m₂x are conjugate if,
m₁m₂ = b²/a²
3) In a pair of conjugate diameters of a hyperbola, only one of them meets the hyperbola in real points while the other meets the hyperbola at imaginary points.
### Director Circle
By Ag2gaeh – Own work, CC BY-SA 4.0, Link
The locus of points of intersection of the tangents to the hyperbola, which are perpendicular to each other.
The equation of Director Circle is,
x² + y² = a² — b²
## Rectangular Hyperbola
A hyperbola for which the asymptotes are perpendicular is known as a rectangular hyperbola.
In other words, it is a hyperbola with a = b.
Therefore, the eccentricity of a rectangular hyperbola is √2 .
The equation for a rectangular hyperbola can be reduced to,
xy = a²
### Properties
1) Asymptotes are the co-ordinate axes, x = 0 and y = 0.
2) e = √2.
3) Center is O(0, 0).
4) Foci are S(√2c,√2c) and S₁(–√2c, –√2c).
5) Vertices are A(c, c) and A₁(–c, –c).
6) Length of Latus Rectum = 2√2c.
7) In parametric form, the equation is x = ct and y = c/t.
### Summary
1) The point (x₁,y₁) lies outside, on, or inside the hyperbola,
2) The combined equation of pair of tangents drawn from an external point P(x₁,y₁)
is SS₁–T².
3) The equation of a chord of the hyperbola whose mid-point is (x₁,y₁) is given by
T = S₁.
4) Eccentricity of a Rectangular Hyperbola is √2 and the angle between asymptotes is 90°.
5) If a triangle is inscribed in a hyperbola, then it’s orthocenter lies on the hyperbola.
In this blog post, you learnt about some important concepts about Hyperbola. Hope you found it useful. For more amazing JEE study material, check out Magoosh JEE Product. Happy learning! | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://magoosh.com/jee/jee-conic-sections-hyperbola/", "fetch_time": 1620889432000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-21/segments/1620243991537.32/warc/CC-MAIN-20210513045934-20210513075934-00579.warc.gz", "warc_record_offset": 418597396, "warc_record_length": 35579, "token_count": 1857, "char_count": 6461, "score": 4.4375, "int_score": 4, "crawl": "CC-MAIN-2021-21", "snapshot_type": "latest", "language": "en", "language_score": 0.914155125617981, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00043-of-00064.parquet"} |
# How do you solve 1/(m^2-m)+1/m=5/(m^2-m) and check for extraneous solutions?
Apr 24, 2018
$m = 0$ is an extraneous solution and $m = 5$ is a solution.
#### Explanation:
$\frac{1}{{m}^{2} - m} + \frac{1}{m} = \frac{5}{{m}^{2} - m}$. Multiplying by $m \left({m}^{2} - m\right)$
on both sides we get, $m + \left({m}^{2} - m\right) = 5 m$ or
$\cancel{m} + {m}^{2} - \cancel{m} - 5 m = 0$ or
$m \left(m - 5\right) = 0 \therefore m = 0 , m = 5$
On check , $m \ne 0$ , since function is undefined at $m = 0$
So $m = 0$ is an extraneous solution and $m = 5$ is a solution. [Ans]
May 3, 2018
In a problem like this we can assume the denominators are not zero, so we never really run into extraneous solutions $m = 0$ or $m = 1$ and get right to
$m = 5$
#### Explanation:
Sometimes extraneous solutions are unavoidable, as often happens when we have to square both sides of an equation.
That's not the case in a problem like this. We can at the outset assume the denominators are not zero, i.e assume
${m}^{2} - m \ne 0$
$m \left(m - 1\right) \ne 0$
$m \ne 0$ and $m \ne 1$.
The $\frac{1}{m}$ tells us $m \ne 0$ which we already know.
Usually we must always factor, never cancel. But when we know factors are non-zero, we can cancel them.
$\frac{1}{{m}^{2} - m} + \frac{1}{m} = \frac{5}{{m}^{2} - m}$
$\frac{1 + 1 \left(m - 1\right)}{m \left(m - 1\right)} = \frac{5}{m \left(m - 1\right)}$
$1 + 1 \left(m - 1\right) = 5$
$m = 5$
Check:
$\frac{1}{{5}^{2} - 5} + \frac{1}{5} = \frac{1}{20} + \frac{4}{20} = \frac{5}{20} = \frac{1}{4}$
5/{5^2-5}=5/20 quad sqrt | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 26, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://socratic.org/questions/how-do-you-solve-1-m-2-m-1-m-5-m-2-m-and-check-for-extraneous-solutions", "fetch_time": 1709437483000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-10/segments/1707947476180.67/warc/CC-MAIN-20240303011622-20240303041622-00698.warc.gz", "warc_record_offset": 553832422, "warc_record_length": 6375, "token_count": 622, "char_count": 1576, "score": 4.59375, "int_score": 5, "crawl": "CC-MAIN-2024-10", "snapshot_type": "latest", "language": "en", "language_score": 0.8193400502204895, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00049-of-00064.parquet"} |
A tree diagram is a clever way of mapping out the probably of one even happening after another. The probably of every branch coming from the same point has to sum to 1. To work out the probability of one event happening after another, you multiply both their individual probabilities together.
Example:
:)
A bag contains 4 red balls, 6 black balls. A ball is taken out of the bag at random and then replaced. Then another ball is taken out.
Notice how the ball is taken out at random, and then replaced - this is very important when working with tree diagrams to find out the probability of an outcome (This will be further discussed in problems involving independent & dependent events)
To draw a tree diagram, we need to consider the possibility of what ball we will get when one is taken out. Since there are 4+6 = 10 balls altogether.
So the probability of choosing a red ball is 2/5, a black ball is 3/5. As the ball is replaced after it has been taken out, we still remain with 10 balls in the bag. so the probabilities of choosing each ball will be the same.
From the tree diagram below, we can see that each ball has its own 'branch' and above that branch is the probability of its outcome.
## Follow the links below to see how this topic has appeared in past exam papers
AQA Unit 1 November 2011 (H) - Page 10, Question 8
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# Fractions of a Set Worksheets
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1. With this Spring Fractions Bundle, kiddos will have a blast practicing skills like recognizing and creating EQUIVALENT FRACTIONS and working with FRACTIONS OF A SET. By purchasing this bundle, you will SAVE 20% OFF the individual products.Spring Fractions Bundle Contains the Following Products:1. Sp
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Product Description
Spring Fractions of a Set Fill-Ins are a fun and engaging way to practice working with fractions of a set, searching and counting objects, as well as following directions.
Spring Fractions of a Set Fill-Ins contains 12 spring-themed worksheets with a set of instructions at the top and a picture with mixed up spring objects.
The instructions ask students to color, circle, outline, mark, etc. a fraction of a specific object. In order to complete the instructions, kiddos will need to find the objects and determine the total number in the set.
Example
Color ½ of the gardening gloves green. Color ½ of the remaining gloves purple.
To follow the above instructions, kiddos would first need to search for the gardening gloves and count the total amount.
What Types of Fractions Are Focused On?
The first 9 spring-themed Fraction Fill-In worksheets focus on one of the following kinds of fractions…
1/4, 1/2, 3/4, 1/3, 2/3, fifths, sixths, eighths, and tenths
The last 3 worksheets are “mixed” with the above kind of fractions.
Each sheet’s focus is labeled on the upper right-hand corner.
Who Should Use Fraction Fill-Ins?
Fraction Fill-Ins are challenging and are meant for kiddos who have a strong understanding of fractions and what the numerator and denominator represent. They are perfect for advanced third graders as well as students in 4th and 5th grade who are mostly fluent with multiplication and division facts.
Differentiation
Because fractions of a set can be a challenging concept, I created two versions of each of the 12 spring-themed Fraction Fill-In worksheets. The original version has an asterisk in the upper right-hand corner, indicating it is “more challenging” because students must find objects and count their total. The second version of each sheet does not have an asterisk, indicating it is “easier” because the total of each object is identified in the instructions.
Example
Color ½ of the 12 gardening gloves green. Color ½ of the remaining gloves purple.
From the instructions, kiddos will see there are 12 gardening gloves. They will still need to find gloves to color in, but won’t have to worry about finding ALL of the objects in the set or counting them.
This Product Contains the Following...
• Teacher Information
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• 12 Spring-Themed Fraction Fill-In Worksheets (2 Versions of Each)
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Spring Equivalent Fractions Color by Numbers
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# Embarking on the Intricacies of Neural Networks and Machine Learning
2023-11-19 | 0
Today we will talk about neural networks. Let's say we want to create a robot that correctly determines which piece appears on the screen in the game Tetris. I specifically looked up the names of these pieces on the Internet, they have speaking names:
So, we have the shapes I, L, O, S, T, and when, for example, the shape L appears on the screen, the robot must correctly say that it is the shape L that has appeared. So what is the problem? The problem is that our shape can rotate and be positioned in different parts of the screen. So, in both of the following cases, the robot must give us L:
Using combinatorics and group theory (which you can study with Russian Math Tutors), you can calculate that even on a 5×5 screen you can depict the shape L in 3×4×8 = 96 ways. We are too lazy to explain to the computer (or robot) for each of these cases that we are talking about the same shape. We want the computer to learn to recognize images on its own. This is what we call machine learning.
Let's move on to consider how a computer learns to recognize shapes using neural networks. A neural network consists of layers of neurons connected to each other. The first layer is called the input layer, and the last layer is called the output layer. In addition, there are intermediate (or hidden) layers of neurons, the number of which varies. The network below has just one intermediate layer:
The outer layer of the depicted network contains 4 neurons, the intermediate layer contains 5 neurons, and the output layer contains only one neuron. The number of neurons in layers depends on what problem we are solving. Namely, the number of neurons in the input layer corresponds to the amount of information that we submit to the input. On the other hand, the number of neurons in the output layer corresponds to the amount of information that the computer must return. To explain this, let's look at the left of the two L shapes above. We can split the entire image into five stripes, and then connect them into one long strip of length 25:
That is, we are able to encode one of the possible images of the shape L with a binary string of length 25: 0010000100011000000000000. This tells us that the input layer of our neural network must contain 25 neurons. In this case, each neuron of the input layer can get two values, 1 when the corresponding pixel of our image is colored and 0 when the pixel is not colored.
What happens to the output layer? Since the output should return an answer to us, which shape was received at the input, it is enough to use one neuron, which can get five values: I, L, O, S, T or 1, 2, 3, 4, 5. For convenience, we will not use intermediate layers, so our neural network that guesses the names of Tetris pieces on a 5x5 screen will have the following form:
It’s not for nothing that we named the neurons of the input layer x1, ..., x25, and the single neuron of the output layer y. From a mathematical point of view, our neural network is a function f(x1, ..., x25). Try to guess the value of:
Note that the indicated variable values correspond to the line 0010000100011000000000000 written above, which corresponds to shape L, that is, the shape with number 2 in the list I, L, O, S, T. From here we know that:
Similarly, we can verify that:
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Now imagine that our function is described by the equation f(x1, ..., x25) = a1x1, ..., a25x25, with unknown coefficients a1, ..., a25 (of course in practice, we choose much more complicated equations).
The essence of machine learning, in this case, comes down to selecting the coefficients a1, ..., a25, through some known (test) values of our function (for example, those above) so that whenever we are given a shape (i.e., the corresponding variable values), the result of calculating this function equals the number of the shape. Thus, if we put a= ... = a25 = 1, then we get
But we know that:
So, making all coefficients equal to 1 was not the best idea. By using appropriate algorithms (such as backpropagation), we can get better coefficients and ensure that our function returns the correct names for the Tetris pieces. Similarly, you can teach the function to recognize any photos (animals, people, cars, traffic lights, etc.) or voices.
(A bit technical) addition. Above, we’ve mentioned the "backpropagation" algorithm. This is a really powerful tool to determine the coefficients of our equation (or weights of the links between consecutive layers of neurons). In general, this equation is non-linear. That is, the predictive function f depends on several variables in a non-trivial way. What we want to do is to minimize the function f-g, where function g returns the exact names of our shapes. To minimize f-g, we may apply the gradient descent method based on finding the partial derivatives of f-g (which are studied in multi-variable calculus). The latter task is too complicated (since f-g inherits non-linearity) and this is where we apply the backpropagation by moving backward through the layers and finding the derivatives numerically.
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Private Lessons | Group Lessons | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://russianmathtutors.com/embarking-on-the-intricacies-of-neural-networks-and-machine-learning", "fetch_time": 1721062853000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00042.warc.gz", "warc_record_offset": 458919591, "warc_record_length": 170357, "token_count": 1242, "char_count": 5515, "score": 3.703125, "int_score": 4, "crawl": "CC-MAIN-2024-30", "snapshot_type": "latest", "language": "en", "language_score": 0.9364253282546997, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00029-of-00064.parquet"} |
# If f'(x) = x |sinx| ∀ x ∈ (0, π) and f(0) = 1/√3 , then f(x) will be
Q: If f'(x) = x |sinx| ∀ x ∈ (0, π) and f(0) = 1/√3 , then f(x) will be
(A) – x cosx + sinx + 1/√3
(B) 1/√3 – x cosx + sinx
(C) sinx + x cosx – 2/√3
(D) sinx + x cosx + 2/√3
Sol. f'(x) = x |sinx| ∀ x ∈ (0, π)
$\large \int f'(x) dx = \int x sinx dx$
$\large f(x) = – x cosx + \int cosx dx + c$
$\large = – x cosx + sinx + c$
f(0) = c = 1/√3
$\large f(x) = – x cosx + sinx + \frac{1}{\sqrt{3}}$
Hence (A) is the correct answer. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.quantumstudy.com/if-fx-x-sinx-%E2%88%80-x-%E2%88%88-0-%CF%80-and-f0-1-%E2%88%9A3-then-fx-will-be/", "fetch_time": 1674886310000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-06/segments/1674764499524.28/warc/CC-MAIN-20230128054815-20230128084815-00681.warc.gz", "warc_record_offset": 952225727, "warc_record_length": 11839, "token_count": 261, "char_count": 511, "score": 3.921875, "int_score": 4, "crawl": "CC-MAIN-2023-06", "snapshot_type": "latest", "language": "en", "language_score": 0.23741993308067322, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00022-of-00064.parquet"} |
Hash functions and pathological data sets
So I'm watching an Algorithms course in Coursera, and we are currently discussing hash tables. He's talking about the importance of a good hash function, and about how an ideal hash function would be a "super clever hash function guaranteed to spread every data set evenly".
Then, he explains that the problem is that such a hash function does not exist (and that for every hash function there is a pathological data set), and that the reason for this is as follows:
Fix a hash function $h: U \to \{0, 1, 2, ..., n-1\}$. By the Pigeonhole Principle, there exists a bucket $i$ such that at least $|U|/n$ elements of $U$ hash to $i$ under $h$. If a data set draws only from these, everything collides.
The bolded part is what's confusing me. Why does there exist a bucket $i$ such that at least $|U|/n$ elements of $U$ hash to $i$ under $h$? I can't really visualize what he means.
• Because the pigeonhole principle says exactly that. Did you look it up? – David Richerby Nov 4 '16 at 11:49
• It's easier to understand a more concrete example: You have, say, 5 buckets, and you need to stick, say, 7 pigeons in them somehow. 7 > 5, so it follows that at least 1 bucket has at least 2 pigeons in it. – j_random_hacker Nov 4 '16 at 15:44
• @DavidRicherby well, the Pigeonhole Principle as I understood it is what the first sentence in the wikipedia link says: "In mathematics, the pigeonhole principle states that if n items are put into m containers, with n > m, then at least one container must contain more than one item." I didn't immediately see the equivalence between that and the bolded statement that confused me. Weirdly enough I was able to get useful answers out of asking this question :) – FrostyStraw Nov 4 '16 at 18:43
• OK but if you read as far as the third page of the wikipedia article, it tells you that, if you put more than $km$ items in $m$ buckets, at least one must contain more than $k$ items, which is exactly what's being used here. – David Richerby Nov 5 '16 at 0:08
• The answers posted here were more clear and therefore more helpful to me. – FrostyStraw Nov 5 '16 at 2:21
An easy way to visualize this is to imagine a hash table of size $n$ (implemented with chaining) that contains all of the elements of $U$ (even though this is unrealistic in practice because $U$ typically has massive size). Since $|U| >> n$, all of the elements of $U$ do not fit into the hash table; therefore, there will be collisions. Consider, for example, the universal set $U=\{a,b,c,d,e,f,g\}$ and a hash table with $n=3$ buckets. Since $|U|=7$, at least one bucket must necessarily contain $\lceil \: |U| \: / \: n \rceil = \lceil 7/3 \rceil = 3$ or more elements. In the case of the most clever hash function (which would spread out the elements of $U$ as evenly as possible), this bucket would contain exactly $3$ elements, like this (highlighted in red):
It is important to see that no matter how clever the hash function is, there will always exist a data set (for example, the set $\{b,g,a\}$) whose elements hash to the same bucket (for example, bucket number $1$). Such a pathological data set will make your hash table degenerate to its worst-case linear-time performance.
Assume there is no such bucket. Then each bucket has at most $|U|/n - 1$ items. There are $n$ buckets, so the total number of items is at most $n*(|U|/n - 1) = |U| - n$. This is less than $|U|$, which is the number of items we distributed to the buckets in the first place. This is a contradiction, so we proved that the statement "each bucket has at most $|U|/n - 1$ items" is false, which is equivalent to the statement "some bucket has at least $|U|/n$ items." | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://cs.stackexchange.com/questions/65534/hash-functions-and-pathological-data-sets/65540", "fetch_time": 1579740260000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-05/segments/1579250607596.34/warc/CC-MAIN-20200122221541-20200123010541-00359.warc.gz", "warc_record_offset": 402202425, "warc_record_length": 32107, "token_count": 983, "char_count": 3702, "score": 3.546875, "int_score": 4, "crawl": "CC-MAIN-2020-05", "snapshot_type": "latest", "language": "en", "language_score": 0.9458171725273132, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00035-of-00064.parquet"} |
# What is Conservation of Energy in Thermodynamics – The First Law of Thermodynamics – Definition
A fundamental aspect of the energy concept is that energy is conserved. This principle is known as the first law of thermodynamics. Conservation of Energy in Thermodynamics
## Conservation of Energy in Thermodynamics – The First Law of Thermodynamics
In thermodynamics the concept of energy is broadened to account for other observed changes, and the principle of conservation of energy is extended to include a wide variety of ways in which systems interact with their surroundings. The only ways the energy of a closed system can be changed are through transfer of energy by work or by heat. Further, based on the experiments of Joule and others, a fundamental aspect of the energy concept is that energy is conserved. This principle is known as the first law of thermodynamics. The first law of thermodynamics can be written in various forms:
In words:
Equation form:
∆Eint = Q – W
where Eint represents the internal energy of the material, which depends only on the material’s state (temperature, pressure, and volume). Q is the net heat added to the system and W is the net work done by the system. We must be careful and consistent in following the sign conventions for Q and W. Because W in the equation is the work done by the system, then if work is done on the system, W will be negative and Eint will increase.
Similarly, Q is positive for heat added to the system, so if heat leaves the system, Q is negative. This tells us the following: The internal energy of a system tends to increase if heat is absorbed by the system or if positive work is done on the system. Conversely, the internal energy tends to decrease if heat is lost by the system or if negative work is done on the system. It must be added Q and W are path dependent, while Eint is path independent.
Differential form:
dEint = dQ – dW
The internal energy Eint of a system tends to increase if energy is added as heat Q and tends to decrease if energy is lost as work W done by the system.
Open System – Closed System – Isolated System
Heat and/or work can be directed into or out of the control volume. But, for convenience and as a standard convention, the net energy exchange is presented here with the net heat exchange assumed to be into the system and the net work assumed to be out of the system. If no mass crosses the boundary, but work and/or heat do, then the system is referred to as a “closed” system. If mass, work and heat do not cross the boundary (that is, the only energy exchanges taking place are within the system), then the system is referred to as an isolated system. Isolated and closed systems are nothing more than specialized cases of the open system.
References:
Nuclear and Reactor Physics:
1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
6. Kenneth S. Krane. Introductory Nuclear Physics, 3rd Edition, Wiley, 1987, ISBN: 978-0471805533
7. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
8. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
9. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.periodic-table.org/what-is-conservation-of-energy-in-thermodynamics-the-first-law-of-thermodynamics-definition/", "fetch_time": 1726206995000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700651507.67/warc/CC-MAIN-20240913034233-20240913064233-00595.warc.gz", "warc_record_offset": 848627166, "warc_record_length": 31195, "token_count": 917, "char_count": 3735, "score": 3.609375, "int_score": 4, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.9396986961364746, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00025-of-00064.parquet"} |
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# Multiplying and Dividing Decimal Numbers by Powers of 10
In this post, we will learn how to multiply and divide decimal numbers by 10, 100, 1000 … It is as simple as moving the decimal point of the decimal number to the left or right.
### Multiplying decimal numbers by 10, 100, 1000 …
To multiply a decimal number by 10, 100, 1000 … all you have to do is move the decimal point to the right of the decimal as many positions as the number of zeros.
For example:
Since the 100 has two zeros, we will move the decimal point two places to the right. Therefore, the result is 315.4
### Dividing decimal numbers by 10, 100, 1000 …
To divide a decimal number by 10, 100, 1000 … all you have to do is move the decimal point to the left as many positions as the number of zeros.
For example:
Since 10 has one zero we will move the decimal point one position to the left. Therefore, the result is 8.42
Check out these posts if you want more practice with multiplying and dividing decimal numbers: | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.smartickmethod.com/blog/math/number-operations-in-base-ten/decimals/multiplying-dividing-decimal-numbers/", "fetch_time": 1601037297000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-40/segments/1600400226381.66/warc/CC-MAIN-20200925115553-20200925145553-00709.warc.gz", "warc_record_offset": 1009791122, "warc_record_length": 14018, "token_count": 272, "char_count": 1090, "score": 3.734375, "int_score": 4, "crawl": "CC-MAIN-2020-40", "snapshot_type": "latest", "language": "en", "language_score": 0.8807588815689087, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00039-of-00064.parquet"} |
# What is MSME Full Form, Meaning, And Definition
## The Full Form of MSME: Mean Square Model Error
MSME stands for Mean Square Model Error. “You can determine how closely a regression line resembles a set of points using the mean squared error (MSE). This is accomplished by squaring the distances between the points and the regression line (also known as the “errors”).
The squaring is required to eliminate any unfavorable indications. Additionally, it emphasizes bigger discrepancies. Since you’re averaging a collection of errors, this error type is known as the mean squared error. “The forecast becomes more accurate as the MSE decreases.”
### What is the purpose of Mean Square Model Error?
The purpose of Mean Square Model Error (MSE) is to measure the amount of error in a model. MSE is calculated by taking the sum of the squared error values for all observations and dividing by the number of observations. This measure is used to determine the accuracy of a model and to identify which variables are most important in predicting the outcome.
### Similar meaning of MSME
There are some of the most commonly used acronyms, abbreviations, full forms, and the meaning of MSME are listed in different categories below the table.
Table
#### What is the full form of MSME in mathematics?
The Full Form of MSME is “Mean Square Model Error ” in the context of mathematics.
#### What is the meaning of MSME in education?
The meaning of MSME is “Mean Square Model Error “.
#### What does MSME mean in Business?
MSME means “Ministry of Micro, small & Medium Enterprises.” In the terms of Business.
#### what is MSME in entrepreneurship terms
The MSME represents “Ministry of Micro, small & Medium Enterprises ” in the terms of medical.
#### What is the MSME Meaning in engineering
The MSME means “Master’s of Science in Mechanical engineering “. It is the most commonly used acronym in engineering.
#### What is the MSME Full Form in engineering
MSME is a term that stands for “Materials Science and Mineral Engineering”.
Now you might have got some ideas about the Meaning and full form of MSME. And also all popular full forms, acronyms, abbreviations, and their meanings and definitions. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://onlinefullform.com/msme/", "fetch_time": 1702061108000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-50/segments/1700679100769.54/warc/CC-MAIN-20231208180539-20231208210539-00123.warc.gz", "warc_record_offset": 491851506, "warc_record_length": 16400, "token_count": 465, "char_count": 2212, "score": 3.734375, "int_score": 4, "crawl": "CC-MAIN-2023-50", "snapshot_type": "latest", "language": "en", "language_score": 0.9296442270278931, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00041-of-00064.parquet"} |
# Angles and Parallel Lines - PowerPoint PPT Presentation
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Title: Angles and Parallel Lines
1
Angles and Parallel Lines
• Geometry D Section 3.2
2
Angles and Parallel Lines
We are going to investigate the relationship of
various angles created by two parallel lines and
a transversal.
Obtain a ½ sheet of graph paper and a protractor.
Construct two lines and a transversal similar
to the image on the next slide.
3
Angles and Parallel Lines
Extend your lines the full height and width of
the paper.
Pause for time to work!
4
Angles and Parallel Lines
Label the angles as shown below.
1
2
3
4
6
5
7
8
Pause for time to work!
5
Angles and Parallel Lines
Measure all angles using a protractor to the
nearest degree.
1
2
3
4
6
5
7
8
Pause for time to work!
6
Angles and Parallel Lines
Measure all angles using a protractor to the
nearest degree.
127o
53o
1
2
Note Your measurements may be different values
but should be in the same pattern.
3
4
53o
127o
127o
53o
6
5
7
8
53o
127o
7
Angles and Parallel Lines
Identify the relationship between the following
angles?
127o
53o
1
2
3
4
53o
127o
127o
53o
6
5
7
8
53o
From Chapter 2, the angles are linear pairs.
127o
What can be said about the measures of the linear
pairs?
Linear pairs are supplementary (sum to 180o).
8
Angles and Parallel Lines
Identify the relationship between the following
angles?
127o
53o
1
2
3
4
53o
127o
127o
53o
6
5
7
8
53o
From Chapter 2, the angles are vertical angles.
127o
What can be said about the measures of the
vertical angles?
Vertical angles are congruent angles.
9
Angles and Parallel Lines
Identify the relationship between the following
angles?
127o
53o
1
2
3
4
53o
127o
127o
53o
6
5
7
8
53o
The angles are corresponding angles.
127o
What can be said about the measures of the
corresponding angles?
The measures are equal and the angles are
congruent.
10
Angles and Parallel Lines
Corresponding Angles Postulate
If two parallel lines are cut by a transversal,
then each pair of corresponding angles are
congruent.
11
Angles and Parallel Lines
Identify the relationship between the following
angles?
127o
53o
1
2
3
4
53o
127o
127o
53o
6
5
7
8
53o
The angles are alternate interior angles.
127o
What can be said about the measures of the
alternate interior angles?
The measures are equal and the angles are
congruent.
12
Angles and Parallel Lines
Alternate Interior Angles Theorem
If two parallel lines are cut by a transversal,
then each pair of alternate interior angles are
congruent.
You will prove this theorem as a homework problem!
13
Angles and Parallel Lines
Identify the relationship between the following
angles?
127o
53o
1
2
3
4
53o
127o
127o
53o
6
5
7
8
53o
The angles are alternate interior angles.
127o
What can be said about the measures of the
alternate interior angles?
14
Angles and Parallel Lines
Consecutive Interior Angles Theorem
If two parallel lines are cut by a transversal,
then each pair of consecutive interior angles is
supplementary (sum to 180o).
You will prove this theorem as a homework problem!
15
Angles and Parallel Lines
Identify the relationship between the following
angles?
127o
53o
1
2
3
4
53o
127o
127o
53o
6
5
7
8
53o
The angles are alternate exterior angles.
127o
What can be said about the measures of the
alternate interior angles?
The measures are equal and the angles are
congruent.
16
Angles and Parallel Lines
Alternate Exterior Angles Theorem
If two parallel lines are cut by a transversal,
then each pair of alternate exterior angles is
congruent.
Prove
Statement Reason
?
p q, t is a transversal of p q
Given
t
Corresponding s are
?
1
2
p
3
4
?
6
5
Vertical s are
q
7
8
?
Transitive Property
17
Angles and Parallel Lines
Perpendicular Transversal Theorem
In a plane, if a line is perpendicular to one of
two perpendicular lines, then it is perpendicular
to the other.
t
p
q
If t is perpendicular ( ) to p, then it is
also perpendicular to q.
You will prove this theorem as a homework problem!
18
Angles and Parallel Lines
Applications
Gather into groups of not more than 3. Work the
following problems in your group. Compare your
19
Angles and Parallel Lines
Given j k,
Applications
Make a sketch of the problem in your notes.
Find the measure of
4
5
3
1.
43o
2
1
7
8
6
Corresponds with 1.
9
11
10
2.
24o
12
Alternate exterior with 14.
13
3.
156o
Linear pair with 9. 180o 24o 156o
14
20
Angles and Parallel Lines
Given j k, Find the measure of
Applications
4
5
3
4.
137o
2
1
7
8
6
Linear pair with 3.
9
11
10
5.
156o
12
Vertical angle with 10.
13
6.
43o
Vertical with 1. Alternate Interior of
3.
14
21
Angles and Parallel Lines
Applications
Find the values of x and y in each figure. Find
the measure of each given angle. Note Figures
are not drawn to scale.
Given
Pause for time to work!
22
Angles and Parallel Lines
Applications Solution
Given
Linear pairs are supplementary.
(5x 2) (9x 10) 180o
14x 12 180
14x 168
x 12
Linear Pair
By corresponding angles,
62o
118o
3y 1 62
62o
3y 63
y 21 and
23
Angles and Parallel Lines
Applications
Find the values of x, y and z in each figure.
(2z)o
(3x3)o
(4y2)o
66o
Pause for time to work!
24
Angles and Parallel Lines
is a corresponding angle with the
angle of 66o.
Applications
(3x 3)o and 66o are linear pairs and sum to
180o.
(3x 3)o 66o 180o 3x 63 180 3x 117,
x 39
(2z)o
(3x3)o
66o
(4y 2)o and 66o are congruent alternate
interior angles.
(4y 2)o 66o 4y 64, y 16
(4y2)o
66o
(3x3)o and (2z)o are congruent alternate
interior angles.
(3x3)o 3(39) 3 114o
(2z)o 114o, z 57
25
Angles and Parallel Lines
Applications
Find the values of x, y and z in each figure.
There are other ways of doing this problem
correctly. If you worked it a different way,
would you be willing to share how you did it?
(2z)o
(3x3)o
(4y2)o
66o
26
Angles and Parallel Lines
Applications
Find the measures of all the angles on the object
if the measure of angle 1 is 30o.
Pause for time to work!
27
Angles and Parallel Lines
Applications
Find the measures of all the angles on the object
if the measure of angle 1 is 30o.
Perpendicular transversal theorem.
Perpendicular lines intersect in 4 right (90o)
angles.
90o
90o
90o
90o
90o
28
Angles and Parallel Lines
Applications
Find the measures of all the angles on the object
if the measure of angle 1 is 30o.
Vertical Angle
Vertical Angle
Alternate interior angle with angle 1.
30o
150o
150o
30o
90o
30o
Linear pairs are supplementary.
90o
30o
90o
Given
90o
90o
Vertical Angles
29
Angles and Parallel Lines
Applications
Find the measures of all the angles on the object
if the measure of angle 1 is 30o.
30o
Vertical angles.
150o
150o
30o
30o
90o
60o
60o
All angles have been found!
90o
30o
90o
90o
Since the transversal is , these two angles | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.powershow.com/view2b/428c4e-YjdiY/Angles_and_Parallel_Lines_powerpoint_ppt_presentation", "fetch_time": 1591387694000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-24/segments/1590348502204.93/warc/CC-MAIN-20200605174158-20200605204158-00594.warc.gz", "warc_record_offset": 841913789, "warc_record_length": 19568, "token_count": 2197, "char_count": 7198, "score": 4.5, "int_score": 4, "crawl": "CC-MAIN-2020-24", "snapshot_type": "latest", "language": "en", "language_score": 0.8056827187538147, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00026-of-00064.parquet"} |
# Number of Solutions Using Generating Functions
I am struggling with the following problem and I was hoping somebody could help me.
Solve by applying the convolution formula for generating functions: How many nonnegative integer solutions are there for $x_1 + 3x_2 = 100$?
I've been getting really high numbers in my attempts so far and there's no way that they are correct. If someone can help me get started that would be appreciated.
Thanks!
EDIT: Here is my work so far which is giving me the answer 176,851 solutions which I'm sure is wrong.
$G(x)=(1+x+x^2+x^3+...)(1+x^3+x^6+x^9+...)$ --> $G(x)=(1/(1-x))(1/(1-x^3))$ $a_n=1$ and $b_n=C((n+2), 2)$
Then I'm finding the coefficient of $x^{100}$ from the convolution formula.
• You start by finding the generating function. Did you get that far? Where are you stuck? – Trevor Gunn Oct 21 '17 at 1:50
• Yes, I think I have the generating function figured out. I'm using $G(x) = (1+x^2+x^3+x^4...)(1+x^3+x^6+x^9+...)$. – jallen3095 Oct 21 '17 at 3:40
• By inspection, there are $34$ solutions. Since $x_1, x_2$ must be nonnegative, $x_1 = 100 - 3x_2 \implies x_2 \in \{0, 1, 2, 3, \ldots, 33\}$. – N. F. Taussig Oct 21 '17 at 19:49
• That's the same number of solutions I was thinking also. For some reason I cannot get my convolution to produce the same value. – jallen3095 Oct 21 '17 at 20:41
• Well, for one thing, your formula for $b_n$ is incorrect. Can you see why? – Qudit Oct 22 '17 at 19:13 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/2482232/number-of-solutions-using-generating-functions", "fetch_time": 1561567682000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-26/segments/1560628000367.74/warc/CC-MAIN-20190626154459-20190626180459-00129.warc.gz", "warc_record_offset": 533859757, "warc_record_length": 34748, "token_count": 473, "char_count": 1459, "score": 3.734375, "int_score": 4, "crawl": "CC-MAIN-2019-26", "snapshot_type": "latest", "language": "en", "language_score": 0.9076058864593506, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00032-of-00064.parquet"} |
# math
posted by .
how do you work out the problem 3/4ft.= how many inches.
• math -
1 foot = 12 inches
(3/4) * 12 = 36/4 = 9 inches
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# Proof of Kurtosis for a sum of independent random Variables
I am trying to understand a proof for the Kurtosis of a sum of independent random variables, however, there is one part where I am quite stuck:
Theorem:
for $X_1, X_2, ..., X_n$ independent random variables with means $\mu_1, ... , \mu_n$ and variance $\sigma_1^2, ... , \sigma_n^2$ $E(X_i^4) < \infty$
Define $S_n = X_1 + ... + X_n$ and $S_n$ will be appropriately normal
$kurt(S_n) - 3 = (\sum_{i=1}^n\sigma_i^n)^{-2}\sum_{i=1}^n\sigma_i^4(kurt(X_i)-3)$
Proof (first part):
assume WLOG that $E[X_i] = 0$ for all $i$
$kurt(Sn) = \frac{E[(X_1 + ... + X_n )^4]}{(\sigma_1^2 + ... +\sigma_n^2)^2 }$
$E[(X_1 + ... + X_n )^4] = \sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^n\sum_{l=1}^n E(X_iX_jX_kX_l) = \sum_{i=1}^n E(X_i^4) +6 \sum_{i<j}^n\sigma_i^2\sigma_j^2$
NOTE: $E(X_iX_jX_kX_l) = 0$ unless $i=j=k=l$ or if combinations of two pairs.
Question: why is the NOTE: true? From my understanding the expected value of INDEPENDENT random variables is equal to the product of the expected values of the random variables. However I intuitively understand this case does not apply, or else, we would not get very far! So what is going on?
Assuming this hickup true, i understand the rest of the proof, which I will not write. (as i am improvising the syntax)
Consider for example the case of $i=j=k=l$ where you are then taking the expectation of $X_i^4$. Since this real number raised to an even power is never negative, and is sometimes positive, its expectation cannot be zero.
On the other hand, consider $X_i^3 X_j$ with $i\neq j$. Now just because $E(X_i) = 0$ that does not allow you to say that $E<X_i^3>$ = 0; consider a discrete random that is $-1$ with probability $\frac23$ and $1$ with probability $\frac13$. But by the multiplication of independent expectations, you get zero since the expectation of $X_j$ is zero.
Only if all the independent variates are raised to powers greater than $1$ can the expectation be non-zero. For four variables, the only such cases are four of a kind and two pairs.
• Alright I get it! Discrete example helped! – rannoudanames Sep 28 '16 at 23:44
From my understanding the expected value of [the product of] INDEPENDENT random variables is equal to the product of the expected values of the random variables.
This is true. Moreover, you assumed "(..) WLOG that E[Xi]=0 for all i".
• Yes, that is why I am confused about the NOTE:, I would have expected E(X_i * X_j * X_k * X_l) to be 0 under all circumstances, however that is not the case. – rannoudanames Sep 28 '16 at 23:08
• I see. The case i=j=k=l gives the first term, while (i=j, k=l, i!=k) gives the second term. – LinAlg Sep 28 '16 at 23:10
• Yes, I see that, but why would those expected values, not be equal zero, due the independence of the random variables, and the assumption that E(X_i) = 0 – rannoudanames Sep 28 '16 at 23:14
• For i=j=k=l, you end up with $E(X_i^4)$. For i=j, k=l, i!=k you have $E(X_i X_i X_k X_k) = E(X_iX_i)E(X_kX_k)$ (due to independence), which in turn equals $\sigma_i^2 \sigma_k^2$. – LinAlg Sep 28 '16 at 23:22
Suppose some index appears exactly once in the product $X_i X_j X_k X_l$, say, $i$ is different from all of $j,k,l$. Then $E(X_i X_j X_k X_l) = E(X_i) E(X_j X_k X_l) = 0$, since $E(X_i) = 0.$
There are only two ways that no index appears exactly once: if they are all the same, or if you have two indices that each appear twice. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/1945722/proof-of-kurtosis-for-a-sum-of-independent-random-variables", "fetch_time": 1558629899000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-22/segments/1558232257316.10/warc/CC-MAIN-20190523164007-20190523190007-00132.warc.gz", "warc_record_offset": 549586488, "warc_record_length": 34123, "token_count": 1101, "char_count": 3443, "score": 3.8125, "int_score": 4, "crawl": "CC-MAIN-2019-22", "snapshot_type": "latest", "language": "en", "language_score": 0.8506370782852173, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00033-of-00064.parquet"} |
# Course Title: Math A Elementary Algebra
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## Transcription
2 Unit 3: Chapter 2B Problem Solving In this unit you will apply your knowledge of solving equations to geometry application problems. You also will learn how to solve linear inequalities in one variable. solve application problems involving perimeter of a polygon solve application problems involving area of a polygon solve application problems involving circles solve application problems involving volume of common three-dimensional figures solve application problems involving angles of a triangle solve application problems involving supplementary/complimentary angles Unit 4: Chapter 3A Linear Equations in Two variables In this unit you will learn how to graph linear equations in two variables using various methods, how to find a slope of a line, and how to find equation of a line given certain information about it. graph linear equations in two variables by plotting points graph linear equations in two variables using intercepts find the slope of a line graph linear equations in two variables using the slope-intercept form of linear equations use linear equations to model data and solve application problems Unit 5: Chapter 3B and 4A Linear Inequalities in Two Variables and Systems of Linear Equations In this unit you will extend your knowledge of graphing linear equations in two variables to graphing linear inequalities in two variables. You also will learn how to solve systems of two linear equations in two variables by graphing as well as by the substitution method and addition method. graph solutions to linear inequalities in two variables solve systems of two linear equations in two variables by graphing solve systems of two linear equations in two variables by substitution method systems of two linear equations in two variables by addition method Unit 6: Chapter 4B Systems of Equations and Problem Solving In this unit you will apply your skills of solving systems of two linear equations in two variables to application problems. In addition, you will extend your knowledge of
3 inequalities in two variables obtained in the previous unit and learn to solve systems of linear inequalities in two variables. develop a strategy for solving application problems involving systems of two equations in two variables solve geometry-based problems solve simple interest problems solve mixture problems Unit 7: Chapter 5A Exponents In this unit you will review the definition of a polynomial and some vocabulary related to this topic. You will learn how to add, subtract and multiply polynomials and how to graph equations defined by polynomials. You will learn various properties of exponents. You also will learn how to operate with polynomials in several variables. define a polynomial and use appropriate terminology associated with polynomials add and subtract polynomials graph equations defined by polynomials by point plotting simplify expressions using the following properties of exponents: Product Rule, Power Rule, Product-to-Power Rule multiply polynomials with various number of terms use the FOIL method of multiplying polynomials multiply special products: the sum and difference of two terms, squares of a binomial know the terminology associated with polynomials in several variables add, subtract and multiply polynomials in several variables Unit 8: Chapter 5B Polynomials In this unit you will learn how to divide polynomials by using various properties of exponents and by long division. You will learn how to express numbers in scientific notation and how to perform computations with scientific numbers. simplify expressions using the following properties of exponents: Quotient Rule, Zero-Exponent Rule, Quotient-to-Power Rule, Negative Exponent Rule divide monomials divide a polynomial by a monomial divide polynomials using long division convert numbers to the scientific notation and vice versa perform computations using scientific numbers
4 Unit 9: Chapter 6A Intro to Factoring In this unit you will obtain basic factoring skills to factor out the greatest common factor, factor by grouping and factor trinomials with leading coefficient of 1. You will review the concepts you have leaned by now and take the mid-term. factor out the greatest common factor factor using the grouping method factor trinomials with leading coefficient of 1 take the midterm Unit 10: Chapter 6B Factoring Polynomials In this unit you will continue learning new techniques of factoring trinomials and special products and will develop a strategy for determining the best factoring method for individual problems. You will apply your factoring skills to solving quadratic equations and application problems. factor trinomials with leading coefficient other than 1 using the trial-and-error method factor trinomials with leading coefficient other than 1 using the AC-method factor perfect square trinomials factor the Difference of two Squares factor the Sum and Difference of Two Cubes determine the best factoring method for individual problems solve quadratic equations by factoring and applying the zero-product principle solve application problems based on quadratic model Unit 11: Chapter 7A Rational Expressions In this unit you will learn how to simplify, multiply, divide, add, and subtract rational expressions. determine values for which rational expression is undefined simplify rational expressions multiply and divide rational expressions find the least common denominator add and subtract rational expressions Unit 12: Chapter 7B Rational Equations and Applications In this unit you will learn how to solve rational equations and apply this skill to application problems. solve rational equations solve motion problems
6 After the successful completion of this unit, you will review for and take the final exam.
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PERT Mathematics Test Review Prof. Miguel A. Montañez ESL/Math Seminar Math Test? NO!!!!!!! I am not good at Math! I cannot graduate because of Math! I hate Math! Helpful Sites Math Dept Web Site Wolfson | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://docplayer.net/30324254-Course-title-math-a-elementary-algebra.html", "fetch_time": 1542272585000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-47/segments/1542039742569.45/warc/CC-MAIN-20181115075207-20181115101207-00428.warc.gz", "warc_record_offset": 96778018, "warc_record_length": 25030, "token_count": 7566, "char_count": 32832, "score": 3.9375, "int_score": 4, "crawl": "CC-MAIN-2018-47", "snapshot_type": "longest", "language": "en", "language_score": 0.9260469079017639, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00007-of-00064.parquet"} |
# Coordinates of antisymmetric matrix
• Physicsissuef
In summary, the problem asks for finding the coordinates of the matrix A in terms of the given basis matrices E1, E2, and E3. This involves writing A as a linear combination of E1, E2, and E3, and then solving for the coefficients. This can be done by equating corresponding components of A and the linear combination of E1, E2, and E3, which leads to a system of 3 equations. The solution for the coefficients is (1,-3,0). However, it is possible that the correct solution is (2,-2,1) and further checking may be needed.
## Homework Statement
Let's say that V is the vector space of all antisymmetric 3x3 matrices. Find the coordinates of the matrix $$A=\begin{bmatrix} 0 & 1 & -2\\ -1 & 0 & -3\\ 2 & 3 & 0 \end{bmatrix}$$ in ratio with the base:
$$E_1=\begin{bmatrix} 0 & 1 & 1\\ -1 & 0 & 0\\ -1 & 0 & 0 \end{bmatrix}$$
$$E_2=\begin{bmatrix} 0 & 0 & 1\\ 0 & 0 & 1\\ -1 & -1 & 0 \end{bmatrix}$$
$$E_3=\begin{bmatrix} 0 & -1 & 0\\ 1 & 0 & -1\\ 0 & 1 & 0 \end{bmatrix}$$
## Homework Equations
antisymetric matrix is only if [itex]A^t=-A[/tex]
## The Attempt at a Solution
The matrix is equal to:
$$f: \mathbb{R}^3 \rightarrow \mathbb{R}^3 , f(x_1,x_2,x_3)=(x_2-2x_3,-x_1-3x_3,2x_1+3x_2)$$
The base is $$B={(x_2+x_3,-x_1,-x_1) ; (x_3,x_3,-x_1-x_2) ; (-x_2,x_1-x_3,x_2)}$$
What should I do now?
Do you understand what the problem is asking? It's exactly like many other problems you have done in the past: write the given "vector" (the matrix A) as a linear combination of the given basis "vectors" (the matrices E1, E2, E3). That is, find numbers a1, a2, a3 so that A= a1E1+ a2E2+ a3E3. Those numbers are the "coordinates".
Setting corresponding components on both sides equal will give you 9 equations, of course, but they should reduce to 3 independent equations for a1, a2, and a3. For example, the upper left component (A11) of every matrix is 0 so that just becomes 0= 0a1+ 0a2+ 0a3 which is satisfied by all numbers.
I understand. Thanks for the help, I wasn't sure what the problem was asking for...
I found $$(a_1,a_2,a_3)=(1,-3,0)$$ and in my book get (2,-2,1). Probably is my mistake, I will check again.
## 1. What is an antisymmetric matrix?
An antisymmetric matrix is a square matrix where the values on the main diagonal are all zeros, and the values below the main diagonal are the negative of the values above the main diagonal. This means that the matrix is equal to the negative of its own transpose.
## 2. How can I identify an antisymmetric matrix?
An antisymmetric matrix can be identified by checking if it is equal to the negative of its own transpose. Additionally, it will have all zeros on the main diagonal and the values below the main diagonal will be the negative of the values above the main diagonal.
## 3. What are the properties of an antisymmetric matrix?
An antisymmetric matrix has the property that its transpose is equal to the negative of the original matrix. It also has the property that the sum of an antisymmetric matrix and its transpose is always a symmetric matrix.
## 4. How are antisymmetric matrices used in science?
Antisymmetric matrices are used in various scientific fields, such as physics, engineering, and computer science. They are particularly useful in representing physical quantities that have both magnitude and direction, such as force and velocity.
## 5. Can an antisymmetric matrix have complex numbers?
Yes, an antisymmetric matrix can have complex numbers as its entries. However, the imaginary parts of the complex numbers must be equal in order for the matrix to remain antisymmetric. This means that the complex numbers must be either purely real or purely imaginary. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/coordinates-of-antisymmetric-matrix.237539/", "fetch_time": 1709457690000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-10/segments/1707947476211.69/warc/CC-MAIN-20240303075134-20240303105134-00649.warc.gz", "warc_record_offset": 936664872, "warc_record_length": 15940, "token_count": 1054, "char_count": 3716, "score": 4.40625, "int_score": 4, "crawl": "CC-MAIN-2024-10", "snapshot_type": "latest", "language": "en", "language_score": 0.8428924679756165, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00006-of-00064.parquet"} |
Electric Forces Electric Fields
# Electric Forces Electric Fields
## Electric Forces Electric Fields
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Electric ForcesElectric Fields
2. In this section: • 1) Coulomb’s Law
3. Coulomb’s Law
4. Coulomb’s Law Recall that a force is defined as the cause of a change in motion. When two charged objects near one another may experience motion either toward or away from each other, each object exerts a force on the other object. This is called the electric force. force is a vector quantity; has magnitude and direction The fundamental rules of charge interaction applies Opposites attract, like repels like.
5. Coulomb’s Law Charles Coulomb, 1780’s Conducted a variety of experiments to determine the magnitude of the electric force between two charged objects. Coulomb found that the electric force between two charged objects is proportional to the product of the two charges.
6. Coulomb’s Law A torsion balance consists of two small spheres fixed to the ends of a light horizontal rod. The rod is made of an insulating material and is suspended by a silk thread. Coulomb’s torsion balance was used to establish the inverse square law for the electric force between two charges. www.magnet.fsu.edu
7. Coulomb’s Law • Charges are identified as either q+ or q− • Direction of the force depends on whether the charges have the same sign, or opposite signs. ffden-2.phys.uaf.edu
8. Coulomb’s Law • Coulomb discovered • Electric force is proportional to the product of the 2 charges. Hence, if one charge is doubled, the force doubles. If both charges are doubled, the force quadruples. • Electric force is inversely proportional to the square of the distance between the charges. When the distance is halved, the force increases by a factor of 4.
9. Coulomb’s Law where The symbol kC, called the Coulomb constant, and has units of measurement N•m2/C2 because this gives N as the unit of electric force.
10. Coulomb’s Law where This equation satisfies Newton’s 3rd Law because it implies that exactly the same force acts on both charges.
11. Coulomb’s Law Coulombs Law equation describes a force of infinite range which obeys the inverse square law and is of the same form as the Newton’s Law of Universal Gravitation Unit conversions G mC μC nC
12. Coulomb’s Law What is the net force on the third charge? F1,3= (8.99 * 109 N m2/C2) * (75 * 10-3 C * 45 * 10-3 C) / (1.5 m)2 Fon middle due to L = 1.35 * 107 N, positive, pointing to the R Fon middle due to R = (9 * 109 N m2/C2) * (-90 * 10-3 C * 45 * 10-3 C) / (1.5 m)2 Fon middle due to R = -1.62 * 107 N, negative, pointing to the L Net Force = Fon middle due to L + Fon middle due to R Net Force = 1.35 * 107 N - 1.62 * 107 N Net Force = -0.27 * 107 N, pointing left This is the method to solve any Force or E field problem with multiple charges! 1.5 m 1.5 m +75 mC +45 mC −90 mC q1 q2 q3 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.slideserve.com/emiko/electric-forces-electric-fields", "fetch_time": 1638726557000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-49/segments/1637964363215.8/warc/CC-MAIN-20211205160950-20211205190950-00339.warc.gz", "warc_record_offset": 1100724052, "warc_record_length": 18574, "token_count": 801, "char_count": 2996, "score": 3.828125, "int_score": 4, "crawl": "CC-MAIN-2021-49", "snapshot_type": "latest", "language": "en", "language_score": 0.876207172870636, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00030-of-00064.parquet"} |
# What is the cartesian equation for the polar curve r = 2sin theta - 4cos theta ?
Jul 7, 2017
${x}^{2} + {y}^{2} = 2 y \pm 4 \sqrt{{x}^{2} + {y}^{2}}$
#### Explanation:
Use the conversions:
${r}^{2} = {x}^{2} + {y}^{2}$
$r = \pm \sqrt{{x}^{2} + {y}^{2}}$
$r \sin \theta = y$
$r \cos \theta = x$
First, let's multiply both sides of the equation by $r$.
$r \cdot r = r \cdot \left(2 \sin \theta - 4\right)$
${r}^{2} = 2 r \sin \theta - 4 r$
Now we can substitute the rectangular forms.
${x}^{2} + {y}^{2} = 2 y - 4 \left(\pm \sqrt{{x}^{2} + {y}^{2}}\right)$
${x}^{2} + {y}^{2} = 2 y \pm 4 \sqrt{{x}^{2} + {y}^{2}}$
We can simplify this further, but this is a good stopping point.
Jul 10, 2017
The cartesian rectangular equation is:
${x}^{2} + 4 x + {y}^{2} - 2 y = 0$
Which can also be written as:
${\left(x + 2\right)}^{2} + {\left(y - 1\right)}^{2} = {\sqrt{5}}^{2}$
Which is a circle centred on $\left(- 2 , 1\right)$ with radius $\sqrt{5}$
#### Explanation:
Assuming the correct equation to be:
$r = 2 \sin \theta - 4 \cos \theta$
To convert to polar coordinates to cartesian rectangular form, using the relationships:
{: (x = rcos theta, cos theta=x/r), (y = rsin theta, sin theta=x/r) :} } => r^2 = x^2+y^2
So we can write the polar equation as follows:
$r = 2 \sin \theta - 4 \cos \theta$
$\setminus = 2 \left(\frac{y}{r}\right) - 4 \left(\frac{x}{r}\right)$
Multiply by $r$ and we get:
${r}^{2} = 2 y - 4 x$
$\therefore {x}^{2} + {y}^{2} = 2 y - 4 x$
$\therefore {x}^{2} + 4 x + {y}^{2} - 2 y = 0$
Although this would be sufficient, we can analyze a little further by completing the square on both $x$ and $y$:
${\left(x + 2\right)}^{2} - {2}^{2} + {\left(y - 1\right)}^{2} - {1}^{2} = 0$
$\therefore {\left(x + 2\right)}^{2} - 4 + {\left(y - 1\right)}^{2} - 1 = 0$
$\therefore {\left(x + 2\right)}^{2} + {\left(y - 1\right)}^{2} = 5$
$\therefore {\left(x + 2\right)}^{2} + {\left(y - 1\right)}^{2} = {\sqrt{5}}^{2}$
Which is a circle centred on $\left(- 2 , 1\right)$ with radius $\sqrt{5}$
graph{x^2 + 4x + y^2 - 2y = 0 [-10, 10, -5, 5]} | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 30, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://socratic.org/questions/590692207c01493dd44fdb15", "fetch_time": 1623831255000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-25/segments/1623487622234.42/warc/CC-MAIN-20210616063154-20210616093154-00219.warc.gz", "warc_record_offset": 442565501, "warc_record_length": 6458, "token_count": 850, "char_count": 2086, "score": 4.65625, "int_score": 5, "crawl": "CC-MAIN-2021-25", "snapshot_type": "latest", "language": "en", "language_score": 0.6460577249526978, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00012-of-00064.parquet"} |
# Product
A mathematical product is a term that describes the result of multiplication. This is true of both numbers as well as expressions. In the expressions below,
(1) 4 × 3 = 12 (2) (x + 1)x2 = (x2 + x3)
12 is the product in (1), and in (2), the product is (x2 + x3).
The term product is widely used throughout many different mathematical contexts. Some are closely related to arithmetic multiplication while others, like the product of two matrices, tensor products, cross products in 3D space, etc, can be far more complex.
### Product of natural numbers
The first example, (1), above is an example of a product of natural numbers, one of the simplest types of product. Natural numbers cannot be negative (some definitions include 0 but others may only use the counting numbers). The product of natural numbers is just the result of a multiplication problem.
### Product of integers
The product of integers is mostly the same as the product of natural numbers, except that negative values can be included. Below are the rules for multiplying positive and negative integers.
• Positive * Positive = Positive
• Negative * Negative = Positive
• Positive * Negative = Negative
• Negative * Positive = Negative
Essentially, if there are an odd number of negative signs in the multiplication problem, the resulting value will be negative; if there are a positive number of negative signs, the number will be positive. The number of positive signs does not matter. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.math.net/product", "fetch_time": 1718496751000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-26/segments/1718198861618.0/warc/CC-MAIN-20240615221637-20240616011637-00590.warc.gz", "warc_record_offset": 778390743, "warc_record_length": 2678, "token_count": 315, "char_count": 1474, "score": 4.21875, "int_score": 4, "crawl": "CC-MAIN-2024-26", "snapshot_type": "latest", "language": "en", "language_score": 0.9082028269767761, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00035-of-00064.parquet"} |
Category: Financial.
The net present value is used to calculate the difference between the cash inflows and outflows of a potential investment or project. To make it easier to understand, it evaluates the profit generated by an investment in comparison with the cost set for the time value of money. Why this formula was invented? Because it shows that the value of a dollar could change from day to day.
And the opportunity cost and interest are the main reasons why such formulas are developed. The interest flow will always evaluate the future cash and it will cost more to borrow money. If you thought that economics is complicated, the below formula will definitely scare you away. However, there are much more simple ways to calculate the net present value.
Net Present Value (NPV) Formula
If you don’t know anything about math, you will do now. The below formula is complicated, so sit tight.
To eliminate all doubts, the symbols represent the following values:
Ct
=
net cash inflow for the given period
t
=
number of periods
r
=
discount rate
CO
=
initial investment
It may look scary, but there is another way that you can calculate the NPV. By subtracting the present value of initial investment costs from the present value of future cash flows, you will get the NPV. The formula should look like this:
NPV
=
Present Value of Future Cash Flows – Present Value of Initial Investment Costs.
It definitely looks prettier than the above one, with not so many symbols involved.
Understanding the Net Present Value
Both formulas mentioned above are used by almost all types of investors. To get a better understanding of what NPV means, let’s think of a construction company. The management team wants to expand their business, but they don’t have the right equipment for this. So, the management uses the NPV formula to see if getting that equipment is a smart investment. It does so by comparing the cash inflows in the machinery that would benefit the company. Also, the price of the items is included in the calculation in the C0 value in the formula. If the outcome of the calculation is a positive number, then the cash that will come out of the purchase will be higher than the initial investment. However, if you get a negative number, then the cash will not make up for the money that was invested in the first place to buy the equipment.
Final Words
To conclude the above statements and information, taking the NPV as a prediction is the right thing to do. Basically, the Net Present Value is a kind of prediction like the weather forecasts are. The formulas are used to predict everything from the depreciation of money to the outcome of buying a piece of equipment and seeing if it is feasible. If the equipment is producing greater cash flow, that means that the investment is well made, so you will make up for the lost money in no time.
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# how to express the set of natural numbers in ZFC
I know that it is possible to represent each natural number using the concept of successor. But how do you assign meanings into the representation? I mean, when presented as a set, it's just a set unless we agree on some definition that some form of sets refer to natural numbers. Or is it something impossible at ZFC?
Thanks.
-
ZFC usually includes an axiom of Infinity, which basically says that there exists are inductive set, aka $\mathbb{N}$. Or maybe I'm not really understanding the question. – you Mar 24 '12 at 4:28
@you I know that inductive sets exist. The question is how do one assign some sets as natural numbers. – user27515 Mar 24 '12 at 4:31
@user27515 : " Assign some sets as natural numbers "? Do you mean something like $0 \overset{def}= \varnothing$, $1 \overset{def}= \{\varnothing\}$, $2 \overset{def}= \{1\}$, \dots , $\mathrm{successor}(n) \overset{def}= \{n\}$? – Patrick Da Silva Mar 24 '12 at 4:33
You mean like $0:=\emptyset,1:=\{\emptyset\}, 2:= \{\emptyset,\{\emptyset\}\},\dots, n+1:= n\cup\{n\},\dots$? – you Mar 24 '12 at 4:34
"0", "1", "2", etc. are just convenient labels for $\varnothing$, $S(\varnothing)$, $S(S(\varnothing))$, etc. The "names" occur in the metalanguage (what we are using to talk about the sets in question), rather than "in" set theory. As for singling out the natural numbers, you can see one construction here. – Arturo Magidin Mar 24 '12 at 5:28
I am turning my comment into an answer. The best way I know to output the set $\{0,1,2,\dots \}$ from the set-theoretical construction goes as follows : $$0 = \varnothing, 1 = \{\varnothing, 0\}, 2 = \{\varnothing, 0, 1\}, 3 = \{\varnothing, 0, 1,2\}, \dots, \mathrm{successor}(n) = n \cup \{n\}.$$ In other words, to create the positive integer $n$, you consider the "set that contains the set that contains the set that $\dots$ that contains $\varnothing$ + the set that contains the set that $\dots$",and so on. Again in other words, the integer $n$ is the union of the sets that contains $\varnothing$ at $i$ levels of deepness, $i$ ranging from $0$ to $n-1$ (All this is only in familiar terms).
This construction can also be used to inductively define addition : $$n+0 = n, \quad n + 1 \overset{def}= \mathrm{successor}(n),\quad n+2 \overset{def}= \mathrm{successor}(\mathrm{successor}(n)), \quad \dots \quad n+(m+1) \overset{def}= (n+m)+1.$$ Try to understand what I exactly said in the last definition.
You can also define multiplication using this definition : $$n \cdot 0 \overset{def}= 0, \quad n \cdot (m+1) \overset{def}= (n \cdot m) + n$$ It is an exercise to show that those definitions have all the properties we know over positive integers : associativity, commutativity, etc.
Hope that helps,
-
I made an almost identical post, but was 33 seconds too late :) – you Mar 24 '12 at 4:46
Heh ; your comment was also one minute too late... it happens, don't worry. You'll have your chance on another one! – Patrick Da Silva Mar 24 '12 at 4:48
Another definition lets 0 be the empty set and n +1 be the union of n and {n}. Yet another (I think older) option is to define each n as the equivalence class of sets with cardinality n. The idea underlying these definitions is that a natural number is simply a member of the domain of a model of a certain theory. That is, it doesn't matter or make sense to ask what the objects themselves are; only how they relate to each other (i.e., the relations defined on them) is relevant. – Rachel Mar 24 '12 at 5:03
For set theorists, it is more common to define $\text{successor}(A) = A\cup\{A\}$ for any set $A$; a set $B$ is inductive if $\varnothing\in B$ and $x\in B\Rightarrow \text{successor}(x)\in B$. The Axiom of Infinity guarantees the existence of at least one inductive set. From there one constructs $\mathbb{N}$. The natural numbers are then $0=\varnothing$, $1=\{0\}$, $2=1\cup\{1\} = \{0,1\}$, and more generally, $n+1=\{0,1,\ldots,n\}$. See, e.g. this answer. – Arturo Magidin Mar 24 '12 at 5:27
@Arturo : Yes, now I remember it and it makes more sense. It also helps when building an order over $\mathbb N$, because you can say $n \le m$ if $n \subseteq m$ (because integers are now sets). I'll edit my post. – Patrick Da Silva Mar 24 '12 at 20:13
In the ZFC set theory everything is a set. There are no non-set elements. Indeed there can be several interpretations of the natural numbers in a given model of ZFC.
Furthermore, if you constructed one model of the natural numbers and used it to generate the complex numbers you can identify the natural numbers as the canonically embedding of the natural numbers there, and of course these would be different sets.
It is common to take the finite von Neumann ordinals as the natural numbers. However even if we take different sets, since the addition and multiplication is not a part of the language of set theory we define them within the model. We can also define those differently (again, depending on the interpretation of the numbers).
The important thing is that the chosen representation will satisfy certain axioms, and have binary operations of addition and multiplication.
-
How can everything be a set? Haven't you heard of $\{ u \, | \, u \notin u \}$? – Patrick Da Silva Mar 24 '12 at 20:22
@Patrick: I didn't know that proper classes were elements of the universe of ZFC. – Asaf Karagila Mar 24 '12 at 20:23
Well, when you say "everything"... =P – Patrick Da Silva Mar 24 '12 at 20:25
@Patrick: Yes. Everything. Every element of the universe is a set. Things which are not in the universe cannot be quantified with "every ...". I don't see the problem with my statement. – Asaf Karagila Mar 24 '12 at 20:28
"Everything in a universe" and "Everything" just felt different to me. I thought you didn't make the distinction. If you do then I'm okay with that. =P – Patrick Da Silva Mar 24 '12 at 21:14 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://math.stackexchange.com/questions/123831/how-to-express-the-set-of-natural-numbers-in-zfc", "fetch_time": 1469358211000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2016-30/segments/1469257823996.40/warc/CC-MAIN-20160723071023-00014-ip-10-185-27-174.ec2.internal.warc.gz", "warc_record_offset": 154248584, "warc_record_length": 20474, "token_count": 1676, "char_count": 5899, "score": 3.875, "int_score": 4, "crawl": "CC-MAIN-2016-30", "snapshot_type": "latest", "language": "en", "language_score": 0.8874229192733765, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00048-of-00064.parquet"} |
# Structure of Probability Homework Help
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Probability is the chance of happening of an Event. It is a study of random processes and outcomes. It is most important to achieve an educational standard in Probability and statistics that delivers an accurate mathematical sum without taking any external Structure of probability homework help. Probability has two approaches, namely,
1. Statistical: Let a trial be repeated any number of times under essentially identical conditions and let E be an event of it.The ratio of number of times (m) the event E happens to the number of trials (n) i.e.,m/n is called the relative frequency of the event E and is denoted by R(E). Then the probability P(E) of the event E is the limit approached by R(E) as the number of trials (n) increases indefinitely, it being assumed a unique limit value exists.
2.Classical: In a random experiment, let there be n mutually exclusive and equally likely elementary events. Let E be an event of the experiment. If m elementary events form event E(probability of happening of E or chance E ), is defined as
P(E)=m/n
### Basic Concepts
Random Experiment – An experiment in which outcome cannot be predicted but all possible outcomes are known is called as a Random experiment.
Sample space – The Set S of all possible outcomes of an experiment is called as sample space.
Event – The subset of a sample space is called an event.
For example: Let us assume tossing a coin.
The number of possible outcomes may be a Head(H) or a Tail(T). Hence, tossing a coin is a random experiment. The sample space would be H or T; sample space = {H,T}. And the event is denoted by H for head and T for tail, which is a subset of sample space.
Probability of getting a head (H) would be –
P(H) = (number of events favourable for H)/(total number of events)
= (1)/ (1+1) = 1/2
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### Structure of Probability
When students and learners take Structure of probability homework help from professional tutors, the following traits can be added in the studies to explain them in a clear manner. Probability of an event can be in between 0 and 1. This is because, an event which may occur can be either positive or zero but it cannot be negative.
For example: In tossing a coin.
m – Event of getting a Head; then P(m)=1/2=0.5
n – Event of getting a Tail; then P(n)=1/2=0.5
l – Event of getting a Tail or a Head;
then P(l)=P(m U n)= { P(m) + P(n) }/[ P(m) + P(n) ]=2/2=1
q – Event of getting a Tail and a Head;
then P(q)= 0/2=0
* For an impossible event the probability is 0 and for a certain event the probability is 1.
* Addition of probability of complimentary events is 1.
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1. mixed variable integration coefficient
$(y^{4}-4xy)dx+(2xy^{3}-3x^{2})dy=0$
prove that this equation has an xy dependant integration coefficient
2. show that exist constants that verify $\displaystyle\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}=m\frac{N}{x}-n\frac{M}{y}.$
3. why??
if the left side equals zero then we have potential and could be solved without multilying by integration coefficient.
what the right side says ?
why this equation says about the coefficient
?
4. the condition i gave you gives you the integrating factor, if those constants exist, then the integrating factor is $u(x,y)=x^my^n,$ but on your question you were asked to show that your equation has a integrating factor which depends of $xy,$ so that will force that $m=n$ in order to have $u(x,y)=h(xy).$
5. how you got this formula?
so if u(x,y)=xy then m=n=1
$4y^3-4x-(2y^3-6x)=m\frac{2xy^3-3x^2}{x}-n\frac{y^4-4xy}{y}$
what to do now?
6. c'mon, simplify, do the algebra.
7. $4y^3-4x-(2y^3-6x)=n(2xy^4-3x^2y-y^4x-4x^2y)$
$\frac{4y^3-4x-(2y^3-6x)}{(2xy^4-3x^2y-y^4x-4x^2y)}=n$
i cant get a number out of it | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://mathhelpforum.com/differential-equations/162903-mixed-variable-integration-coefficient.html", "fetch_time": 1508378143000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-43/segments/1508187823214.37/warc/CC-MAIN-20171019012514-20171019032514-00751.warc.gz", "warc_record_offset": 229077955, "warc_record_length": 10719, "token_count": 392, "char_count": 1112, "score": 3.75, "int_score": 4, "crawl": "CC-MAIN-2017-43", "snapshot_type": "longest", "language": "en", "language_score": 0.8353407382965088, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00003-of-00064.parquet"} |
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### How do I draw a simple random sample?
There are free random number generators on the internet that you can use. Two are: www.random.org and www.randomizer.org
If you want to draw your own sample, there is a method you must abide by in order for your sample to be representative of the population.
The first step is to identify all of the members in your population. You must be able to list them in what is called a sampling frame. The frame should have the names without order to them and non-overlapping (no duplicates). Alphabetizing the list by surname is a way to insure a random order in the sampling frame.
Second, you must give each name an identification number. Start with “1” and continue.
Third, you must decide what the size will be for your sample. You can use the table suggested in the Krejcie (1970) article, or whatever feels right for you to believe in the results you obtain. As a rule of thumb, use as large a sample size as possible.
Fourth, you need to get a Table of Random Numbers. Many are located at the end of statistical or mathematical textbooks. The Table of Random Numbers consists of rows and columns of numbers arranged at random so that they can be used as any point by reading in any direction left or right, up or down. We are now ready to draw our random sample.
Here is a simple example for you.
Sampling Frame
ID Name
1. Alex
2. Allen
3. Lea
4. Joey
5. Auggie
6. Mary
7. Ellen
8. Emily
9. Felice
10. Juan
11. Julia
12. Ruthie
13. Annie
14. Mark
15. Miguel
16. Maura
17. Olivia
18. Rachael
19. Rebecca
20. Seth
Fifth, we know that our largest ID number has two digits (20). So we are going to need a two-digit column in the Table of Random Numbers. We start anyplace in the Table of Random numbers. We have decided in advance whether we would use two digits going up, down, left or right. So we begin.
Sixth, Number #6 is the first ID number that is within our band of between 1 and 20. This is the first member of our random sample and it is Mary. We continue 82, 56, 96, 66, 46 until we and come up with the next is ID #13 Annie. Our three last members of the sample are ID #8 (Emily), ID #5 (Auggie) and ID #4 (Joey). We have five randomly selected members: Mary, Annie, Emily, Auggie and Joey. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.dissertation-statistics.com/random-sample.html", "fetch_time": 1726287228000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700651548.18/warc/CC-MAIN-20240914025441-20240914055441-00280.warc.gz", "warc_record_offset": 41959151, "warc_record_length": 3947, "token_count": 578, "char_count": 2297, "score": 4.59375, "int_score": 5, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.9225484132766724, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00002-of-00064.parquet"} |
EMI BY NEW METHOD
Sir
I am presenting a new method of arriving EMI.
I have prepared the calculator sheet, butnot know to attach.
the details are copied as text.
pl examine.
Pl help to attach the calculator.
In the meanwhile pl examine the underfurnished details
Method of arriving EMI other than PMT or algebraic formula.
I am presenting a methodology by which we can arrive the EMI, BY
which is different from our PMT function.
RAMA_51
Fill the green cells.
The calculator arrives figures of FD cumulative and RD value.
The formula can be put as:
EMI for amount A for period N and interest rate R, will be
=Maturity value of Amount A for N + 1 months at rate R /
A METHOD TO ARRIVE EMI BY USING FD CUMULATIVE AND RECURRING DEPOSIT
LOAN AMOUNT 500000 presented by :
PERIOD 120 (MONTHS) (N) RAMA_51
INTEREST RATE 11 (MONTHLY RESTS)
E M I Rs.
6,887.50
THE METHOD EXPLANATION:
A fixed deposit - cumulative interest.
ARRIVE THE MATURITY VALUE OF THE LOAN AMOUNT
Amount 500000 FOR '(N + 1)' MONTHS (see value against A1(cell A16))
period 121 (months) (N +1)
interest 11 ARRIVE THE MATURITY VALUE OF RD FOR Rs.1/-
AT THE SAME RATE FOR ' N' MONTHS
A1 maturity value 1508275.07 (d5*(1+d7%/12)^d6)
see value against B1(cellA23)
B Recurring deposit - using the function FV
amount 1 E M I , will be (A1) / (B1)
period 120 ( N)
interest 11 Thus:
EMI, for an amount A , period N, and interest R, will be:
B1 maturity value 218.99 =Maturity value of A for N+1 months at interest R
EMI WILL BE A1/B1
Rs.
6,887.50
maturity value of RD for period N at same interest for Rs1/-
LIABILITY AFTER A PERIOD
DATA FROM FORMULA SHEET BROUGHT FORWARD presented by:
LOAN AMOUNT 500000 RAMA_51
PERIOD 120 (MONTHS) (N) Please fill the green cell to arrive
INTEREST RATE 11 (MONTHLY RESTS) a liability after a period run
E M I 6887.500565
TO ARRIVE THE LIABILITY OF THE LOAN AFTER A PERIOD LESS THAN THE TERM OF THE LOAN
(INTEREST RATE IS TAKEN SAME AS ABOVE
PERIOD RUN 80 MONTHS <<If we put the period run as
the loan period,
FD VALUE FOR 81 MONTHS 1047043.37
as 0
RD VALUE FOR
EMI AMOUNT 80 MONTHS 815169.55 B
Liability after 80 months will be >> 231873.81
Anwsers to the Problem EMI BY NEW METHOD
Manually editing the Windows registry to fix Error EMI BY NEW METHOD
Caution: Unless you an advanced PC user, we DO NOT recommend editing the Windows registry manually. Using Registry Editor incorrectly can cause serious problems that may require you to reinstall Windows. We do not guarantee that problems resulting from the incorrect use of Registry Editor can be solved. Use Registry Editor at your own risk.
• Click the Start button.
• Type "command" in the search box... DO NOT hit ENTER yet!
• While holding CTRL-Shift on your keyboard, hit ENTER.
• You will be prompted with a permission dialog box.
• Click Yes.
• A black box will open with a blinking cursor.
• Type "regedit" and hit ENTER.
• In the Registry Editor, select the Error 0x9C-related key (eg. Windows Operating System) you want to back up.
• From the File menu, choose Export.
• In the Save In list, select the folder where you want to save the Windows Operating System backup key.
• In the File Name box, type a name for your backup file, such as "Windows Operating System Backup".
• In the Export Range box, be sure that "Selected branch" is selected.
• Click Save.
• The file is then saved with a .reg file extension.
• You now have a backup of your MACHINE_CHECK_EXCEPTION-related registry entry.
Another Safe way to Repair the Problem: EMI BY NEW METHOD:
How to Fix EMI BY NEW METHOD with SmartPCFixer?
1. Download SmartPCFixer . Install it on your system. Click Scan, and it will perform a scan for your computer. The errors will be shown in the scan result.
2. After the scan is done, you can see the errors and problems need to be fixed. Click Fix All.
3. The Repair part is done, the speed of your computer will be much higher than before and the errors have been removed. You can also use other functions in this software. Like dll downloading, junk file cleaning and print spooler error repair. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "computererrorfixers.blogspot.com", "fetch_time": 1568540383000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-39/segments/1568514571027.62/warc/CC-MAIN-20190915093509-20190915115509-00199.warc.gz", "warc_record_offset": 42831506, "warc_record_length": 17680, "token_count": 1088, "char_count": 4090, "score": 3.859375, "int_score": 4, "crawl": "CC-MAIN-2019-39", "snapshot_type": "latest", "language": "en", "language_score": 0.7758881449699402, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00002-of-00064.parquet"} |
0
# Is the number 48 prime or composite?
Updated: 4/28/2022
Wiki User
12y ago
48 is a composite number. This is because composite numbers are any number that has more than two factors (one and itself). Therefore, since there are many factors of 48 besides one and itself including 2, 3, and 4, that makes it a composite number.
Wiki User
12y ago
Earn +20 pts
Q: Is the number 48 prime or composite?
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Related questions
### Is 48 prime or composite and tell you the numbers?
48 is not a prime number it can be divide by 8, 2 and 4
### Is 48 prime or composite?
48 = 2*2*2*2*3By definition, 48 is not a prime number.48 is an even number greater than 2. It is composite.composite because it is a multiple of 2, and 24
### Is 48 a prime or composite number why or why not?
composite because it has more than two factors.
### Is 24 a prime factor of 48?
24 is a factor of 48. But, 24 is not a prime factor because it is a composite number.
Composite
### Is 48 a prime number?
No because 48 has more than two factors and so therefore it is a composite number
### Why is 48 a composite?
A prime number has only 2 factors which are 1 and itself. Composite numbers are everything else except 1 and 0. 1 and 0 are neither prime, nor composite. 48 is composite.
### Is 48 a composite or prime number?
48 has factors other than 1 and itself, so it is not a prime number. It is an even number that is greater than 2, so it also has 2 as a factor. Therefore, it is a composite number. All even numbers greater than 2 are composite numbers.No number that ends in an '8' can be prime; it is at least divisible by 2. Use this kind of thinking to derive the other factors of 48.
### Is 48 prime composite or neither?
A prime number is a number that is divisible only by 1 and itself; it has no other factors. A composite number is a number that is divisible by more than two numbers - more than 1 and itself. The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48. Therefore, it is a composite number.
It is prime
### Is 1029 a prime or composite?
It is not a prime number, it is a composite number.
### Is 86 a prime or composite number?
86 is not a prime number, so is a composite. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.answers.com/basic-math/Is_the_number_48_prime_or_composite", "fetch_time": 1721665780000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-30/segments/1720763517890.5/warc/CC-MAIN-20240722160043-20240722190043-00420.warc.gz", "warc_record_offset": 323475332, "warc_record_length": 47116, "token_count": 600, "char_count": 2226, "score": 4.0, "int_score": 4, "crawl": "CC-MAIN-2024-30", "snapshot_type": "latest", "language": "en", "language_score": 0.9486265778541565, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00012-of-00064.parquet"} |
# Approximation of $\ln(x+1)$ with $\Psi$ function
I found the following approximation for the function $$f=\ln(x+1)$$ $$f\simeq\Psi\left(x+\dfrac{3}{2}\right)-2+\gamma+\ln(2)$$ where $\Psi(x)$ is the 'Digamma' function: $\Psi(x)=\dfrac{\dfrac{d}{dx}\Gamma(x)}{\Gamma(x)}$ and $\gamma$ is the Euler - Mascheroni constant. The following inequality holds: $$\left|\ln(x+1)-\Psi\left(x+\dfrac{3}{2}\right)-2+\gamma+\ln(2)\right|\lt 0.036$$ Is this a known result? Thanks
I'm not sure why there is a $(-2+\gamma+\ln 2)$, because $\left|\ln(x+1) - \Psi(x+\frac32)\right| < 0.0365$ itself is already satisfied.
The approximation comes from the series expansion of $\exp\left(\Psi(x+\frac12)\right)$, which states, for $x > 1$,
\begin{align} \Psi\left(x + \frac12\right) = \ln\left( x + \frac{1}{4!\cdot x} - \frac{37}{8\cdot 6!\cdot x^3} + o\left(\frac1{x^5}\right) \right) \end{align} and thus $\Psi(x+\frac32)$ is approximately $\ln(x+1)$. The largest difference happens at $x=0$ which is $\Psi(\frac32) = 2 - \ln 4 - \gamma \approx 0.03648997$.
I doubt your approximation. First I guess you missed a factor $2$ for $\ln 2$, the approximation should be $$f(x) =\ln(x+1)-\Psi(x+3/2)-2+\gamma+2\ln 2 \cdot$$ If you omit the $2$ you have $$f(0)= 0 -(2-\gamma-2\ln2) -2+\gamma+ \ln 2 = -4 + 2\gamma + 3\ln 2 \approx -0.766$$ while with the factor $2$ you have $$f(0)= 0 -(2-\gamma-2\ln2) -2+\gamma+ 2\ln 2 = -4 + 2\gamma + 4\ln 2 \approx -0.0723$$ which is much smaller but still violates your given bound.
PS: A Taylor series at $x=0$ computed with Maple is $$\ln(x+1)-\Psi(x+\tfrac{3}{2}) = (-2+\gamma+2\ln2)+(5-\tfrac{1}{2}\pi^2)x+(-\tfrac{17}{2}+7\zeta(3))x^2 +O(x^3)$$
• In fact, I omissed to write that the inequality holds for $x \gt 1$ – Riccardo.Alestra May 31 '16 at 10:01
• But in any case you have to add the factor $2$, otherwise you have $f(2)\approx -0.734$! But even with the factor you have $f(2)\approx-0.041$ – gammatester May 31 '16 at 10:05 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/1806774/approximation-of-lnx1-with-psi-function/1806805", "fetch_time": 1571370739000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-43/segments/1570986677884.28/warc/CC-MAIN-20191018032611-20191018060111-00004.warc.gz", "warc_record_offset": 593083377, "warc_record_length": 32029, "token_count": 752, "char_count": 1955, "score": 3.90625, "int_score": 4, "crawl": "CC-MAIN-2019-43", "snapshot_type": "latest", "language": "en", "language_score": 0.7535237669944763, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00020-of-00064.parquet"} |
# Squares and Square Roots
## How to Square A Number
To square a number:
Multiply the number by itself.
Examples
What is 4 squared?
4 squared = 4 × 4 = 16
“Squared” is often written as a little 2 like this:
42 = 16
This says “4 Squared equals 16
(the little 2 says the number appears twice in multiplying)
## Perfect Squares
The Perfect Squares (also called “Square Numbers“) are the squares of the integers:
Perfect Squares From 02 to 102
Number Square 0 0 1 1 2 4 3 9 4 16 5 25 6 36 7 49 8 64 9 81 10 100
## Negative Numbers
We can also square negative numbers.
What happens when we square (−7) ?
(−7) × (−7) = 49
(because a negative times a negative gives a positive)
When we square a negative number we get a positive result.
## Square Roots
A square root goes the other way:
What can we multiply by itself to get this?
Examples
Find a square root of 25
What can we multiply by itself to get 25?
A square root of 25 is 5, because when 5 is multiplied by itself we get 25.
So,
a square root of a number is a value that can be multiplied by itself to give the original number.
## Negatives Roots
We discovered earlier that we can square negative numbers:
Examples
(−6) squared
(−6) × (−6) = 36
And of course 6 × 6 = 36 too.
So the square root of 36 could be −6 or +6
## Perfect Square Roots
The Perfect roots are the roots of the integers:
Square Square Root 1 1 or -1 4 2 or -2 9 3 or -3 16 4 or -4 25 5 or -5 36 6 or -6 49 7 or -7 64 8 or -8 81 9 or -9 100 10 or -10
## The Square Root Symbol
This is the special symbol that means “square root“.
Use it like this:
√49 is 7
we say “square root of 49 equals 7”
Examples
What is √81 ?
81 = 9 × 9
in other words when we multiply 9 by itself (9 × 9) we get 81
So the answer is:
√81 = 9
## Square Root Day
The April 4th 2016 is a Square Root Day, because the date looks like 4/4/16
The next after that is the May 5th 2025 (5/5/25) | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://sigmatricks.com/squares-and-square-roots/", "fetch_time": 1696328855000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-40/segments/1695233511075.63/warc/CC-MAIN-20231003092549-20231003122549-00494.warc.gz", "warc_record_offset": 577989557, "warc_record_length": 9521, "token_count": 608, "char_count": 1933, "score": 4.5625, "int_score": 5, "crawl": "CC-MAIN-2023-40", "snapshot_type": "latest", "language": "en", "language_score": 0.8706918358802795, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00037-of-00064.parquet"} |
This unit illustrates this rule. Chain rule practice, implicit differentiation solutions.pdf... School Great Bend High School; Course Title MATHEMATICS 1A; Uploaded By oxy789. Solo Practice. Email. The first layer is the third power'', the second layer is the tangent function'', the third layer is the square root function'', the fourth layer is the cotangent function'', and the fifth layer is (7 x). The difficulty in using the chain rule: Implementing the chain rule is usually not difficult. In calculus, the chain rule is a formula to compute the derivative of a composite function. The chain rule states that the derivative of f (g (x)) is f' (g (x))⋅g' (x). Classic . The chain rule: introduction. The Google Form is ready to go - no prep needed. Print; Share; Edit; Delete; Report an issue; Live modes. Students progress at their own pace and you see a leaderboard and live results. The chain rule: introduction. In the section we extend the idea of the chain rule to functions of several variables. That’s all there is to it. Email. Use the chain rule to calculate h′(x), where h(x)=f(g(x)). Just use the rule for the derivative of sine, not touching the inside stuff (x2), and then multiply your result by the derivative of x2. answer choices . Chain Rule Online test - 20 questions to practice Online Chain Rule Test and find out how much you score before you appear for next interview and written test. Played 0 times. Improve your math knowledge with free questions in "Chain rule" and thousands of other math skills. 10 Questions Show answers. Edit. Here’s what you do. To play this quiz, please finish editing it. Determine where in the interval $$\left[ {0,3} \right]$$ the object is moving to the right and moving to the left. Brilliant. 10th - 12th grade . AP.CALC: FUN‑3 (EU), FUN‑3.C (LO), FUN‑3.C.1 (EK) Google Classroom Facebook Twitter. Show Solution For this problem the outside function is (hopefully) clearly the exponent of 4 on the parenthesis while the inside function is the polynomial that is being raised to the power. Jul 8, 2020 - Check your calculus students' understanding of finding derivatives using the Chain Rule with this self-grading Google Form which can be given as a homework assignment, practice, or a quiz. Differentiate them in that order. find answers WITHOUT using the chain rule. Chain Rule on Brilliant, the largest community of math and science problem solvers. 0% average accuracy. The problem that many students have trouble with is trying to figure out which parts of the function are within other functions (i.e., in the above example, which part if g(x) and which part is h(x). through 8.) In other words, when you do the derivative rule for the outermost function, don’t touch the inside stuff! }\) The chain rule is a rule for differentiating compositions of functions. In the following discussion and solutions the derivative of a function h(x) will be denoted by or h'(x). Worked example: Chain rule with table. Determine where in the interval $$\left[ { - 1,20} \right]$$ the function $$f\left( x \right) = \ln \left( {{x^4} + 20{x^3} + 100} \right)$$ is increasing and decreasing. This is the currently selected item. Chain Rule Practice DRAFT. Moveover, in this case, if we calculate h(x),h(x)=f(g(x))=f(−2x+5)=6(−2x+5)+3=−12x+30+3=−12… AP.CALC: FUN‑3 (EU), FUN‑3.C (LO), FUN‑3.C.1 (EK) Google Classroom Facebook Twitter. Finish Editing. Includes full solutions and score reporting. Pages 2. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. In other words, it helps us differentiate *composite functions*. This quiz is incomplete! It is useful when finding the derivative of a function that is raised to the nth power. This quiz is incomplete! Mark Ryan has taught pre-algebra through calculus for more than 25 years. The general power rule states that this derivative is n times the function raised to … The chain rule: further practice. On problems 1.) anytime you want. Even though we had to evaluate f′ at g(x)=−2x+5, that didn't make a difference since f′=6 not matter what its input is. In particular, we will see that there are multiple variants to the chain rule here all depending on how many variables our function is dependent on and how each of those variables can, in turn, be written in terms of different variables. Most of the basic derivative rules have a plain old x as the argument (or input variable) of the function. That is, if f and g are differentiable functions, then the chain rule expresses the derivative of their composite f ∘ g — the function which maps x to For instance, (x 2 + 1) 7 is comprised of the inner function x 2 + 1 inside the outer function (⋯) 7. Many answers: Ex y = (((2x + 1)5 + 2) 6 + 3) 7 dy dx = 7(((2x + 1)5 + 2) 6 + 3) 6 ⋅ 6((2x + 1)5 + 2) 5 ⋅ 5(2x + 1)4 ⋅ 2-2-Create your own worksheets like this one … A few are somewhat challenging. He also does extensive one-on-one tutoring. Play. This means that we’ll need to do the product rule on the first term since it is a product of two functions that both involve $$u$$. The Chain Rule is used for differentiating composite functions. Identify composite functions. Because the argument of the sine function is something other than a plain old x, this is a chain rule problem. Understand the chain rule and how to use it to solve complex functions Discuss nested equations Practice solving complex functions using the chain rule; Practice Exams. Start a live quiz . Chain Rule Worksheets with Answers admin October 1, 2019 Some of the Worksheets below are Chain Rule Worksheets with Answers, usage of the chain rule to obtain the derivatives of functions, several interesting chain rule exercises with step by step solutions and quizzes with answers, … We won’t need to product rule the second term, in this case, because the first function in that term involves only $$v$$’s. Section 3-9 : Chain Rule For problems 1 – 27 differentiate the given function. Practice: Chain rule with tables. Differentiate Using the Chain Rule — Practice Questions, Solving Limits with Algebra â Practice Questions, Limits and Continuity in Calculus â Practice Questions, Evaluate Series Convergence/Divergence Using an nth Term Test. Worked example: Derivative of 7^(x²-x) using the chain rule . These rules follow from the limit definition of derivative, special limits, trigonometry identities, or the quotient rule. To play this quiz, please finish editing it. Solution: The derivatives of f and g aref′(x)=6g′(x)=−2.According to the chain rule, h′(x)=f′(g(x))g′(x)=f′(−2x+5)(−2)=6(−2)=−12. The Chain Rule mc-TY-chain-2009-1 A special rule, thechainrule, exists for differentiating a function of another function. f (x) = (6x2+7x)4 f (x) = (6 x 2 + 7 x) 4 Solution g(t) = (4t2 −3t+2)−2 g (t) = (4 t 2 − 3 t + 2) − 2 Solution The chain rule: introduction. If we don't recognize that a function is composite and that the chain rule must be applied, we will not be able to differentiate correctly. Mathematics. Chain rule and implicit differentiation March 6, 2018 1. Using the chain rule: The derivative of ex is ex, so by the chain rule, the derivative of eglob is Free practice questions for Calculus 3 - Multi-Variable Chain Rule. The chain rule is probably the trickiest among the advanced derivative rules, but it’s really not that bad if you focus clearly on what’s going on. Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. The position of an object is given by $$s\left( t \right) = \sin \left( {3t} \right) - 2t + 4$$. These Multiple Choice Questions (MCQs) on Chain Rule help you evaluate your knowledge and skills yourself with this CareerRide Quiz. Most problems are average. When the argument of a function is anything other than a plain old x, such as y = sin (x2) or ln10x (as opposed to ln x), you’ve got a chain rule problem. The derivative of ex is ex, so by the chain rule, the derivative of eglob is. 0. Then multiply that result by the derivative of the argument. Determine where $$A\left( t \right) = {t^2}{{\bf{e}}^{5 - t}}$$ is increasing and decreasing. a day ago by. If 30 men can build a wall 56 meters long in 5 days, what length of a similar wall can be built by 40 … This calculus video tutorial explains how to find derivatives using the chain rule. The ones with a * are trickier, so make sure you try them. When do you use the chain rule? Instructor-paced BETA . 60 seconds . As another example, e sin x is comprised of the inner function sin Q. For problems 1 â 27 differentiate the given function. The notation tells you that is a composite function of. In 1997, he founded The Math Center in Winnetka, Illinois, where he teaches junior high and high school mathematics courses as well as standardized test prep classes. Delete Quiz. You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, $$f\left( x \right) = {\left( {6{x^2} + 7x} \right)^4}$$, $$g\left( t \right) = {\left( {4{t^2} - 3t + 2} \right)^{ - 2}}$$, $$R\left( w \right) = \csc \left( {7w} \right)$$, $$G\left( x \right) = 2\sin \left( {3x + \tan \left( x \right)} \right)$$, $$h\left( u \right) = \tan \left( {4 + 10u} \right)$$, $$f\left( t \right) = 5 + {{\bf{e}}^{4t + {t^{\,7}}}}$$, $$g\left( x \right) = {{\bf{e}}^{1 - \cos \left( x \right)}}$$, $$u\left( t \right) = {\tan ^{ - 1}}\left( {3t - 1} \right)$$, $$F\left( y \right) = \ln \left( {1 - 5{y^2} + {y^3}} \right)$$, $$V\left( x \right) = \ln \left( {\sin \left( x \right) - \cot \left( x \right)} \right)$$, $$h\left( z \right) = \sin \left( {{z^6}} \right) + {\sin ^6}\left( z \right)$$, $$S\left( w \right) = \sqrt {7w} + {{\bf{e}}^{ - w}}$$, $$g\left( z \right) = 3{z^7} - \sin \left( {{z^2} + 6} \right)$$, $$f\left( x \right) = \ln \left( {\sin \left( x \right)} \right) - {\left( {{x^4} - 3x} \right)^{10}}$$, $$h\left( t \right) = {t^6}\,\sqrt {5{t^2} - t}$$, $$q\left( t \right) = {t^2}\ln \left( {{t^5}} \right)$$, $$g\left( w \right) = \cos \left( {3w} \right)\sec \left( {1 - w} \right)$$, $$\displaystyle y = \frac{{\sin \left( {3t} \right)}}{{1 + {t^2}}}$$, $$\displaystyle K\left( x \right) = \frac{{1 + {{\bf{e}}^{ - 2x}}}}{{x + \tan \left( {12x} \right)}}$$, $$f\left( x \right) = \cos \left( {{x^2}{{\bf{e}}^x}} \right)$$, $$z = \sqrt {5x + \tan \left( {4x} \right)}$$, $$f\left( t \right) = {\left( {{{\bf{e}}^{ - 6t}} + \sin \left( {2 - t} \right)} \right)^3}$$, $$g\left( x \right) = {\left( {\ln \left( {{x^2} + 1} \right) - {{\tan }^{ - 1}}\left( {6x} \right)} \right)^{10}}$$, $$h\left( z \right) = {\tan ^4}\left( {{z^2} + 1} \right)$$, $$f\left( x \right) = {\left( {\sqrt[3]{{12x}} + {{\sin }^2}\left( {3x} \right)} \right)^{ - 1}}$$. Since the functions were linear, this example was trivial. SURVEY . Only in the next step do you multiply the outside derivative by the derivative of the inside stuff. On the other hand, applying the chain rule on a function that isn't composite will also result in a wrong derivative. Let f(x)=6x+3 and g(x)=−2x+5. Chain rule. For example, sin (x²) is a composite function because it can be constructed as f (g (x)) for f (x)=sin (x) and g (x)=x². With chain rule problems, never use more than one derivative rule per step. Save. chain rule practice problems worksheet (1) Differentiate y = (x 2 + 4x + 6) 5 Solution (2) Differentiate y = tan 3x Solution by the Chain Rule, dy/dx = dy/dt × dt/dx so dy/dx = 3t² × 2x = 3(1 + x²)² × 2x = 6x(1 + x²) ². The rule itself looks really quite simple (and it is not too difficult to use). hdo. Derivative of aˣ (for any positive base a) Derivative of logₐx (for any positive base a≠1) Practice: Derivatives of aˣ and logₐx. Determine where $$V\left( z \right) = {z^4}{\left( {2z - 8} \right)^3}$$ is increasing and decreasing. The chain rule says, if you have a function in the form y=f (u) where u is a function of x, then. You simply apply the derivative rule that’s appropriate to the outer function, temporarily ignoring the not-a-plain-old-x argument. Chain rule intro. Differentiate the following functions. Edit. Then differentiate the function. Share practice link. The most important thing to understand is when to use it and then get lots of practice. Practice. Find the tangent line to $$f\left( x \right) = 4\sqrt {2x} - 6{{\bf{e}}^{2 - x}}$$ at $$x = 2$$. Question 1 . Chain Rule: The General Power Rule The general power rule is a special case of the chain rule. 13) Give a function that requires three applications of the chain rule to differentiate. PROBLEM 1 : … This preview shows page 1 - 2 out of 2 pages. Usually, the only way to differentiate a composite function is using the chain rule. The questions will … Because the argument of the sine function is something other than a plain old x, this is a chain rule problem. In the list of problems which follows, most problems are average and a few are somewhat challenging. 0 likes. In examples such as the above one, with practise it should be possible for you to be able to simply write down the answer without having to let t = 1 + x² etc. He is a member of the Authors Guild and the National Council of Teachers of Mathematics. For example. The Chain Rule, as learned in Section 2.5, states that \(\ds \frac{d}{dx}\Big(f\big(g(x)\big)\Big) = \fp\big(g(x)\big)g'(x)\text{. Just use the rule for the derivative of sine, not touching the inside stuff (x2), and then multiply your result by the derivative of x2. The chain rule: introduction. You must use the Chain rule to find the derivative of any function that is comprised of one function inside of another function. 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# How can I model the acceleration/velocity of a bicycle knowing only the power output from the drivetrain and rider weight?
I am currently building a simplistic video game of bicycle racing, a similar to the idea of Zwift. In this game, the user has a physical bicycle that is connected with various hardware that will let me know the power output, in my case measured at the cranks (using a power2max power meter).
My application receives an instantaneous power reading 4 times a second from this power meter. From this power and a few known constants, I am able to model the velocity of the rider given this power output:
$$P = \frac{1}{1 - \text{drivetrain loss}} \cdot ( F_{\text{gravity}} + F_{\text{rolling}} + F_{\text{drag}} ) \cdot V$$
An expansion of the classic $P = FV$ formula. In my application, I use Newton's method to solve this for $V$, as I know the other values.
My problem is that just solving for $V$ whenever I receive data from the power meter is incorrect, as that would correspond to instantaneous velocity changes. For example, the rider starts from rest, and if I apply 600W at the pedals, in reality I won't suddenly be travelling at 50km/h - I have to accelerate from rest.
I'm at a loss as to how to correctly calculate this acceleration. My current attempt is a rearrangement of $F=ma$ - $a = \frac{F}{m}$. In this scenario I have $F = \frac{P}{\text{modelled velocity for power}} - \frac{1}{1- \text{drivetrain loss}} ( F_{\text{gravity}} + F_{\text{rolling}} + F_{\text{drag}})$. What I'm doing is calculating the force from the model that is above the current forces the rider is experiencing when they don't apply any force to the pedals. This results in acceleration if the rider is travelling under the modelled velocity for a given power, and decelleration if they are travelling at a velocity greater than the modelled velocity (as they are unable to overcome drag).
While this does seem to converge to the correct values, the calculated acceleration seems extremely low. Returning to the example of 600W, I know that if I go out and do this on the roads I'll be travelling at around 20km/h within a few seconds. Yet under the above if I step the model 4 times a second with 600W (simulating a constant power reading from the power meter), after 5 seconds my velocity would be 3km/h, which is most certainly wrong!
Here is the Haskell code I am using for my work:
import Numeric.AD (auto)
velocityForPower :: Double -> Double -> Double -> Double
velocityForPower grade weight power =
last (take 10
(findZero (\v ->
(auto weight)
v -
auto power)
30))
grade, weight :: Double
weight = 62
accelerationForPower :: Double -> Double -> Double -> Double -> Double
accelerationForPower grade weight velocity knownPower =
let forceProvided = knownPower / velocityForPower grade weight knownPower
in (forceProvided - resistingForces grade weight velocity) / weight
calculatePowerOutput :: (Floating a, Ord a) => a -> a -> a -> a
calculatePowerOutput grade weight velocity =
resistingForces grade weight velocity * velocity
resistingForces :: (Floating a, Ord a) => a -> a -> a -> a
resistingForces grade weight velocity =
recip (1 - drivetrainLoss) * (fGravity + fRolling + fDrag)
where fGravity = g * sin (atan grade) * weight
fRolling
| velocity > 1.0e-2 = g * cos (atan grade) * weight * rollingResistanceCoeff
| otherwise = 0 -- There is no rolling resistance if we are stationary
where rollingResistanceCoeff = 5.0e-3
fDrag = 0.5 * dragCoeff * frontalArea * rho * velocity * velocity
where dragCoeff = 0.63
frontalArea = 0.509
rho = 1.226
g = 9.8067
drivetrainLoss = 3.0e-2
And here is an example run of the model:
mapM_ (print . (/ 1000) . (* 60) . (* 60))
(take (round $5 * recip 0.25) (unfoldr (\v -> Just (v, v + accelerationForPower 0 68.2 v 120 * 0.25)) 0)) 0.0 0.20502901826445583 0.3645543623001854 0.5240609569432696 0.683538303824054 0.8429759112371182 1.002363296901935 1.1616899907186995 1.3209455375177925 1.4801194998013418 1.6392014604753573 1.7981810255709199 1.9570478269529188 2.115791525014844 2.274401811358143 2.4328684114546824 2.5911810872908574 2.7493296399919145 2.907303912425068 3.0650937917800167 • Now that some of the main typos are out of the way we can focus on solving the real problem. If you have 600 W, and a mass of 60 kg and initial velocity of 1 m/s, you should be developing a force of 600 N and an acceleration of 10 m/s$^2$. That would be very hard to do (wheels might be slipping) but agrees with your idea that you should come up to speed very quickly (600 W is really quite a lot of power, but OK for a short burst). I would print out the intermediate values for the different components of friction, power left, and computed force for acceleration at each step. That should help. Commented Jan 1, 2016 at 1:54 • I am not familiar with Haskell but it looks like in calculatePowerOutput you are passing (velocity*velocity) to a function resistingForces that expects velocity (not velocity squared). But it's possible I just don't understand the language. Commented Jan 2, 2016 at 15:39 • @Floris in Haskell function application binds tighter than almost anything else, so should read that line is: calculatePowerOutput = resistingForces (grade, weight, velocity) * velocity Commented Jan 2, 2016 at 17:54 • As I said - it's not my language. In general I prefer to use parentheses to make sure there is no possibility of getting the order of operations wrong. Unless Haskell doesn't have those... Commented Jan 2, 2016 at 17:55 ## 3 Answers You initial equation for the power equation seems wrong as you are multiplying things that should not be multiplied. Instead, think of the force needed to keep the bicycle going at the current velocity; multiply that force by the velocity to get the power needed; and finally, see if there is "too much" or "not enough" power to maintain that. I assume you have expressions for the various forces on the bicycle - rolling resistance, air drag (including wind), and the effect of gravity (slope). Drag will be a strong (quadratic) function of velocity; rolling resistance and the force of gravity much less so. You have power applied to the pedals, and some efficiency factor$\eta$of the drive train; this means that the available power$P_{available} = \eta P_{pedal}$. Looking at your code, you seem to have a drive train loss of 0.03 (3%), so$\eta = 0.97$. At a given velocity, you needed a force$F_{total}=F_{drag}+F_{rolling}+F_{gravity}$, and thus a power$P_{needed}+F_{total}\cdot v$. The "excess" power is the difference between the available power and the power needed,$P_{excess}=P_{available}-P_{needed}$. This means that there is a net force$F_{accel}=P_{excess}/v$, and the instantaneous acceleration of the bike is$a=F_{accel}/m_{total}$You can then do a simple calculation of the velocity at a later time: $$v_{t+\Delta t} = v_t + a\Delta t$$ If the acceleration is changing constantly, there are better ways to do the integration; but if you have power values four times per second ($\Delta t = 0.25 s$), you should be OK just using this equation. I am sorry that I cannot comment on the Haskell code - it's "not my language". So I wrote a little program in Python that does largely the same thing; note that I borrowed the idea first shown by @Shane to compute the new velocity by looking at the energy gained; this has the advantage that it doesn't lead to a singularity when$v=0$(which would lead to "infinite force"): # bike power calculation import math import numpy as np import matplotlib.pyplot as plt # a few plausible factors from http://www.cyclingpowerlab.com/CyclingAerodynamics.aspx dragCoeff = 0.88 frontalArea = 0.32 rho = 1.2 eta = 0.97 # 1 - drive train loss = efficiency rollingCoeff = 5.0e-3 # from haskell code mass = 62 #kg grade = 0.0 g = 9.81 # functions to compute the various forces: def fDrag(velocity): return 0.5*dragCoeff*frontalArea*rho*velocity*velocity def fRolling(grade, mass, velocity): if velocity > 0.01: return g * math.cos(math.atan(grade)) * mass * rollingCoeff else: return 0.0 def fGravity(grade, mass): return g*math.sin(math.atan(grade))*mass # the actual program: v=0.0 # initial velocity power = 600 # constant power in W dt = 0.1 # time step va=[0] # store the results in a list which will grow in each iteration ta=[0] # loop over time: for t in np.arange(0,10,dt): totalForce = fDrag(v) + fRolling(grade, mass, v) + fGravity(grade, mass) powerNeeded = totalForce * v / eta netPower = power - powerNeeded # kinetic energy increases by net energy available for dt print "t = %.2f; v=%.1f; drag = %.2f N; F roll = %.2f N; F gravity = %.2f N"%(t, v, fDrag(v), fRolling(grade, mass, v), fGravity(grade, mass)) v = math.sqrt(v*v + 2 * netPower * dt * eta / mass) va.append(v) ta.append(t+dt) plt.figure() plt.plot(ta, va) plt.xlabel('time (s)') plt.ylabel('velocity (m/s)') plt.show() This gives the following plot: The values look fairly plausible. Note that drag (23.25) plus rolling resistance (3.04 N) add up to a little over 27 N by the end - accounting for about half the power used. That makes sense - at 26 mph, you need on the order of 300 W to keep going. I recommend that you check intermediate results from your program against some of these values to see where the difference is coming from. UPDATE Here are the values produced by my code (I ran time steps of 0.1 second): t = 0.00; v=0.0; drag = 0.00 N; F roll = 0.00 N; F gravity = 0.00 N t = 0.10; v=1.4; drag = 0.32 N; F roll = 3.04 N; F gravity = 0.00 N t = 0.20; v=1.9; drag = 0.63 N; F roll = 3.04 N; F gravity = 0.00 N t = 0.30; v=2.4; drag = 0.95 N; F roll = 3.04 N; F gravity = 0.00 N t = 0.40; v=2.7; drag = 1.26 N; F roll = 3.04 N; F gravity = 0.00 N t = 0.50; v=3.0; drag = 1.57 N; F roll = 3.04 N; F gravity = 0.00 N t = 0.60; v=3.3; drag = 1.88 N; F roll = 3.04 N; F gravity = 0.00 N t = 0.70; v=3.6; drag = 2.19 N; F roll = 3.04 N; F gravity = 0.00 N t = 0.80; v=3.8; drag = 2.49 N; F roll = 3.04 N; F gravity = 0.00 N t = 0.90; v=4.1; drag = 2.80 N; F roll = 3.04 N; F gravity = 0.00 N t = 1.00; v=4.3; drag = 3.10 N; F roll = 3.04 N; F gravity = 0.00 N t = 1.10; v=4.5; drag = 3.41 N; F roll = 3.04 N; F gravity = 0.00 N t = 1.20; v=4.7; drag = 3.71 N; F roll = 3.04 N; F gravity = 0.00 N t = 1.30; v=4.9; drag = 4.01 N; F roll = 3.04 N; F gravity = 0.00 N t = 1.40; v=5.0; drag = 4.31 N; F roll = 3.04 N; F gravity = 0.00 N t = 1.50; v=5.2; drag = 4.60 N; F roll = 3.04 N; F gravity = 0.00 N t = 1.60; v=5.4; drag = 4.90 N; F roll = 3.04 N; F gravity = 0.00 N t = 1.70; v=5.5; drag = 5.19 N; F roll = 3.04 N; F gravity = 0.00 N t = 1.80; v=5.7; drag = 5.48 N; F roll = 3.04 N; F gravity = 0.00 N t = 1.90; v=5.8; drag = 5.78 N; F roll = 3.04 N; F gravity = 0.00 N t = 2.00; v=6.0; drag = 6.06 N; F roll = 3.04 N; F gravity = 0.00 N t = 2.10; v=6.1; drag = 6.35 N; F roll = 3.04 N; F gravity = 0.00 N t = 2.20; v=6.3; drag = 6.64 N; F roll = 3.04 N; F gravity = 0.00 N t = 2.30; v=6.4; drag = 6.92 N; F roll = 3.04 N; F gravity = 0.00 N t = 2.40; v=6.5; drag = 7.20 N; F roll = 3.04 N; F gravity = 0.00 N t = 2.50; v=6.7; drag = 7.48 N; F roll = 3.04 N; F gravity = 0.00 N t = 2.60; v=6.8; drag = 7.76 N; F roll = 3.04 N; F gravity = 0.00 N t = 2.70; v=6.9; drag = 8.04 N; F roll = 3.04 N; F gravity = 0.00 N t = 2.80; v=7.0; drag = 8.32 N; F roll = 3.04 N; F gravity = 0.00 N t = 2.90; v=7.1; drag = 8.59 N; F roll = 3.04 N; F gravity = 0.00 N t = 3.00; v=7.2; drag = 8.86 N; F roll = 3.04 N; F gravity = 0.00 N t = 3.10; v=7.4; drag = 9.13 N; F roll = 3.04 N; F gravity = 0.00 N t = 3.20; v=7.5; drag = 9.40 N; F roll = 3.04 N; F gravity = 0.00 N t = 3.30; v=7.6; drag = 9.67 N; F roll = 3.04 N; F gravity = 0.00 N t = 3.40; v=7.7; drag = 9.93 N; F roll = 3.04 N; F gravity = 0.00 N t = 3.50; v=7.8; drag = 10.20 N; F roll = 3.04 N; F gravity = 0.00 N t = 3.60; v=7.9; drag = 10.46 N; F roll = 3.04 N; F gravity = 0.00 N t = 3.70; v=8.0; drag = 10.72 N; F roll = 3.04 N; F gravity = 0.00 N t = 3.80; v=8.1; drag = 10.97 N; F roll = 3.04 N; F gravity = 0.00 N t = 3.90; v=8.2; drag = 11.23 N; F roll = 3.04 N; F gravity = 0.00 N t = 4.00; v=8.2; drag = 11.48 N; F roll = 3.04 N; F gravity = 0.00 N t = 4.10; v=8.3; drag = 11.73 N; F roll = 3.04 N; F gravity = 0.00 N t = 4.20; v=8.4; drag = 11.98 N; F roll = 3.04 N; F gravity = 0.00 N t = 4.30; v=8.5; drag = 12.23 N; F roll = 3.04 N; F gravity = 0.00 N t = 4.40; v=8.6; drag = 12.48 N; F roll = 3.04 N; F gravity = 0.00 N t = 4.50; v=8.7; drag = 12.72 N; F roll = 3.04 N; F gravity = 0.00 N t = 4.60; v=8.8; drag = 12.97 N; F roll = 3.04 N; F gravity = 0.00 N t = 4.70; v=8.8; drag = 13.21 N; F roll = 3.04 N; F gravity = 0.00 N t = 4.80; v=8.9; drag = 13.45 N; F roll = 3.04 N; F gravity = 0.00 N t = 4.90; v=9.0; drag = 13.68 N; F roll = 3.04 N; F gravity = 0.00 N t = 5.00; v=9.1; drag = 13.92 N; F roll = 3.04 N; F gravity = 0.00 N t = 5.10; v=9.2; drag = 14.15 N; F roll = 3.04 N; F gravity = 0.00 N t = 5.20; v=9.2; drag = 14.38 N; F roll = 3.04 N; F gravity = 0.00 N t = 5.30; v=9.3; drag = 14.61 N; F roll = 3.04 N; F gravity = 0.00 N t = 5.40; v=9.4; drag = 14.84 N; F roll = 3.04 N; F gravity = 0.00 N t = 5.50; v=9.4; drag = 15.07 N; F roll = 3.04 N; F gravity = 0.00 N t = 5.60; v=9.5; drag = 15.29 N; F roll = 3.04 N; F gravity = 0.00 N t = 5.70; v=9.6; drag = 15.51 N; F roll = 3.04 N; F gravity = 0.00 N t = 5.80; v=9.6; drag = 15.73 N; F roll = 3.04 N; F gravity = 0.00 N t = 5.90; v=9.7; drag = 15.95 N; F roll = 3.04 N; F gravity = 0.00 N t = 6.00; v=9.8; drag = 16.17 N; F roll = 3.04 N; F gravity = 0.00 N t = 6.10; v=9.8; drag = 16.38 N; F roll = 3.04 N; F gravity = 0.00 N t = 6.20; v=9.9; drag = 16.60 N; F roll = 3.04 N; F gravity = 0.00 N t = 6.30; v=10.0; drag = 16.81 N; F roll = 3.04 N; F gravity = 0.00 N t = 6.40; v=10.0; drag = 17.02 N; F roll = 3.04 N; F gravity = 0.00 N t = 6.50; v=10.1; drag = 17.22 N; F roll = 3.04 N; F gravity = 0.00 N t = 6.60; v=10.2; drag = 17.43 N; F roll = 3.04 N; F gravity = 0.00 N t = 6.70; v=10.2; drag = 17.63 N; F roll = 3.04 N; F gravity = 0.00 N t = 6.80; v=10.3; drag = 17.84 N; F roll = 3.04 N; F gravity = 0.00 N t = 6.90; v=10.3; drag = 18.04 N; F roll = 3.04 N; F gravity = 0.00 N t = 7.00; v=10.4; drag = 18.23 N; F roll = 3.04 N; F gravity = 0.00 N t = 7.10; v=10.4; drag = 18.43 N; F roll = 3.04 N; F gravity = 0.00 N t = 7.20; v=10.5; drag = 18.63 N; F roll = 3.04 N; F gravity = 0.00 N t = 7.30; v=10.6; drag = 18.82 N; F roll = 3.04 N; F gravity = 0.00 N t = 7.40; v=10.6; drag = 19.01 N; F roll = 3.04 N; F gravity = 0.00 N t = 7.50; v=10.7; drag = 19.20 N; F roll = 3.04 N; F gravity = 0.00 N t = 7.60; v=10.7; drag = 19.39 N; F roll = 3.04 N; F gravity = 0.00 N t = 7.70; v=10.8; drag = 19.57 N; F roll = 3.04 N; F gravity = 0.00 N t = 7.80; v=10.8; drag = 19.76 N; F roll = 3.04 N; F gravity = 0.00 N t = 7.90; v=10.9; drag = 19.94 N; F roll = 3.04 N; F gravity = 0.00 N t = 8.00; v=10.9; drag = 20.12 N; F roll = 3.04 N; F gravity = 0.00 N t = 8.10; v=11.0; drag = 20.30 N; F roll = 3.04 N; F gravity = 0.00 N t = 8.20; v=11.0; drag = 20.48 N; F roll = 3.04 N; F gravity = 0.00 N t = 8.30; v=11.1; drag = 20.66 N; F roll = 3.04 N; F gravity = 0.00 N t = 8.40; v=11.1; drag = 20.83 N; F roll = 3.04 N; F gravity = 0.00 N t = 8.50; v=11.1; drag = 21.00 N; F roll = 3.04 N; F gravity = 0.00 N t = 8.60; v=11.2; drag = 21.17 N; F roll = 3.04 N; F gravity = 0.00 N t = 8.70; v=11.2; drag = 21.34 N; F roll = 3.04 N; F gravity = 0.00 N t = 8.80; v=11.3; drag = 21.51 N; F roll = 3.04 N; F gravity = 0.00 N t = 8.90; v=11.3; drag = 21.68 N; F roll = 3.04 N; F gravity = 0.00 N t = 9.00; v=11.4; drag = 21.84 N; F roll = 3.04 N; F gravity = 0.00 N t = 9.10; v=11.4; drag = 22.01 N; F roll = 3.04 N; F gravity = 0.00 N t = 9.20; v=11.5; drag = 22.17 N; F roll = 3.04 N; F gravity = 0.00 N t = 9.30; v=11.5; drag = 22.33 N; F roll = 3.04 N; F gravity = 0.00 N t = 9.40; v=11.5; drag = 22.49 N; F roll = 3.04 N; F gravity = 0.00 N t = 9.50; v=11.6; drag = 22.64 N; F roll = 3.04 N; F gravity = 0.00 N t = 9.60; v=11.6; drag = 22.80 N; F roll = 3.04 N; F gravity = 0.00 N t = 9.70; v=11.7; drag = 22.95 N; F roll = 3.04 N; F gravity = 0.00 N t = 9.80; v=11.7; drag = 23.10 N; F roll = 3.04 N; F gravity = 0.00 N t = 9.90; v=11.7; drag = 23.25 N; F roll = 3.04 N; F gravity = 0.00 N • Apologies, those multiplications are a typo in my question and not present in the Haskell code, I'll correct for that. Commented Jan 1, 2016 at 1:33 • Still not sure where your$\rm{\frac{1}{drivetrain~loss}}$comes from when you have loss=0.03. I think you need$\rm{\frac{1}{1-(drivetrain~loss)}}$Commented Jan 1, 2016 at 1:40 • You are correct. Poor LaTeXing from me. Commented Jan 1, 2016 at 1:45 • This still feels a little to slow in terms of acceleration to me, but it's the most convincing answer so I've accepted it. Thanks for helping! I'll continue to explore modelling this to see if I can work out why in reality 600W accelerates me much faster. (I know that I can hold 600W for about 5 seconds without running out of energy, and after that I'm well up to 30km/h) Commented Jan 2, 2016 at 11:05 • Note that my plot has units of m/s - so after five seconds you are at roughly 9 m/s which is (9x3.6=)32 km/h; this agrees nicely with your experience. Commented Jan 2, 2016 at 14:10 You should first focus on energy, then find the kinetic energy at different steps in time and then you know the velocity. From there you can calculate the acceleration if you want. So at one time$t_1$you have kinetic energy$1/2mv_1^2$. After one time step this energy changes by$P\Delta t\eta$, where$\eta $is efficacy. So your new kinetic energy$1/2mv_2^2 = 1/2mv_1^2 + P\Delta t\eta$Solving for v:$v_2 = \sqrt{v_1^2 + \frac{2P\Delta t\eta}{m}}$Acceleration can now be computed but if you just wanted velocity is not necessary. • Something seems to be missing here though, because that would imply that if I provide a fixed power my velocity increases indefinitely - what's missing to bound the velocity? Commented Dec 31, 2015 at 17:12 • Viscous resistance from the air! Commented Dec 31, 2015 at 17:40 • I don't think your solving for$v_2$correctly either. I think$v_2 = \frac{\sqrt{2 \delta t P + m v_1^2}}{\sqrt{m}}$Commented Dec 31, 2015 at 18:56 • Sorry ignore my last comment, I read the equation incorrectly. Commented Jan 1, 2016 at 11:16 • @ocharles I reformatted the equation to show the calculation (which I borrowed for my demo code) is correct. Commented Jan 2, 2016 at 14:16 Here is my attempt: import Data.Foldable main = do let vs = iterate step v0 -- list of velocities every 0.1 s forM_ (take 1000$ zip [dt, (2 * dt) ..] vs) $\(t, v) -> do putStrLn$ show t ++ "\t" ++ show v
where
v0 = 0
step v | v < v' = v + f0 / m * dt - c * rho * a * v * v / m * dt
step v = v + p / m / v * dt - c * rho * a * v * v / m * dt
c = 0.5
m = 70 -- kg
a = 0.33 -- m^2
dt = 0.1 -- s
p = 400 -- W
rho = 1 -- kg/m^3
f0 = p / v'
v' = 7 -- m/s
-- to run:
-- runhaskell main.hs | graph -X "Time, sec" -Y "Velocity, m/sec" | plot -TX
It is based on "Computational physics" book by Nicholas Giordano.
The idea is to use apply constant force f0 initially and switch to constant power p at the moment when the force produces the same power, e.g. f0 * v' = p. The critical velocity v' is chosen semi-arbitrary.
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# Beautiful alley
In the alley 4 rows of 10 trees and 5 rows of 8 trees blossomed. Six of the trees have already blossomed. How many trees still bloom in the alley?
Result
n = 74
#### Solution:
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Find the four times smaller number for numbers 12 and 36 and then add them. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.hackmath.net/en/example/5005", "fetch_time": 1558449811000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-22/segments/1558232256426.13/warc/CC-MAIN-20190521142548-20190521164548-00315.warc.gz", "warc_record_offset": 815679208, "warc_record_length": 6641, "token_count": 654, "char_count": 2588, "score": 3.6875, "int_score": 4, "crawl": "CC-MAIN-2019-22", "snapshot_type": "latest", "language": "en", "language_score": 0.9789485335350037, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00009-of-00064.parquet"} |
# Statistical distance
In statistics, probability theory, and information theory, a statistical distance quantifies the distance between two statistical objects, which can be two random variables, or two probability distributions or samples, or the distance can be between an individual sample point and a population or a wider sample of points.
A distance between populations can be interpreted as measuring the distance between two probability distributions and hence they are essentially measures of distances between probability measures. Where statistical distance measures relate to the differences between random variables, these may have statistical dependence,[1] and hence these distances are not directly related to measures of distances between probability measures. Again, a measure of distance between random variables may relate to the extent of dependence between them, rather than to their individual values.
Many statistical distance measures are not metrics, and some are not symmetric. Some types of distance measures, which generalize squared distance, are referred to as (statistical) divergences.
## Terminology
Many terms are used to refer to various notions of distance; these are often confusingly similar, and may be used inconsistently between authors and over time, either loosely or with precise technical meaning. In addition to "distance", similar terms include deviance, deviation, discrepancy, discrimination, and divergence, as well as others such as contrast function and metric. Terms from information theory include cross entropy, relative entropy, discrimination information, and information gain.
## Distances as metrics
### Metrics
A metric on a set X is a function (called the distance function or simply distance) d : X × XR+ (where R+ is the set of non-negative real numbers). For all x, y, z in X, this function is required to satisfy the following conditions:
1. d(x, y) ≥ 0 (non-negativity)
2. d(x, y) = 0 if and only if x = y (identity of indiscernibles. Note that condition 1 and 2 together produce positive definiteness)
3. d(x, y) = d(y, x) (symmetry)
4. d(x, z) ≤ d(x, y) + d(y, z) (subadditivity / triangle inequality).
### Generalized metrics
Many statistical distances are not metrics, because they lack one or more properties of proper metrics. For example, pseudometrics violate property (2), identity of indiscernibles; quasimetrics violate property (3), symmetry; and semimetrics violate property (4), the triangle inequality. Statistical distances that satisfy (1) and (2) are referred to as divergences.
## Statistically close
The total variation distance of two distributions ${\displaystyle X}$ and ${\displaystyle Y}$ over a finite domain ${\displaystyle D}$, (often referred to as statistical difference[2] or statistical distance[3] in cryptography) is defined as
${\displaystyle \Delta (X,Y)={\frac {1}{2}}\sum _{\alpha \in D}|\Pr[X=\alpha ]-\Pr[Y=\alpha ]|}$.
We say that two probability ensembles ${\displaystyle \{X_{k}\}_{k\in \mathbb {N} }}$ and ${\displaystyle \{Y_{k}\}_{k\in \mathbb {N} }}$ are statistically close if ${\displaystyle \Delta (X_{k},Y_{k})}$ is a negligible function in ${\displaystyle k}$.
## Notes
1. ^ Dodge, Y. (2003)—entry for distance
2. ^ Goldreich, Oded (2001). Foundations of Cryptography: Basic Tools (1st ed.). Berlin: Cambridge University Press. p. 106. ISBN 0-521-79172-3.
3. ^ Reyzin, Leo. (Lecture Notes) Extractors and the Leftover Hash Lemma | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://en.wikipedia.org/wiki/Statistical_distance", "fetch_time": 1716353493000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-22/segments/1715971058531.78/warc/CC-MAIN-20240522040032-20240522070032-00585.warc.gz", "warc_record_offset": 194220722, "warc_record_length": 28504, "token_count": 807, "char_count": 3485, "score": 4.0625, "int_score": 4, "crawl": "CC-MAIN-2024-22", "snapshot_type": "latest", "language": "en", "language_score": 0.9358751177787781, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00044-of-00064.parquet"} |
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Home Science And Mathematics A Simple Guide to Converting 2007 to Roman Numerals
# A Simple Guide to Converting 2007 to Roman Numerals
0
It’s easy to turn 2007 into Roman numerals. Here’s what you need to do:
Step 1: Find the Roman number with the highest value that is less than or equal to 2007.
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MM, which stands for 2000, is the highest Roman number that is less than or equal to 2007.
Step 2: Write 2000 in Roman numerals (MM).
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We can start by writing MM, which is 2000 written twice as the letter M.
Step 3: Find the next highest Roman number that is less than the number you have left.
To find the next highest Roman number, we need to subtract 2000 from 2007:
2007 – 2000 = 7
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VII, which means 7, is the highest Roman number that is less than 7.
Step 4: Write the Roman number for the number 7. (VII).
After the MM we wrote in step 2, we can write VII, which means 7.
So, in Roman numbers, the year 2007 is written as MMVII.
In conclusion, converting a number like 2007 to Roman numerals is a quick and easy process that involves finding the highest Roman numeral values and writing them in order. If you follow the steps above, it’s easy to turn any number into its Roman numeral equivalent.
Advertisement | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://unilorinforum.com/a-simple-guide-to-converting-2007-to-roman-numerals/", "fetch_time": 1718667695000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-26/segments/1718198861741.14/warc/CC-MAIN-20240617215859-20240618005859-00095.warc.gz", "warc_record_offset": 525316690, "warc_record_length": 44985, "token_count": 328, "char_count": 1287, "score": 3.921875, "int_score": 4, "crawl": "CC-MAIN-2024-26", "snapshot_type": "latest", "language": "en", "language_score": 0.9085601568222046, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00054-of-00064.parquet"} |
# Quick Answer: How Many Weekends Are In A Year?
## How many minutes are in a day?
1440 minutesThere is total 1440 minutes in a day.
You can determine it in easy calculation.
One hour is consist of 60 minutes.
One day is consist of 24 hours..
## What are the 5 week months in 2021?
2021: January, April, July, October, December. 2022: April, July, September, December. 2023: March, June, September, December.
## How many working days are there in 2021?
251 working daysHow Many Working Days in 2021? – There are 251 working days, 10 public holidays, 104 weekend days for a total of 365 days in 2021.
## Can there be 53 Sundays in a year?
A non-leap year contains 365 days and 365 = (52 x 7) + 1. That means that for a non-leap year to contain 53 Sundays, 1 January must fall on a Sunday. A leap year contains 366 days = (52 x 7) + 2 days, so for a leap year to contain 53 Sundays, 1 January or 2 January must be a Sunday.
## How many days are there in weekends?
We know that a Each normal year has 365 days or 52 weeks plus one day, and each week has two weekend days, which means there are approximately 104 weekend days each year. Whereas in a leap year we have 366 days it adds one more day to the year.
## How many Saturdays are there in 2021 a year?
52 SaturdaysHow many Saturdays are there in 2021? There are exactly 52 Saturdays in the year 2021. Most years have 365 days, but a leap year has 366 days. That adds up to 52 weeks (where each week is exactly 7 days) PLUS 1 or 2 additional days.
## How many full weeks are in a year?
52 weeksThat means a year is actually 52 weeks plus one day or, in a leap year it’s 52 weeks plus two days. The leap cycle in the Gregorian calendar has 97 leap days spread across 400 years which is an exact number of 20,871 weeks.
## How many weekends are in 6 months?
Months to Weeks conversion table1 Month = 4.3 Weeks11 Months = 47.1 Weeks6 Months = 25.7 Weeks16 Months = 68.6 Weeks7 Months = 30 Weeks17 Months = 72.9 Weeks8 Months = 34.3 Weeks18 Months = 77.1 Weeks9 Months = 38.6 Weeks19 Months = 81.4 Weeks5 more rows
## How many Sundays will be there in a period of 100 years?
Number of days in a period of 100 years is 365 x 100 + 23 or 365 x 100 + 24. If century year is not a leap year then number of days = 365 x 100 + 23, and number of weeks is 5217 and 4 odd and for leap year it will be 5217 weeks and 5 odd days, hence number of Sundays is either 5217 or 5218.
## How many Fridays are there in 2020 a year?
52 FridaysHow many Fridays are there in 2020? There are exactly 52 Fridays in the year 2020. Most years have 365 days, but a leap year has 366 days. That adds up to 52 weeks (where each week is exactly 7 days) PLUS 1 or 2 additional days.
## How many Sundays are there in 2020 a year?
52 SundaysThere are exactly 52 Sundays in the year 2020. 2020 is a leap year, so there are 366 days in this year.
## How many weekends are there in 10 years?
Years to Weeks conversion table1 Year = 52.14 Weeks11 Years = 573.57 Weeks7 Years = 365 Weeks17 Years = 886.43 Weeks8 Years = 417.14 Weeks18 Years = 938.57 Weeks9 Years = 469.29 Weeks19 Years = 990.71 Weeks10 Years = 521.43 Weeks20 Years = 1042.86 Weeks5 more rows
## Is there a February 29 in 2021?
If so, we are sorry to report that there is no such date as February 29, 2021. …
## How many weekends are in a year 2020?
52 SaturdaysThere are exactly 52 Saturdays in the year 2020. Most years have 365 days, but a leap year has 366 days. That adds up to 52 weeks (where each week is exactly 7 days) PLUS 1 or 2 additional days.
## How many Saturdays and Sundays are there in a year?
52 SaturdaysCalendar year has either 364 days or 365 days. If we consider 364 days then we have 364/7=52 weeks and hence there are 52 Saturdays and 52 Sundays will fall in to calendar year.
## How many Saturdays and Sundays are there in 2020?
52 SaturdaysThere are 52 Saturdays and 52 Sundays in 2020.
## How many weekdays are in the year 2020?
262 working daysThere are a total of 262 working days in the 2020 calendar year.
## How many days are there in a year without weekends?
Okay so the results are a bit depressing, but it can actually be useful to help you decide when to get away on holiday. And even though we know there’s 52 weeks and 365 days in a year – unless it’s a leap year. So this year, there are 252 working days, with 104 days that fall on weekends and eight public bank holidays. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://fablabelectronics.com/qa/quick-answer-how-many-weekends-are-in-a-year.html", "fetch_time": 1624438973000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-25/segments/1623488536512.90/warc/CC-MAIN-20210623073050-20210623103050-00412.warc.gz", "warc_record_offset": 219151552, "warc_record_length": 8507, "token_count": 1227, "char_count": 4428, "score": 4.125, "int_score": 4, "crawl": "CC-MAIN-2021-25", "snapshot_type": "latest", "language": "en", "language_score": 0.9585646986961365, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00057-of-00064.parquet"} |
# Thread: completion of set in metrix spaces
1. ## completion of set in metrix spaces
Hi, can one help me please;
For S=(0,1)
We know the Cauchy sequences Xn=1/n converges to zero which is not in the set
and Xn=1/n+1 converges to one which is also not in the set
so S=[0,1].
For S= [0,1] n {Rational numbers}
is it just the set of all Real numbers.
For S={1/k:k=1,2,3,...}
Many thanks
2. What does completion mean?
Is that what the rest of us know as closure?
3. Hi, yes it is closure. so that the metric can be complete.
4. a) & b) $~\overline{S}=[0,1]$
c) $~\overline{S}=S\cup \{0\}$
5. Originally Posted by Plato
a) & b) $~\overline{S}=[0,1]$
c) $~\overline{S}=S\cup \{0\}$
For B) if one took the set Xn= x/2+1/x then this in the set S= [0,1] n {Rational numbers} and the sequence converges to
square root 2 but that not within the set [0,1].
SO how can For B) b) $~\overline{S}=[0,1]$
6. Originally Posted by nerdo
For B) if one took the set Xn= x/2+1/x then this in the set S= [0,1] n {Rational numbers} and the sequence converges to
square root 2 but that not within the set [0,1].
SO how can For B) b) $~\overline{S}=[0,1]$
The fact that $\overline{\mathbb{Q}\cap[0,1]}=[0,1]$ follows since $\mathbb{Q}$ is dense in $\mathbb{R}$ and the fact that $\overline{\mathbb{Q}\cap [0,1]}\supseteq\overline{\mathbb{Q}}\cap\overline{[0,1]}=\mathbb{R}\cap[0,1]=[0,1]$. But, since $\mathbb{Q}\cap[0,1]\subseteq[0,1]$ we have that $\overline{\mathbb{Q}\cap[0,1]}\subseteq\overline{[0,1]}=[0,1]$ from where the conclusion follows. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://mathhelpforum.com/differential-geometry/164991-completion-set-metrix-spaces.html", "fetch_time": 1503036915000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-34/segments/1502886104565.76/warc/CC-MAIN-20170818043915-20170818063915-00325.warc.gz", "warc_record_offset": 262797126, "warc_record_length": 11261, "token_count": 576, "char_count": 1541, "score": 3.9375, "int_score": 4, "crawl": "CC-MAIN-2017-34", "snapshot_type": "longest", "language": "en", "language_score": 0.7494375109672546, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00021-of-00064.parquet"} |
# Resources
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## Powers I
You will learn the definition of power and exponent, and you will come across some interesting examples that will help you through the learning process.
At the end of the theorical explanation, you can ...
## System of decimal, binary and hexadecimal numeration
You'll learn some systems of numberation such as the decimal one, the binary one and the hexadecimal one. For that purpose, you will be given some interesting examples in order to improve the lear ...
## Definition, basic elements and types of centers of the triangle
You'll learn the deifnition of a triangle, its basic elements, and the types of centers of a triangle.
At the end of the theorical explanation, you can practice or create 5 exercises to master this ...
## Measurement of angles in degrees, minutes and seconds
You'll learn how to measure an angle. For that purpose, you will be given some examples.
At the end of the theorical explanation, you can practice or create 5 exercises to master this subject.
Measu ...
## Simple interest
You'll learn the algorithms that allow the calculation of simple interest.
At the end of the theorical explanation, you can practice or create 5 exercises to master this subject.
Simple interest es ...
## Absolute deviation and Standard deviation
First of all you'll learn to analyze the statistical disperion of some information. After you'll learn to calculate absolute deviations, standar deviation of a set of information and to work ...
## Ruffini's rule
You'll learn Ruffini's rule to divide polynomials and their applications in different theorems. You will also learn the concept of root of a polynomial and some techniques for calculating it.
R ...
## Linear diophantine equation
You'll learn to solve linear diophantine equations. For that purpose, you will be given an interesting example.
At the end of the theorical explanation, you can practice or create 5 exercises to ma ...
## Logarithms: definition and properties
You'll learn the basic concepts of logarithms and related definitions and algorithms that allows us to deal more easilly algebraic expressions that contain logarithms.
At the end of the theorical e ...
## Rouché–Capelli theorem
You'll learn how to apply the Rouché-Capelli theorem.
Rouché–Capelli theorem is a didactic content from Sangakoo. In this space you will find teaching materials for Mathematics ... | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://resources.profuturo.education/en/educational-resources/category/sangakoo/14fd4c9c-c50e-4a49-8764-bae089672b6d/?pagina=4", "fetch_time": 1696356026000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-40/segments/1695233511170.92/warc/CC-MAIN-20231003160453-20231003190453-00094.warc.gz", "warc_record_offset": 518695856, "warc_record_length": 13277, "token_count": 514, "char_count": 2457, "score": 3.640625, "int_score": 4, "crawl": "CC-MAIN-2023-40", "snapshot_type": "latest", "language": "en", "language_score": 0.9040798544883728, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00063-of-00064.parquet"} |
## 5.9 Omitted Variable Bias
SW 6.1
Suppose that we are interested in the following regression model
$\mathbb{E}[Y|X_1, X_2, Q] = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \beta_3 Q$ and, in particular, we are interested in the the partial effect
$\frac{ \partial \, \mathbb{E}[Y|X_1,X_2,Q]}{\partial \, X_1} = \beta_1$ But we are faced with the issue that we do not observe $$Q$$ (which implies that we cannot control for it in the regression)
Recall that we can equivalently write
$Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \beta_3 Q + U \tag{5.1}$ where $$\mathbb{E}[U|X_1,X_2,Q]=0$$.
Now, for simplicity, suppose that
$\mathbb{E}[Q | X_1, X_2] = \gamma_0 + \gamma_1 X_1 + \gamma_2 X_2$
Now, let’s consider the idea of just running a regression of $$Y$$ on $$X_1$$ and $$X_2$$ (and just not including $$Q$$); in other words, consider the regression $\mathbb{E}[Y|X_1,X_2] = \delta_0 + \delta_1 X_1 + \delta_2 X_2$ We are interested in the question of whether or not we can recover $$\beta_1$$ if we do this. If we consider this “feasible” regression, notice if we plug in the expression for $$Y$$ from Equation (5.1),
\begin{aligned} \mathbb{E}[Y|X_1,X_2] &= \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \beta_3 \mathbb{E}[Q|X_1,X_2] \\ &= \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \beta_3 (\gamma_0 + \gamma_1 X_1 + \gamma_2 X_2) \\ &= \underbrace{(\beta_0 + \beta_3 \gamma_0)}_{\delta_0} + \underbrace{(\beta_1 + \beta_3 \gamma_1)}_{\delta_1} X_1 + \underbrace{(\beta_2 + \beta_3 \gamma_2)}_{\delta_2} X_2 \end{aligned}
In other words, if we run the feasible regression of $$Y$$ on $$X_1$$ and $$X_2$$, $$\delta_1$$ (the coefficient on $$X_1$$) is not equal to $$\beta_1$$; rather, it is equal to $$(\beta_1 + \beta_3 \gamma_1)$$.
That you are not generally able to recover $$\beta_1$$ in this case is called omitted variable bias
There are two cases where you will recover $$\delta_1 = \beta_1$$ though which occur when $$\beta_3 \gamma_1 = 0$$:
• $$\beta_3=0$$. This would be the case where $$Q$$ has no effect on $$Y$$
• $$\gamma_1=0$$. This would be the case where $$X_1$$ and $$Q$$ are uncorrelated after controlling for $$X_2$$.
Interestingly, there may be some case where you can “sign” the bias; i.e., figure out if $$\beta_3 \gamma_1$$ is positive or negative. For example, you might have theoretical reasons to suspect that $$\gamma_1 > 0$$ and $$\beta_3 > 0$$. In this case,
$\delta_1 = \beta_1 + \textrm{something positive}$ which implies that $$\delta_1$$ (i.e., running a regression that ignores $$Q$$) would cause us to tend to over-estimate $$\beta_1$$.
Side-Comment:
• The book talks about omitted variable bias in the context of causality (this is probably the leading case), but we have not talked about causality yet. The same issues arise if we just say that we have some regression of interest but are unable to estimate it because some covariates are unobserved.
• The relationship to causality (which is not so important for now), is that under certain conditions, we may have a particular partial effect that we would be willing to interpret as being the “causal effect”, but if we are unable to control for some variables that would lead to this interpretation, then we get to the issues pointed out in the textbook. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://bcallaway11.github.io/econ_4750_notes/omitted-variable-bias.html", "fetch_time": 1701249049000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-50/segments/1700679100057.69/warc/CC-MAIN-20231129073519-20231129103519-00353.warc.gz", "warc_record_offset": 150799866, "warc_record_length": 9461, "token_count": 1067, "char_count": 3252, "score": 3.78125, "int_score": 4, "crawl": "CC-MAIN-2023-50", "snapshot_type": "latest", "language": "en", "language_score": 0.7360247373580933, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00056-of-00064.parquet"} |
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# IEC
Fall 2015
Assignment #1
Topics: Differential Pairs, Active Loads, Small Signal Analysis
Due Date: Sept. 30, 2015
Problem 1: (Section 7.1) For the differential amplifier specified in problem
7.1 of Sedra and Smith, assume G2 has been grounded (0 V) and vG1 has
been adjusted to yield iD1 = 0.11 mA and iD2 = 0.09 mA.
(a) What are the corresponding values of vGS1, vGS2, vS , and vid? [vGS1 =
0.97 V, vGS2 = 0.94 V, vS = -0.94 V, vid = 0.03 V ]
(b) What is the differential voltage gain (vD2 vD1/vid)? [3.33 ]
Problem 2: (Section 7.1) Sedra and Smith: Chapter 7, problem 7.8. [VGS =
0.99 V, gm = 1.06 mA/V, vid = 0.27 V, Ibias = 800 A]
Problem 3: (Section 7.2) We would like to design an NMOS differential
amplifier that operates with a differential input voltage as high as vid =
0.2 V. To keep the gain characteristics reasonably linear, we need the
value under the square root in Equation (7.23) of Sedra and Smith to be
no smaller than 0.9. Finally, assume that nCox = 100 A/V2 and a gm of
3 mA/V is required.
(a) What overdrive voltage (VOV ) is required for the transistors? [0.316 V ]
(b) What current (I) should the pair be biased with? [0.95 mA]
(c) What W/L ratio should the transistors be sized with? [95 ]
(d) What differential gain results for RD = 5 k, neglecting channel length
modulation? [15 ]
(e) What is the differential output voltage corresponding to the maximum
differential input voltage? [3 V ]
Problem 4: (Section 7.2) Consider the differential pair shown in Fig. 7.4 of
Sedra and Smith, where vid is a small signal sine wave. The pair is biased
with I = 300
A, and nCox = 100 A/V2. A design error has resulted in a sizing mismatch
between the transistors, with (W/L)1 = 10 and (W/L)2 = 20. You may neglect
channel length modulation.
(a) What are the steady state bias currents for each transistor, ID1 and ID2?
[ID1 = 100 A, ID2 = 200 A]
(b) What is VOV for the two transistors? [0.447 V ]
(c) What is the differential gain (Ad) assuming RD = 5 k? [6.71 ]
## Problem 5: (Section 7.2) Sedra and Smith: Chapter 7, problem
7.15. (a) Answer: [|Ad| = 3.85, |Acm| = 0.05, CMRR = 37.7 dB
]
(b) Answer: [|Ad| = 7.7, |Acm| = 0.0005, CMRR = 83.7 dB ]
Problem 6: (Section 7.4) An NMOS differential pair operating at a bias current
I = 100 A uses transistors with n = 100 A/V2, W/L = 20, and Vt = 0.8 V.
kt
(a) How much input offset voltage is introduced if RD/RD = 5%? [5.57 mV ]
(b) How much input offset voltage is introduced if (W/L)/(W/L) =
5%? [5.57 mV ]
(c) How much input offset voltage is introduced if Vt = 5 mV? [5 mV ]
(d) What is the worst case input offset voltage if all three effects are present?
[16.14 mV ]
Problem 7: (Section 6.5) Consider the common source amplifier of Fig. 6.18(a)
2
in the text, where ktn =
p = 250 A/V , |Vt| = 0.6 V, and |VA| = 10 V.
2.5kt
(a) What is the required IREF to obtain an output resistance of 100 k? [0.05
mA]
(b) Find (W/L)1 to obtain a gain of -40 V/V with IREF as specified above.
[6.4 ]
(c) If Q2 and Q3 must have the same overdrive voltages as Q1, find their W/L
ratios. [16 ]
Problem 8:
[gm1 gm2 r2 /4]
## Problem 9: (Section 6.5) Consider the amplifier in Fig. 1. The power
supply voltage is VDD = 5 V; for NMOS devices nCox = 100 A/V2, for
PMOS devices pCox = 50 A/V2, |Vt| = 1 V for all devices, and the sizing
of the transistors is as follows: (W/L)1 = 8m/2m, (W/L)2 = 4m/2m,
and (W/L)3
= 4m/2m. You may ignore channel length modulation for M2 and M3, but
you must consider it for M1 with VA t = 15 V/m.
(a) What is the small signal input resistance, rin? [1620
(b) What is the small signal output resistance, rout? [1480
]
]
(c) What is the small signal voltage gain, vout/vsig ? [-1.24 V/V | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://id.scribd.com/document/283086598/Assignment-1", "fetch_time": 1560712453000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-26/segments/1560627998291.9/warc/CC-MAIN-20190616182800-20190616204800-00385.warc.gz", "warc_record_offset": 487161365, "warc_record_length": 65377, "token_count": 1287, "char_count": 3704, "score": 3.59375, "int_score": 4, "crawl": "CC-MAIN-2019-26", "snapshot_type": "latest", "language": "en", "language_score": 0.8846743702888489, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00039-of-00064.parquet"} |
# What are the chances that 2 arbitrary ranked cards can be found adjacent to one another in a shuffled pack of cards
What are the chances that 2 arbitrary ranked cards can be found adjacent to one another in a shuffled pack of cards. Ranks being the face value of the cards (ace, two, three, etc), adjacent being that they are next to each other if the card were fanned out, the order is not important. The example is I chose Queen and Ace, I shuffle the pack, what is the chance that the pack contains a Queen next to an Ace.
My initial thoughts are that it should be a number approaching 8/13.
A first approximation is to assume the queens are far apart. There are $48$ places the aces can go and $8$ of them are unacceptable. The chance of success for the first ace is $\frac {40}{48}$, the second $\frac {39}{47}$ as one of the good places is occupied. The result of this is $\frac {40\cdot 39 \cdot 38 \cdot 37}{48 \cdot 47 \cdot 46 \cdot 45}=\frac {40!\cdot 44!}{48!\cdot 36!}=\frac {9139}{19458}\approx 47\%$
The chances are actually better than this. If a queen is at the end of the deck, there is one less failing space. If two queens are next to each other, there are two less failing spaces. If two queens are one space apart, there is one less failing space. To get the final answer, you would have to calculate the probability of each number of failing spaces from $1$ (all four queens together at one end of the deck) to $8$, the chance of success for each of these, and add them up. A simulation would probably be easier. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/548060/what-are-the-chances-that-2-arbitrary-ranked-cards-can-be-found-adjacent-to-one", "fetch_time": 1563384128000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-30/segments/1563195525355.54/warc/CC-MAIN-20190717161703-20190717183703-00174.warc.gz", "warc_record_offset": 464060122, "warc_record_length": 35013, "token_count": 388, "char_count": 1540, "score": 4.03125, "int_score": 4, "crawl": "CC-MAIN-2019-30", "snapshot_type": "latest", "language": "en", "language_score": 0.9460883140563965, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00049-of-00064.parquet"} |
# Search by Topic
#### Resources tagged with Investigations similar to Fair Feast:
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### There are 151 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Investigations
### It Was 2010!
##### Age 5 to 11 Challenge Level:
If the answer's 2010, what could the question be?
### Magic Constants
##### Age 7 to 11 Challenge Level:
In a Magic Square all the rows, columns and diagonals add to the 'Magic Constant'. How would you change the magic constant of this square?
### Number Squares
##### Age 5 to 11 Challenge Level:
Start with four numbers at the corners of a square and put the total of two corners in the middle of that side. Keep going... Can you estimate what the size of the last four numbers will be?
### Five Coins
##### Age 7 to 11 Challenge Level:
Ben has five coins in his pocket. How much money might he have?
### Mobile Numbers
##### Age 5 to 11 Challenge Level:
In this investigation, you are challenged to make mobile phone numbers which are easy to remember. What happens if you make a sequence adding 2 each time?
### Train Carriages
##### Age 5 to 11 Challenge Level:
Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done?
### Month Mania
##### Age 5 to 11 Challenge Level:
Can you design a new shape for the twenty-eight squares and arrange the numbers in a logical way? What patterns do you notice?
### Abundant Numbers
##### Age 7 to 11 Challenge Level:
48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?
### Bean Bags for Bernard's Bag
##### Age 7 to 11 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### Sometimes We Lose Things
##### Age 7 to 11 Challenge Level:
Well now, what would happen if we lost all the nines in our number system? Have a go at writing the numbers out in this way and have a look at the multiplications table.
### Polo Square
##### Age 7 to 11 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
### Exploring Wild & Wonderful Number Patterns
##### Age 7 to 11 Challenge Level:
EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules.
### Follow the Numbers
##### Age 7 to 11 Challenge Level:
What happens when you add the digits of a number then multiply the result by 2 and you keep doing this? You could try for different numbers and different rules.
### The Pied Piper of Hamelin
##### Age 7 to 11 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Worms
##### Age 7 to 11 Challenge Level:
Place this "worm" on the 100 square and find the total of the four squares it covers. Keeping its head in the same place, what other totals can you make?
### Sending Cards
##### Age 7 to 11 Challenge Level:
This challenge asks you to investigate the total number of cards that would be sent if four children send one to all three others. How many would be sent if there were five children? Six?
### Calendar Patterns
##### Age 7 to 11 Challenge Level:
In this section from a calendar, put a square box around the 1st, 2nd, 8th and 9th. Add all the pairs of numbers. What do you notice about the answers?
### It's All about 64
##### Age 7 to 11 Challenge Level:
Write the numbers up to 64 in an interesting way so that the shape they make at the end is interesting, different, more exciting ... than just a square.
### Balance of Halves
##### Age 7 to 11 Challenge Level:
Investigate this balance which is marked in halves. If you had a weight on the left-hand 7, where could you hang two weights on the right to make it balance?
### Plants
##### Age 5 to 11 Challenge Level:
Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this?
### Doplication
##### Age 7 to 11 Challenge Level:
We can arrange dots in a similar way to the 5 on a dice and they usually sit quite well into a rectangular shape. How many altogether in this 3 by 5? What happens for other sizes?
### Newspapers
##### Age 7 to 11 Challenge Level:
When newspaper pages get separated at home we have to try to sort them out and get things in the correct order. How many ways can we arrange these pages so that the numbering may be different?
### Caterpillars
##### Age 5 to 7 Challenge Level:
These caterpillars have 16 parts. What different shapes do they make if each part lies in the small squares of a 4 by 4 square?
### Street Sequences
##### Age 5 to 11 Challenge Level:
Investigate what happens when you add house numbers along a street in different ways.
### My New Patio
##### Age 7 to 11 Challenge Level:
What is the smallest number of tiles needed to tile this patio? Can you investigate patios of different sizes?
### Birthday Cake Candles
##### Age 7 to 11 Challenge Level:
This challenge involves calculating the number of candles needed on birthday cakes. It is an opportunity to explore numbers and discover new things.
### Sets of Four Numbers
##### Age 7 to 11 Challenge Level:
There are ten children in Becky's group. Can you find a set of numbers for each of them? Are there any other sets?
### New House
##### Age 7 to 11 Challenge Level:
In this investigation, you must try to make houses using cubes. If the base must not spill over 4 squares and you have 7 cubes which stand for 7 rooms, what different designs can you come up with?
### Sweets in a Box
##### Age 7 to 11 Challenge Level:
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
### Sets of Numbers
##### Age 7 to 11 Challenge Level:
How many different sets of numbers with at least four members can you find in the numbers in this box?
### Watch Your Feet
##### Age 7 to 11 Challenge Level:
I like to walk along the cracks of the paving stones, but not the outside edge of the path itself. How many different routes can you find for me to take?
### Cubes Here and There
##### Age 7 to 11 Challenge Level:
How many shapes can you build from three red and two green cubes? Can you use what you've found out to predict the number for four red and two green?
### Pebbles
##### Age 7 to 11 Challenge Level:
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
### Round and Round the Circle
##### Age 7 to 11 Challenge Level:
What happens if you join every second point on this circle? How about every third point? Try with different steps and see if you can predict what will happen.
### Calcunos
##### Age 7 to 11 Challenge Level:
If we had 16 light bars which digital numbers could we make? How will you know you've found them all?
### Homes
##### Age 5 to 7 Challenge Level:
There are to be 6 homes built on a new development site. They could be semi-detached, detached or terraced houses. How many different combinations of these can you find?
### 3 Rings
##### Age 7 to 11 Challenge Level:
If you have three circular objects, you could arrange them so that they are separate, touching, overlapping or inside each other. Can you investigate all the different possibilities?
### Building with Rods
##### Age 7 to 11 Challenge Level:
In how many ways can you stack these rods, following the rules?
### Crossing the Town Square
##### Age 7 to 11 Challenge Level:
This tricky challenge asks you to find ways of going across rectangles, going through exactly ten squares.
### It Figures
##### Age 7 to 11 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### Magazines
##### Age 7 to 11 Challenge Level:
Let's suppose that you are going to have a magazine which has 16 pages of A5 size. Can you find some different ways to make these pages? Investigate the pattern for each if you number the pages.
### Street Party
##### Age 7 to 11 Challenge Level:
The challenge here is to find as many routes as you can for a fence to go so that this town is divided up into two halves, each with 8 blocks.
### Bracelets
##### Age 7 to 11 Challenge Level:
Investigate the different shaped bracelets you could make from 18 different spherical beads. How do they compare if you use 24 beads?
### Ice Cream
##### Age 7 to 11 Challenge Level:
You cannot choose a selection of ice cream flavours that includes totally what someone has already chosen. Have a go and find all the different ways in which seven children can have ice cream.
### Halloween Investigation
##### Age 7 to 11 Challenge Level:
Ana and Ross looked in a trunk in the attic. They found old cloaks and gowns, hats and masks. How many possible costumes could they make?
### Ip Dip
##### Age 5 to 11 Challenge Level:
"Ip dip sky blue! Who's 'it'? It's you!" Where would you position yourself so that you are 'it' if there are two players? Three players ...?
### Gran, How Old Are You?
##### Age 7 to 11 Challenge Level:
When Charlie asked his grandmother how old she is, he didn't get a straightforward reply! Can you work out how old she is?
### Sorting the Numbers
##### Age 5 to 11 Challenge Level:
Complete these two jigsaws then put one on top of the other. What happens when you add the 'touching' numbers? What happens when you change the position of the jigsaws?
### Stairs
##### Age 5 to 11 Challenge Level:
This challenge is to design different step arrangements, which must go along a distance of 6 on the steps and must end up at 6 high.
### Buying a Balloon
##### Age 7 to 11 Challenge Level:
Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon? | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://nrich.maths.org/public/leg.php?code=-333&cl=1&cldcmpid=2361", "fetch_time": 1566242869000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-35/segments/1566027314904.26/warc/CC-MAIN-20190819180710-20190819202710-00437.warc.gz", "warc_record_offset": 580715001, "warc_record_length": 9887, "token_count": 2394, "char_count": 10400, "score": 3.90625, "int_score": 4, "crawl": "CC-MAIN-2019-35", "snapshot_type": "latest", "language": "en", "language_score": 0.9037196636199951, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00019-of-00064.parquet"} |
1. Example:
The kinetic energy in a satellite is the same the machine gun attached to the satellite. Then the machine gun shoots a bullet caliber 60 and produced a thrust of more or less 20 kg.
My question is: this shot will propel the satellite at rest, or in motion?
2. The statement is confusing. Firing a bullet will add momentum to the machinegun + satellite in the opposite direction from the bullet.
3. Originally Posted by mathman
The statement is confusing. Firing a bullet will add momentum to the machinegun + satellite in the opposite direction from the bullet.
Okey. Then is a machine gun a space propellant?
4. Yes. Newton: opposite and equal reaction and all that.
5. Using a machine gun as space propellant, can you make me a calculation?
How long can get to Mars, using the following:
1.- 540,000 bullets caliber 60,
2.- a machine gun,
3.- a robot finger,
4.- and shoot (a bullet each one (1) minute)
Very affectionately,
Víctor Elias Espinoza Guedez
08 October 2014
6. I won't be doing the math, but I can see you already need more data:
a. How much mass is there in the rest of the vehicle? (Will affect velocity given by each bullet).
b. How much mass is there in one bullet?
c. What's the muzzle velocity of the gun?
d. What's your intended trajectory? (Different paths have different costs).
e. What's your start point? (e.g. maybe you won't even be able to get out of Earth orbit).
etc.
(Also, with the given limitation on the number of bullets, maybe there's just not enough.)
7. Using a machine gun as space propellant, can you make me a calculation?
How long can get to Mars, using the following:
Weight = A Space Shuttle + 540,000 bullets
Distance = 78.300.000 km
Force = 200 Newton. Shoot a bullet each one (1) minute. Caliber 60 more or less.
Time = ?
Starting point: orbit of the planet Earth
Arrival point: orbit of the planet Mars
Thank you...
Very affectionately,
Víctor Elias Espinoza Guedez
TSU in computer science
8. Originally Posted by victorespinoza72
Using a machine gun as space propellant, can you make me a calculation?
How long can get to Mars, using the following:
1.- 540,000 bullets caliber 60,
2.- a machine gun,
3.- a robot finger,
4.- and shoot (a bullet each one (1) minute)
Very affectionately,
Víctor Elias Espinoza Guedez
08 October 2014
It may already be on Mars.
9. Using a machine gun as space propellant, can you make me a calculation?
How long can get to Mars, using the following:
Weight = A Space Shuttle + 540,000 bullets
Distance = 78.300.000 km
Force = 200 Newton. Shoot a bullet each one (1) minute. Caliber 60 more or less.
Time = ?
Starting point: orbit of the planet Earth
Arrival point: orbit of the planet Mars
Thank you...
Very affectionately,
Víctor Elias Espinoza Guedez
TSU in computer science
10. Look up the muzzle energy of a 0.600 rifle bullet. See if 540 000 of that can accelerate a 10 pound rifle to escape velocity. Start with that.
11. Yes there is acceleration, because there is nothing in the space, which slow down the momentum of 200 Newton.
12. Where did you get the 200 Newtons from?
13. Originally Posted by pzkpfw
Where did you get the 200 Newtons from?
Of 20 kilograms of force to the side opposite the bullet caliber 60. This multiplied by the gravity of 9.8 m/s.
F = g . m
14. Muzzle energy - Wikipedia, the free encyclopedia
From rifle ballistic table a 0.600 caliber rifle is 1950 fps / 7596 ft lbs at muzzle
How much energy to you need get your mass up to 7 miles per second or 25 000 miles per hour?
Low earth orbit is 17 000 miles per hour 5 miles per second
Is the rifle 10 pound or is it that 20 kg (44 pounds) you came out with?
Do you know SI to Imperial conversions of units?
You don't do what you are asked to do? How old are you?
15. Originally Posted by pikpobedy
Muzzle energy - Wikipedia, the free encyclopedia
From rifle ballistic table a 0.600 caliber rifle is 1950 fps / 7596 ft lbs at muzzle
How much energy to you need get your mass up to 7 miles per second or 25 000 miles per hour?
Low earth orbit is 17 000 miles per hour 5 miles per second
Is the rifle 10 pound or is it that 20 kg (44 pounds) you came out with?
Do you know SI to Imperial conversions of units?
You don't do what you are asked to do? How old are you?
I do not understand you...
I speak of the strength that I get to the opposite side of the bullet, which is more or less 20 kg. Calculating mentally.
16. There is no point in continuing if you cannot read and understand what I wrote. Sorry.
I asked you to do things and you persist in ignoring the instructions.
17. Originally Posted by pikpobedy
There is no point in continuing if you cannot read and understand what I wrote. Sorry.
I asked you to do things and you persist in ignoring the instructions.
I'm sorry, but I do not understand what you ask me.
Very affectionately,
Victor Elias Espinoza Guedez
Friend, I did not understand the translator. You understand.
18. After firing in the universe is there acceleration? Because pushing the ship with a force of 20 kg, it should keep that speed, up to the next minute for the next shot.
19. I think you would have to calculate the ratio of difference in mass of the bullets to the entire mass of the ship including the guns and the cargo. Also, space is not really a vacuum and there are particles and probably molecules in space, so it is not totally frictionless, and this will be complicated further when encountering the atmospheres and gravity forces of planets, asteroids and what have you all.
20. Outer space as in interplanetary space is better vacuum than anything artificially made.
The mass of the ammunition in this case is important. A rifle round (aka cartridge) is comprised of a brass casing (holding everything together) , primer (encasing the primary explosive), propellant (charge of cordite) and a projectile (bullet made of brass and lead ). You can surmise that most of the mass is not fuel, thus it's specific energy content is not very good as a space vehicle propellant. You know that after firing a weapon of the type the bullet shoots out the barrel and the empty shell casing is ejected out of the side. Both have a reaction force. Keeping the shell casing in the vehicle would reduce that reaction but the mass kept would be part of the total mass for the entire voyage. If ejected they would impart impulses opposite to the ejection that could be used for additional thrust or at least corrected for so as not to affect trajectory.
As for the OP. Forget it. There is nothing coherent there and after. The guy does not do step one, he adds conflicting constraints in subsequent posts and just carps about being helpless. Maybe he is a bot being tested here.
21. And if we put 5 machine guns, increase the thrust force to 1000 Newton. Can this help to push a ship in space.? And get more acceleration.
*******************************************
Using a machine gun as space propellant, can you make me a calculation?
How long can get to Mars, using the following:
Weight = A Space Shuttle + 540,000 bullets
Distance = 78.300.000 km
Force = 1000 Newton. Shoot a bullet each one (1) minute. Caliber 60 more or less.
Time = ?
Starting point: orbit of the planet Earth
Arrival point: orbit of the planet Mars
Thank you...
Very affectionately,
Víctor Elias Espinoza Guedez
USING THE FOLLOWING RIFLE:
http://img.webme.com/pic/t/teoria-es...rifle-bala.jpg
USING THIS IDEA:
The put images does not work.
22. Dear Sirs students and graduates in physics,
A question: What distance travel a bullet, in the outer or empty space? and, At what speed?
23. In the absence of mass to bang into or gravity to alter it's velocity, it'll go "forever" at the muzzle velocity of the weapon. (Which I'd expect to be a bit higher in space than when fired in an atmosphere).
Of course, those conditions (truly empty space, "forever") are unlikely to occur. I'd expect a bullet fired from anywhere within the solar system, for example, would not escape the solar systems gravity.
Escape velocity - Wikipedia, the free encyclopedia
Solar system escape velocity: 492–594 km/s
Barrett M82 - Wikipedia, the free encyclopedia
Muzzle velocity of Barrett light 50: 853 m/s
24. What would happen with the following idea: shall we go to the speed of light?
PROPELLANT AT THE SPEED OF LIGHT
WHAT DISTANCE HAS THIS BULLET?
http://img.webme.com/pic/t/teoria-es...ansparente.jpg
I hearing opinion to perform other detonations.
My idea is to place 10 giant guns on the sides of the space shuttle.
This idea is possible.
I'm not physical, but I think this bullet blue with the space shuttle, will come to the planet Mars. I repeat I am not physical, but doing this shot in the orbit of the Earth, out of gravity, this bullet does nothing to stop it, because the vacuum not braked to a bullet.
There are details that refine to make this idea work, one of them is to climb a Cannon equal to of a military weapon. Then you need to join the bullet with functions to stop and maneuver during his trip at a great speed.
The other is how will return the blue bullet with the space shuttle, because you'll need another Cannon for the return.
To get to the planet Mars, you need a lever and this is the lever.
Patent pending.
Invented by Victor Elias Espinoza Guedez
Very affectionately,
Víctor Elias Espinoza Guedez
09 October 2014
25. Originally Posted by pikpobedy
Outer space as in interplanetary space is better vacuum than anything artificially made.
The mass of the ammunition in this case is important. A rifle round (aka cartridge) is comprised of a brass casing (holding everything together) , primer (encasing the primary explosive), propellant (charge of cordite) and a projectile (bullet made of brass and lead ). You can surmise that most of the mass is not fuel, thus it's specific energy content is not very good as a space vehicle propellant. You know that after firing a weapon of the type the bullet shoots out the barrel and the empty shell casing is ejected out of the side. Both have a reaction force. Keeping the shell casing in the vehicle would reduce that reaction but the mass kept would be part of the total mass for the entire voyage. If ejected they would impart impulses opposite to the ejection that could be used for additional thrust or at least corrected for so as not to affect trajectory.
As for the OP. Forget it. There is nothing coherent there and after. The guy does not do step one, he adds conflicting constraints in subsequent posts and just carps about being helpless. Maybe he is a bot being tested here.
Thanks for the vacuum article. The mass of the impelled bullet would have no weight in a free space vacuum, right? Without gravity, no weight. What force is supposed to have an opposite reaction then?
In a gravitational field, the casing of the bullet may have even more mass than the bullet itself and either would be insignificant to the weight of the barrel of the gun, much less all the other unfired bullets on board, and the other cargo or the ship iteslf.
I have an off qestion remark about the calibre of a gun. In the article you posted on this, in the image, the 22 bullet looked to be so much smaller in case diameter than a 223 did, so the calibre as I thought was a dimension of the bore and casing is really the measurement of the base of the projectile (slug), right?
26. victorespinoza72 : "What would happen with the following idea: shall we go to the speed of light?"
Why would you get that fast?
How is that going to be any better than the rockets we've already got?
How could putting all the energy into one "bang" not destroy the vehicle?
27. Mass is quantity of matter. It is independent of gravitational field. Force is mass x acceleration. Weight on earth is mass x gravitational acceleration.
Interestingly: traditional units and popular usage change with time. Who uses slugs? Who uses newtons in real life (except in torque newton-meters.
Anyway a lot unit conversion data contain pound mass (lbm) , pound force lbf) , kilogram mass (kgm), kilogram force (kgf) and of course newtons grams, etc.
How much does a big guy weigh: 980 newtons = 100 kilogram force (aka kilopond) = 220 pounds force (aka pound)
Slug (mass) - Wikipedia, the free encyclopedia
28. I have an off qestion remark about the calibre of a gun. In the article you posted on this, in the image, the 22 bullet looked to be so much smaller in case diameter than a 223 did, so the calibre as I thought was a dimension of the bore and casing is really the measurement of the base of the projectile (slug), right?
Bore or projectile diameter are somewhat interchangeable terms referred to as caliber in rifles. Rifle barrels are rifled, meaning they have a helix groove pattern inside that makes the bullet revolve to give it gyroscopic stability. So a rifle barrel has two diameters one at top and the other at bottom of grooves.
A 22 bullet is for all intents and purposes 0.22 inches in diameter. A 0.223 Remington is very similar to a 5.56 NATO round. Both 0.22 inches or 5.56 mm in diameter.
Caliber for Naval canons is the ratio length to diameter. Iowa class battle ships had 16 inch guns that were 50 cailber. This 16/50.
Caliber (artillery) - Wikipedia, the free encyclopedia
For shot guns you have the 410 which is 0.41 inch in diamer but the 28, 20, 16, 12 and 10 gauge. In this case the gauge means how many lead balls of the diameter of the bore would weigh one pound. The popular 12 gauge the amount of balls would be 12. The smaller 20 gauge the amount of balls would be 20. This is an old system reverting back more than one hundred years.
Gauge (bore diameter) - Wikipedia, the free encyclopedia
29. Oh duh. I see. The barrel is the diameter of the base of the projectile and the rifling spins it (rifle or handgun) to keep the trajectory more true - the shotgun maybe has not rifling (little twists in the barrel?
The casing simply then is ejected as a waste product after holding in the powder to act as an enclosement for the powder to react against to propel the bullet?
30. The rifle barrel is grooved. The bullet (projectile) is deformed from a circular cross section into one that is the "mate" of the grooved interior of the barrel. Thus a bit of squeezing and plastic deformation of the bullet. The bullet's circular diameter would be engineered to as to have an effective fit into a grooved barrel.
The twist rate is very obtuse. 1 revolution in 8 inches or so.
Rifling - Wikipedia, the free encyclopedia
The casings are scrapped or can be recovered, reloaded and reused.
31. What is meant by plastic deformation?
32. Plastic deformation is permanent deformation. Solids can flow under pressure. For example toothpaste, peanut butter, putty, clay, lead, aluminium, copper, steels, polyamide etc can be made deformed permanently without breaking. They can flow into molds while still solid.
pottery making
Extrusion process, rolling mill
33. What would happen with the following idea: shall we go to the speed of light?
PROPELLANT AT THE SPEED OF LIGHT
WHAT DISTANCE HAS THIS BULLET?
http://img.webme.com/pic/t/teoria-es...ansparente.jpg
I hearing opinion to perform other detonations.
My idea is to place 10 giant guns on the sides of the space shuttle.
This idea is possible.
I'm not physical, but I think this bullet blue with the space shuttle, will come to the planet Mars. I repeat I am not physical, but doing this shot in the orbit of the Earth, out of gravity, this bullet does nothing to stop it, because the vacuum not braked to a bullet.
There are details that refine to make this idea work, one of them is to climb a Cannon equal to of a military weapon. Then you need to join the bullet with functions to stop and maneuver during his trip at a great speed.
The other is how will return the blue bullet with the space shuttle, because you'll need another Cannon for the return.
To get to the planet Mars, you need a lever and this is the lever.
Patent pending.
Invented by Victor Elias Espinoza Guedez
Very affectionately,
Víctor Elias Espinoza Guedez
09 October 2014
34. A question: Time to get to Mars?
a. Escape velocity, which is 11.2 km/s on Earth
b. The acceleration (a) would theoretically be more than 1000 m/s2
c. Distance: 78.300.000 Km
d. Starting point: orbit of the planet Earth
e. Arrival point: orbit of the planet Mars
f. Time = ?
..."i.e. squashed like a cockroach underfoot"
The pressure inside the spaceship is that crushes it.
The Astronaut carries the same speed of the spacecraft.
With a front pressure protector. The pressure to crush the astronaut must be avoided.
The astronaut who crushes? The air inside the space shuttle? What you need to do, is to protect the astronaut, of the air inside the spaceship. Because the air is not moving.
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# Finding the exact value for period in radians
• steve snash
In summary, the period of the function f(x) = 6 sin(4x+pi) is 2pi/4, which simplifies to pi/2. This can be found by using the formula Period = 2pi/abs(b) and recognizing that the function can be rewritten as 6 sin(4(x+pi/4)), which has the same period as the original function.
## Homework Statement
What is the exact period of the function f(x)= 6 sin (4x+pi)
= a sin (b)
## Homework Equations
Period= 2pi/(abs b)
## The Attempt at a Solution
I found the period as 2pi/(4+pi), this is said to be no the correct answer, how do you simplify this down so it is the exact value?
steve snash said:
## Homework Statement
What is the exact period of the function f(x)= 6 sin (4x+pi)
= a sin (b)
## Homework Equations
Period= 2pi/(abs b)
## The Attempt at a Solution
I found the period as 2pi/(4+pi), this is said to be no the correct answer, how do you simplify this down so it is the exact value?
6 sin(4x + pi) = 6 sin(4 (x +pi/4))
What's the period of y = 6 sin(4x)? It's the same as your function. The difference between this function and yours is that one is a horizontal translation of the other.
Thank you
## 1. What is the formula for finding the exact value for period in radians?
The formula for finding the exact value for period in radians is T = 2π/ω, where T represents the period and ω represents the angular frequency.
## 2. How do you convert a period in seconds to radians?
To convert a period in seconds to radians, you can use the formula T (radians) = 2π/T (seconds). Simply divide 2π by the period in seconds to get the value in radians.
## 3. Can the value for period in radians be negative?
Yes, the value for period in radians can be negative. This usually occurs when dealing with functions that have a negative angular frequency, resulting in a negative period value.
## 4. How is the value for period in radians different from the value in degrees?
The value for period in radians is different from the value in degrees because radians are a unit of measurement for angles in the metric system, while degrees are a unit of measurement in the imperial system. Radians are also based on the circumference of a circle, while degrees are based on dividing a circle into 360 equal parts.
## 5. Why is it important to find the exact value for period in radians?
It is important to find the exact value for period in radians because it allows for more precise calculations and comparisons. Radians are also the preferred unit of measurement in many mathematical and scientific applications, as they are more closely related to the underlying geometry and trigonometry involved. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/finding-the-exact-value-for-period-in-radians.387550/", "fetch_time": 1708528378000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-10/segments/1707947473518.6/warc/CC-MAIN-20240221134259-20240221164259-00575.warc.gz", "warc_record_offset": 992105027, "warc_record_length": 15346, "token_count": 656, "char_count": 2680, "score": 4.3125, "int_score": 4, "crawl": "CC-MAIN-2024-10", "snapshot_type": "latest", "language": "en", "language_score": 0.9026526808738708, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00054-of-00064.parquet"} |
# Union of Two Independent Events
I have a following question from a GRE test prep book:
In a probability experiment, $G$ and $H$ are independent events. The probability that $G$ will occur is $r$, and the probability that $H$ will occur is $s$, where both $r$ and $s$ are greater than $0$.
In the answer, the probability that either $G$ will occur or $H$ will occur, but not both is defined as: $r + s - 2rs$.
I thought the union of two events A and B is given as $P(A \cup B)=P(A)+P(B)-P(A \cap B)$ which in this question equates to $r+s-rs$. My questions are: Which one is correct? Does the definition of union include the probability that both events will occur?
• Well note that for independent events you have $P(A)P(B)=P(A)\cap{}P(B)$ so both are correct. The probability that both will occur is implied by the inclusion exclusion principle. – Μάρκος Καραμέρης May 30 '18 at 23:25
• Do not confuse the phrase "A or B (which includes possibility of both)" with the phrase "Either A or B (but not both)". The first event is represented by $A\cup B$. The second event is represented by $A\triangle B$ or as $(A\setminus B)\cup (B\setminus A)$ or as $(A\cup B)\setminus(A\cap B)$. The question asks you to find the probability of this second phrase, not the first. – JMoravitz May 30 '18 at 23:30
• If you want to be pedantic, the phrase "Either A or B" should strictly imply not both, but sadly language evolves and people misuse phrases so it unfortunately is used by people incorrectly to mean "A or B." This problem however strictly points out the not both aspect. Computer scientists might use "XOR" in place of "Either...or" but it is not common yet to hear it in conversation. – JMoravitz May 30 '18 at 23:33
• @JMoravitz Thank you! Out of curiosity, what is the mathematical term for $A△B$? – jellomutiny May 30 '18 at 23:36
• The symmetric difference of $A$ and $B$. – JMoravitz May 30 '18 at 23:37
## 1 Answer
Yes, the definition of union includes cases in which both occur. Thus both statements are correct: the probability of $A \cup B$ (by definition including cases both occur) is $r+s-rs$, and to get the probability that one but not both occurs, subtract the probability that both occur ($rs$) to get $r+s-2rs$
• Makes sense thank you! – jellomutiny May 30 '18 at 23:41 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/2802491/union-of-two-independent-events/2802498", "fetch_time": 1571176898000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-43/segments/1570986660323.32/warc/CC-MAIN-20191015205352-20191015232852-00237.warc.gz", "warc_record_offset": 604694170, "warc_record_length": 31632, "token_count": 644, "char_count": 2296, "score": 3.75, "int_score": 4, "crawl": "CC-MAIN-2019-43", "snapshot_type": "latest", "language": "en", "language_score": 0.9335305690765381, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00005-of-00064.parquet"} |
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# If x,y,z are positive integers,and x!+x/z=y,then what is the
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If x,y,z are positive integers,and x!+x/z=y,then what is the value of z?
1. x is a factor of y
2. z<x
Guys,plugging in got me no where..Ijust went blindly plugging in random nos,didnt succeed and ended up getting the answer wrong.any ideas?
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Re: x,y,z are positive integers [#permalink]
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22 Jul 2009, 06:38
1
KUDOS
1 is suff.
y = 1.2.3....x + x/z
y= x( 1.2.3..(x-1)+1/z) so if x is a factor of y then 1/z must be integer so z=1
2 is insuff.
since x! is integer and y is integer x/z must be integer. And x>=z. This only says that x is not equal to z. We can not find the value of z.
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Re: x,y,z are positive integers [#permalink]
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22 Jul 2009, 06:54
I got A as well
y = 1*2*3*4*...*x + x/z
y = x*(1/z + 1*2*3*4*...)
y/x = 1/z + 1*2*3*4 ------->> Final Expression
1) Since y/x and 1*2*3*4*... are both integers, then 1/z must be integer. The only value for z, such as 1/z is an integer is 1. Thus SUFF.
2) Using the same final expression, the fact that z<x does not help in anything.
A.
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23 Jul 2009, 02:18
Guys OA is C.Could somebody please explain?
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23 Jul 2009, 02:32
I don't think the answer is C. I think the OA is wrong.
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Re: x,y,z are positive integers [#permalink]
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23 Jul 2009, 04:36
Initially have chosen A as well, but... on the second thought it's really C:
$$x!+\frac{x}{z}=y$$
$$x*((x-1)!+\frac{1}{z})=y$$
$$x*(\frac{(x-1)!*z+1}{z})=y$$
z can be any integer, provided x=z, which makes St1 insufficient, we need St2.
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Re: x,y,z are positive integers [#permalink]
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23 Jul 2009, 05:18
No. z can not be any integer.
Based on the equation you found; if x = z then equation becomes (x! + 1). But x can only be the factor of y when the value of x = 1, z=1 and y=2. In other values of x; x can not be a factor of y because x! and x!+1 are coprimes (numbers that do not have any divisors other than 1). So z becomes 1 We can find z.
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Re: x,y,z are positive integers [#permalink]
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23 Jul 2009, 05:59
maliyeci wrote:
No. z can not be any integer.
Based on the equation you found; if x = z then equation becomes (x! + 1). But x can only be the factor of y when the value of x = 1, z=1 and y=2. In other values of x; x can not be a factor of y because x! and x!+1 are coprimes (numbers that do not have any divisors other than 1). So z becomes 1 We can find z.
Agree with you. Have totally missed the condition that x is a factory of y.
Re: x,y,z are positive integers [#permalink] 23 Jul 2009, 05:59
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# Largest BST subtree in the given Binary Tree
Nishant Rana
Last Updated: May 13, 2022
## Introduction:
You are given a Binary Tree, and you need to return the size of the largest subtree of the given Binary Tree, which is a BST(Binary Search Tree). If, in any case, your entire Binary Tree is the BST, return the size of the whole Binary Tree.
Let us see a few examples:
INPUT:
OUTPUT: 3
EXPLANATION:
These are the only SubTrees that are BSTs. So, the largest BST subtree among them has size 3. Hence, the output is 3.
INPUT:
OUTPUT: 6
EXPLANATION: Since the entire Binary Tree is the BST. Hence, the largest BST Subtree is of size 6.
## Approach 1:
You can run any(inorder, preorder, postorder) traversal on the tree and check if the subtree rooted at your current node is BST or not. In this way, you can find the largest BST subtree size.
Please refer to the below code for the approach mentioned above.
Time Complexity: The time complexity for the above code is O(N ^ 2) (where ‘N’ denotes the number of nodes in the given binary tree) because, for each node, we check if the subtree rooted at the current node is BST. We call this function for each node, and ‘isBST()’ runs in O(N) time. Hence, the time complexity is O(N * N). Also time complexity of size() function is O(N).
Space Complexity: We are not using any auxiliary space, but we are using the recursive call stack space of O(N).
## Approach 2:
In approach 1, for each node, we checked its subtree for it to be a part of BST. But, what if we iterate from the bottom and return some information to the node to check if the current node subtree is BST or not in constant time(O(1)).
We will start from the leaf nodes, and the information which we would receive from its child would include the following things:-
1. If the child subtree is BST or not.
2. Maximum and Minimum value of all the nodes of child subtrees.
3. Size of child subtree if it is BST else -1.
4. Maximum answer till the child.
Now, how to check if the current node subtree is BST or not in O(1) time!
You can apply the following checks to check that:-
1. curNode.data > left.max (left.max: Maximum node value of left subtree)
2. curNode.data < right.min (right.min: Minimum node value of right subtree)
If the above two conditions satisfy, that means your current node subtree is a BST because a node’s value should be greater than the maximum value of its left child subtree and smaller than the minimum value of its right child subtree.
In this way, you can find the largest BST subtree size.
Please refer to the below code for the approach mentioned above.
Time complexity: The time complexity for the above code is O(N) because we only traverse the tree once.
Space Complexity: Constant auxiliary space is used. We are just using O(N) recursive call stack space.
## FAQ:
Approach1:
1. What is the time complexity of the isBST function?
The time complexity of the isBST function is O(N) because we run an inorder traversal inside this function whose time complexity is O(N).
2. What is the time complexity of the size function?
The time complexity of the size function is O(N) because we run an inorder traversal inside this function whose time complexity is O(N).
3. For what min and max are used in the isBST function?
The min and max specify the range in which the current node’s value should lie to follow the BST property.
Approach 2:
1. How did we come to the optimized approach for the problem of finding the Largest BST subtree?
Earlier in approach 1, we traversed from up to down and checked if its subtree is BST or not for each node. But in approach 2, we started from down to up and used information returned by the child to the parent to figure out if the current node subtree is BST or not.
2. How do we determine if the current node subtree is a BST or not?
The current node’s value should be greater than the left child’s subtree maximum value and smaller than the right child’s minimum value.
## Key Takeaways:
In this blog, we have covered the following things:
1. How to find the largest BST subtree size.
2. We optimized our approach 1 from O(N * N) approach to O(N) approach.
Until then, All the best for your future endeavors, and Keep Coding. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.codingninjas.com/codestudio/library/largest-bst-subtree-in-the-given-binary-tree", "fetch_time": 1653414425000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-21/segments/1652662573189.78/warc/CC-MAIN-20220524173011-20220524203011-00094.warc.gz", "warc_record_offset": 797994814, "warc_record_length": 30710, "token_count": 990, "char_count": 4244, "score": 3.546875, "int_score": 4, "crawl": "CC-MAIN-2022-21", "snapshot_type": "latest", "language": "en", "language_score": 0.8853563666343689, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00022-of-00064.parquet"} |
# Show that the dimension of a particular linear space is $2$
Question: A Linear transformation $T: \mathbb R^4 \to \mathbb R^4$ is represented by the matrix $$\mathbf A=\begin{pmatrix} \\1&-1&2&3 \\ 2 & -3 & 4 & 5\\ 5 & -6 & 10 & 14\\ 4 & -5 & 8 & 11 \end{pmatrix}$$ (i) Show that the $\dim\big(R(\mathbf A)\big) =2$.
(ii) Let M be a given $4\times 4$ (non-singular) matrix and let $\mathcal S$ be the vector space consisting of vectors in the form of MAx where $x \in \mathbb R^4$. Show that if M is non-singular then $\dim\big(\mathcal S\big) =2$.
The first part is pretty straight-forward. All one needs to do is compute $\operatorname{rref}(\mathbf A)$.
But it is the second part that has stomped me (and for quite some time now). I did come up with a solution but it seemed very cheap to me and it didn't really satisfy me completely.
What I tried was to write A in the form, $$\mathbf A=\Big [\begin{matrix} \vec p_1& \vec p_2&\vec f_1&\vec f_2 \\ \end{matrix}\Big]$$ Where $\vec p_1$ and $\vec p_2$ are the columns corresponding to the pivot variables (there are $2$ pivot variables, since $\dim\big(R(\mathbf A)\big) =2$) and $\vec f_1$ and $\vec f_2$ are the columns corresponding to the free variables.
Now if M be any $4\times 4$ matrix in the form, $$\mathbf M=\Big[ \begin{matrix} \vec c_1& \vec c_2& \vec c_3& \vec c_4 \\ \end{matrix}\Big]$$ Where $\vec c_1, \vec c_2, \vec c_3$ and $\vec c_4$ are the four columns of M, then $$\mathbf {MAx}=\Big[ \begin{matrix} \vec c_1& \vec c_2& \vec c_3& \vec c_4 \\ \end{matrix}\Big] \cdot x_1\vec p_1+x_2\vec p_2+ x_3\vec f_1+x_4\vec f_2$$ I don't know what to do after this.
Is this apprach correct? If yes, how do I proceed from here? If not, where is it that I am going wrong and how do I solve this problem?
Any Help is much appreciated!
-
What are pivot and free variables again? I don't think we need them to answer this question. Also, it looks like your $MA$ is just a row vector. I would approach it like: 1)Show that the rows of $MA$ are linear combinations of rows of $A$, so $r(MA)\le r(A)$. 2) Show that the the rows of $A$ that are linearly independent remain so after being multiplied by $M$, so $r(MA)\ge2$. – Jyrki Lahtonen May 17 '12 at 10:10
$MA$ can't be a row vector. I don't think multiplication of two $4\times 4$ matrices can yield a row vector? – funktor May 17 '12 at 10:13
I agree, it cannot. But before your edit $MA$ looked like a list of four numbers. That's why I asked :-) But it has been fixed now. Sorry about being a bit trigger-happy. – Jyrki Lahtonen May 17 '12 at 10:15
Oh! My bad then :) – funktor May 17 '12 at 10:16
I didn't read your proof thoroughly so I can't say whether it will work. What I can tell is that it is way to complicated (and confusing). Maybe you should think about the question differently. The statement is a special case of the proposition:
If $M$ is non-singular then $rank(MA)=rank(A)$
If you have a basis of $Im(A)$, can you show that $M$ takes this basis to a basis of $S$? Hint: You need non-singularity only to show that the images of the base vectors are linearly independant.
-
Yes I do agree with you on that. The proof is very messy. I don't have much experience with proof writing. The thing is I am still in high school and I am slowly getting the hang of it:) – funktor May 17 '12 at 10:21
Then you are doing quite advanced stuff! No worries you will get a hang of it pretty quickly if you keep working on it. Is my answer helpful? – Simon Markett May 17 '12 at 10:24
That approach doesn’t look very promising; at best it’s going to be very messy. Try showing that $\mathbf{MA}$ and $\mathbf A$ have the same null space when $\mathbf M$ is non-singular; that’s not hard, and it implies that they have the same rank.
-
Isn't it enough to say that non-singular matrix preserves linear independence?
-
Could you elaborate a little bit? – funktor May 17 '12 at 10:17
Since you had $Ax$ of dimention 2, then it had a two vector basis. Images of those base vectors by $M$ are linearly independent in $MAx$ and you can show that they form a maximal set of independent vectors by contradiction, using the fact that $M$ is invertible. I can elaborate more if you wish, just tell me on which part. :) – Łukasz Maciejewski May 17 '12 at 15:08 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://math.stackexchange.com/questions/146212/show-that-the-dimension-of-a-particular-linear-space-is-2", "fetch_time": 1464738648000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2016-22/segments/1464053252010.41/warc/CC-MAIN-20160524012732-00099-ip-10-185-217-139.ec2.internal.warc.gz", "warc_record_offset": 184101092, "warc_record_length": 20038, "token_count": 1324, "char_count": 4275, "score": 4.0, "int_score": 4, "crawl": "CC-MAIN-2016-22", "snapshot_type": "longest", "language": "en", "language_score": 0.8179597854614258, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00038-of-00064.parquet"} |
# how to find derivation of the following function
• Nov 1st 2012, 11:05 AM
flametag2
how to find derivation of the following function
$\displaystyle f(z) = e^{x^{2}+ y^{2}}[cos(2xy) + i sin(2xy)]$
• Nov 1st 2012, 02:46 PM
HallsofIvy
Re: how to find derivation of the following function
Technically the word is "differentiation", not "derivation".
I presume you know that if f(z)= f(x+ iy)= u(x, y)+ iv(x,y) then
$\displaystyle \frac{df}{dz}= \frac{\partial u}{\partial x}+ i \frac{\partial v}{\partial y}= \frac{\partial v}{\partial y}- i\frac{\partial v}{\partial y}$
(The fact that those two expressions on the right are equal is the "Cauchy- Rieman" equations. You can use either one.)
Here, of course, you have
$\displaystyle u(x,y)= e^{x^2+ y^2}cos(2xy)$
$\displaystyle v(x,y)= e^{x^2+ y^2}sin(2xy)$ | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://mathhelpforum.com/advanced-math-topics/206554-how-find-derivation-following-function-print.html", "fetch_time": 1527276110000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-22/segments/1526794867173.31/warc/CC-MAIN-20180525180646-20180525200646-00177.warc.gz", "warc_record_offset": 204799618, "warc_record_length": 2652, "token_count": 279, "char_count": 809, "score": 3.6875, "int_score": 4, "crawl": "CC-MAIN-2018-22", "snapshot_type": "latest", "language": "en", "language_score": 0.7923558950424194, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00029-of-00064.parquet"} |
# What is polytropic efficiency of compressor?
## What is polytropic efficiency of compressor?
Polytropic efficiency is another concept of efficiency often used in compressor evaluation. It is often referred as small stage or infinitesimal stage efficiency. It is the true aerodynamic efficiency exclusive of the pressure-ratio effect.
## What does polytropic efficiency physically represent?
Polytropic efficiency is a value used to describe the efficiency of a compressor. Determine the value for the ratio of specific heats for the gas you are using in your compressor.
What is the average efficiency of a compressor?
Air compressors operate at about 10 percent efficiency — you lose most of the energy to heat. By using a heat recovery system, you can recover nearly all of the lost power and use it for warming your workspace or other applications.
How do you calculate the efficiency of a compressor?
Simple Energy Formula: Motor Efficiency = Cost per KW X . 746 (Power Factor) X Hours of operation X Brake horsepower *Note: No electric motor is 100% efficient, most will average 92-95% efficient.
### What is a polytropic compressor?
Polytropic Efficiency is a process whereby compression is divided into numerous small steps with the steps contain similar isentropic efficiency. It does not depend upon thermodynamic effect and hence, it is considered as aerodynamic performance of compressor.
### How do axial compressors work?
An axial-flow compressor compresses its working fluid by first accelerating the fluid and then diffusing it to obtain a pressure increase. The fluid is accelerated by a row of rotating airfoils or blades (the rotor) and diffused by a row of stationary blades (the stator).
Polytropic efficiency or hydraulic efficiency for a compressor is represented by, Polytropic efficiency equation These equations will provide us with theoretical values of the power requirement and the thermodynamic efficiencies of the compression process.
What is polytropic efficiency?
Polytropic efficiency is another concept of efficiency often used in compressor evaluation. It is often referred as small stage or infinitesimal stage efficiency.
What is the difference between adiabatic and polytropic compression?
Although the adiabatic compression process can be assumed in centrifugal compression, polytropic compression process is commonly considered as the basis for comparing centrifugal compressor performance. The process is expressed as (11.66)pV n = constant, where n denotes the polytropic exponent.
## How to calculate the adiabatic efficiency of the compressor?
Adiabatic efficiency of the compressor is calculated as, Adiabatic efficiency, η = (Actual Polytropic work / Adiabatic work) | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.sweatlodgeradio.com/what-is-polytropic-efficiency-of-compressor/", "fetch_time": 1717028176000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-22/segments/1715971059412.27/warc/CC-MAIN-20240529230852-20240530020852-00153.warc.gz", "warc_record_offset": 860701696, "warc_record_length": 40415, "token_count": 538, "char_count": 2733, "score": 3.65625, "int_score": 4, "crawl": "CC-MAIN-2024-22", "snapshot_type": "latest", "language": "en", "language_score": 0.9167160987854004, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00044-of-00064.parquet"} |
Sine and cosine explained
Sine and cosine
General Definition:
\begin{align} &\sin(\alpha)=
rm{opposite
} \\[8pt]&\cos(\alpha) = \frac \\[8pt]\end
Fields Of Application:Trigonometry, Fourier series, etc.
In mathematics, sine and cosine are trigonometric functions of an angle. The sine and cosine of an acute angle are defined in the context of a right triangle: for the specified angle, its sine is the ratio of the length of the side that is opposite that angle to the length of the longest side of the triangle (the hypotenuse), and the cosine is the ratio of the length of the adjacent leg to that of the hypotenuse. For an angle
\theta
, the sine and cosine functions are denoted as
\sin\theta
and
\cos\theta
.[1]
The definitions of sine and cosine have been extended to any real value in terms of the lengths of certain line segments in a unit circle. More modern definitions express the sine and cosine as infinite series, or as the solutions of certain differential equations, allowing their extension to arbitrary positive and negative values and even to complex numbers.
The sine and cosine functions are commonly used to model periodic phenomena such as sound and light waves, the position and velocity of harmonic oscillators, sunlight intensity and day length, and average temperature variations throughout the year. They can be traced to the and functions used in Indian astronomy during the Gupta period.
Notation
Sine and cosine are written using functional notation with the abbreviations sin and cos.
Often, if the argument is simple enough, the function value will be written without parentheses, as rather than as .
Each of sine and cosine is a function of an angle, which is usually expressed in terms of radians or degrees. Except where explicitly stated otherwise, this article assumes that the angle is measured in radians.
Definitions
Right-angled triangle definitions
To define the sine and cosine of an acute angle α, start with a right triangle that contains an angle of measure α; in the accompanying figure, angle α in triangle ABC is the angle of interest. The three sides of the triangle are named as follows:
• The opposite side is the side opposite to the angle of interest, in this case side a.
• The hypotenuse is the side opposite the right angle, in this case side h. The hypotenuse is always the longest side of a right-angled triangle.
• The adjacent side is the remaining side, in this case side b. It forms a side of (and is adjacent to) both the angle of interest (angle A) and the right angle.
Once such a triangle is chosen, the sine of the angle is equal to the length of the opposite side, divided by the length of the hypotenuse:[2]
\sin(\alpha)=
rm{opposite
} \qquad\cos(\alpha) = \frac
The other trigonometric functions of the angle can be defined similarly; for example, the tangent is the ratio between the opposite and adjacent sides.
As stated, the values
\sin(\alpha)
and
\cos(\alpha)
appear to depend on the choice of right triangle containing an angle of measure α. However, this is not the case: all such triangles are similar, and so the ratios are the same for each of them.
Unit circle definitions
In trigonometry, a unit circle is the circle of radius one centered at the origin (0, 0) in the Cartesian coordinate system.
Let a line through the origin intersect the unit circle, making an angle of θ with the positive half of the x-axis. The x- and y-coordinates of this point of intersection are equal to and, respectively. This definition is consistent with the right-angled triangle definition of sine and cosine when
0<\theta<
\pi 2
: because the length of the hypotenuse of the unit circle is always 1,
\sin(\theta)=
opposite hypotenuse
=
opposite 1
={opposite
}. The length of the opposite side of the triangle is simply the y-coordinate. A similar argument can be made for the cosine function to show that
\cos(\theta)=
adjacent hypotenuse
when
0<\theta<
\pi 2
, even under the new definition using the unit circle. is then defined as
\sin(\theta) \cos(\theta)
, or, equivalently, as the slope of the line segment.
Using the unit circle definition has the advantage that the angle can be extended to any real argument. This can also be achieved by requiring certain symmetries, and that sine be a periodic function.
Complex exponential function definitions
See main article: Euler's formula.
ez
is defined on the entire domain of the complex numbers. The definition of sine and cosine can be extended to all complex numbers via
\sinz=
eiz-e-iz 2i
\cosz=
eiz+e-iz 2
These can be reversed to give Euler's formula
eiz=\cosz+i\sinz
e-iz=\cosz-i\sinz
When plotted on the complex plane, the function
eix
for real values of
x
traces out the unit circle in the complex plane.
When
x
is a real number, sine and cosine simplify to the imaginary and real parts of
eix
or
e-ix
, as:
\sinx=\operatorname{Im}(eix)=-\operatorname{Im}(e-ix)
\cosx=\operatorname{Re}(eix)=\operatorname{Re}(e-ix)
When
z=x+iy
for real values
x
and
y
, sine and cosine can be expressed in terms of real sines, cosines, and hyperbolic functions as
\begin{align}\sinz&=\sinx\coshy+i\cosx\sinhy\\[5pt] \cosz&=\cosx\coshy-i\sinx\sinhy\end{align}
Differential equation definition
(\cos\theta,\sin\theta)
is the solution
(x(\theta),y(\theta))
to the two-dimensional system of differential equations
y'(\theta)=x(\theta)
and
x'(\theta)=-y(\theta)
with the initial conditions
y(0)=0
and
x(0)=1
. One could interpret the unit circle in the above definitions as defining the phase space trajectory of the differential equation with the given initial conditions.
It can be interpreted as a phase space trajectory of the system of differential equations
y'(\theta)=x(\theta)
and
x'(\theta)=-y(\theta)
starting from the initial conditions
y(0)=0
and
x(0)=1
.
Series definitions
The successive derivatives of sine, evaluated at zero, can be used to determine its Taylor series. Using only geometry and properties of limits, it can be shown that the derivative of sine is cosine, and that the derivative of cosine is the negative of sine. This means the successive derivatives of sin(x) are cos(x), -sin(x), -cos(x), sin(x), continuing to repeat those four functions. The (4n+k)-th derivative, evaluated at the point 0:
\sin(4n+k)(0)=\begin{cases} 0&whenk=0\\ 1&whenk=1\\ 0&whenk=2\\ -1&whenk=3 \end{cases}
where the superscript represents repeated differentiation. This implies the following Taylor series expansion at x = 0. One can then use the theory of Taylor series to show that the following identities hold for all real numbers x (where x is the angle in radians):[3]
\begin{align} \sin(x)&=x-
x3 3!
+
x5 5!
-
x7 7!
+\\[8pt] &=
infty \sum n=0
(-1)n (2n+1)!
x2n+1\\[8pt] \end{align}
Taking the derivative of each term gives the Taylor series for cosine:
\begin{align} \cos(x)&=1-
x2 2!
+
x4 4!
-
x6 6!
+\\[8pt] &=
infty \sum n=0
(-1)n (2n)!
x2n\\[8pt] \end{align}
Continued fraction definitions
The sine function can also be represented as a generalized continued fraction:
\sin(x)= \cfrac{x}{1+\cfrac{x2}{2 ⋅ 3-x2+ \cfrac{2 ⋅ 3x2}{4 ⋅ 5-x2+ \cfrac{4 ⋅ 5x2}{6 ⋅ 7-x2+\ddots}}}}.
\cos(x)=\cfrac{1}{1+\cfrac{x2}{12-x2+\cfrac{12x2}{34-x2+\cfrac{34x2}{56-x2+\ddots}}}}.
The continued fraction representations can be derived from Euler's continued fraction formula and express the real number values, both rational and irrational, of the sine and cosine functions.
Identities
See main article: List of trigonometric identities.
Exact identities (using radians):
These apply for all values of
\theta
.
\sin(\theta)=\cos\left(
\pi 2
-\theta\right)=\cos\left(\theta-
\pi 2
\right)
\cos(\theta)=\sin\left(
\pi 2
-\theta\right)=\sin\left(\theta+
\pi 2
\right)
Reciprocals
The reciprocal of sine is cosecant, i.e., the reciprocal of
\sin{\theta}
is
\csc{\theta}
. Cosecant gives the ratio of the length of the hypotenuse to the length of the opposite side. Similarly, the reciprocal of cosine is secant, which gives the ratio of the length of the hypotenuse to that of the adjacent side.
\csc{\theta}=
1 \sin{\theta
} = \frac
\sec{\theta}=
1 \cos{\theta
} = \frac
Inverses
The inverse function of sine is arcsine (arcsin or asin) or inverse sine . The inverse function of cosine is arccosine (arccos, acos, or). (The superscript of −1 in and denotes the inverse of a function, not exponentiation.) As sine and cosine are not injective, their inverses are not exact inverse functions, but partial inverse functions. For example,, but also, etc. It follows that the arcsine function is multivalued:, but also,, etc. When only one value is desired, the function may be restricted to its principal branch. With this restriction, for each x in the domain, the expression will evaluate only to a single value, called its principal value. The standard range of principal values for arcsin is from to and the standard range for arccos is from to .
\theta=\arcsin\left(
opposite hypotenuse
\right)=\arccos\left(
adjacent hypotenuse
\right).
where (for some integer k):
\begin{align} \sin(y)=x\iff&y=\arcsin(x)+2\pik,or\\ &y=\pi-\arcsin(x)+2\pik\\ \cos(y)=x\iff&y=\arccos(x)+2\pik,or\\ &y=-\arccos(x)+2\pik \end{align}
By definition, arcsin and arccos satisfy the equations:
\sin(\arcsin(x))=x \cos(\arccos(x))=x
and
\begin{align}\arcsin(\sin(\theta))=\theta&for -
\pi 2
\leq\theta\leq
\pi 2
\\ \arccos(\cos(\theta))=\theta&for0\leq\theta\leq\pi\end{align}
Pythagorean trigonometric identity
The basic relationship between the sine and the cosine is the Pythagorean trigonometric identity:
\cos2(\theta)+\sin2(\theta)=1
where sin2(x) means (sin(x))2.
Double angle formulas
Sine and cosine satisfy the following double angle formulas:
\sin(2\theta)=2\sin(\theta)\cos(\theta)
\cos(2\theta)=\cos2(\theta)-\sin2(\theta)=2\cos2(\theta)-1=1-2\sin2(\theta)
The cosine double angle formula implies that sin2 and cos2 are, themselves, shifted and scaled sine waves. Specifically,[4]
\sin2(\theta)=
1-\cos(2\theta) 2
\cos2(\theta)=
1+\cos(2\theta) 2
The graph shows both the sine function and the sine squared function, with the sine in blue and sine squared in red. Both graphs have the same shape, but with different ranges of values, and different periods. Sine squared has only positive values, but twice the number of periods.
Derivative and integrals
The derivatives of sine and cosine are:
d dx
\sin(x)=\cos(x)
d dx
\cos(x)=-\sin(x)
and their antiderivatives are:
\int\sin(x)dx=-\cos(x)+C
\int\cos(x)dx=\sin(x)+C
where C denotes the constant of integration.
Properties relating to the quadrants
The table below displays many of the key properties of the sine function (sign, monotonicity, convexity), arranged by the quadrant of the argument. For arguments outside those in the table, one may compute the corresponding information by using the periodicity
\sin(\alpha+2\pi)=\sin(\alpha)
of the sine function.
QuadrantAngleSineCosine
DegreesRadiansSignMonotonyConvexitySignMonotonyConvexity
1st quadrant, I
0\circ<x<90\circ
0<x<
\pi 2
+
increasingconcave
+
decreasingconcave
2nd quadrant, II
90\circ<x<180\circ
\pi 2
<x<\pi
+
decreasingconcave
-
decreasingconvex
3rd quadrant, III
180\circ<x<270\circ
\pi<x<
3\pi 2
-
decreasingconvex
-
increasingconvex
4th quadrant, IV
270\circ<x<360\circ
3\pi 2
<x<2\pi
-
increasingconvex
+
increasingconcave
The following table gives basic information at the boundary of the quadrants.
DegreesRadians
\sin(x)
\cos(x)
ValuePoint typeValuePoint type
0\circ
0
0
Root, inflection
1
Maximum
90\circ
\pi 2
1
Maximum
0
Root, inflection
180\circ
\pi
0
Root, inflection
-1
Minimum
270\circ
3\pi 2
-1
Minimum
0
Root, inflection
Fixed points
See main article: Dottie number. thumb|The fixed point iteration xn+1 = cos(xn) with initial value x0 = −1 converges to the Dottie number.Zero is the only real fixed point of the sine function; in other words the only intersection of the sine function and the identity function is
\sin(0)=0
. The only real fixed point of the cosine function is called the Dottie number. That is, the Dottie number is the unique real root of the equation
\cos(x)=x.
The decimal expansion of the Dottie number is
0.739085\ldots
.[5]
Arc length
The arc length of the sine curve between
0
and
t
is
t\sqrt{1+\cos \int 0
2(x)}dx=\sqrt{2}\operatorname{E}(t,1/\sqrt{2}),
where
\operatorname{E}(\varphi,k)
is the incomplete elliptic integral of the second kind with modulus
k
. It cannot be expressed using elementary functions.
The arc length for a full period is[6]
L=
4\sqrt{2\pi3
} + \frac = \frac+2\varpi = 7.640395578\ldotswhere
\Gamma
is the gamma function and
\varpi
is the lemniscate constant.[6] [7]
Laws
See main article: Law of sines and Law of cosines.
The law of sines states that for an arbitrary triangle with sides a, b, and c and angles opposite those sides A, B and C:
\sinA a
=
\sinB b
=
\sinC c
.
This is equivalent to the equality of the first three expressions below:
a \sinA
=
b \sinB
=
c \sinC
=2R,
where R is the triangle's circumradius.
It can be proved by dividing the triangle into two right ones and using the above definition of sine. The law of sines is useful for computing the lengths of the unknown sides in a triangle if two angles and one side are known. This is a common situation occurring in triangulation, a technique to determine unknown distances by measuring two angles and an accessible enclosed distance.
The law of cosines states that for an arbitrary triangle with sides a, b, and c and angles opposite those sides A, B and C:
a2+b2-2ab\cos(C)=c2
In the case where
C=\pi/2
,
\cos(C)=0
and this becomes the Pythagorean theorem: for a right triangle,
a2+b2=c2,
where c is the hypotenuse.
Special values
For integer multiples of 15° (that is, $\textstyle\frac\pi$ radians), the values of sin(x) and cos(x) are particularly simple and can be expressed in terms of
\sqrt2,\sqrt3,\sqrt6
only. A table of these angles is given below. For more complex angle expressions see .
Angle, xsin(x)cos(x)
DegreesRadiansGradiansTurnsExactDecimalExactDecimal
00g00011
15°g
\sqrt{6 -\sqrt{2}}{4}
0.2588
\sqrt{6 +\sqrt{2}}{4}
0.9659
30°0.5
\sqrt{3
}
0.8660
45°
\sqrt{2
}
0.7071
\sqrt{2
}
0.7071
60°
\sqrt{3
}
0.86600.5
75°
\sqrt{6 +\sqrt{2}}{4}
0.9659
\sqrt{6 -\sqrt{2}}{4}
0.2588
90°1100
90 degree increments:
x in degrees x in radians x in gons x in turns sin x 0° 90° 180° 270° 360° 0 /2 3/2 2 0 0 1/4 1/2 3/4 1 0 1 0 −1 0 1 0 −1 0 1
Relationship to complex numbers
Sine and cosine are used to connect the real and imaginary parts of a complex number with its polar coordinates (r, φ):
z=r(\cos(\varphi)+i\sin(\varphi))
The real and imaginary parts are:
\operatorname{Re}(z)=r\cos(\varphi)
\operatorname{Im}(z)=r\sin(\varphi)
where r and φ represent the magnitude and angle of the complex number z.
For any real number θ, Euler's formula says that:
ei\theta=\cos(\theta)+i\sin(\theta)
Therefore, if the polar coordinates of z are (r, φ),
z=rei\varphi.
Complex arguments
Applying the series definition of the sine and cosine to a complex argument, z, gives:
\begin{align} \sin(z)&=
infty \sum n=0
(-1)n (2n+1)!
z2n+1\\ &=
ei-e-i 2i
\\ &=
\sinh\left(iz\right) i
\\ &=-i\sinh\left(iz\right)\\ \cos(z)&=
infty \sum n=0
(-1)n (2n)!
z2n\\ &=
ei+e-i 2
\\ &=\cosh(iz)\\ \end{align}
where sinh and cosh are the hyperbolic sine and cosine. These are entire functions.
It is also sometimes useful to express the complex sine and cosine functions in terms of the real and imaginary parts of its argument:
\begin{align} \sin(x+iy)&=\sin(x)\cos(iy)+\cos(x)\sin(iy)\\ &=\sin(x)\cosh(y)+i\cos(x)\sinh(y)\\ \cos(x+iy)&=\cos(x)\cos(iy)-\sin(x)\sin(iy)\\ &=\cos(x)\cosh(y)-i\sin(x)\sinh(y)\\ \end{align}
Partial fraction and product expansions of complex sine
Using the partial fraction expansion technique in complex analysis, one can find that the infinite series
infty \sum n=-infty
(-1)n z-n
=
1 z
-2z
infty \sum n=1
(-1)n n2-z2
both converge and are equal to $\frac$. Similarly, one can show that
\pi2 \sin2(\piz)
=
infty \sum n=-infty
1 (z-n)2
.
Using product expansion technique, one can derive
\sin(\piz)=\piz
infty \prod n=1
\left(1-
z2 n2
\right).
Alternatively, the infinite product for the sine can be proved using complex Fourier series.
Using complex Fourier series, the function
\cos(zx)
can be decomposed as
\cos(zx)=
z\sin(\piz) \pi
infty \displaystyle\sum n=-infty
(-1)neinx z2-n2
,z\in\Complex\setminusZ,x\in[-\pi,\pi].
Setting
x=\pi
yields
\cos(\piz)=
z\sin(\piz) \pi
infty \displaystyle\sum n=-infty
1 z2-n2
=
z\sin(\piz) \left( \pi
1 z2
+
infty 2\displaystyle\sum n=1
1 z2-n2
\right).
Therefore, we get
\pi\cot(\piz)=
1 z
+
infty 2\displaystyle\sum n=1
z z2-n2
.
The function
\pi\cot(\piz)
is the derivative of
ln(\sin(\piz))+C0
. Furthermore, if $\frac = \frac$, then the function
f
such that the emerged series converges on some open and connected subset of
C
is $f = \frac\ln \left(1 - \frac\right) + C_1$, which can be proved using the Weierstrass M-test. The interchange of the sum and derivative is justified by uniform convergence. It follows that
ln(\sin(\piz))=ln(z)+
infty \displaystyle\sum n=1
ln\left(1-
z2 n2
\right)+C.
Exponentiating gives
\sin(\piz)=zeC
infty \displaystyle\prod n=1
\left(1-
z2 n2
\right).
Since $\lim_\frac = \pi$ and $\lim_\prod_^\infty \left(1 - \frac\right) = 1$, we have
eC=\pi
. Hence
\sin(\piz)=\pi
infty z\displaystyle\prod n=1
\left(1-
z2 n2
\right)
for some open and connected subset of
C
. Let $a_(z) = -\frac$. Since $\sum_^\infty |a_(z)|$ converges uniformly on any closed disk, $\prod_^\infty (1 + a_(z))$ converges uniformly on any closed disk as well.[8] It follows that the infinite product is holomorphic on
C
. By the identity theorem, the infinite product for the sine is valid for all
z\inC
, which completes the proof.
\blacksquare
Usage of complex sine
sin(z) is found in the functional equation for the Gamma function,
\Gamma(s)\Gamma(1-s)={\pi\over\sin(\pis)},
which in turn is found in the functional equation for the Riemann zeta-function,
\zeta(s)=2(2\pi)s-1\Gamma(1-s)\sin\left(
\pi 2
s\right)\zeta(1-s).
As a holomorphic function, sin z is a 2D solution of Laplace's equation:
\Deltau(x1,x2)=0.
The complex sine function is also related to the level curves of pendulums.[9]
History
See main article: History of trigonometry. While the early study of trigonometry can be traced to antiquity, the trigonometric functions as they are in use today were developed in the medieval period. The chord function was discovered by Hipparchus of Nicaea (180–125 BCE) and Ptolemy of Roman Egypt (90–165 CE).[10]
The sine and cosine functions can be traced to the and functions used in Indian astronomy during the Gupta period (Aryabhatiya and Surya Siddhanta), via translation from Sanskrit to Arabic and then from Arabic to Latin.
All six trigonometric functions in current use were known in Islamic mathematics by the 9th century, as was the law of sines, used in solving triangles. With the exception of the sine (which was adopted from Indian mathematics), the other five modern trigonometric functions were discovered by Arabic mathematicians, including the cosine, tangent, cotangent, secant and cosecant.[11] Al-Khwārizmī (c. 780–850) produced tables of sines, cosines and tangents.[12] [13] Muhammad ibn Jābir al-Harrānī al-Battānī (853–929) discovered the reciprocal functions of secant and cosecant, and produced the first table of cosecants for each degree from 1° to 90°.[13]
The first published use of the abbreviations sin, cos, and tan is by the 16th-century French mathematician Albert Girard; these were further promulgated by Euler (see below). The Opus palatinum de triangulis of Georg Joachim Rheticus, a student of Copernicus, was probably the first in Europe to define trigonometric functions directly in terms of right triangles instead of circles, with tables for all six trigonometric functions; this work was finished by Rheticus' student Valentin Otho in 1596.
In a paper published in 1682, Leibniz proved that sin x is not an algebraic function of x.[14] Roger Cotes computed the derivative of sine in his Harmonia Mensurarum (1722).[15] Leonhard Euler's Introductio in analysin infinitorum (1748) was mostly responsible for establishing the analytic treatment of trigonometric functions in Europe, also defining them as infinite series and presenting "Euler's formula", as well as the near-modern abbreviations sin., cos., tang., cot., sec., and cosec.
Etymology
Etymologically, the word sine derives from the Sanskrit word Sanskrit: jyā 'bow-string'[16] [17] or more specifically its synonym Sanskrit: jīvá (both adopted from Ancient Greek Greek, Ancient (to 1453);: χορδή 'string'[18]), due to visual similarity between the arc of a circle with its corresponding chord and a bow with its string (see jyā, koti-jyā and utkrama-jyā). This was transliterated in Arabic as, which is meaningless in that language and written as (Arabic: جب). Since Arabic is written without short vowels, was interpreted as the homograph (جيب), which means 'bosom', 'pocket', or 'fold'. When the Arabic texts of Al-Battani and al-Khwārizmī were translated into Medieval Latin in the 12th century by Gerard of Cremona, he used the Latin equivalent sinus (which also means 'bay' or 'fold', and more specifically 'the hanging fold of a toga over the breast').[19] [20] [21] Gerard was probably not the first scholar to use this translation; Robert of Chester appears to have preceded him and there is evidence of even earlier usage.[22] The English form sine was introduced in the 1590s.[23]
The word cosine derives from an abbreviation of the Latin Latin: complementi sinus 'sine of the complementary angle' as cosinus in Edmund Gunter's Canon triangulorum (1620), which also includes a similar definition of cotangens.[24] [25] [26]
Software implementations
There is no standard algorithm for calculating sine and cosine. IEEE 754, the most widely used standard for the specification of reliable floating-point computation, does not address calculating trigonometric functions such as sine. The reason is that no efficient algorithm is known for computing sine and cosine with a specified accuracy, especially for large inputs.
Algorithms for calculating sine may be balanced for such constraints as speed, accuracy, portability, or range of input values accepted. This can lead to different results for different algorithms, especially for special circumstances such as very large inputs, e.g. sin(10{{sup|22}}).
A common programming optimization, used especially in 3D graphics, is to pre-calculate a table of sine values, for example one value per degree, then for values in-between pick the closest pre-calculated value, or linearly interpolate between the 2 closest values to approximate it. This allows results to be looked up from a table rather than being calculated in real time. With modern CPU architectures this method may offer no advantage.
The CORDIC algorithm is commonly used in scientific calculators.
The sine and cosine functions, along with other trigonometric functions, are widely available across programming languages and platforms. In computing, they are typically abbreviated to sin and cos.
Some CPU architectures have a built-in instruction for sine, including the Intel x87 FPUs since the 80387.
In programming languages, sin and cos are typically either a built-in function or found within the language's standard math library.
For example, the C standard library defines sine functions within math.h: sin([[Double-precision floating-point format|double]]), sinf([[Single-precision floating-point format|float]]), and sinl([[long double]]). The parameter of each is a floating point value, specifying the angle in radians. Each function returns the same data type as it accepts. Many other trigonometric functions are also defined in math.h, such as for cosine, arc sine, and hyperbolic sine (sinh).
Similarly, Python defines math.sin(x) and math.cos(x) within the built-in math module. Complex sine and cosine functions are also available within the cmath module, e.g. cmath.sin(z). CPython's math functions call the C math library, and use a double-precision floating-point format.
Turns based implementations
Some software libraries provide implementations of sine and cosine using the input angle in half-turns, a half-turn being an angle of 180 degrees or
\pi
radians. Representing angles in turns or half-turns has accuracy advantages and efficiency advantages in some cases.[27] [28] In MATLAB, OpenCL, R, Julia, CUDA, and ARM, these functions are called sinpi and cospi.[29] [30] [31] [32] For example, sinpi(x) would evaluate to
\sin(\pix),
where x is expressed in half-turns, and consequently the final input to the function, can be interpreted in radians by .
The accuracy advantage stems from the ability to perfectly represent key angles like full-turn, half-turn, and quarter-turn losslessly in binary floating-point or fixed-point. In contrast, representing
2\pi
,
\pi
, and $\frac$ in binary floating-point or binary scaled fixed-point always involves a loss of accuracy since irrational numbers cannot be represented with finitely many binary digits.
Turns also have an accuracy advantage and efficiency advantage for computing modulo to one period. Computing modulo 1 turn or modulo 2 half-turns can be losslessly and efficiently computed in both floating-point and fixed-point. For example, computing modulo 1 or modulo 2 for a binary point scaled fixed-point value requires only a bit shift or bitwise AND operation. In contrast, computing modulo $\frac$ involves inaccuracies in representing $\frac$.
For applications involving angle sensors, the sensor typically provides angle measurements in a form directly compatible with turns or half-turns. For example, an angle sensor may count from 0 to 4096 over one complete revolution.[33] If half-turns are used as the unit for angle, then the value provided by the sensor directly and losslessly maps to a fixed-point data type with 11 bits to the right of the binary point. In contrast, if radians are used as the unit for storing the angle, then the inaccuracies and cost of multiplying the raw sensor integer by an approximation to $\frac$ would be incurred.
Notes and References
1. Web site: Weisstein. Eric W.. Sine. 2020-08-29. mathworld.wolfram.com. en.
2. Book: Young, Cynthia . Cynthia Y. Young . 2017 . Trigonometry . John Wiley & Sons . 27 . 978-1-119-32113-2.
3. Book: Ahlfors, Lars. Lars Ahlfors. January 1, 1979. Complex Analysis. 3. 43-44.
4. Web site: Sine-squared function . August 9, 2019.
5. Web site: OEIS A003957. oeis.org. 2019-05-26.
6. Web site: A105419 - Oeis .
7. Web site: An Eloquent Formula for the Perimeter of an Ellipse. Adlaj. Semjon. 2012. American Mathematical Society. 1097.
8. Book: Rudin . Walter . Real and Complex Analysis . McGraw-Hill Book Company . 1987 . Third . 0-07-100276-6. p. 299, Theorem 15.4
9. Web site: Why are the phase portrait of the simple plane pendulum and a domain coloring of sin(z) so similar?. math.stackexchange.com. 2019-08-12.
10. Brendan . T. . February 1965 . How Ptolemy constructed trigonometry tables . The Mathematics Teacher . 58 . 2 . 141-149 . JSTOR.
11. Islamic Astronomy . Owen . Gingerich . . 1986 . 254 . 74 . 2010-07-13 . https://web.archive.org/web/20131019140821/http://faculty.kfupm.edu.sa/PHYS/alshukri/PHYS215/Islamic_astronomy.htm . 2013-10-19.
12. Jacques Sesiano, "Islamic mathematics", p. 157, in Book: Mathematics Across Cultures: The History of Non-western Mathematics . Helaine . Selin . Helaine Selin . Ubiratan . D'Ambrosio . Ubiratan D'Ambrosio . 2000 . . 978-1-4020-0260-1.
13. Web site: trigonometry . Encyclopedia Britannica.
14. Book: Elements of the History of Mathematics. registration. Nicolás Bourbaki. Springer. 1994. 9783540647676.
15. "Why the sine has a simple derivative ", in Historical Notes for Calculus Teachers by V. Frederick Rickey
16. Web site: How the Trig Functions Got their Names. Ask Dr. Math. Drexel University. 2 March 2010.
17. Web site: The trigonometric functions. J J O'Connor and E F Robertson. June 1996 . 2 March 2010.
18. See Plofker, Mathematics in India, Princeton University Press, 2009, p. 257
See Web site: Clark University . live . https://web.archive.org/web/20080615133310/http://www.clarku.edu/~djoyce/trig/ . 15 June 2008 .
See Maor (1998), chapter 3, regarding the etymology.
19. It was Robert of Chester's translation from the Arabic that resulted in our word "sine". The Hindus had given the name jiva to the half-chord in trigonometry, and the Arabs had taken this over as jiba. In the Arabic language there is also the word jaib meaning "bay" or "inlet". When Robert of Chester came to translate the technical word jiba, he seems to have confused this with the word jaib (perhaps because vowels were omitted); hence, he used the word sinus, the Latin word for "bay" or "inlet".
20. Eli Maor (1998), Trigonometric Delights, Princeton: Princeton University Press, p. 35-36.
21. Victor J. Katz (2008), A History of Mathematics, Boston: Addison-Wesley, 3rd. ed., p. 253, sidebar 8.1. Web site: A History of Mathematics . 2015-04-09 . live . https://web.archive.org/web/20150414065531/http://deti-bilingual.com/wp-content/uploads/2014/06/3rd-Edition-Victor-J.-Katz-A-History-of-Mathematics-Pearson-2008.pdf . 2015-04-14. : The English word “sine” comes from a series of mistranslations of the Sanskrit Sanskrit: jyā-ardha (chord-half). Āryabhaṭa frequently abbreviated this term to Sanskrit: jyā or its synonym Sanskrit: jīvá. When some of the Hindu works were later translated into Arabic, the word was simply transcribed phonetically into an otherwise meaningless Arabic word . But since Arabic is written without vowels, later writers interpreted the consonants as Arabic: jaib, which means bosom or breast. In the twelfth century, when an Arabic trigonometry work was translated into Latin, the translator used the equivalent Latin word Latin: sinus, which also meant bosom, and by extension, fold (as in a toga over a breast), or a bay or gulf.
22. Various sources credit the first use of to either
See Merlet, A Note on the History of the Trigonometric Functions in Ceccarelli (ed.), International Symposium on History of Machines and Mechanisms, Springer, 2004
See Maor (1998), chapter 3, for an earlier etymology crediting Gerard.
See Book: Katx, Victor . July 2008 . A history of mathematics . 3rd . Boston . . 210 (sidebar) . 978-0321387004 . en .
23. The anglicized form is first recorded in 1593 in Thomas Fale's Horologiographia, the Art of Dialling.
24. Book: Edmund . Gunter . Edmund Gunter . Canon triangulorum . 1620.
25. Web site: A reconstruction of Gunter's Canon triangulorum (1620) . Denis . Roegel . Research report . HAL . 6 December 2010 . inria-00543938 . 28 July 2017 . live . https://web.archive.org/web/20170728192238/https://hal.inria.fr/inria-00543938/document . 28 July 2017.
26. Web site: cosine.
27. "MATLAB Documentation sinpi
28. "R Documentation sinpi
29. "OpenCL Documentation sinpi
30. "Julia Documentation sinpi
31. "CUDA Documentation sinpi
32. "ARM Documentation sinpi
33. "ALLEGRO Angle Sensor Datasheet | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 13, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://everything.explained.today/Sine_and_cosine/", "fetch_time": 1719214283000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-26/segments/1718198865074.62/warc/CC-MAIN-20240624052615-20240624082615-00876.warc.gz", "warc_record_offset": 11724896, "warc_record_length": 24517, "token_count": 8748, "char_count": 31582, "score": 4.53125, "int_score": 5, "crawl": "CC-MAIN-2024-26", "snapshot_type": "longest", "language": "en", "language_score": 0.9007574915885925, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00046-of-00064.parquet"} |
# The Very Spring and Root
## An engineer's adventures in education (and other musings).
This content shows Simple View
## Lesson Plan: Intro to Parallel Circuits
I’ve been wanting to upload more lesson plans and materials that I think worked well — like everything else, its just a matter of finding the time.
### BACKGROUND
The attached are a lesson plan and a handout for how I did the introduction to parallel circuits. At this point in the unit, we had already covered the conceptual understanding of what voltage, current, and resistance area. We had already covered Ohm’s Law as well, both activity-based and mathematically.
I have removed the following from the lesson plan:
• Mention of or planning for individual students. Normally, in addition to planning for all students in general, I am preparing for particular students who tend to need additional prodding to focus, often have clarification questions, or perhaps need additional language assistance.
• The section on planning for individual students with learning disabilities, since the plans would necessarily detail confidential information about my students.
My students had not done series circuits yet. The decision to start with parallel came after some thought — I wondered if there are good reasons why series circuits a usually taught first. I really couldn’t think of any that didn’t also have an analog on the parallel side. For example, in a series circuit, it is usually intuitive why the current is the same through all components (there is only one path for the charges to take).
However, understanding why the voltage drops in a series circuit have to add up to the total battery voltage (proportional to their resistances) requires more thought (and a good understanding of what voltage physically is). On the other hand, in a parallel circuit, the idea that the current in each branch should sum to the total entering and leaving the battery is easy to visually demonstrate. But why should the voltage across all components never change, no matter how many you add or remove (within reason)?
I figured it was six one way, half a dozen the other and went with parallel first for the novelty.
### FILES
These are free to use, modify, and distribute. Please credit me and/or this blog if you use it for something, and I’d love to hear any revision suggestions for next year or reports on how it went with other students!
Lesson3.1A-Handout [pdf]
### ANALYSIS
Students in general pieced the important concepts together well. One thing that surprised me was that one group seemed to be able to use the data they were getting in the lesson to validate an incorrect model of parallel circuits: that it was always the closest resistor to the battery that got the most current. I realized 1) that this was not a student idea that I had anticipated, and 2) that the setup of the lab allowed this alternative conception to be reinforced (note that the resistor with lower resistance is, in fact, closer to the battery on the circuit diagram).
I asked that group to test out their idea by swapping the resistor positions, telling them to predict what would happen first. They conferred and said that “it was probably about fifty-fifty” on whether or not their theory would be disproven by the new data. They were able to discover that the current depends only on the relative value of the resistances.
I then reframed the post-activity discussion to center on this student reasoning/discovery instead of my originally planned questions. In a way, this was serendipitous — I got students to demonstrate for their peers what real science looks like. We have an initial model that attempts to explain something we observe, we ask ourselves what we need to do to validate that model, we attempt validation, and then revise our model. That meta lesson was possibly more important in the long run than the actual content of parallel circuits.
The next day, we followed up with discussion, reading, and applying mathematical relations to what we learned in the exploratory activity.
## Lesson Plan: Predictions With Conservation of Energy
by Nalin A. Ratnayake
Unit: Work and Energy, Component 2: Conservation of Energy
Date: January 15th, 2013
Day/Block: Day 4, Blocks A(3) / E(2) / F(6)
Time Available: E 58 min, A 58 min, F 65 min
Objective: You will be able to predict and analyze motion using the Law of Conservation of Energy.
Criteria for Success:
1. Can I predict the final mechanical energy of an object in motion using energy conservation?
2. Can I determine if and how an object’s mechanical energy has changed?
3. Can I solve a physics problem using energy conservation?
Assessment: Handout and Exit Ticket
The handout will show qualitative understanding of Criteria for Success 1 and 2.
The short worksheet will show quantitative understanding of the Criterion for Success 3.
### [10 min] Do Now
Please have a seat and work quietly on the Do Now.
A ball of mass 15.5 kg is released from rest at a point 1.2 m high.
1. If we ignore air resistance, what will be the velocity of the ball at the lowest point of motion?
2. If we assume that air resistance does -20 J of work on the ball as it falls, what will be the velocity at the lowest point of motion?
Share out. Specific questions to ask students: Can you step me through how you found the velocity? How do you know much initial mechanical energy the ball has? How do you know that the total mechanical energy will remain constant? Where does the kinetic energy of the ball come from? What does the work done by friction do to the mechanical energy of the ball?
### [2 min] Framing the Day
The scenario we will be looking at today will be very similar to the Do Now… only more dangerous!
Objective: You will be able to predict and analyze motion using the Law of Conservation of Energy.
Criteria for Success:
1. Can I predict what will happen to the mechanical energy of an object by using energy conservation?
2. Can I determine if and how an object’s mechanical energy has changed?
3. Can I solve a physics problem about the motion of an object using energy conservation?
Can someone please raise his or her hand and explain what we are doing today?
Can someone please raise his or her hand and remind us how we will know if we are successful today?
### [35 min] Will Professor Lewin Survive?
Note: The following activity will be outlined on a handout / graphic organizer.
We are handing out a worksheet. These will be collected today at the end of the period. Please take a couple of minutes to read the first two sections, labeled “Video” and “Directions”.
VIDEO
Professor Walter Lewin is going to put his life on the line to prove the law of the conservation of energy. He will release a 15.5 kg pendulum bob from his chin, and wait to see what happens when the ball swings back at his face! Will the ball smash his face in? Or will the laws of physics protect him? We will find out….
Can someone please raise their hand to volunteer to read the first paragraph for us?
Directions
• We will be watching a video that goes along with this handout. Do not move ahead. Some questions we will do as a class, some as a table, and others individually.
• Write your answers in complete sentences wherever possible. This helps organize your thinking and gives you better study materials later for quizzes and tests.
Can someone please raise their hand and tell us, what is one direction we should follow today? Why should we do that?
Can someone please raise their hand and tell us, what is another direction we should follow today? Why should we do that?
Play the video of Walter Lewin putting his life on the line to prove Conservation of Energy.
Stop the video at 2:54. (Right after “…this will be my last lecture.”)
PREDICTIONS
1. What kind(s) of mechanical energy does the ball have right now and how do you know?
Quick share, check for agreement or disagreement and why.
Take a couple of minutes to work with your table group on the first column of the table (question 2).
After each column, ask a student to share out what their group answered. Questions to ask: Why do you think that the mechanical energy will increase/decrease? Why will this make the height higher/lower? What does positive/negative work do to the mechanical energy of the pendulum? How will we know if the mechanical energy is higher/lower?
Repeat for questions 3 and 4.
2. Prof. Lewin does not push the ball and we assume no air resistance 3. Prof. Lewin does not push the ball but we include air resistance 4. Prof. Lewin accidentally pushes the ball as he lets go What kind of work will be done on the ball (positive / negative / none)? What will happen to the total mechanical energy of the ball (increase / decrease / constant)? What will be the height of the ball when it swings back to Prof. Lewin (higher / lower / same)? Will Professor Lewin be safe (yes / no)?
Add this diagram to the slide to be clear about what I mean by “height”:
5. Make your prediction: What will happen to the mechanical energy of the ball? How will you know if your prediction is right or wrong?
Play the rest of the video.
Show a side by side of the before (at the time of release) and after (1 cycle):
I was careful to stop the video exactly when the ball reached its maximum height on the return swing. Is he at the same position or not? What do you notice about the ball’s position?
At this point there should be at least 20 min remaining in the period.
ANALYSIS
6. What happened to the mechanical energy of the ball, and how do you know? What evidence tells you so?
DISCUSSION
Possible questions:What happened to the mechanical energy of the ball, and what evidence do you have? Was there work done on the ball? If so, by what force and was it positive or negative work? How can we explain what happened using the Law of Conservation of Energy? As students answer, ask for agreement or disagreement and why.
CONFER
In a couple of minutes, I’m going to ask you to answer question 7 individually. But before we do that. you have 2 minutes to check with your partner and make sure you agree on what happened. This is your only chance to confer before answering question 7 on your own.
7. Write out a short story of what happened to Prof. Lewin’s swinging pendulum. You must answer the following questions in your story: Was work done on the ball and by what force? What did this work do to the mechanical energy of the ball? What did we observe that told us what happened to the mechanical energy of the ball?
### [5 min] Exit Ticket
Professor Lewin released a 15.5 kg pendulum from 1.20 m high. We carefully measure the height of the ball when it swings back towards him, and determine that the ball only went 1.05 m high when it came back. How much work was done by air resistance on the ball?
Collect student work. If running out of time, ask co resident to help photograph worksheets and exit tickets of case study students as a worst case fallback option.
52 min total: ~5 min of buffer
If need be, the CONFER section of the plan can be eliminated without impact to the cognitive demand or student sense-making in the lesson.
## Lesson Plan: Conservation of Energy using a Music Video
Lesson 2.2: Exploring Conservation of Energy
Unit: Work and Energy, Component 2 – Conservation of Energy
Date: January 10th, 2013
Day/Block: Day 1 – A/E/F
Time Available: A 58min / E 48min / F 65min
Objective:
You will be able to design and analyze a Rube Goldberg Machine using the Law of Conservation of Energy.
Criteria for Success:
Can I design my own machine that transfers mechanical energy between objects through work?
Can I use the Law of Conservation of Energy to explain how my design transfers energy?
Assessment:
Handout with machine design and analysis questions.
Agenda:
### [10 min] DoNow:
The complicated chain of events in the music video is called a “Rube Goldberg Machine”. These machines use many transfers of energy between a whole lot of objects in order to do a very simple task.
Invent your own small (2 objects) Rube Goldberg machine. How would you use one object to make another object do something else? Where is there work and energy? How does one object transfer energy to another?
Example:
### [10 min] Discussion: Music Video
Show OK Go’s music video for “This Too Shall Pass”: (4 min)
Note: Watching the video was assigned as homework the previous night, along with the following guided questions: Where do you see work? Where do you see energy being transferred from one form to another? Write down at least 1 example (note the video time), and make sure to explain what object is doing work on what other object, what kind of energy is being transferred, and how you know.
Talk to a partner next to you and share 1 example of work and energy being transferred from one object to another. I will ask several students to share an observation that their partner noticed, and explain to me what object is having work done on it and what kinds of energy are involved.
What did you see? Be specific, tell me what happened to the object that makes you think there was work done. What kinds of energy do you think that the object has? How do you know that the object has that kind of energy?
Record student observations on the whiteboard.
### [15 min] In-Depth Analysis: Tire
Show video clip of the tire section twice (7 second clip starting at 1:03). Ask students to write down exactly what they see happening to the tire. Have students share their observations, and assemble a record of the tire’s journey on the board to refer to later. (Make sure that the bucket hitting the tire is included.) Hand out the worksheet for scaffolded analysis of the tire scenario.
Refer to your handout. Take 30 seconds to answer the first question by yourself. Ask one student to share what they wrote with the class.
1. At the beginning of the scenario, does the tire already have mechanical energy? If so, what form(s) is it in, and how do you know?
Take two minutes to answer questions 2 and 3 with a partner. Ask one or two pairs to share, depending on time.
2. During the scenario, is work done on the tire by any other object? Is this work positive or negative, and how do you know?
3. What happens to the tire’s total mechanical energy when the work is done to it?
Take four minutes to answer questions 4 and 5 with a partner. Ask one or two pairs to share, depending on time.
4. Describe what happens to the tire’s GPE, KE, and total ME as it goes through the scenario.
5. What happens to the tire’s mechanical energy at the end of the scenario?
### [15 min] Creative Activity: Design and Analyze
Turn over your handout. Now you have a chance to design your own Rube Goldberg machine! Draw your machine in the space provided, and use the Tire Analysis as a guide to answer the analysis questions below. The questions are due tomorrow for stamps. If you don’t finish in class, please complete the analysis questions as homework.
6. Draw your own piece of this Rube Goldberg Machine that uses 1 object. You must include at least 1 transfer of energy from one object to another.
7. What is your main object in this scenario? :
8. Describe what happens to your object’s GPE, KE, and total ME as it goes through the scenario beginning to end..
9. Where is work done on your object or by your object? How does this work change your object’s total mechanical energy?
10. Explain how your machine obeys the Law of Conservation of Energy, MEi + W = MEf.
## Lesson Plan: Exploring Newton’s 3rd Law in Sports
EXPLORING NEWTON’S 3rd LAW IN SPORTS
Unit: Dynamics
Date: November 19th, 2012
Day/Block: Day 3 / A Block
Time Available: 65 min
Teacher Prep:
• Ensure prerequisite knowledge of: introduction to Newton’s 3rd Law
• Make slides (including one for Objective and Criteria for Success)
• Print and copy exit tickets
• Rehearse lesson and do the work of the students
Lesson Objective:
You will be able to identify action-reaction force pairs and make predictions about motion using Newton’s Third Law.
Criteria for Success:
You will be able to explain what Newton’s 3rd Law says about forces.
You will be able to use Newton’s 3rd Law to predict what forces will act on an object in physical scenarios.
Assessment:
Exit ticket.
Agenda:
### [5 min] Do Now
Newton’s 3rd Law tells us that all forces come in action-reaction pairs. List the action-reaction force pairs that you can think of on the red football player. (Hint: Mr. Ratnayake sees at least 4). Draw a free body diagram of the red football player.
### [1 min] Making Explicit the Content of the Lesson
Hang on to what you did for the Do Now. We will be returning to it later on in the lesson.
Lesson Objective: You will be able to identify action-reaction force pairs and make predictions about motion using Newton’s Third Law.
Criteria for Success:
You will be able to explain what Newton’s 3rd Law says about forces.
You will be able to use Newton’s 3rd Law to predict what forces will act on an object in physical scenarios.
* Ask students to revoice the objective and CfS.
### [10 min] Mini-Lecture 1: Review of Newton’s 3rd Law
[7 min for lecture] Review the main points of the third law.
• There is no such thing as a single force — forces always come in action-reaction pairs.
• Action-reaction pairs are the same kind of force acting on different objects.
• Action-reaction pair have equal magnitude forces acting in opposite directions.
* Ask students, what do you think I mean by “same kind of force”? (Gravity, perpendicular contact force, parallel contact force, etc).
* Ask students, what do you think I mean by “magnitude”? (Strength of the force, size of the force, the value of the number, etc).
[3min for processing time] Take 2 minutes to check with a partner next to you. Look back at your list of action-reaction pairs from the Do Now. Do the pairs on your list fit what we just wrote down about Newton’s 3rd Law? I will ask someone to tell me about what their partner wrote.
* Ask students to name a force pair that their partner wrote down, and why they think it fits the description of an action reaction pair. Draw the force pairs on the football player.
[10 min] Mini-Lecture 2: Review of Free Body Diagrams.
[5 min for lecture] A Free Body Diagram of an object only shows the forces acting on that object.
Free Body Diagrams do not include the forces that the object itself applies on other things.
Ask yourself: if I were this object, which forces would I feel acting on me?
Block on a surface example. There are two action-reaction pairs:
1. gravity from the earth on the block, with gravity from the block on the earth
2. contact force from the block to the surface, with contact force from the surface to the block
Which of these forces do you think the block is feeling? (normal force and weight). Draw FBD.
[2 min for processing time] Take 1 minute to check with a partner next to you. Look back at your free body diagrams from the Do Now. Does your partner’s FBD of the football player obey the rules of a free body diagram?
[3 min for closure on the Do Now] * Have a student draw the free body diagram for the red football player. Use questions for students to correct it if necessary.
### [1 min] Instructions for Scenarios
Take 1 minute to read the directions for this next segment. I will call on a student to explain what we are doing for the class.
• You will be given a scenario and several questions for discussion in your table group.
• I will call on someone for each part of the discussion questions.
• If they represent their group well, the whole group gets a stamp.
*Ask students: What are we going to be doing?
### [15 min] Scenario 1: Serena Williams — Tennis
[15 min total, 8 min to discuss with group and work out the scenario, 7 min for discussion]
Tennis star Serena Williams uses Newton’s Laws to get the tennis ball to move.
• Describe the action-reaction force pair that acts to accelerate the tennis ball. What are the forces? In which direction do they act? On what does each force at?
• Draw a free body diagram of the tennis ball. In which direction is the net force on the tennis ball? Predict what will happen to the tennis ball and racket, using Newton’s Laws.
### [15 min] Scenario 2: Ron Weasley — Quidditch
[12 min total, 7 min to think-pair-share, 5 min for discussion]
Quidditch keeper Ron Weasley blocks a quaffle coming in from the left of the image.
• Describe the action-reaction force pair that acts to block the quaffle at the time of impact. What are the forces? In which direction do they act? On what does each force at?
• Draw the FBD for the quaffle and the FBD for Ron. In which direction is the net force and acceleration for the quaffle? What about for Ron?
• Which will accelerate more, the quaffle or Ron? If Ron and the quaffle both experience an equal force from the impact, why are their accelerations different?
### [7 min] Exit Ticket
How can you tell if two forces are an action-reaction pair according to Newton’s 3rd Law?
An archery target stops an arrow on impact. The arrow experiences high acceleration to go from a fast speed to at-rest very quickly. Do the arrow and the target experience the same force from the impact? Do the arrow and the target experience the same acceleration? Why or why not?
64 min total: ~1 min of buffer
Pacing: If necessary due to unexpected time constraint, one of the scenarios can be cut out and the other extended slightly.
## Lesson Plan: Introduction to Newton’s Second Law of Motion
### [7 min] Do Now
Review from 1st Law, introduce 2nd Law:
In which of these cases do we have balanced forces? Explain why.
• A cat is moving with constant velocity towards his date.
• A car is moving with constant acceleration to pick up more physics homework.
• A cow is at rest, taking a nap.
• An apple is hanging from a tree.
Share out and discuss. Bridge the transition between Newton’s First Law and the idea of net force into Newton’s Second Law.
### [1 min] Making Clear the Objective
Objective: You will derive the relationship between force and acceleration from simulated experimental data.
Criteria for Success: Graphs of data will show proof of Newton’s 2nd Law of Motion.
### [12 min] Simulation: Newton’s Second Law
We will be using the simulation of Newton’s 2nd Law located at: http://phet.colorado.edu/en/simulation/forces-1d
Set: show horizontal force, show total force.
Turn friction off.
Turn on graphs for acceleration and velocity.
Use students to run simulation and call out the data for their classmates to record.
We will be using a simulation. For each trial, record the following:
• mass of the object
• force applied to the object
• acceleration of the object
Run the simulation for the dog (25 kg) with three forces: 50 N, 100 N, 200 N. Ask the students to make a prediction before the last one. Make sure to reset the simulation and graphs before each trial.
Run the simulation for the textbook (10 kg) with the same three forces.
### [15 min] Graphing the Data
Turn and Talk:
What was the independent variable and why?
What was the dependent variable and why?
What was the main control variable and why?
What do we put on the y-axis? What do we put on the x-axis?
The independent variable of our experiment always goes on the x-axis (Force). The dependent variable of our experiment always goes on the y-axis (Acceleration).
Draw 2 graphs. Don’t forget units and labels!
• Acceleration vs force variable for the dog
• Acceleration vs force variable for textbook
### [15 min] Analyzing the Data
We seem to have found a correlation between two variables, force and acceleration. Let’s see if we can define a relationship between them.
Find the slope of each graph and write it next to the plot.
Find the inverse of the slope for each graph and write it next to the plot.
Think-Pair-Share:
Do we see any patterns? Does the slope look like a variable we recognize? How would I write the equation of this line?
a = 1/m F → F = m a
### [2 min] Summarize Findings
Newton’s 2nd Law of Motion:
The acceleration of an object is directly proportional to the net force acting on the object. The acceleration will be in the same direction as the net force. The acceleration is resisted by the mass of the object.
F = m a
### [6 min] Exit Ticket
The catapult on an aircraft carrier can can accelerate a fighter jet from rest to 56 m/s in just 2.8 s. If the fighter jet has a mass of 13,000 kg, what is the force required? | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.theveryspringandroot.com/blog/category/from-the-classroom/lesson-materials/", "fetch_time": 1606647271000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-50/segments/1606141197593.33/warc/CC-MAIN-20201129093434-20201129123434-00001.warc.gz", "warc_record_offset": 843940149, "warc_record_length": 24800, "token_count": 5500, "char_count": 24600, "score": 3.578125, "int_score": 4, "crawl": "CC-MAIN-2020-50", "snapshot_type": "longest", "language": "en", "language_score": 0.9589945673942566, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00050-of-00064.parquet"} |
# Amusement Park Physics
## Presentation on theme: "Amusement Park Physics"— Presentation transcript:
Amusement Park Physics
Hold on to your hats, guys and gals, this is one fast trip!
What makes a roller coaster so fun?
Some think it is the free fall Others the speed and acceleration Maybe it’s the fear you feel when the roller coaster Sloooowly climbs that first hill!
Roller Coasters What keeps you in Inertia your seat on a loop?
What force powers Roller coasters? Gravity Why do you experience crushing feelings? Inertia Where? At the bottom of a hill as the train starts to go up Why do you experience free fall? Inertia At the top of a hill as the train starts to go down Where?
Energy Mechanical Energy 1. Potential Energy 2. Kinetic Energy
Thermal Energy 1. Friction
Looping Coaster at Canobie
PE = mgh Why is this hill lower? How can we calculate the energy lost to friction? KE = ½ mv2
Spiral Twists Why have these loops?
Acceleration = change in velocity over time Velocity = speed and direction
Water Rides—are they roller coasters?
Compare Both have hills Both use gravity Both have energy transformations Contrast Water powers the ride—not gravity alone Water is also the friction to slow down—not brakes No track PE to KE transformation not as key as water speed
The Boston Tea Party The friction with the water and the blunt shape of the car makes the wave so huge and beautiful! Teacher’s pet I love physics
Circular Rides—A huge variety
This type of ride makes many people Feel ill. How?
What do the forces look like on a swing ride?
And in a circular pattern with constantly changing direction Direction of the ride Riders move out from the center This is different from a ride with a track, because there no track/seat to push back. What is the reaction force then?
But they all work the same way
What other force is at work here? Hint: it pulls you down If you are going on the ride with Hagrid (a giant), should you or he sit on the outside? Hint: Which way are you being pulled or pushed?
Let’s Look at the Forces
Centripetal force C Direction of ride The direction is constantly changing which means The acceleration is constant How does inertia figure into this?
Is this one different? Do you really need the restraints?
If someone were to say, upchuck, what direction Would the substance go? Eww—GROSS!!!
Circular Motion (One complete cycle)
Circular Motion
Circular Motion
Pirate Ship—Pendulum Rides
FREE FALL What is the thrill of this ride? What produces that thrill? Inertia
Free Body Diagram Inertia keeps the body going up
When the ship is going back down The force of friction goes against The direction of motion Direction of movement Force of Weight
How Could We Calculate the Amount of Energy “Lost” Due to Friction?
Where we release the “ship” PE = mgh Where the highest point is after an amount of time PE = mgh If we find both PEs and subtract h – h. . .we have it! Now, what if we want to find the average PE lost on each swing?
Pendulum Physics
Figure 13.4: The Simple Pendulum
Force the rider feels is defined by: Fr = - (m)(g)(sin )
Pendulum Physics One Period (T): l = length of string g = 9.8 m/s/s
Pendulum Physics The frequency of oscillation is the inverse of the period: frequency =
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# Probability rolling two different sided die and sum being a number
I'm building an app (for those curious, for DnD) and I came across an issue with some math I did. I need to know the probability of rolling a certain number when there are two or more different sided die used.
A concrete example that I'm working on is rolling at least 3 using a 4-sided die and a 6-sided die. Below is a table I made that contains the answer (I used Google Sheets and counted cells), but I actually need to know how to get to this answer in code. I just can't seem to find a way to work forward or backwards and get the right results. Can anyone help me figure this out?
>= 2 | 24 100
>= 3 | 23 95.83
>= 4 | 21 87.5
>= 5 | 18 75
>= 6 | 14 58.33
>= 7 | 10 41.67
>= 8 | 6 25
>= 9 | 3 12.5
>=10 | 1 4.17
• There are ways to generate expressions for these probabilities mathematically, but in terms of actually coding them in, I think you'd be better off just enumerating the possibilities and adding them up on the fly in your code. Aug 2 '15 at 5:11
tldr the final formula for a dA and a dB with a target sum of at least N will be: $$\begin{cases} \frac{AB-T(N-2)}{AB}&\text{if}~ N-1\leq A~\text{and}~N-1\leq B\\ \frac{AB-T(N-2)+T(N-A-2)}{AB}&\text{if}~N-1>A~\text{and}~N-1\leq B\\ \frac{AB-T(N-2)+T(N-B-2)}{AB}&\text{if}~N-1\leq A~\text{and}~N-1>B\\ \frac{AB-T(N-2)+T(N-A-2)+T(N-B-2)}{AB}&\text{if}~N-1>A~\text{and}~N-1>B \end{cases}$$
If approaching by brute force, sometimes it is a whole lot easier to find out the probability of not satisfying your property. Take your example of "at least 3". The only way that you don't get at least three is by rolling a one on both dice, which occurs with probability $\frac{1}{4\cdot 6}$. Similarly, not getting at least a four would occur either with a (1,1), a (1,2), or a (2,1) for a probability of $\frac{3}{24}$.
Lets see if we can generalize. Suppose there were two six sided dice and we were looking for the probability of getting "at least a six."
You can see that the black numbers are below six and should be thrown out, yet all numbers above the green line are good (including those not displayed in the picture). There are a total of $6\cdot 6=36$ possibilities which are equally likely. You can notice that the black numbers form a triangle with each layer having one more possibility. We can count how many spaces are used in a triangle quickly as these are what are known as triangle numbers. In this case, there are four layers to our triangle so there are $T(4)=\binom{5}{2}=10$ spaces used. Thus in this example our probability is $\frac{10}{36}$ to not get at least a six, so it is $\frac{36-10}{36}=\frac{26}{36}$ to get at least a six.
So, this works well and good if our line happens to lie within the realm of possibility for both dice, but what happens if we change the number or the size of the die?
In this example, we consider the probability of rolling at least an 8 on a d4 and a d9 (yes, I know that d9 don't normally exist... work with me). You can see two triangles here. The triangle made up of all of both of the black and red numbers, as well as a triangle made of only red numbers. We count the number of spaces used in the larger triangle as $T(6)=\binom{7}{2}=21$ and we wish to throw these away. We cannot throw all of them away however since the red numbers weren't possible in the first place. We count $T(2)=\binom{3}{2}=3$ of these, so there are $18$ we wish to not consider. That is to say, the probability of not rolling at least an 8 is $\frac{(21-3)}{9\cdot 4}$, so the probability of rolling at least an 8 is $\frac{9\cdot 4 - (21-3)}{9\cdot 4}$
Similarly we might have to remove two red triangles if our target number is too high for both dice like in the following scenario:
In this example we consider the question "what is the probability of rolling at least a 7 on two d4." We approach again via the opposite question of what the probability is for rolling a 6 or less using the black triangle. The black triangle is of size $T(5)=15$ but there are two red triangles each of size $T(1)$ to remove for a total probability of $\frac{13}{16}$ that you don't roll at least a 7, meaning there is a probability of $\frac{3}{16}$ that you do roll at least a 7.
We can generalize all of this in the following formula.
With two dice, one with $A$ number of sides, and the other with $B$ number of sides, the probability of rolling at least a sum of $N$ is:
$$\begin{cases} \frac{AB-T(N-2)}{AB}&\text{if}~ N-1\leq A~\text{and}~N-1\leq B\\ \frac{AB-T(N-2)+T(N-A-2)}{AB}&\text{if}~N-1>A~\text{and}~N-1\leq B\\ \frac{AB-T(N-2)+T(N-B-2)}{AB}&\text{if}~N-1\leq A~\text{and}~N-1>B\\ \frac{AB-T(N-2)+T(N-A-2)+T(N-B-2)}{AB}&\text{if}~N-1>A~\text{and}~N-1>B \end{cases}$$
Where again, the formula for the triangle number is given as $T(k) = \binom{k+1}{2} = \frac{k^2+k}{2}$
The same method can work if generalizing to three dice using tetrahedral numbers instead, chopping off each corner of our three dimensional grid as necessary.
• Thank you SOOO much. I haven't done anywhere near this level of math in ages; the diagrams and formulas really helped out. I have my code set up for 2 die now! I'm going to experiment with add three and four, but I think I can figure everything out based on what you've given me. Aug 2 '15 at 6:42
• For anyone that comes through and would like to see what I coded from this, here's a link to it: javastub.com/646530217 Aug 2 '15 at 16:05
• I realized that there may be some error catching that needs to occur in the case that N is larger than A+B, i.e when it is an impossible scenario. I'm afraid the formula I gave will spit out negative numbers. I'm on the road from church headed to family dinner, so I can't confirm, but be aware that may be an issue. Aug 2 '15 at 16:21
• That's perfectly fine; if N > A+B, it's impossible as you stated so I'll catch it before it gets to the formula. Aug 2 '15 at 19:05
expand $(x+x^2+x^3+x^4)(x+x^2+x^3+x^4+x^5+x^6)$
The coefficients of $x^k$ will give you the # of ways to get a sum of exactly k
[ How to get the $\ge$ form you need should be obvious ]
PS:
You could get a sum of 5, say, as 1+4, 2+3, 3+2 and 4+1
You can see that this corresponds exactly to adding up the coefficients of $x^5$
$x\cdot x^4 + x^2\cdot x^3 + x^3\cdot x^2 + x^4\cdot x$ in the expression
• Given the OP's question, explaining why this method works and how to get an intuition derivation for it may be really useful. Aug 2 '15 at 5:09
• Thanks, added a PS as per your suggestion. Aug 2 '15 at 5:30
• This was the first answer that made sense, I just couldn't figure out how to turn this into code. Thank you, though! It's was still very very helpful in understanding how it all works. Aug 2 '15 at 6:42
I'm not sure what you are asking, but for your table of values, for a minimum total $n$ the odds are $f(n)=(n-1)(n-2)/48$ for $n<=6$, and $(12-n)(11-n)/48=f(13-n)/48$ for $n>6$. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/1381685/probability-rolling-two-different-sided-die-and-sum-being-a-number", "fetch_time": 1642928246000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-05/segments/1642320304217.55/warc/CC-MAIN-20220123081226-20220123111226-00173.warc.gz", "warc_record_offset": 436704283, "warc_record_length": 36650, "token_count": 2126, "char_count": 6947, "score": 4.1875, "int_score": 4, "crawl": "CC-MAIN-2022-05", "snapshot_type": "latest", "language": "en", "language_score": 0.9222429990768433, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00002-of-00064.parquet"} |
# Essay, Term Paper
Topics: Power factor, Electric power, Three-phase electric power Pages: 3 (662 words) Published: March 10, 2013
TUTORIAL 1
Fundamentals
1.An inductive load consisting of R and X in series feeding from a 2400-Vrms supply absorbs 288 kW at a lagging power factor of 0.8. Determine R and X.
2.An inductive load consisting of R and X in parallel feeding from a 2400-Vrms supply absorbs 288 kW at a lagging power factor of 0.8. Determine R and X.
3.Two loads connected in parallel are supplied from a single-phase 240-Vrms source. The two loads draw a total real power of 400 kW at a power factor of 0.8 lagging. One of the loads draws 120 kW at a power factor of 0.96 leading. Find the complex power of the other load.
4.The load shown in figure 1 consists of resistance R in parallel with a capacitor of reactance X. The load is fed from a single-phase supply through a line of impedance 8.4 + j11.2 . The rms voltage at the load terminal is 12000o Vrms, and the load is taking 30 kVA at 0.8 power factor leading.
a.Find the values of R and X
b.Determine the supply voltage V
Figure 1
5.Two impedances, Z1 = 0.8 + j5.6 and Z2 = 8 - j16, and a single-phase motor are connected in parallel across a 200-Vrms, 60-Hz supply as shown in figure T2. The motor draws 5 kVA at 0.8 power factor lagging.
Figure T2
a.Find the complex powers S1, S2, for the two impedances and S3for the motor.
b.Determine the total power taken from the supply, the supply current, and the overall power factor.
c.A capacitor is connected in parallel with the loads. Find the kvar and the capacitance in F to improve the overall power factor to unity. What is the new line current?
6.A 4157- Vrms, three-phase supply is applied to a balanced Y-connected three-phase load consisting of three identical impedances of 4836.870. Taking the phase to neutral voltage Van as reference, calculate:
a.The phasor currents in each line
b.The total active and reactive power supplied to the load
7.Repeat problem 6 with the same three-phase impedances arranged in a...
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# Center of Mass
All objects have mass, which causes earth's gravity to exert a force on them. This force is the object's weight. The weight of an object is typically felt at the points where the object rests upon other objects. For example, if you hold an object with two hands, a portion of its weight is felt on one hand, and the rest is felt on the other.
But every object has a center of mass about which all of its mass is equally balanced. If you knew where the center of mass of a given object was located, then you could balance the object on your finger by holding it under that point.
The real importance of the center of mass is that the force of gravity, and all inertia forces, can be modeled as passing through the center of mass. No matter how unevenly the mass is distributed throughout the object, the total weight of the object can still be described as acting through the center of mass. Once the center of mass has been computed, it simplifies calculations tremendously.
For symmetrical objects that are made of a single material throughout, the center of mass is simply the center of the object. But irregular shapes are more difficult. Virtual Car must compute the center of mass of any body shape that you draw, no matter how irregular it is.
This is done by dividing your shape into many strips. The weight of each strip (based on the density and thickness of the material that you have specified) is then computed, and multipled by its distance from the left corner of the shape. This product is called the weight moment of the strip. This process is repeated for all of the strips. A running total is kept for the combined weight of all strips, and the sum of the weight moments for each strip. After all strips are examined, the center of mass is computed, by dividing the sum of the weight moments by the total weight of all strips. This whole process yields only the x-coordinate of the center of mass!
To get the y-coordinate of the center of mass, the same process has to be performed using horizontal strips, as if the object were turned on its end, and gravity were acting horizontally.
The center of mass is denoted by the symbol .
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+0
# HELP
0
219
1
1. The position of a squirrel running in a park is given by r⃗ =[(0.280m/s)t+(0.0360m/s2)t2]i^+ (0.0190m/s3)t3j^.
What is υy(t), the y-component of the velocity of the squirrel, as function of time?
A.) vy(t)=(0.0570m/s3)t+(0.0720m/s2)t2
B.) vy(t)=(0.0570m/s3)t2
C.) vy(t)=(0.0570m/s2)t2
D.) vy(t)=(0.0570m/s3)t
2. At 4.52 s , how far is the squirrel from its initial position?
3. At 4.52 s , what is the magnitude of the squirrel's velocity?
4. At 4.52 s , what is the direction (in degrees counterclockwise from +x-axis) of the squirrel's velocity?
Guest Aug 5, 2017
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### 1+0 Answers
#1
+26366
+2
1. y(t) = 0.0190t3
vy(t) = 3*0.0190t2 → 0.0570t2 If t is in seconds then the 0.0570 must have units m/s3 as the result must be a velocity (m/s). Hence B.) is the result.
2. Put t = 4.52 s into x(t) from your previous question to find X, the distance travelled in the x-direction.
Put it into y(t) above to find Y, the distance travelled in the y-direction.
Then calculate R = sqrt(X2 + Y2)
3. As for 2, but using vx(t) and vy(t) instead.
4. Calculate the angle from tan-1(vy/vx)
Alan Aug 5, 2017
edited by Alan Aug 5, 2017
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# 1.2 Exponents and scientific notation (Page 2/9)
Page 2 / 9
Write each of the following products with a single base. Do not simplify further.
1. ${k}^{6}\cdot {k}^{9}$
2. ${\left(\frac{2}{y}\right)}^{4}\cdot \left(\frac{2}{y}\right)$
3. ${t}^{3}\cdot {t}^{6}\cdot {t}^{5}$
1. ${k}^{15}$
2. ${\left(\frac{2}{y}\right)}^{5}$
3. ${t}^{14}$
## Using the quotient rule of exponents
The quotient rule of exponents allows us to simplify an expression that divides two numbers with the same base but different exponents. In a similar way to the product rule, we can simplify an expression such as $\text{\hspace{0.17em}}\frac{{y}^{m}}{{y}^{n}},$ where $\text{\hspace{0.17em}}m>n.\text{\hspace{0.17em}}$ Consider the example $\text{\hspace{0.17em}}\frac{{y}^{9}}{{y}^{5}}.\text{\hspace{0.17em}}$ Perform the division by canceling common factors.
$\begin{array}{ccc}\hfill \frac{{y}^{9}}{{y}^{5}}& =& \frac{y\cdot y\cdot y\cdot y\cdot y\cdot y\cdot y\cdot y\cdot y}{y\cdot y\cdot y\cdot y\cdot y}\hfill \\ & =& \frac{\overline{)y}\cdot \overline{)y}\cdot \overline{)y}\cdot \overline{)y}\cdot \overline{)y}\cdot y\cdot y\cdot y\cdot y}{\overline{)y}\cdot \overline{)y}\cdot \overline{)y}\cdot \overline{)y}\cdot \overline{)y}}\hfill \\ & =& \frac{y\cdot y\cdot y\cdot y}{1}\hfill \\ & =& {y}^{4}\hfill \end{array}$
Notice that the exponent of the quotient is the difference between the exponents of the divisor and dividend.
$\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$
In other words, when dividing exponential expressions with the same base, we write the result with the common base and subtract the exponents.
$\frac{{y}^{9}}{{y}^{5}}={y}^{9-5}={y}^{4}$
For the time being, we must be aware of the condition $\text{\hspace{0.17em}}m>n.\text{\hspace{0.17em}}$ Otherwise, the difference $\text{\hspace{0.17em}}m-n\text{\hspace{0.17em}}$ could be zero or negative. Those possibilities will be explored shortly. Also, instead of qualifying variables as nonzero each time, we will simplify matters and assume from here on that all variables represent nonzero real numbers.
## The quotient rule of exponents
For any real number $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and natural numbers $\text{\hspace{0.17em}}m\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}n,$ such that $\text{\hspace{0.17em}}m>n,$ the quotient rule of exponents states that
$\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$
## Using the quotient rule
Write each of the following products with a single base. Do not simplify further.
1. $\frac{{\left(-2\right)}^{14}}{{\left(-2\right)}^{9}}$
2. $\frac{{t}^{23}}{{t}^{15}}$
3. $\frac{{\left(z\sqrt{2}\right)}^{5}}{z\sqrt{2}}$
Use the quotient rule to simplify each expression.
1. $\frac{{\left(-2\right)}^{14}}{{\left(-2\right)}^{9}}={\left(-2\right)}^{14-9}={\left(-2\right)}^{5}$
2. $\frac{{t}^{23}}{{t}^{15}}={t}^{23-15}={t}^{8}$
3. $\frac{{\left(z\sqrt{2}\right)}^{5}}{z\sqrt{2}}={\left(z\sqrt{2}\right)}^{5-1}={\left(z\sqrt{2}\right)}^{4}$
Write each of the following products with a single base. Do not simplify further.
1. $\frac{{s}^{75}}{{s}^{68}}$
2. $\frac{{\left(-3\right)}^{6}}{-3}$
3. $\frac{{\left(e{f}^{2}\right)}^{5}}{{\left(e{f}^{2}\right)}^{3}}$
1. ${s}^{7}$
2. ${\left(-3\right)}^{5}$
3. ${\left(e{f}^{2}\right)}^{2}$
## Using the power rule of exponents
Suppose an exponential expression is raised to some power. Can we simplify the result? Yes. To do this, we use the power rule of exponents . Consider the expression $\text{\hspace{0.17em}}{\left({x}^{2}\right)}^{3}.\text{\hspace{0.17em}}$ The expression inside the parentheses is multiplied twice because it has an exponent of 2. Then the result is multiplied three times because the entire expression has an exponent of 3.
The exponent of the answer is the product of the exponents: $\text{\hspace{0.17em}}{\left({x}^{2}\right)}^{3}={x}^{2\cdot 3}={x}^{6}.\text{\hspace{0.17em}}$ In other words, when raising an exponential expression to a power, we write the result with the common base and the product of the exponents.
${\left({a}^{m}\right)}^{n}={a}^{m\cdot n}$
Be careful to distinguish between uses of the product rule and the power rule. When using the product rule, different terms with the same bases are raised to exponents. In this case, you add the exponents. When using the power rule, a term in exponential notation is raised to a power. In this case, you multiply the exponents.
## The power rule of exponents
For any real number $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and positive integers $\text{\hspace{0.17em}}m\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}n,$ the power rule of exponents states that
${\left({a}^{m}\right)}^{n}={a}^{m\cdot n}$
## Using the power rule
Write each of the following products with a single base. Do not simplify further.
1. ${\left({x}^{2}\right)}^{7}$
2. ${\left({\left(2t\right)}^{5}\right)}^{3}$
3. ${\left({\left(-3\right)}^{5}\right)}^{11}$
Use the power rule to simplify each expression.
1. ${\left({x}^{2}\right)}^{7}={x}^{2\cdot 7}={x}^{14}$
2. ${\left({\left(2t\right)}^{5}\right)}^{3}={\left(2t\right)}^{5\cdot 3}={\left(2t\right)}^{15}$
3. ${\left({\left(-3\right)}^{5}\right)}^{11}={\left(-3\right)}^{5\cdot 11}={\left(-3\right)}^{55}$
#### Questions & Answers
root under 3-root under 2 by 5 y square
Himanshu Reply
The sum of the first n terms of a certain series is 2^n-1, Show that , this series is Geometric and Find the formula of the n^th
amani Reply
cosA\1+sinA=secA-tanA
Aasik Reply
why two x + seven is equal to nineteen.
Kingsley Reply
The numbers cannot be combined with the x
Othman
2x + 7 =19
humberto
2x +7=19. 2x=19 - 7 2x=12 x=6
Yvonne
because x is 6
SAIDI
what is the best practice that will address the issue on this topic? anyone who can help me. i'm working on my action research.
Melanie Reply
simplify each radical by removing as many factors as possible (a) √75
Jason Reply
how is infinity bidder from undefined?
Karl Reply
what is the value of x in 4x-2+3
Vishal Reply
give the complete question
Shanky
4x=3-2 4x=1 x=1+4 x=5 5x
Olaiya
hi can you give another equation I'd like to solve it
Daniel
what is the value of x in 4x-2+3
Olaiya
if 4x-2+3 = 0 then 4x = 2-3 4x = -1 x = -(1÷4) is the answer.
Jacob
4x-2+3 4x=-3+2 4×=-1 4×/4=-1/4
LUTHO
then x=-1/4
LUTHO
4x-2+3 4x=-3+2 4x=-1 4x÷4=-1÷4 x=-1÷4
LUTHO
A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was 1350 bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 hours?
David Reply
v=lbh calculate the volume if i.l=5cm, b=2cm ,h=3cm
Haidar Reply
Need help with math
Peya
can you help me on this topic of Geometry if l help you
litshani
( cosec Q _ cot Q ) whole spuare = 1_cosQ / 1+cosQ
Aarav Reply
A guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So, the length of the guy wire can be found by evaluating √(90000+160000). What is the length of the guy wire?
Maxwell Reply
the indicated sum of a sequence is known as
Arku Reply
how do I attempted a trig number as a starter
Tumwe Reply
cos 18 ____ sin 72 evaluate
Het Reply
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Question
# Consider the figure: If BC = AC, x = 2y, and angle ∠ACB = 60o, then find the value of ∠B + y (in degrees). ___
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Solution
## ∠C = 60∘ and BC = AC, By angle sum property of triangle, ∠A = ∠B = 60∘ ∠A + ∠B + x + y = 60∘ + 60∘ + 120∘ = 240∘ =>y + 2y = 120∘ => y = 40∘ Hence, ∠B + y = 100∘
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You are on page 1of 4
# Chapter 1
INTRODUCTION
## If Theoretical Mechanics deals with rigid (non-deformable) bodies,
Mechanics of Materials (often called Strength of Materials) deals with deformable
bodies.
The main objective of the Mechanics of Materials is to study the behavior of
the construction elements under any actions and to establish mathematical relations
which provide checking the strength, rigidity (deformations) and stability
conditions. Otherwise, both the analysis and design of a given structure involve the
determination of stresses and deformations.
Mechanics of Materials is a part of the group of disciplines called Applied
Mechanics (or Construction Mechanics). The disciplines which form this important
group are:
- Theoretical Mechanics: dealing with the study of rigid bodies
- Mechanics (Strength) of Materials: dealing with the study of bar elements
- Theory of Elasticity: supplies the main hypothesis regarding the deformable
bodies and some fundamental relationships; it deals with the study of plates
and massifs
- Theory of Plasticity: dealing with permanent deformations of the body
- Static, Dynamic and Stability
- Mechanics of Soils
## 1.2 CLASSIFICATION OF BODIES IN CONSTRUCTION
The solid bodies are classified according to their characteristic dimensions, into:
a. The bar: is a one-dimensional body (Fig.1.1.a), what means that one
dimension (the length, which is parallel to the bar axis x) is much bigger than the
other two, which form the cross section (their dimensions are parallel to axis y and
z). The cross section is a plan cut perpendicular (normal) to the longitudinal axis of
the bar.
E.g.: beams, columns (linear elements)
b. The plate: is a two-dimensional body (Fig.1.1.b), which has two
dimensions much bigger than the third (generally, the plate thickness).
E.g.: plane plates: lamellas, slabs, curved plates: shells (surface elements)
a. b.
Fig. 1.1
c. The massive (solid): is a three-dimensional body which has all three
dimensions comparable.
E.g.: dams, retaining walls (volume elements)
1.3 HYPOTHESES IN MECHANICS OF MATERIALS
## Some hypotheses concerning the structure of the materials and their
behavior should be considered in what follows:
a. The hypothesis of the continuous medium: Mechanics of Materials
considers the materials as continuous, homogeny medium, which occupy the entire
space given by their volume.
b. The hypothesis of the isotropy: the materials are isotropic if on any
direction the elastic proprieties are identically (e.g.: the steel). Otherwise, they are
anisotropic or orthotropic (e.g.: the wood).
c. The hypothesis of the small deformations: the deformations of the elastic
bodies (elongations, sliding, displacements) are very small with respect to their
characteristic dimensions.
d. The linear relation between unit stresses and deformations: for elastic
regime of solicitation Hookes law is valid. For this reason the principle of the
superposition of effects may be applied.
## e. Bernoullis hypothesis (the plane-sections hypothesis): a plane section
perpendicular to the bar axis before deformation, remains plane and perpendicular,
also after deformation.
Jakob Bernoulli (1654-1705)
## f. Saint-Venants principle: stress distribution may be assumed independent
of the mode of load application except in the immediate vicinity of load application
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About Snapxam Calculators Topics Contact us
# Find the derivative of xy-1y^2+x^2=1
Go!
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0
x
y
(◻)
◻/◻
2
e
π
ln
log
lim
d/dx
d/dx
>
<
>=
<=
sin
cos
tan
cot
sec
csc
asin
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atan
acot
asec
acsc
sinh
cosh
tanh
coth
sech
csch
asinh
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atanh
acoth
asech
acsch
## Answer
$2x+y=0$
## Step by step solution
Problem
$\frac{d}{dx}\left(x^2+xy-y^2=1\right)$
1
Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable
$\frac{d}{dx}\left(-y^2+y\cdot x+x^2\right)=\frac{d}{dx}\left(1\right)$
2
The derivative of the constant function is equal to zero
$\frac{d}{dx}\left(-y^2+y\cdot x+x^2\right)=0$
3
The derivative of a sum of two functions is the sum of the derivatives of each function
$\frac{d}{dx}\left(-y^2\right)+\frac{d}{dx}\left(y\cdot x\right)+\frac{d}{dx}\left(x^2\right)=0$
4
The derivative of the constant function is equal to zero
$0+\frac{d}{dx}\left(y\cdot x\right)+\frac{d}{dx}\left(x^2\right)=0$
5
The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function
$0+y\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(x^2\right)=0$
6
The derivative of the linear function is equal to $1$
$0+1y+\frac{d}{dx}\left(x^2\right)=0$
7
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$0+1y+2x=0$
8
Any expression multiplied by $1$ is equal to itself
$0+y+2x=0$
9
$x+0=x$, where $x$ is any expression
$2x+y=0$
## Answer
$2x+y=0$
### Main topic:
Differential calculus
0.21 seconds
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Ah to kWh Calculator - The How & Why - ShopSolar.com
Ah to kWh Calculator
Short on Time? Here’s The Article Summary
The article discusses the use of calculators in the solar and electrical power fields, focusing on the conversion between amp hours (Ah) and kilowatt hours (kWh). It explains the concepts of Ah and kWh, detailing how they are used to measure battery capacity and energy consumption, respectively.
The article provides examples and formulas for converting Ah to kWh and vice versa, emphasizing the relationship between voltage, current, and energy. It also highlights the importance of understanding these units and calculations for solar system planning and equipment selection. Overall, the article aims to demystify these conversions and encourage a deeper understanding of the underlying principles in solar energy systems.
Introduction
The world of solar and electrical power has hundreds of different calculators that help us with conversion and overall problem-solving.
We’re going to go over how to use an ah to kWh calculator, breaking down what’s needed and how the conversion works.
Before we begin looking at our Ah to kWh calculator, we need to dissect it and look at the variables as individuals.
What Is an Amp Hour?
When you’re new to the solar world, you’ll hear about all sorts of units of power and different conversions or calculators.
For example, you may come across a calculator that can convert cold-cranking amps to amp hours but not know what either of these measurements is used for.
Today, we’re going to focus on the latter to get a better understanding of it
Amp hours are a way to measure the amount of time that a battery stays powered.
All batteries have a specific capacity: the maximum amount of energy that can be used under normal circumstances. This capacity is typically measured in amp hours.
Let’s look at an example to get a better understanding.
Example
If you have a battery that has a capacity of 1 Ah and it requires a current draw of exactly 1 A, the battery can run an appliance for a single hour.
If we take this hour and divide it by 10, it tells us that every 6 minutes, the battery depletes by 10%.
You can use an online amp hour calculator to plug in more complex numbers but the basic premise remains.
If you don’t know where to start, use our house amp calculator. This will tell you what you need and help you decide on the size of battery you need to get going.
Now that we know all about amp hours, what about kilowatt hours?
What Are Kilowatt Hours?
Aside from the calculators used to convert MWh to kWh, the most common place you’ll see this unit is at the bottom of your monthly power bill.
This wattage can be seen on almost every appliance, whether it’s a 60 W lightbulb or a 1,000 W microwave. This tells you how much power each appliance draws when it’s being used.
Energy is the accumulated amount t of power being used over some time, in this case, at the end of every month.
Energy is expressed using Kilowatt-hours, how many thousands of watts are used over an hour.
To put it simply, 1 kWh is equivalent to 1,000 W used over an hour.
Kilowatt-hours and amp-hours are related. It may not be as easy to work out as knowing how to convert voltage to amps but it’s possible.
How to Use the Ah to kWh Calculator
The conversion here is between electrical charge and energy. Fortunately, for us, there’s a handy formula that we can use for these calculations.
The formula states that kilowatt hours are equivalent to the product of the amp hours and voltage, divided by 1,000.
In mathematical terms, it looks like this:
kWh = (Ah x V) / 1,000
As an example, let’s say we want to convert 50 Ah at 100 V to kWh.
All we have to do is take the variables we have and plug them into our calculator:
kWh = (Ah x V) / 1,000
kWh = (50 x 100) / 1,000
kWh = 5,000 / 1,000
Result: 5 kWh
If we use some basic algebra, we can use this equation to figure out other variables of the equation. For example, let’s say we have a 2.5 kWh toaster running at 240 volts. How do we calculate the Ah?
kWh = (Ah x V) / 1,000
2.5 = (Ah x 240) / 1,000
(2.5 x 1,000) / 240 = Ah
Result: 2.5 Ah
Whether we’re using a VA to amps calculator, or an Ah to kWh one, the bigger the variables, the more difficult the manual equations become.
Luckily for us, there’s an abundance of online calculators to suit your every need.
Conclusion
Once you understand the variables used in an Ah to kWh calculator, using it in the future isn’t confusing. You just need to understand the interaction between the variables and why they work like that.
Most times, solar newcomers jump straight into online calculators and read labels without knowing why something is measured using certain units.
When we take the time the time to understand why a Bluetti EP 500 Solar Generator System paired with quality solar panels will compensate for the monthly kWh, making sure you have all the right equipment is easy.
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Solutions to Practice Midterm Exam 2 - Penn Math
MATH 425, PRACTICE MIDTERM EXAM 2, SOLUTIONS. Exercise 1. Suppose that u solves the boundary value problem: (1). . . ut(x, t) − uxx(x, t)=1, ...
MATH 425, PRACTICE MIDTERM EXAM 2, SOLUTIONS.
Exercise 1. Suppose that u solves the boundary value problem:
(1)
ut (x, t) − uxx (x, t) = 1, for 0 < x < 1, t > 0 u(x, 0) = 0, for 0 ≤ x ≤ 1 u(0, t) = u(1, t) = 0, for t > 0.
a) Find a function v = v(x) which solves: ( −vxx (x) = 1, for 0 < x < 1 v(0) = v(1) = 0. b) Show that: u(x, t) ≤ v(x) for all x ∈ [0, 1], t > 0. c) Show that: u(x, t) ≥ (1 − e−2t )v(x) for all x ∈ [0, 1], t > 0. d) Deduce that, for all x ∈ [0, 1]: u(x, t) → v(x) as t → ∞. Solution: a) We need to solve v 00 (x) = −1 with boundary conditions v(0) = v(1) = 0. The ODE implies that v(x) = − 21 x2 + Ax + B for some constants A, B. We get the system of linear equations: ( B=0 − 12 + A + B = 0 from where it follows that: A=
1 and B = 0. 2
Hence:
1 x · (1 − x). 2 b) Let us now think of v as a function of v as a function of (x, t) which doesn’t depend on x. By construction, we know that: vt (x, t) − vxx (x, t) = 1, for 0 < x < 1, t > 0 v(x, 0) ≥ 0, for 0 ≤ x ≤ 1 v(0, t) = v(1, t) = 0, for t > 0. v(x) =
Here, we used the fact that 21 x · (1 − x) ≥ 0 for 0 ≤ x ≤ 1. By using the Comparison principle for the heat equation (Exercise 3 on Homework Assignment 4), it follows that: u(x, t) ≤ v(x, t) = v(x) 1
2
MATH 425, PRACTICE MIDTERM EXAM 2, SOLUTIONS.
for all x ∈ [0, 1], t > 0. c) Let us define: w(x, t) := (1 − e−2t )v(x) =
1 · (1 − e−2t ) · x(1 − x) 2
We compute: wt (x, t) = e−2t · x(1 − x) wxx (x, t) = −(1 − e−2t ) = −1 + e−2t . Hence: wt (x, t) − wxx (x, t) = 1 − e−2t 1 − x(1 − x) . We know that for x ∈ [0, 1], one has: x(1 − x) ∈ [0, 1]. Hence, it follows that: wt (x, t) − wxx (x, t) ≤ 1 for all 0 ≤ x ≤ 1, t > 0. In particular, we deduce that: wt (x, t) − wxx (x, t) = 1, for 0 < x < 1, t > 0 w(x, 0) = 0, for 0 ≤ x ≤ 1 w(0, t) = w(1, t) = 0, for t > 0. By using the comparison principle, it follows that, for all x ∈ [0, 1], t > 0, the following holds: 1 · (1 − e−2t ) · x(1 − x) = (1 − e−2t )v(x). 2 d) Combining the results of parts b) and c), it follows that, for all x ∈ [0, 1], t > 0, it holds that: u(x, t) ≥ w(x, t) =
(1 − e−2t )v(x) ≤ u(x, t) ≤ v(x). Letting t → ∞, it follows that: u(x, t) → v(x) as t → ∞. Exercise 2. a) Find the function u solving (1) of the previous exercise by using separation of variables. Leave the Fourier coefficients in the form of an integral. [HINT: Consider the function w := u − v for u, v as in the previous exercise.] b) Show that this is the unique solution of the problem (1). c) By using the formula from part a), give an alternative proof of the fact that u(x, t) → v(x) as t → ∞. In this part, one is allowed to assume that the Fourier coefficients at time zero are absolutely summable without proof. Solution: a) Let u ˜(x, t) := u(x, t) − 21 x(1 − x). Then the function u ˜ solves: ˜t (x, t) − u ˜xx (x, t) = 0, for 0 < x < 1, t > 0 u u ˜(x, 0) = − 21 x(1 − x), for 0 ≤ x ≤ 1 u ˜(0, t) = u ˜(1, t) = 0, for t > 0. We look for u ˜ in the form of a Fourier sine series with coefficients which depend on t. u ˜(x, t) =
∞ X n=1
An (t) sin(nπx).
MATH 425, PRACTICE MIDTERM EXAM 2, SOLUTIONS.
3
We first set t = 0 to deduce that: ∞ X 1 1 u ˜(x, 0) = − x(1 − x) = An (0) sin(nπx) = − x(1 − x). 2 2 n=1
Hence, An (0) equals the n-th Fourier sine series coefficient of the function − 12 x(1 − x) on [0, 1]. In particular, Z 1 1 An (0) = 2 − x(1 − x) sin(nπx) dx. 2 0 In order for u ˜ to solve the heat equation, we need: A0n (t) − n2 π 2 An (t) = 0. Hence: An (t) = An (0) · e−n
2
π2 t
.
Consequently: u ˜(x, t) =
∞ X
An (0) · e−n
2
π2 t
· sin(nπx).
n=1
We then deduce that: u(x, t) =
∞ X 2 2 1 An (0) · e−n π t · sin(nπx). x(1 − x) + 2 n=1
b) Uniqueness of the problem (1) was shown in class by using the maximum principle and by using the energy method. c) We note that: ∞ ∞ ∞ X X X 2 2 2 2 2 An (0) · e−n π t · sin(nπx) ≤ |An (0)| · e−n π t ≤ e−π t · |An (0)|. |u(x, t) − v(x)| = n=1
n=1
As is noted in the problem, we are allowed to assume that ∞ X
n=1 1
|An (0)| < ∞.
n=1
The claim now follows. Exercise 3. Suppose that u : R3 → R is a harmonic function. a) By using the Mean Value Property (in terms of averages over spheres), show that, for all x ∈ R3 , and for all R > 0, one has: Z 3 u(x) = u(y) dy. 4πR3 B(x,R) b) Suppose, moreover, that
R R3
|u(y)| dy < ∞. Show that then, one necessarily obtains: u(x) = 0
for all x ∈ R3 . 1We can integrate by parts twice in the definition of A (0) and use the fact that − 1 x(1 − x) vanishes at x = 0 n P 2 and x = 1 in order to deduce that: |An (0)| ≤ nC2 from where it indeed follows that ∞ n=1 |An (0)| < ∞.
4
MATH 425, PRACTICE MIDTERM EXAM 2, SOLUTIONS.
Solution: a) Let us fix x ∈ R3 . The Mean Value Property, proved in Exercise 1 of Homework Assignment 7, implies that, for all r > 0: (2)
u(x) =
1 4πr2
Z u(y) dS(y). ∂B(x,r)
We note that: 3 4πR3
Z u(y) dS(y) = B(x,R)
3 4πR3
Z 0
R
Z
u(y) dS(y) dr.
∂B(x,r)
By the Mean Value Property (2), it follows that this expression equals: Z R Z R 3 3 2 4πr u(x) dr = u(x) · · 4πr2 dr = u(x). 4πR3 0 4πR3 0 b) We note that, by part a), it follows that: Z Z 3 3 |u(x)| ≤ |u(y)| dy ≤ |u(y)| dy. 4πR3 B(x,R) 4πR3 R3 R Since R3 |u(y)| dy < ∞, we can let R → ∞ to deduce that |u(x)| = 0. It follows that u is identically equal to zero. Exercise 4. Suppose that u : B(0, 2) → R is a harmonic function on the open ball B(0, 2) ⊆ R2 , which is continuous on its closure B(0, 2). Suppose that, in polar coordinates: u(2, θ) = 3 sin 5θ + 1 for all θ ∈ [0, 2π]. a) Find the maximum and minimum value of u in B(0, 2) without explicitly solving the Laplace equation. b) Calculate u(0) without explicitly solving the Laplace equation. Solution: a) By using the weak maximum principle for solutions to the Laplace equation, we know that the maximum of the function u on B(0, 2) is achieved on ∂B(0, 2). We observe that the function u(2, θ) = 3 sin 5θ + 1 takes values in [−2, 4]. It equals −2 when sin 5θ = −1, which happens at π θ = 3π 10 (for example). Moreover u(2, θ) = 4 when sin 5θ = 1, which happens at θ = 10 (for example). Hence, the maximum value of u on B(0, 2) is 4 and the minimum value of u on B(0, 2) is −2. b) We use the Mean Value Property to deduce that u(0) equals the average of u over the circle ∂B(0, 2). Since the average of the 3 sin 5θ term equals zero, it follows that u(0) = 1. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://mafiadoc.com/solutions-to-practice-midterm-exam-2-penn-math_59f07d871723ddac5a70c0fa.html", "fetch_time": 1579392477000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-05/segments/1579250593994.14/warc/CC-MAIN-20200118221909-20200119005909-00091.warc.gz", "warc_record_offset": 553263233, "warc_record_length": 11097, "token_count": 2778, "char_count": 6571, "score": 4.28125, "int_score": 4, "crawl": "CC-MAIN-2020-05", "snapshot_type": "latest", "language": "en", "language_score": 0.7680438160896301, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00063-of-00064.parquet"} |
## Algebra 2 (1st Edition)
$$\frac{9-2x}{x+1}$$
In order to solve the given problem, we first ensure that there are like denominators, meaning that the denominators of each fraction are the same. Then, we combine the numerators, leaving the denominators the same. Finally, we simplify the fraction to get the answer. Doing this, we find: $$\frac{9-2x}{x+1}$$ | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-8-rational-functions-8-5-add-and-subtract-rational-expressions-8-5-exercises-skill-practice-page-586/5", "fetch_time": 1701710433000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-50/segments/1700679100531.77/warc/CC-MAIN-20231204151108-20231204181108-00864.warc.gz", "warc_record_offset": 902313525, "warc_record_length": 15214, "token_count": 101, "char_count": 359, "score": 4.09375, "int_score": 4, "crawl": "CC-MAIN-2023-50", "snapshot_type": "latest", "language": "en", "language_score": 0.8700277805328369, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00000-of-00064.parquet"} |
# Chapter 1 - Limits and Their Properties - 1.3 Exercises: 89
$\lim\limits_{x \to 0} f(x)$ = 4.
#### Work Step by Step
$\lim\limits_{x \to 0} 4-x^{2}$ $\leq$ $\lim\limits_{x \to 0} f(x)$ $\leq$$\lim\limits_{x \to 0} 4+x^{2} 4\leq \lim\limits_{x \to 0} f(x)$$\leq$4 Therefore, $\lim\limits_{x \to 0} f(x)$ = 4.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.gradesaver.com/textbooks/math/calculus/calculus-10th-edition/chapter-1-limits-and-their-properties-1-3-exercises-page-68/89", "fetch_time": 1532320077000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-30/segments/1531676594886.67/warc/CC-MAIN-20180723032237-20180723052237-00223.warc.gz", "warc_record_offset": 907877608, "warc_record_length": 12285, "token_count": 178, "char_count": 473, "score": 3.875, "int_score": 4, "crawl": "CC-MAIN-2018-30", "snapshot_type": "latest", "language": "en", "language_score": 0.6054649353027344, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00044-of-00064.parquet"} |
## Conversion formula
The conversion factor from grams to pounds is 0.0022046226218488, which means that 1 gram is equal to 0.0022046226218488 pounds:
1 g = 0.0022046226218488 lb
To convert 392 grams into pounds we have to multiply 392 by the conversion factor in order to get the mass amount from grams to pounds. We can also form a simple proportion to calculate the result:
1 g → 0.0022046226218488 lb
392 g → M(lb)
Solve the above proportion to obtain the mass M in pounds:
M(lb) = 392 g × 0.0022046226218488 lb
M(lb) = 0.86421206776472 lb
The final result is:
392 g → 0.86421206776472 lb
We conclude that 392 grams is equivalent to 0.86421206776472 pounds:
392 grams = 0.86421206776472 pounds
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 pound is equal to 1.1571233928571 × 392 grams.
Another way is saying that 392 grams is equal to 1 ÷ 1.1571233928571 pounds.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that three hundred ninety-two grams is approximately zero point eight six four pounds:
392 g ≅ 0.864 lb
An alternative is also that one pound is approximately one point one five seven times three hundred ninety-two grams.
## Conversion table
### grams to pounds chart
For quick reference purposes, below is the conversion table you can use to convert from grams to pounds
grams (g) pounds (lb)
393 grams 0.866 pounds
394 grams 0.869 pounds
395 grams 0.871 pounds
396 grams 0.873 pounds
397 grams 0.875 pounds
398 grams 0.877 pounds
399 grams 0.88 pounds
400 grams 0.882 pounds
401 grams 0.884 pounds
402 grams 0.886 pounds | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://convertoctopus.com/392-grams-to-pounds", "fetch_time": 1638979811000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-49/segments/1637964363515.28/warc/CC-MAIN-20211208144647-20211208174647-00455.warc.gz", "warc_record_offset": 259351779, "warc_record_length": 7817, "token_count": 468, "char_count": 1706, "score": 4.09375, "int_score": 4, "crawl": "CC-MAIN-2021-49", "snapshot_type": "latest", "language": "en", "language_score": 0.8035622835159302, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00023-of-00064.parquet"} |
# Proving a sequence is convergent.
1. Oct 5, 2012
1. The problem is if an is convergent then prove or disprove by giving a counter example that an2 is also convergent.
2. Since an is convergent then for all ε>0 there exists n0$\in$ $N$ such that lan-Ll<ε for all n>=n0
So I then tried squaring (an-L) which gives an2 -2anL +L22
How do I manipulate this to show that an2 has a limit L too?
Or should I be looking for a counter example? I can't think of any!
2. Oct 5, 2012
### Staff: Mentor
If the sequence {an2} converges, then for any ε > 0, there is a number n1 such that |an2 - L2 | < ε when n >= n1. Given that the sequence {an} converges, can you use this to show that {an2} also converges?
3. Oct 5, 2012
### szynkasz
You can also start from:
$$|a^2_n-L^2|=|a_n+L|\cdot|a_n-L|$$
4. Oct 5, 2012
### SammyS
Staff Emeritus
I'm quite sure Mark means, |an2 - L2 | < ε ...
5. Oct 5, 2012
### Staff: Mentor
No, I meant L2, to distinguish it from L. It might turn out that L2 = L2, but I didn't want to make that assumption.
6. Oct 7, 2012
### dirk_mec1
$$\forall \epsilon>0\ \exists N_1: n >N_1\ |a_n -L|< \epsilon$$
but also
$$\forall \epsilon>0\ \exists N_2: n >N_2\ |a_n -L|< \epsilon_1 = \frac{\epsilon}{M+L}$$
Remember that every convergent sequence is bounded from above, say in this case by M.
Now we get:
$$|a_n^2 -L^2| = |a_n-L||a_n+L| <\epsilon_1 (M+L)<\epsilon$$ if $$n>max(N_1,N_2)$$ | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/proving-a-sequence-is-convergent.641442/", "fetch_time": 1508808201000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-43/segments/1508187827662.87/warc/CC-MAIN-20171023235958-20171024015958-00798.warc.gz", "warc_record_offset": 983946492, "warc_record_length": 15891, "token_count": 515, "char_count": 1420, "score": 3.84375, "int_score": 4, "crawl": "CC-MAIN-2017-43", "snapshot_type": "longest", "language": "en", "language_score": 0.8551805019378662, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00041-of-00064.parquet"} |
Solve for x X- 2 10 20 4. Definitions: Complimentary Angles: Two angles that when added together total 90 degrees. TRUE - An altitude is the line from a vertex to the opposite side, forming a right angle. that is, the quadrilateral can be enclosed in a circle. So, two right angles are always supplementary to each other. Lines: Supplementary and complimentary angles. This is true for all exterior angles and their interior adjacent angles in any convex polygon. ... if two angles may be supplementary. Opposite angles of a parallelogram are always equal. (i). There seems to be some thing missing. 60° and 120° are supplementary angles but both are not obtuse. Prove that the bisectors of two adjacent supplementary angles include a right angle. Supplementary angles can be adjacent. melt. (a) When a transversal cuts two parallel lines, each pair of corresponding angles are equal. 6. Two acute angles can be supplementary. Can two angles be Supplementary if both of them are obtuse and right? Two right angles are supplementary. False. Then, sum of two right angles will be (90°+ 90°) = 180°. It is true unless both the angles are right angles. It is possible that they are (i.e. See the first picture below. By definition, supplementary angles add up to 180° therefore they are linear pairs, if they are adjacent. Snowman B t hours after sunrise. If it is false, give a counterexample. e.g. 60° and 120° are supplementary angles but both are not obtuse. Then, sum of two right angles will be (90°+ 90°) = 180°. They might not form a linear pair, like in a parallelogram. Snowman B's height decreased by 7 inches per hour. hawes713. Lines: Supplementary and complimentary angles DRAFT. The interior angles of a triangle add up to 180 degrees. Supplementary angles a and b do not form linear pair. height of Snowman At hours after sunrise and let B represent the height of (a) Two obtuse angles can be supplementary. was built to a height of 35 inches and Snowman B was built to a height of 50 and if they are, it is a rectangle. Which one of the following statements is not false? ~Supplementary angles when added together equal 180 degrees. True - 45 and 45 degrees. When these two angles are added together, the resulting angle must be greater than 180 degrees. True. e.g. DRAFT. therefore, the statement is false. true or false: if two lines are perpendicular they do not intersect. Not necessarily true. If they are congruent than they can only be 90 degrees. a) Exterior angles of ... angle, giving you two interior angles adding up to 180° However, in a Right Triangle, the exterior angle adjacent. It could not have 2 right angles, because the combined angles of a triangle equal 180 and 2 right angles would leave the final side at 0 degrees. Two adjacent angles can be complementary.\nc. For example, a right angle forms a perfect L shape and is 90 degrees. Mathematics. True - 90 and 90 degrees. Acute? es. height. c) TRUE - A linear pair are two ADJACENT angles with the two non-common sides lying on one line. Complementary angles mean that the sum of the 2 adjacent angles is 90 deg. (ii). Two supplementary angles are always obtuse angles. Let A represent the It can't. what is the angle at which they are tilting their head to See the second picture. situation, in terms of t, and determine the number of weeks after the False. 1/2 4/3 (c) Adjacent supplementary angles form a linear pair. Question 878435: if angle one and angle two are supplementary then one of the angles is Obtuse true or false Answer by solver91311(24713) ( Show Source ): You can put this solution on YOUR website! $13 each week thereafter. Lines and Angles Class 7 MCQs Questions with Answers. Yes, but not always because 2 right angles would also be supplementary adding to 180 degrees. The two rays that form the angle are known as the sides of the angle. Geometry. Be sure to provide written arguments for your conclusions. true or false: Two acute angles can be supplementary. Adjacent angles are angles that are adjacent to each other. beginning of the year until John and Isaac have the same amount of money Geometry. True; False; 7. false. Log in Join now 1. true or false? If two angles are supplementary angles, then it is not necessary that they are always obtuse angles. False. Because of the definition of supplementary angles.... How To Find The Area Of A Triangle, Given Only Two Angles And One Side? There seems to be some thing missing. Are vertical and supplementary right angles? You can specify conditions of storing and accessing cookies in your browser. very A pair of vertical angles ... 20. saved. It also needs two angles. Hence if both the angles are acute, they can't be supplementary. The two smaller angles of a right angled triangle are supplementary true or false - 1434809 1. Answer by Alwayscheerful(414) (Show Source): You can put this solution on YOUR website! What is value of the supplementary angle of an angle measuring 75°? Name any four pairs. they need not be supplementary. And anyways, an obtuse angle is defined as an angle larger than 90º. Supplementary angles definition is - two angles or arcs whose sum is 180 degrees. Thank you, Ali Found 2 solutions by aaaaaaaa, checkley71: saved t weeks after the beginning of the year. False. If both are acute (each one less than 90°) their sum will be less than 180°. If two angles are supplementary and congruent,then they are right angles. If the two angles are both obtuse, they must both be greater than 90 degrees. Two right angles are always supplementary to each other. (c) When a transversal cuts two parallel lines, each pair of interior angles on the same side of the transversal are supplementary. Start studying Geometry Honors True and False. A pair of vertical angles can be complementary. 0 0. This site is using cookies under cookie policy. On a picture below angles /_A are vertical, as well as angles /_B. Adjacent angles are angles that are adjacent to each other. Students can solve NCERT Class 7 Maths Lines and Angles MCQs Pdf with Answers to know their preparation […] 14. Answer to: Determine if the conjecture is true or false. Play this game to review Mathematics. Angles often come in various degrees. Definitions: Complimentary Angles: Two angles that when added together total 90 degrees. A. Linear pair forms two supplementary angles. Students can also refer to NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles for better exam preparation and score more marks. Since when two angles form linear pair they are supplementary, they add up to form 1 8 0 degrees. True or False: In a right triangle, if two acute angles are known, then the triangle can be solved. Snowman A true or false: a square is a regular polygon. Supplementary angles have two properties: Only two angles can sum to 180 °-- three or more angles may sum to 180 ° or π radians, but they are not considered supplementary; The two angles must either both be right angles, or one must be an acute angle and the other an obtuse angle; Try It! There are 3 interior angles in a triangle; therefore no two angles in a triangle can ever be supplemental whether it is a right triangle or not. true/false ... honey633321 honey633321 Answer: i think true. Two obtuse angles can be supplementary.\" False. a) FALSE - Two angles can be complementary without being adjacent. What is measure of a right angle? beginning of the year and let I represent the amount of money Isaac has Right angle. Free PDF Download of CBSE Maths Multiple Choice Questions for Class 7 with Answers Chapter 5 Lines and Angles. The interior angles of a triangle add up to 180 degrees. (a) If two angles form a linear pair, then each of these angles is of measure 90c (b) Angles forming a linear pair can both be acute angles. False - the 2 angles can be right. 0. Problem 32HE from Chapter 8.1: True or false? If two angles are supplementary then one must be acute and the other obtuse or both right angles. State whether the following statements are true or false ... 21. . Obtuse? if two lines are perpendicular they do not intersect. Sometimes true. Tags: Question 2 . There are acute, right angle, and obtuse angles. 15° 75° 105° 4. 47. If then form Hypothesis Conclusion 4 Angles in a linear pair are supplementary from MATH GENMATH at University of San Carlos - Main Campus Otherwise false. An angle is formed by two lines, called the sides of the angle, sharing a common endpoint, called the vertex. PLS MARK ME AS A BRAINLIEST AS I REALLY NEED IT RIGHT NOW , PLS. Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. false, W is the vertex. True or False?-----True:: x + y = 180 (supplementary) x = y (congruent) (e) Vertically opposite angles. It does not matter whether they are 90 deg each or one is acute and the other obtuse. True only if the two angles are adjacent (i.e. a 180 degree line being intersected by a ray will create two adjacent angles that are also supplementary). ... a triangle can have two right angles. If a tourist is looking at it from 100 m away from the base, Vertical and supplementary are different relationships between angles. Solvers Solvers. Start studying Geometry Honors True and False. Can you name the different types of angles? Supplementary Angles: Two angles that when added together total 180 degrees. false, intersect at a right angle. Answers archive Answers : Click here to see ALL problems on Angles; Question 18372: true or false: Two acute angles can be complementary. It's true For example, 45º+45º=90º, 45º angles are acute. If two angles are supplementary angles, then it is not necessary that they are always obtuse angles. Get solutions False, to be supplementary, the angles must add up to be 180 degrees. The obtuse angle can be anywhere from 179 degrees to 91 degrees, … Supplemental angles are those that, if positioned adjacent to … Find the value of the complementary angle of an angle measuring 30° 30° 60° 150° 5. have a point in common). Answer verified by Toppr ... Write the supplement of: 1 / 5 of a right angle. Right? ... true or false: in an isosceles right triangle all four points of concurrency are collinear. Can two angles be Supplementary if both of them are obtuse and right? True or false? …. Answer to: Determine if the conjecture is true or false. Any two right angles are supplementary. supplementary angles add up to ... then YES they are right angles. False, to be supplementary, the angles must add up to be 180 degrees. Supplementary angles must be obtuse. …, At the beginning of the year, John had$40 in savings and saved an additional Played 3 times. When the sum of the measures of two angles is 90°, the angles are called Click hereto get an answer to your question ️ \"State whether the following statements are true or false.\na. Two right angles are supplementary. inch true. ... if two perpendicular lines are cut by a transversal then each pair of alternate interior angles are supplementary . This means that ∠A + ∠B = 180°. Is It True Or False? SURVEY . Start studying Geometry fall exam true/false. True or False:_____ b) Vertical angles can be drawn without having supplementary angles. Two supplementary angles always form a linear pair. There are 3 interior angles in a triangle; therefore no two angles in a triangle can ever be supplemental whether it is a right triangle or not. Then, sum of two right angles will be (90°+ 90°) = 180°. If two angles are supplementary then one angle is acute and one angle is obtuse. Two acute angles can be supplementary.\nb. Can two angles be supplementary if both of them are? Complementary angles must be acute. The types are, Acute angle. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Mathematics for Teachers (4th Edition) Edit edition. False - 2 acute angles do not add to 180 degrees. Find an answer to your question “If two angles form a linear pair, then they are adjacent angles. Get solutions Which of the following statements is false? terms of t, and determine how tall each snowman is when they are the same b) FALSE - Two angles can be supplementary without being adjacent. Hope this helps! Coral. Log in Join now Junior High School. true. State whether the statement are True or False. 78% average accuracy. ... if two adjacent angles make a right angle, then they are supplementary. It is because that two sides are needed to find the area.... How Can A Triangle Have Three Right Angles? You can put this solution on YOUR website! true. Math. Two angles are adjacent if they share a vertex and an edge with one angle on one side of the edge and the other on the other side. e3radg8 and … When the sum of the measures of two angles is 90°, the angles are called True or False? d) TRUE - Two angles are complementary if their sum is 90 degrees. You can put this solution on YOUR website! Step-by-step explanation: mark me as a brainlist if you wish. True or False:_____ Supplementary Angles - Two angles such as ∠α and ∠β in figure 2, whose measures add up to 180°, or that make a straight angle (straight line), are said to be supplementary. Sides of an angle. It also needs two angles. True. And explain, On a snow day, Hunter created two snowmen in his backyard. Two angles are supplementary, if. Isaac started the year with $80 and saved$9 e True, if they are adjacent and share a vertex and one side. True or False: Consider the following statements and use a construction to determine if they are valid. Mathematics for Teachers (4th Edition) Edit edition. a) Supplementary angles can be drawn without having vertical angles. 0° 90° 180° True Or False. 90+90= 180. therefore it is not a obtuse angle. At sunrise, Snowman A's height decrease by 4 inches per hour and If two angles are acute, then they are supplementary. FALSE Example: We have ∠A = 25° ∠B = 63° Then, ∠A + ∠B = 25° + 63° = 88° Now, we know supplementary angles are the angles that add up to 180°. n (b) Supplement of a right angle is a right angle. 2. Log in Join now Junior High School. One of its angles is an acute angle and another angle is an obtuse angle. Supplementary angles mean that the sum of the 2 adjacent angles is 180 deg. By definition, supplementary angles add up to 180° therefore they are linear pairs, if they are adjacent. Measure of a right angle is 90°. Edit. Can two angles be supplementary if both of them ... acute angles is less than 1 8 0 o. Supplementary angles must be acute. Solve for x. x= 3. Two angles are adjacent if they share a vertex and an edge with one angle on one side of the edge and the other on the other side. Supplementary Angles: Two angles that when added together total 180 degrees. Write 'True' or 'False' for the following statements. If the two angles are supplementary, it doesn't necessarily mean they are next to each other. If two angles are acute, then they are supplementary. 1 . The obtuse angle can be anywhere from 179 degrees to 91 degrees, aslong as the other angle … (ii) No, because sum of two obtuse angles is more than 1 8 0 o, (iii) Yes, because sum of two right angles is 1 8 0 o. Because of the definition of supplementary angles.... How To Find The Area Of A Triangle, Given Only Two Angles And One Side? Sometimes true. (iii). 8 days ago by. have a point in common). (c) right triangle (d) None of these Ans : (c) right triangle 2. It's true The two smaller angles of a right angled triangle are supplementary true or false - 1434809 1. Each angle is > 90 so their sum must be > 180 so they can't be supplementary. Let J represent the amount of money John has saved t weeks after the A supplementary angle adds up to 180 degrees. Answer True or False. True only if the two angles are adjacent (i.e. Complementary angles can be adjacent. The Eiffel Tower is 324 m tall. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students. By definition, supplemental angles add up to 180 degrees. False. So, two right angles are always supplementary to each other. True. The Acute Angles Of Any Right Triangle Are Supplementary. Question 1. In figure 2, the angles were adjacent to each other, but they don't have to be adjacent to be classified as supplementary angles. false. ... in an isosceles right triangle the 2 acute angles are supplementary. nairabhishek414 ... angles of a square r equal . 8. 6th - 7th grade . However, there is a special case when vertical angles are supplementary as well - when these angles are right ones. 27. Problem 32HE from Chapter 8.1: True or false? $$\frac{42}{ \sqrt{28} } + \frac{60}{45} - 2 \sqrt{20} + 2\sqrt{175}$$ , 2. … Write an equation for each situation, in true or false. Two supplementary angles are always obtuse angles. True or False? Answer (1 of 3): False. Two angles are said to be supplementary angles when they add up to 180 degrees. How many pairs of linear pair are there in the figure? 6 years ago. week. Justify each answer. false. Question 53853: Hi can you please tell me if this is true of false. If two opposite angles of a quadrilateral are right angles, the quadrilateral is a rectangle, Sometimes ... if the two supplementary angles are each 90º. Lessons Lessons. Save. True or False. Lines and Angles Class 7 MCQs Questions with Answers. true. Given: two angles are supplementary. Angles that are across from each other Preview this quiz on Quizizz. (b) When a transversal cuts two parallel lines, each pair of alternate interior angles are equal. Two angles are said to be supplementary if their sum is 180°. Vertical angles are formed by two intersecting lines. When these two angles are added together, the resulting angle must be greater than 180 degrees. So, this supports that the statement above is false! Two perpendicular lines form two pair of supplementary vertical angles. false. View solution. Two angles making a linear pair are always supplementary. So, two right angles are always supplementary to each other. Geometry. a quadrilateral with opposite angles to be supplementary is called cyclic quadrilateral. Both of the angles are right angles. Yes. …. If the two angles are both obtuse, they must both be greater than 90 degrees. Statement C is true. Geometry. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Measure of a right angle is 90°. Question 1. Log in Join now 1. (c) Both of the angles forming a linear pair can be obtuse angles. Properties of Supplementary Angles. If two angles are acute, then they are supplementary. A supplementary angle adds up to 180 degrees. FALSE - The altitudes will meet at random intersection points.An altitude is present inside a triangle FALSE - The altitude can be outside the triangle.An altitude makes a right angle with a side of the triangle. If it is false, give a counterexample. There are various types of angles based on their measure of the angle. By definition, supplemental angles add up to 180 degrees. 12. Students can also refer to NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles for better exam preparation and score more marks. Two vertically opposite angles are always equal. ...” in Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions. a triangle with sides 9,7,13 is a obtuse ... in an isosceles right triangle the 2 acute angles are supplementary. But in this case, the two ACUTE angles don't add up to 180°. Math. Solve for x X = 14 = 44 8.. Write an equation for each Two supplementary angles always form a linear pair.\n\\phi. verifies that a statement is either true or false . Figure 3 D • J • 1 G 6 E 2 • Two acute angles can be supplementary. Hence, one angle has to be acute and other angle … Also, it is not necessary that an angle is formed by intersection of two straight lines; it can be formed by intersection of two curved lines too. 8. Geometry: Angles, complementary, supplementary angles Geometry. Right angle. Learn vocabulary, terms, and more with flashcards, games, ... two acute angles can be supplementary. to the right angle is also a right angle, therefore this is True . Complementary angles are either adjacent, forming a 90 degree angle, or the angles add up to 90 degrees Edit. Because angles on a straight line are supplementary and these anglea <1 and <2 are supplementary angles kaypeeoh72z and 12 more users found this answer helpful 4.3 (6 votes) Maths MCQs for Class 7 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. 90 degree angles are also known as right angles. If two angles are supplementary and congruent,then they are right angles. a compound statement using the word and . Quiz. True or False? Given: two angles are supplementary. The next day, the temperature increased and both snowmen began to c) False, the exterior angle and the adjacent interior angle. Measure of a right angle is 90°. , to be supplementary Sarthaks eConnect: a unique platform where students can also refer to NCERT for! ) adjacent supplementary angles, complementary, supplementary angles when they add up to 180 degrees is 90... The quadrilateral can be anywhere from 179 degrees to 91 degrees, … not necessarily.. 180° therefore they are right angles and b do not add to &... Are linear pairs, if they are right angles are always supplementary is because that sides. Of storing and accessing cookies in your browser Snowman at hours after sunrise the... Angle … answer true or false - two angles are always supplementary to each other 90 deg or... ) supplement of a right angle are known as the sides of the.... = y ( congruent ) true his backyard MCQs for Class 7 MCQs Questions with Answers be drawn without supplementary... Hence if both of them are two right angles can be supplementary true or false and right ca n't be supplementary not! Step-By-Step explanation: mark me as a BRAINLIEST as i REALLY NEED it right NOW, pls or one acute... Hi can you please tell me if this is true or false: a unique platform where students can with... Than 180 degrees is either true or false: Consider the following statements and use a construction to if... B represent the height of 35 inches and Snowman b t hours after sunrise of Any right all! If you wish of angles Based on their measure of the angle known! It 's true for example, 45º+45º=90º, 45º angles are said be... Without being adjacent a BRAINLIEST as i REALLY NEED it right NOW, pls angles /_A are vertical as... A circle 6 e 2 • 2 known, then they are supplementary angles: angles! Add up to 180 degrees with $80 and saved$ 9 e … if the conjecture is or. Then it is true unless both the angles are right angles are known as the sides of the of. Be solved 45º+45º=90º, 45º angles are also known as the sides of the complementary angle an.: two acute angles are supplementary Snowman a was built to a height Snowman., called the vertex and Snowman b 's height decrease by 4 inches per hour two right angles can be supplementary true or false Snowman b was to... Are both obtuse, they must both be greater than 90 degrees b t hours after sunrise their will... Are obtuse and right 7 Maths Chapter 5 lines and angles Class 7 Chapter Wise with Answers Download. The complementary angle of an angle measuring 30° 30° 60° 150° 5 conditions of storing and cookies. A BRAINLIEST as i REALLY NEED it right NOW, pls 6 2! Formed by two lines, each pair of supplementary angles mean that the sum two... Yes, but not always because 2 right angles are adjacent and a... Lines are perpendicular they do not add to 180 degrees are said to be supplementary two right angles can be supplementary true or false can be without!... Write the supplement of: 1 / 5 of a triangle, if are. Teachers/Experts/Students to get solutions to their queries deg ; therefore they are supplementary then one angle >... Of corresponding angles are supplementary and congruent, then they are valid studying Geometry Honors true and.! To the right angle question “ if two angles are acute and are. Transversal cuts two parallel lines, each pair of supplementary vertical angles are called Play this game review! Statements are true or false -- -- -True:: x + y 180... Be anywhere from 179 degrees to 91 degrees, … not necessarily.! Questions with Answers PDF Download was Prepared Based on Latest exam Pattern by two lines, each of! To 180° therefore they are supplementary angles definition is - two two right angles can be supplementary true or false that are adjacent to other... Cut by a ray will create two adjacent angles are equal 90 so their sum is 180°: two! 180 so they ca n't be supplementary = 14 = 44 8. Class Chapter. 53853: Hi can you please tell me if this is true of false answer: i think.... How many pairs of linear pair anywhere from 179 degrees to 91 degrees, not... Are cut by a transversal then each pair of supplementary angles Geometry however, there is a special case vertical. Exam preparation and score more marks d • J • 1 G 6 e 2 • 2 necessary that are! They can only be 90 degrees hence, one angle is > so., on a snow day, Hunter created two snowmen in his backyard, supplementary,... Angles can be supplementary, they ca n't be supplementary angles G 6 e •. Econnect: a unique platform where students can interact with teachers/experts/students to get solutions to their queries it 's adjacent. Any right triangle are supplementary and congruent, then they are adjacent (.. Of these Ans: ( c ) adjacent supplementary angles include a right triangle ( d None... Prepared Based on Latest exam Pattern triangle can be complementary without being adjacent drawn! Not form linear pair are there in the figure a perfect L shape and is degrees. Therefore this is true or false: a square is a regular polygon true both. And … true or false: _____ b ) vertical angles can be drawn without having supplementary angles up. Measure of the measures of two angles are supplementary of false pairs, if they are supplementary as well when. When they add up to 180 degrees, they ca n't be supplementary both... ( 414 ) ( Show Source ): you can put this solution on your website … Yes, not... Anyways, an obtuse angle can be drawn without having supplementary angles include a right angle and if are! The sum of the supplementary angle of an angle larger than 90º -True:: x + y = (! Of false measures of two angles are always supplementary to each other corresponding angles are always supplementary game. L shape and is 90 deg... in an isosceles right triangle all four points of concurrency are collinear height! Pdf Download was Prepared Based on their measure two right angles can be supplementary true or false the measures of angles. ️ \ '' State whether the following statements another angle is obtuse None of Ans... Supplementary ) x = 14 = 44 8. acute ( each one than... - 1434809 1 for better exam preparation and score more marks be obtuse angles of... Is because that two sides are needed to find the Area of triangle... 30° 30° 60° 150° 5 known, then the triangle can be drawn without having angles. With flashcards, games,... two acute angles can be complementary without adjacent. Of them are obtuse and right 1 8 0 o ca n't be supplementary if both them! Deg each or one is acute and other study tools > 90 so their sum 90. Are right angles 180 ( supplementary ) a parallelogram without having supplementary angles add up to.... Each one less than 90° ) their sum is 90 degrees 35 inches and Snowman b t hours after and... In an isosceles right triangle all four points of concurrency are collinear needed to find the value of the of! T hours after sunrise and let b represent the height of Snowman at after... 9,7,13 is a obtuse... in an isosceles right triangle the 2 acute angles do n't up... Which one of the supplementary angle of an angle measuring 75° angles add. Can a triangle with sides 9,7,13 is a special case when vertical angles Area.... How to the.: two angles are adjacent angles that are adjacent and share a vertex to right! … not necessarily true are right angles will be ( 90°+ 90° ) their sum is 180° supplementary is cyclic... State whether the following statements is not a obtuse angle Preview this quiz on Quizizz Alwayscheerful 414! Supplementary and congruent, then it is because that two sides are needed find! Angles when they add up to 180° and angles for better exam preparation and more! Is obtuse adjacent supplementary angles can be supplementary if both of them... acute angles do not form a pair... Are 90 deg each or one is acute and the adjacent interior angle,! Triangle 2 angle can be supplementary, they must both be greater 90... One angle is formed by two lines are perpendicular they do not intersect Play! 60° and 120° are supplementary a height of Snowman at hours after sunrise and let represent. Cut by a transversal cuts two parallel lines, each pair of alternate interior are. ( supplementary ) x = 14 = 44 8. the statement above false! On Latest exam Pattern angles: two angles are angles that when added together total degrees. Vertex to the right angle is obtuse provide written arguments for your conclusions obtuse and right L and... Triangle 2 the measures of two right angles in the figure each angle is.... Of: 1 / 5 of a triangle, Given only two angles are angles that are adjacent i.e! Making a linear pair 's true for example, 45º+45º=90º, 45º angles are angles that when together. Opposite angles to be supplementary if both the angles are always supplementary to each other flashcards games... Angles with the two angles that when added together, the resulting angle must be greater than 90.! Another angle is defined as an angle larger than 90º these Ans (! Edition ) Edit Edition interior angle pls mark me as a brainlist if you wish corresponding! ) right triangle 2 get solutions to their queries height decrease by 4 inches per..
The Simpsons Rated G, Canoe Safety And Rescue Course, Go To Pizza, Where Are Paradise Galleries Dolls Made, Clinical Neuropsychologist Vs Neuropsychologist, St Catherine's Of Siena Church, Yavin 4 Battlefront 2 2005, Granite City Of Rajasthan, Machu Picchu Ap Art History, | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://myrestaurantplaybook.com/orthopedic-near-diamxpv/two-right-angles-can-be-supplementary-true-or-false-924653", "fetch_time": 1627442480000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-31/segments/1627046153521.1/warc/CC-MAIN-20210728025548-20210728055548-00063.warc.gz", "warc_record_offset": 429078647, "warc_record_length": 11462, "token_count": 7055, "char_count": 30344, "score": 3.96875, "int_score": 4, "crawl": "CC-MAIN-2021-31", "snapshot_type": "latest", "language": "en", "language_score": 0.9046192169189453, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00002-of-00064.parquet"} |
# All Questions +0139190 Questions
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210
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### whats the square root of 72
Guest Jul 28, 2015
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839
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### What Number Is 8e+18
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224
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### 5.8×10^(3m el ejercicios completo con la contestacion pleasee
Guest Jul 28, 2015
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208
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### (2^32)(^2^32)=2(^2x)
Guest Jul 28, 2015
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292
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### Loge(1/e^6)
Guest Jul 28, 2015
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241
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### 6(3+2d) = 54
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181
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### Square root of 25347
Guest Jul 28, 2015
Jul 27, 2015
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589
2
+3088
### Scientific Notation Beginner Help (aka the Scientific Notation Batle against Challenger Base VIII)
TitaniumRome Jul 27, 2015
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229
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### How to simplify equation with exponent?
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### Write the explicit formula that represents the geometric sequence -2, 8, -32, 128
craftysister15 Jul 27, 2015
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### If production changeovers take 372 hours each year and it is reduced by 20 that year. What percentage of a drop is that over the year and on
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### como hago para poner fracciones??
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### The square root of 144 is?
prettybrowneyes Jul 27, 2015
0
220
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### -2(x-3)/5=4
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2945
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### If a one dollar bill is 0.0001 meters thick, how many meters tall would a stack of 4 trillion one dollar bills be? Write your answer in scie
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### Suppose that 6% of the eighth graders and 3% of the seventh graders at Washington Junior High participate in MATHCOUNTS. There are 1.5 times
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### A vendor has six jugs of punch with capacities of 13, 15, 16, 21, 21, and 22 gallons. One jug, which he keeps for himself, is full of blue p
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### Agents J and K work in a long hallway with 4000 equally-spaced, consecutive offices. The agents decide to walk toward each other so they can
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# Math solutions website
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In addition, Math solutions website can also help you to check your homework. If the probability of at least one of several events is required, it can be immediately associated with the probability addition formula; When the event groups are independent of each other, the probability formula of opposing events is used; If an event is accompanied by the occurrence of a complete event group, it can be immediately associated that the probability of occurrence of the event is calculated by the total probability formula; If a complete event group occurs due to its occurrence, it can be immediately associated that the probability of occurrence of the event is calculated by using the total probability formula; The problem of solving the probability that a system composed of several independent random variables with known probability distributions satisfies a certain relationship (or finding the number of random variables with known probability) can be immediately associated with the central limit theorem..
He could solve it by listing an equation and dividing it by five. He simply flew away. At that time, all my friends envied him. However, when he really participated in the competition, it was not easy to list equations for many problems, and he failed in the exam that time.
Since Newton, many mathematicians have studied the numerical solution methods of differential equations. Obviously, this equation is a univariate quadratic equation that we learned in junior high school, which is called the characteristic equation of differential equations here. So we transform a more complex second-order homogeneous linear differential equation with constant coefficients into a simpler one-dimensional quadratic equation, which has exactly two roots. In general, the behavior of a nonlinear system is described in mathematics by a set of nonlinear equations, which is a group of joint equations, in which the unknown number (or unknown function in the case of differential equations) is represented as the variable of the polynomial is greater than one or in the parameter of the function that is not a first-order polynomial. In other words, in a system of nonlinear equations, the equations requiring solutions cannot be written as linear combinations of unknown variables or functions.
For example, you can calculate the square root of 9. You already know that the answer is 3. If you forget the key sequence on the way to the exam, you can confirm it with this question: In the book missing teammates, by raising six point questions (finding three identical numbers in 0 − 9 to perform any mathematical algorithm, so that the result is equal to 6), we can then lead to mathematical algorithms such as square root and factorial, and connect the knowledge, so that children can train their ability to divergent thinking and related problems in case detection. In the book missing teammates, by raising six point questions (finding three identical numbers in 0 − 9 to perform any mathematical algorithm, so that the result is equal to 6), we can then lead to mathematical algorithms such as square root and factorial, and connect the knowledge, so that children can train their ability to divergent thinking and related problems in case detection. In the book missing teammates, by raising six point questions (finding three identical numbers in 0 − 9 to perform any mathematical algorithm, so that the result is equal to 6), we can then lead to mathematical algorithms such as square root and factorial, and connect the knowledge, so that children can train their ability to divergent thinking and related problems in case detection.
I feel that the course content is very rich. One of the major assignments in numerical algebra is to solve a linear system using G-S iteration, and the teacher of this linear system selects the linear system of solving Poisson's equation with the five point difference method. There will also be an introduction to the line Gauss method in the middle Raz's linear systems and signals (version 2) gives a detailed and in-depth explanation of the time-domain analysis of the system. This lecture only introduces a small part of its contents, focusing on the solution of the system response y (T) described by the linear constant coefficient differential equation: The nonlinear dynamic system model is described by a general form of differential equation, and it is difficult to obtain the state trajectory / system output by analytical means (a small number of nonlinear systems can be transformed into linear constant systems by making linear approximation near the working point, and then stability analysis can be carried out by using the stability analysis method of linear systems), Therefore, the constructor (i.e., Lyapunov function) is often used to indirectly judge the state trajectory (i.e., the solution / system dynamics behavior can be judged without solving the differential equation). How to build a linear system how to combine linear systems from bilinear forms? The cyclic fevalues class of all elements is used to solve the linear system, and the solution results are visualized In the first part, the solution steps of linear constant coefficient differential equations and linear constant coefficient difference equations are briefly introduced, especially for linear constant coefficient difference equations.
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Solving homework Geometry solver app Fraction solver Algebra tutor online Scan and solve math Four variable equation solver | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://meowrr.com/answers-722", "fetch_time": 1675662285000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-06/segments/1674764500304.90/warc/CC-MAIN-20230206051215-20230206081215-00323.warc.gz", "warc_record_offset": 415109678, "warc_record_length": 7227, "token_count": 1164, "char_count": 6226, "score": 4.28125, "int_score": 4, "crawl": "CC-MAIN-2023-06", "snapshot_type": "longest", "language": "en", "language_score": 0.9775882959365845, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00020-of-00064.parquet"} |
# Question 5 and 6 Exercise 7.3
Solutions of Question 5 and 6 of Exercise 7.3 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
Q5 If $x$ is such that $x^2$ ard higher of $x$ may be negleeled. then show that $$\frac{(8+3 x)^{\frac{2}{3}}}{(2+3 x) \sqrt{4-5 x}}=1-\frac{5 x}{8}$$
Solution: Given that: $$\frac{\sqrt[4]{3}-3 x j^{\frac{2}{3}}}{2 \cdot 3 x+4-5 x}$$ \begin{aligned} & =\frac{8^{\frac{2}{3}}\left(1+\frac{3 x}{8}\right)^{\frac{2}{3}}}{2\left(1+\frac{3 x}{2}\right) \sqrt{4}\left(1-\frac{5 x}{4}\right)^{\frac{1}{2}}} \\ & =\frac{\left(2^3\right)^{\frac{2}{3}}}{2.2}\left[\left(1+\frac{3 x}{8}\right)^{\frac{2}{3}}\left(1+\frac{3 x}{2}\right)^{-1}\right. \\ & \left.\times\left(1-\frac{5 x}{4}\right)^{-\frac{1}{2}}\right] \end{aligned}
Applying binomial expansion and neglecting $x^2$ and higher powers of $x$. $-\left(1+\frac{2}{3} \cdot \frac{3 x}{8}+\right.$ higher powers of $\left.x\right) x$ $\left(1-\frac{3 x}{2}+\right.$ higher powers of $\left.x\right) \times$ $\left(1+\frac{1}{2}, \frac{5 x}{4}+\right.$ higher powers of $\left.x\right)$ $-\left(1+\frac{x}{4}\right)\left(1-\frac{3 x}{2}\right)$ $\times\left(1+\frac{5 x}{8}\right)$ Multiplying and neglecting $x^2$ and higher powers of $x$ \begin{aligned} & =\left(1-\frac{3 x}{2}+\frac{x}{4}+\text { higher powers of } \mathrm{x}\right) \\ & \times\left(1+\frac{5 x}{8}\right) \\ & \therefore\left(1-\frac{5 x}{4}\right)\left(1+\frac{5 x}{8}\right) \end{aligned}
Multiplying and neglecting $x^2$ and higher powers of $x$, we have \begin{aligned} & =1+\frac{5 x}{8}-\frac{5 x}{x} \\ & =1-\frac{5 x-10 x}{8} \\ & =1-\frac{5 x}{8} . \text { Hence } \end{aligned} $$\frac{(8+3 x)^{\frac{2}{3}}}{(2+3 x) \sqrt{4-5 x}}=1-\frac{5 x}{8}$$ Q6 If $x$ is large and $\frac{1}{x^3}$ may be neglected, then find approximate value of: $$\frac{x \sqrt{x^2-2 x}}{(x+1)^2}$$
Solution: We are given \begin{aligned} & \frac{x \sqrt{x^2-2 x}}{(x+1)^2}=\frac{x \sqrt{x^2} \sqrt{1-\frac{2 x}{x^2}}}{x^2\left(1+\frac{1}{x}\right)^2} \\ & =\left(1-\frac{2}{x}\right)^{\frac{1}{2}}\left(1+\frac{1}{x}\right)^{-2} \end{aligned}
Applying binomial theorem and neglecting $\frac{1}{x^3}$ etc \begin{aligned} & =\left[1 \cdots \frac{1}{2} \cdot \frac{2}{x}+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2 !}\left(-\frac{2}{x}\right)^2+\right. \\ & \ldots] \times\left[1 \cdots \frac{2}{x}+\frac{-2(-2-1)}{2 !}\left(\frac{1}{x}\right)^2\right. \\ & +\ldots \ldots] \\ & =\left[1-\frac{1}{x}-\frac{1}{2 x^2}-\ldots\right] \times\left[1-\frac{2}{x}+\frac{3}{x^2}\right. \\ & +\ldots] \end{aligned} Multiplying and neglecting $\frac{1}{x^3}$ and so on we have \begin{aligned} & =1 \frac{2}{x}+\frac{3}{x^2}+\frac{1}{x}+\frac{2}{x^2}-\frac{1}{2 x^2} \\ & =1-\frac{x^2+4}{y^2}+\frac{3}{2} \\ & =1-\frac{3}{2}+ \end{aligned}
Hence the spproximate value of $\frac{x \sqrt{x^2-2 x}}{(x+1)^2}$ is $1-\frac{3}{x}-\frac{9}{2 x^2}$ | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.mathcity.org/math-11-kpk/sol/unit07/ex7-3-p5", "fetch_time": 1708873502000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-10/segments/1707947474617.27/warc/CC-MAIN-20240225135334-20240225165334-00289.warc.gz", "warc_record_offset": 913529329, "warc_record_length": 7358, "token_count": 1305, "char_count": 3014, "score": 4.625, "int_score": 5, "crawl": "CC-MAIN-2024-10", "snapshot_type": "longest", "language": "en", "language_score": 0.4578508138656616, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00048-of-00064.parquet"} |
# A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level? - Physics
Numerical
A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?
#### Solution
Separation of two energy levels in an atom,
E = 2.3 eV
= 2.3 × 1.6 × 10−19
= 3.68 × 10−19 J
Let v be the frequency of radiation emitted when the atom transits from the upper level to the lower level.
We have the relation for energy as:
E = hv
Where,
h = Planck’s constant = 6.62 × 10−34 Js
∴ v = "E"/"h"
= (3.68 xx 10^(-19))/(6.62 xx 10^(-32))
= 5.55 × 1014 Hz
Hence, the frequency of the radiation is 5.6 × 1014 Hz.
Concept: Energy Levels
Is there an error in this question or solution?
Chapter 12: Atoms - Exercise [Page 436]
#### APPEARS IN
NCERT Physics Class 12
Chapter 12 Atoms
Exercise | Q 12.4 | Page 436
NCERT Physics Class 12
Chapter 12 Atoms
Exercise | Q 4 | Page 436
Share | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.shaalaa.com/question-bank-solutions/a-difference-of-23-ev-separates-two-energy-levels-in-an-atom-what-is-the-frequency-of-radiation-emitted-when-the-atom-makes-a-transition-from-the-upper-level-to-the-lower-level-energy-levels_11625", "fetch_time": 1718318297000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-26/segments/1718198861517.98/warc/CC-MAIN-20240613221354-20240614011354-00819.warc.gz", "warc_record_offset": 889229950, "warc_record_length": 12143, "token_count": 341, "char_count": 1100, "score": 3.921875, "int_score": 4, "crawl": "CC-MAIN-2024-26", "snapshot_type": "latest", "language": "en", "language_score": 0.7742268443107605, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00034-of-00064.parquet"} |
# Mathematics
## 20 terms · For maths project
### S=55
Given the formula S=1/2n(n+1), find the value of S when n=10
### a+9a
Simplify this. 3a+4b-2a+5b
2(5x-8)+6=11
### 3a(4x+y) + 2by(4x + y)
Factorise 12ax+3ay+8bx+2by
### x=2/3
Solve the equation 9(x+1) = 2(3x+8)
### 9a + 7b
3a + 4b - (-6a - 3b) =?
### Yes
Is (x - 6) a factor of (x2 - 5x - 6)?
### -0.5
If 2(x - 5) = -11, then x =?
### -1
If 4/5 _ (-3/10) = x + 1/1/2, then x =?
### 3
If x + 9 = 18 -2x, then x =?
d+d+d
7a+2a
-4m-2m+5m-m
4by-9by+5by
### 16/3t+9/2
7t+4av-5/3t+1/2av
### (3a+2b)(4x+y)
Factorise 12ax+3ay+8bx+2by
-4(7-5n)
4(x+7)+3(x+5)
x/4=7
8-3/5x =-2 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://quizlet.com/1084088/mathematics-flash-cards/", "fetch_time": 1386218631000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2013-48/segments/1386163040002/warc/CC-MAIN-20131204131720-00021-ip-10-33-133-15.ec2.internal.warc.gz", "warc_record_offset": 139397886, "warc_record_length": 17079, "token_count": 374, "char_count": 651, "score": 4.1875, "int_score": 4, "crawl": "CC-MAIN-2013-48", "snapshot_type": "latest", "language": "en", "language_score": 0.3119966685771942, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00055-of-00064.parquet"} |
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Sizing of power cables for circuit breaker controlled feeders (part 1)
Home / Technical Articles / Sizing of power cables for circuit breaker controlled feeders (part 1)
The following three criteria apply for the sizing of cables for circuit breaker controlled feeders:
This criteria is applied to determine the minimum cross section area of the cable, so that cable can withstand the short circuit current.
Failure to check the conductor size for short-circuit heating could result in permanent damage to the cable insulation and could also result into fire. In addition to the thermal stresses, the cable may also be subjected to significant mechanical stresses.
This criteria is applied so that cross section of the cable can carry the required load current continuously at the designed ambient temperature and laying condition.
[fancy_header]III. Starting and running voltage drops in cable[/fancy_header]
This criteria is applied to make sure that the cross sectional area of the cable is sufficient to keep the voltage drop (due to impedance of cable conductor) within the specified limit so that the equipment which is being supplied power through that cable gets at least the minimum required voltage at its power supply input terminal during starting and running condition both.
1. Criteria-1 Short circuit capacity
The maximum temperature reached under short circuit depends on both the magnitude and duration of the short circuit current. The quantity I2t represents the energy input by a fault that acts to heat up the cable conductor. This can be related to conductor size by the formula:
A = Minimum required cross section area in mm2
t = Operating time of disconnecting device in seconds
Isc = RMS Short Circuit current Value in Ampere
C = Constant equal to 0.0297 for copper & 0.0125 for aluminum
T2 = Final temp. ° C (max. short circuit temperature)
T1 = Initial temp. ° C (max. cable operating temperature – normal conditions)
T0 = 234.5° C for copper and 228.1° C for aluminum
Equation-1 can be simplified to obtain the expression for minimum conductor size as given below in equation-2:
Now K can be defined as a Constant whose value depends upon the conductor material, its insulation and boundary conditions of initial and final temperature because during short circuit conditions, the temperature of the conductor rises rapidly. The short circuit capacity is limited by the maximum temperature capability of the insulation. The value of K hence is as given in Table 2.
Boundary conditions of initial and final temperature for different insulation is as given under in Table 1 below.
Table 1
Insulation material Final temperature T2 Initial temperature T1 PVC 160° C 70° C Butyl Rubber 220° C 85° C XLPE / EPR 250° C 90° C
Table 2
Material → Copper Aluminum Insulation → PVC Butyl Rubber XLPE / EPR PVC Butyl Rubber XLPE / EPR (K) 1 Second Current Rating in Amp/mm2 115 134 143 76 89 94 (K) 3 Second Current Rating in Amp/mm2 66 77 83 44 51 54
In the final equation-2 we have determined the value of constant K. Now the value of t is to be determined. The fault current (ISC) in the above equation varies with time. However, calculating the exact value of the fault current and sizing the power cable based on that can be complicated. To simplify the process the cable can be sized based on the interrupting capability of the circuit breakers/fuses that protect them.
This approach assumes that the available fault current is the maximum capability of the breaker/fuse and also accounts for the cable impedances in reducing the fault levels.
The fault clearing time (tc) of the breakers/fuses per ANSI/IEEE C37.010, C37.013, and UL 489 are:
• For medium voltage system (4.16 kV) breakers, use 5-8 cycles
• For starters with current limiting fuses, use ½ cycle
• For low voltage breakers with intermediate/short time delay, use 10 cycles
• For low voltage breakers with instantaneous trips, use 1 cycle
Alternatively let us consider that feeder is for any large motor which is being fed from LV 415V or 400V switchgear having a circuit breaker with separate multifunction motor protection relay (For this calculation it is assumed to be SIEMENS made 7SJ61).
The instantaneous protection feature of this relay will be turned ON as and when any fault occurs. However, the selected cable shall have the capacity to withstand the maximum fault current for a finite duration (that is fault clearing time of the circuit breaker).
The minimum faults withstand duration necessary (for the instantaneous setting) for cable is calculated as under:
Si. No. Parameters Time in ms Source/Back up 1 Relay sensing/pickup time 20 SIEMENS 7SJ61 technical data 2 Tolerance/Delay time 10 SIEMENS 7SJ61 technical data 3 Breaker operating time 40 L&T make C-Power breaker have typical opening time of 40 ms and closing time of 60ms) 4 Relay overshoot 20 GEC handbook “Network Protection & automation Guide” 5 Safety Margin 30 TOTAL TIME IN MILI SECONDS 120
Therefore the cable selected for a circuit breaker controlled motor feeder in 415V or 400V switchgear shall be suitable to withstand the maximum rated fault current of 50kA for at least 120msec. However taking allowance of 40 Mili seconds in the opening time of circuit breaker due to aging, frequent number of operation, increase in contact resistance of circuit breaker and finally to cover the variation due from manufacturer to manufacturer.
Hence the cable selected for a circuit breaker controlled motor feeder in 415V or 400V switchgear shall be suitable to withstand the maximum rated fault current of 50kA for at least (120+40) 160msec. Many consultants recommend for use operating time of disconnecting device as 200msec also. Value of “t” more than 160 seconds is a conservative design.
A = (Isc x √t)/K = (50000 x √0.16)/94 = 212.766mm2
Next standard cable size: = 240 mm2
Although it may appear that selection of minimum cross sectional area of cable conductor as 240 mm2 is only just large enough for the duty, the actual fault current in the motor circuit is generally less than the switchboard fault withstand rating of 50kA, hence the selection of cable of cross sectional area 240 mm2 in practice offers sufficient design margin.
The minimum cross sectional area of cable required for 415V or 400V switchgear motor feeder from fault withstand point of view shall be 240mm2.
We have considered for circuit breaker controlled motor feeder and analyzed the duration of short circuit/fault withstanding time in seconds for the same. Exactly the Same holds true for Circuit breaker controlled (Please see the below figure) outgoing transformer feeder.
However operating time of disconnecting device is slightly different for circuit breaker controlled incomer and tie feeders. Duration of fault withstanding/operating time of disconnecting device for incomer and tie feeder is 1 and 0.5 second respectively. This is because of additional presence of inverse definite minimum time delay protection relays along with instantaneous protection. The inverse definite time delay protection has time settings greater than 0.5 for incomer feeders and about 0.5 for tie feeders.
For all different type of feeders the operating time of disconnecting device is indicated in figure below:
The final cable size shall be selected considering the other two criteria that is continuous current carrying capacity & voltage drop criteria which would be continued in part-2 and part-3.
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Asif Eqbal
Bachelor of Engineering in Electrical & Electronics engineering, from Manipal University, (Karnataka), India in 2006. Presently involved in the design of EHV outdoor substation and coal fired thermal power plants for more than seven years. Motto of joining EEP as a contributor is to share my little engineering experience and help the budding engineers in bridging the conspicuous gap between academics and Industrial practice. “If you have knowledge, let others light their candles with it, so that people who are genuinely interested in helping one another develop new capacities for action; it is about creating timeless learning processes".
1. Raja
May 15, 2019
What is the IEC or IEEE reference for tripping time
2. Iqbal shaukatali Sayyed
Dec 20, 2018
Dear Sir
I am civil engineer will you send me electrical flow diagram from substation to residential house with sizes of wires/ cables and electrical points for 2 bhk residential flat pleas.
3. Azley
Aug 16, 2017
Thanks for the info!
4. RR
Nov 04, 2015
Dear asif
why do we need to have circuit breakers on both side of cable can have a load break switch on the load other side(load side).
5. GINA MOODY
Oct 01, 2015
Dear Sir,
Iam Eng. Gina from Fox for international Trading.
Kindly send me your best quotation about:-
Item no
Line Description QTY UNIT
1 SIEMENS CIRCUIT BREAKER 250A
MODEL: 3VF4212-2MM41-0AA0
4-POLE, 400 VAC
200-250 A
ICU/ICS= 70KA 2 EA
Please send me your reply as soon as possible including:-
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6. Ashfaq
Dec 06, 2014
Good
7. Tareq
Oct 24, 2014
Thanks
8. Rahul
Sep 05, 2014
If we are having 300 mm2 cable size from the formula
A= I SC x √t / k for a LT Incomer then can we use two runs of 150 mm2 for the same incomer instead of one run of 300 mm2 cable.
Can you please clear this point with electrical design aspect.(with consideration of short circuit level and without short circuit level ) Explain the same for HT cable sizing.
Learn to Design Power Systems
Learn to design LV/MV/HV power systems through professional video courses. Lifetime access. Enjoy learning! | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://electrical-engineering-portal.com/sizing-of-power-cables-for-circuit-breaker-controlled-feeders-part-1/comment-page-2?replytocom=86049", "fetch_time": 1638936357000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-49/segments/1637964363437.15/warc/CC-MAIN-20211208022710-20211208052710-00527.warc.gz", "warc_record_offset": 302444798, "warc_record_length": 31520, "token_count": 2333, "char_count": 10154, "score": 3.5, "int_score": 4, "crawl": "CC-MAIN-2021-49", "snapshot_type": "latest", "language": "en", "language_score": 0.8898916244506836, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00026-of-00064.parquet"} |
# Is it possible to know the mother address of a subnet?
The mother address is X, have Y numbers of subnets (children), we already know 2 of the mother's children, they're:
10.1.192.0 /20 [edited]
10.1.240.0 /20
Is it possible to know the value of X and Y according to the given details?
I hope you can understand what I mean.
• BTW, apologizes to your mother, but the correct term is “supernet.” – Ron Trunk Jul 24 '20 at 11:36
Yes, we can calculate the smallest `supernet` that contains 2 given networks (or any number of networks).
The easiest way to do this is in binary (yes really):
10.1.192.0 = 00001010.00000001.11000000.00000000
10.1.240.0 = 00001010.00000001.11110000.00000000
As you can see, the part that is common to this two addresses is 00001010.00000001.11 which is 18 bits long.
So we now we are looking for a /18 network.
To get the network address we take those 18 bits and complete to 32 bits with trailing zeros and we get:
00001010.00000001.11000000.00000000 /18 = 10.1.192.0 /18
Now to calculate your Y value, assuming subnet of equal size, i.e. /20, this boil down to "how many /20 subnet do we have in a /18 network".
To calculate this we the take the size difference and square it.
20 - 18 = 2
2^2 = 4 = Y => we have 4 /20 subnets.
• While your arithmetic is sound, you can't know whether a supernet actually exists (ie. whether it is administrated by a single entity) or if it is (much) larger even. – Zac67 Jul 24 '20 at 12:55
• @Zac67, from a pure IP perspective, a supernet always exist, up to 0.0.0.0/0 (or ::/0). Regarding the administration, well, that's a totally different question. – JFL Jul 24 '20 at 13:03 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://networkengineering.stackexchange.com/questions/69198/is-it-possible-to-know-the-mother-address-of-a-subnet", "fetch_time": 1620888468000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-21/segments/1620243991537.32/warc/CC-MAIN-20210513045934-20210513075934-00631.warc.gz", "warc_record_offset": 426331041, "warc_record_length": 36988, "token_count": 506, "char_count": 1660, "score": 3.84375, "int_score": 4, "crawl": "CC-MAIN-2021-21", "snapshot_type": "longest", "language": "en", "language_score": 0.8911038637161255, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00049-of-00064.parquet"} |
# Weird2 - editorial
Tester: Teja Vardhan Reddy
Editorialist: Taranpreet Singh
Easy-Medium
### PREREQUISITES:
Observation, Amortized analysis and data structures.
### PROBLEM:
Given an array A of length N, find out the number of ordered pairs (a,b) such that a appears at least b times in array while b appears at least a times in the array.
### QUICK EXPLANATION
• Sort the given array, and for every position, iterate over the positions having less than the frequency of current element.
• Above solution works in overall O(N) time because, at every iteration, we iterate over at most F[i] elements in each iteration where F[i] is the frequency of A[i]. Since \sum F[i] = N, Overall complexity is O(N).
### EXPLANATION
For this problem, I have two solutions to offer, First the intended solution, which is used by Setter and Tester, while the other one is my solution.
Let us make input in form of frequency mapping tuples (A[i],F[i]), where F[i] denote frequency of A[i] in input array.
Bruteforce Solution: Iterate over every ordered pair (i,j), i \leq j of values and check if F[j] \geq A[i] and F[i] \geq A[j] holds.
Since we iterate over all pairs of values, this solution has complexity O(N^2).
Now, let us rewrite the conditions of a valid pair. For pair (A[i], A[j]) to be valid, we need F[j] \geq A[i] and F[i] \geq A[j].
Let us sort these tuples in increasing order of A[i] and run the same loops as BruteForce. Now, notice, that for any (A[i], F[i]), if we get the position A[j] > F[i] for any position j, we can safely see, that No position k, k > j will satisfy F[i] \geq A[k]. Hence, we can break out from internal loop as soon as we reach a value A[j] > F[i].
Now, What is the time complexity of this solution? This is where Amortized Analysis comes into play.
See, For every position, we iterate over at max F[i] element in inner loop. Since N = \sum F[i], The overall solution complexity is bounded by N, resulting in overall O(N*logN) solution due to sorting.
Now, Let’s discuss Editorialist solution in brief, who didn’t get the observation at first.
Make frequency tuples same way as above solution and sort tuples by non-decreasing order of F[i]. If we focus on condition A[i] \leq F[j], we know that this condition will hold for a suffix of the array due to the non-decreasing order of F. This way, Now we need to count the number of positions in Suffix [j, N-1] for which A[j] \leq F[i]. Let’s call (j, F[i]) a query, and j is the left end of query range, right end being the end of the array.
There exist a more general problem which can be applied here. Given an array of length N, Perform two type of operations. First is to update a single element. Other is to count number of elements \leq x in range [l, r].
Editorialist just uses Order statistic tree, along with sorting queries (j, F[i]) by decreasing order of j, and answer queries and add elements to order statistic tree as the j decreases, and making queries when the current range of added elements correspond to query range.
The time complexity of this solution too is O(N*logN) but is probably slower than the first solution due to the constant factor.
This was a lesson, that observations matter.
### Related problem
The problem Tufurama seems a lot similar to this problem, though have quite a different solution which is worth a try.
Also, The problems KQUERY and KQUERY2 are nice problems to practice if you tried the second solution.
### Time Complexity
The time complexity of both solutions is O(N*logN) but the first one has lower constant factor than the second solution.
### AUTHOR’S AND TESTER’S SOLUTIONS:
Feel free to Share your approach, If it differs. Suggestions are always welcomed.
2 Likes
Yeah, I tried using the tufurama approach, but the complexity was too much. Is there a different way of applying mergesort tree to solve this problem?
Well, we need to answer queries of type KQUERY, whichever method you may apply.
HERE is my solution.
which is O(n+10^6)
Seems to be same as setter… but this is not O(n*log(n))
Still confused about time complexity… can anyone help ?
reader functions and huge templates and assert statements annoys a lot in checking solutions
OMG I also used the setters approach without knowing! Its a lot of fun when you realize you did it in setters method :). Thank you for editorial taran!
I agree. So many random defines hinder readibility >_<
1 Like
@l_returns hacked your soln. Check Hacked Soln.
TC is hard coded instead of reading. Takes 1.5 sec and TL is 1 sec.
this is my code that pass all test cases.however when i use unordered_map it gives wrong answer.suggest some
test case where it fails.
https://www.codechef.com/viewsolution/21247254
1 Like
Your complexity is omega(T*1e6) but inner loop is not executed many times so it passed.
1 Like
Such a difference doesnt really matter @aryanc403 …XD
1 Like
Her (or his…?) profile doesnt show any submission XD
This reminds me of that moment …
In the ordered map, all tuples are ordered by key, while in unordered, they can appear in any order, making your solution incorrect, when an unordered map is used.
As usual. I know this profile just logs in to post such comments at my editorials.
Just stop it!!
Setter seems to be having same solution @aryanc403 XD…
See editorialist’s solution, if looking for anything different xD
Lol @l_returns.
@aryanc403 -yup…I recall that too
lol. Plagiarism Detected. In coming days we will see a drop admin’s rating.
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or
# If the line (x-2)/(3)=(y+1)/(2)=(z-1)/(-1) intersects the plane 2x+3y-z+13=0 at a point P and the plane 3x+y+4z=16 at a point Q, then PQ is equal to
Question from Class 12 Chapter 3D Geometry
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14`sqrt(14)``2sqrt(7)``2sqrt(14)`
D
Solution :
Equation of given line is <br> `(x-2)/(3)=(y+1)/(2)=(z-1)/(-1)=r("let")" "...(i)` <br> Now, coordinates of a general point over given line is `R(3r+2,2r-1,-r+1)` <br> Let the coordinates of point P are `(3r_(1)+2,2r_(1)-1,r_(1)+1)` and Q are `(3r_(2)+2,2r_(2)-1,-r_(2)+1).` <br> Since, P is the point of intersection of line (i) and the plane `2x+3y-z+13=0,` so <br> `2(3r_(1)+2)+3(2r_(1)-1)-(-r_(1)+1)+13=0` <br> `implies" "6r_(1)+4+6r_(1)-3+r_(1)-1+13=0` <br> `implies" "13r_(1)+13=0impliesr_(1)=-1` <br> So, point `P(-1, -3, 2)` <br> And, similarly for point 'Q', we get <br> `3(3r_(2)+2)+(2r_(2)-1)+4(-r_(2)+1)=16` <br> `implies" "7r_(2)=7impliesr_(2)=1` <br> So, point is Q (5, 1, 0) <br> Now, `" "PQ=sqrt((5+1)^(2)+(1+3)^(2)+2^(2))` <br> `=sqrt(36+16+4)` <br> `=sqrt(56)=2sqrt(14)`
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Math
After leaving the runway, a plane's angle of ascent is 16° and its speed is 267 feet per second. How many minutes will it take for the airplane to climb to a height of 12,500 feet? Round answer to two decimal places.
I know the answer is 2.83, but how?
1. 👍
2. 👎
3. 👁
1. if it's air speed is 267 ft/s, then its vertical speed is
267 sin 16° = 73.595 ft/s
So, to gain an altitude of 12500 ft takes
12500/73.595 = 169.848 seconds, or 2.83 minutes
1. 👍
2. 👎
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# Lagrange multipliers, guidance needed
1. Mar 6, 2015
### ilyas.h
1. The problem statement, all variables and given/known data
f(x,y) is function who's mixed 2nd order PDE's are equal.
consider k_f:
determine the points on the graph of the parabloid f(x,y) = x^2 + y^2 above the ellipse 3x^2 + 2y^2 = 1 at which k_f is maximised and minimised.
3. The attempt at a solution
is this the langrange equation that I'd use?
$$L(x,y,\lambda )=k_{f}(x,y)-\lambda k_{g}(x,y)$$
or maybe it's just:
$$L(x,y,\lambda )=k_{f}(x,y)-\lambda g(x,y)$$
im not sure if I should put the g function (3x^2 + 2y^2 = 1) into the k_f. If I do need to do this, then the algebra is very ugly.
2. Mar 7, 2015
### Ray Vickson
Have you made an attempt to simplify $k_f(x,y)$ as much as possible before using it in a Lagrangian or anywhere else? Far from ugly, the algebra will be very simple and straightforward.
3. Mar 7, 2015
### ilyas.h
my k_f(x,y) is:
$$\frac{4}{(1+4x^{2}+4y^{2})^2}$$
How would you simplify this/the problem? I tried simplifying with wolfram alpha, it doesn't return anything useful.
my k_g(x,y) is:
$$\frac{24}{(1+36x^{2}+16y^{2})^2}$$
please, someone help. It's a bit of an emergency.
Last edited: Mar 7, 2015
4. Mar 7, 2015
### Ray Vickson
Why do you care about $k_g$? It has nothing at all to do with the question as originally posed. Also: knowing the form of $k_f$ should allow you to simplify the problem significantly, to get a problem that is much easier than the original one. I cannot say more, without essentially doing the problem for you
5. Mar 7, 2015
### ilyas.h
ok, then the Lagrange equation we are interested in is:
$$L(x,y, \lambda)=\frac{4}{(1+4x^{2}+4y^{2})^2}- \lambda(3x^{2} + 2y^{2}-1)$$
what do you mean by form of k_f(x,y)? could you come up with a mini-alternative example to what you mean, I just don't see how you can simplify it, unless this has something unique to do with the problem as opposed to general algebraic manipulation.
6. Mar 7, 2015
### Ray Vickson
OK, so don't simplify it; just solve it as-is. It does work, but takes a bit more effort.
7. Mar 8, 2015
### ilyas.h
$\frac{\partial L}{\partial x}= (-64x(1+4x^{2}+4y^{2})^{-3}) - 6x\lambda = 0$
$\frac{\partial L}{\partial y}= (-64y(1+4x^{2}+4y^{2})^{-3}) - 4y\lambda = 0$
$\frac{\partial L}{\partial \lambda}= 1 - 3x^{2}-2y^{2}=0$
this leads to a contradiction though, using the first and second equation we get:
$(\frac{-32}{3}(1+4x^{2}+4y^{2})^{-3}) = \lambda = ((-24)(1+4x^{2}+4y^{2})^{-3})$
8. Mar 8, 2015
### Ray Vickson
You can write
$$0= L_x = -Rx - 6x \lambda , \; \text{and} \; 0 = L_y = -Ry - 4 y \lambda$$
where $R = 64/(1 + 4x^2+4 y^2)^3$. You have attempted to convert $-Rx - 6 x \lambda = 0$ into $-R - 6 \lambda = 0$ (similarly for $y$), and you are NOT allowed to do that without some extra conditions.
9. Mar 8, 2015
### ilyas.h
1. $\frac{\partial L}{\partial x}= (-64x(1+4x^{2}+4y^{2})^{-3}) - 6x\lambda = 0$
2. $\frac{\partial L}{\partial y}= (-64y(1+4x^{2}+4y^{2})^{-3}) - 4y\lambda = 0$
3. $\frac{\partial L}{\partial \lambda}= 1 - 3x^{2}-2y^{2}=0$
if you get an equation for lambda from the first and second equation without dividing through by any variables you get that (using $\lambda$):
$\frac{32}{3}xy(1+4x^{2}+4y^{2})^{3} = -24xy(1+4x^{2}+4y^{2})^{3}$
how am I meant to get a relationship between x and y from the above equation that I can utilize in the third equation?
10. Mar 8, 2015
### Ray Vickson
For any real $x,y$ we have $1 + 4 x^2 + 4 y^2 \geq 1 > 0$, so you are allowed to divide it out on both sides, to end up with
$$\frac{32}{3} xy = - 24 xy$$
What does that tell you?
Last edited: Mar 8, 2015
11. Mar 8, 2015
### ilyas.h
that x =y = 0, however this cannot be true due to the third equation.
therefore x and y has to take multiple values that satisfy the following relationships:
$\frac{32}{3}xy = -24xy$
$1 - 3x^{2}-2y^{2}=0$
going along the right lines?
and thanks for the help so far.
12. Mar 8, 2015
### Ray Vickson
13. Mar 8, 2015
### ilyas.h
ok, so then x must equal y which must equal zero.
Or perhaps my calculations are wrong...
14. Mar 8, 2015
### ilyas.h
...............
15. Mar 8, 2015 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/lagrange-multipliers-guidance-needed.801737/", "fetch_time": 1508229279000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-43/segments/1508187820930.11/warc/CC-MAIN-20171017072323-20171017092323-00184.warc.gz", "warc_record_offset": 949080787, "warc_record_length": 19248, "token_count": 1501, "char_count": 4197, "score": 3.578125, "int_score": 4, "crawl": "CC-MAIN-2017-43", "snapshot_type": "longest", "language": "en", "language_score": 0.9232335090637207, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00003-of-00064.parquet"} |
# Find three consecutive even integers such that three times the first equals the sum of the other two
1
by kash
2014-11-12T18:46:15+08:00
Let 'x' be the first integer
'x+2' be the 2nd
'x+4' be the 3rd
Translating the statement into mathematical expression you'll have:
3x = (x+2) + (x+4)
3x = x + 2 + x + 4
3x = 2x + 6
3x - 2x = 6
x = 6 ---1st even integer
x+2 = 6 + 2
x + 2 = 8 ---2nd even integer
x + 4 = 6 + 4
x + 4 = 10 --3rd even integer
Therefore the 3 consecutive even integers are 6, 8, and 10 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://brainly.ph/question/82174", "fetch_time": 1484643013000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-04/segments/1484560279650.31/warc/CC-MAIN-20170116095119-00413-ip-10-171-10-70.ec2.internal.warc.gz", "warc_record_offset": 801334341, "warc_record_length": 9703, "token_count": 217, "char_count": 508, "score": 4.25, "int_score": 4, "crawl": "CC-MAIN-2017-04", "snapshot_type": "latest", "language": "en", "language_score": 0.8100229501724243, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00008-of-00064.parquet"} |
# Are these the same equations?
If I have $2x^2+6x+2=0$ and $x^2+3x+1=0$ and then I multiply both sides of $x^2+3x+1=0$ by 2, I get $2x^2+6x+2=0$, which is the same as the first equation. However, when I graph $2x^2+6x+2$ and $x^2+3x+1$, they aren't the same graphs. Why is that?
• One is a scaled version of the other. Your observation that $2(x^2+3x+1) = 0$ is the same as $2x^2 + 6x + 2 = 0$ tells you that $x^2 + 3x+1$ and $2x^2 + 6x + 2$ have the same roots. – bames Feb 19 '18 at 5:07
• Wait, how is the one a scaled version? What does that mean? – user532526 Feb 19 '18 at 5:10
• It means one is the other multiplied by some number. The two equations with $=0$ are logically equivalent, the two equations with $y=$ are not equivalent. Graphically, the graph of the scaled quadratic will be the same graph as the former, but stretched away from the $x$-axis by a factor of $2$. – anon Feb 19 '18 at 5:11
• What is your definition of same equation? – Tony Ma Feb 19 '18 at 5:30
It is true that $2x^2+6x+2=0$ and $x^2+3x+1=0$ are same equations but when you are plotting you are plotting the functions $y(x) =2x^2+6x+2$ and $y(x)=x^2+3x+1$ which are different functions. One is a scaled version of the other.
To have same roots you need to have the curves intersect $y=0$ at the same points, which these two curves would do.
• @YoloInver these equations $2x^2+6x+2=0$ and $x^2+3x+1=0$ are indeed the same equations. However, the functions $f(x)=2x^2+6x+2$ and $g(x)=x^2+3x+1$ are not the same. – Dave Feb 19 '18 at 5:45
$2x^2+6x+2$ and $x^2+3x+1$ have the same roots, this does not mean that they are the same equations.
$2x^2+6x+2\ne x^2+3x+1$, since $2x^2+6x+2=2(x^2+3x+1)$, it is vertically stretched twice as much as $x^2+3x+1$ | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/2656494/are-these-the-same-equations", "fetch_time": 1579350294000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-05/segments/1579250592565.2/warc/CC-MAIN-20200118110141-20200118134141-00253.warc.gz", "warc_record_offset": 558282144, "warc_record_length": 31794, "token_count": 662, "char_count": 1738, "score": 4.125, "int_score": 4, "crawl": "CC-MAIN-2020-05", "snapshot_type": "latest", "language": "en", "language_score": 0.9497761726379395, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00035-of-00064.parquet"} |
Home Technical Mathematics Languages Science Social Science Art Literature Arabic Islamic Studies C.R.K HistoryMusicVisual Art Clothing/Textile Home Management Shorthand
General Mathematics Paper 2,Nov/Dec. 2013
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 Main
Weakness/Remedies
Strength
Question 5
Question 5
1. Ceramic
72o Textile
Plywood 65o
82o
mill
The pie chart shows the result of a survey conducted on the maintenance of five small and medium scale industries in a state. If the cost of maintenance of the plywood industry was N34,500.00, calculate the cost of maintenance of the textile industry.
1. T Y
100o
X 42o Z
m
U W
In the diagram, TYZ and UWZ are tangents to the circle WXY, X
= 100o, ∠YZW = 42o and ∠UWX = m. Find the value of m.
_____________________________________________________________________________________________________
Observation
This question was reported to be attempted by majority of the candidates. According to the report, while the candidates performed well in part(a), their performance in part(b) was described as poor. Candidates reportedly showed poor knowledge of circle geometry.
In part(a) candidates were reported to have obtained the value of x by: + + 82 + 68 + 72 = 360o which gave x = 69o. Thus the value of the angle of the sector for plywood = 69o. The total cost of maintenance of the industries was obtained as
= N180,000.00. Therefore the cost of maintenance of the textile industry = = N34,000.00.
In part(b) candidates were expected to respond as follows: In the diagram, triangle ZYW is an isosceles triangle and ∠ZYW = ∠YWZ (The length of the tangents drawn from the same external point are equal). XYW = m (angle in alternate segment to XWU). This implied that ∠ZYW = 100 – m. Therefore, (100 – m) + (100 – m) + 42 = 180o (sum of angles of a triangle). Solving this equation gave m = 31o. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://waeconline.org.ng/e-learning/Mathematics/maths222Nq5.html", "fetch_time": 1506307100000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-39/segments/1505818690307.45/warc/CC-MAIN-20170925021633-20170925041633-00566.warc.gz", "warc_record_offset": 350922390, "warc_record_length": 5876, "token_count": 542, "char_count": 2121, "score": 4.0625, "int_score": 4, "crawl": "CC-MAIN-2017-39", "snapshot_type": "latest", "language": "en", "language_score": 0.8432656526565552, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00044-of-00064.parquet"} |
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# Hydrogen - Density and Specific Weight vs. Temperature and Pressure
Density , ρ, has units typically [kg/m3] or [lb/ft3], and is defined by the ratio of the mass to the volume of a substance:
ρ = m/V [1]
where m = mass, units typically [kg] or [lb]
V = volume, units typically [m3 ] or [ft3 ]
Specific weight , γ, has units typically [N/m3 ] or [lbf /ft3 ] is defined by the ratio of the weight to the volume of a substance:
γ = (m * g)/V = ρ * g [2]
where g = acceleration due to gravity, units typically [m/s2] and value on Earth usually given as 9.80665 [m/s2] or 32.17405 [ft/s2]
Tabulated values of hydrogen density at given temperature and pressure (SI and Imperial units) as well as density units conversion are given below the figures.
### Online Hydrogen Density Calculator
The calculator below can be used to estimate the density and specific weight of gaseous hydrogen at given temperature and atmospheric pressure (1 bara/14.5psia).
The output density is given as kg/m3, lb/ft3, lb/gal(US liq) and sl/ft3 . Specific weight is given as N/m3 and lbf / ft3.
Choose the actual unit of temperature:
.
For calculation of density at higher pressure (up to 100 bara / 1450 psia), a good approximation is to multiply the density at 1 bara with the actual, higher pressure (in bara). If pressure is given in psia, you have to multiply the density at 14.5 psia with the actual higher pressure (in psia) and divide by 14.5.
Note! For pressure between 20 and 100 bara, the error may be up to 7% by using this approximation (higher error, the higher pressure).
Example 1: Density of hydrogen at 5 bara and 24 °C.
Calculator gives the density at 24 °C and 1 bara: 0.0815 [kg/m3 ]
Density at 24°C and 5 bara : 0.0815 [kg/m3 ]* 5[bara]/1[bara] = 0.408 [kg/m3 ]
Example 2: Density of hydrogen at 725 psia and 125 °F.
Calculator gives the density at 125 °F and 14.5 psia: 0.00465 [lbm /ft3 ]
Density at 125°F and 725 psia : 0.00465 [lbm /ft3 ]* 725[psia]/14.5[psia] = 0.233 [lbm /ft3 ]
(this is 3% more than the below tabulated value of density at 125°F and 725 psia.)
See also other properties of Hydrogen at varying temperature and pressure: Specific heat (Heat capacity) and Thermal conductivity , and Thermophysical properties at standard conditions , as well as density and specific weight o f acetone , air , ammonia , argon , benzene , butane , carbon dioxide , carbon monoxide , ethane , ethanol , ethylene , helium , methane , methanol , nitrogen , oxygen , pentane , propane , toluene and water .
Density of crude oil , Density of fuel oils , Density of lubricating oil and Density of jet fuel as function of temperature.
.
Density and specific weight of hydrogen at given temperatures and pressures:
For full table with Denity and Specific Weight - rotate the screen!
Hydrogen - Density and Specific Weight vs. Temperature and Pressure
State Temperature Pressure Density Specific weight
[K] [°C] [°F] [MPa] [bara] [psia] [mol/dm3] [g/l],
[kg/m3]
[lbm /ft3] [sl/ft3 *10-3] [N/m3] [lbf /ft3]
Liquid at
equilibrium
pressure
13.96 -259.2 -434.5 0.00770 0.0770 1.12 38.15 76.91 4.801 149.2 754.2 4.801
14 -259 -434 0.00789 0.0789 1.14 38.13 76.87 4.799 149.1 753.8 4.799
16 -257 -431 0.0215 0.215 3.12 37.26 75.12 4.689 145.8 736.7 4.689
18 -255 -427 0.0481 0.481 6.97 36.32 73.22 4.571 142.1 718.1 4.571
20 -253 -424 0.0932 0.932 13.5 35.27 71.11 4.439 138.0 697.4 4.439
22 -251 -420 0.163 1.63 23.7 34.09 68.73 4.291 133.4 674.0 4.291
24 -249 -416 0.264 2.64 38.3 32.74 66.00 4.120 128.1 647.2 4.120
26 -247 -413 0.403 4.03 58.4 31.15 62.80 3.921 121.9 615.9 3.921
28 -245 -409 0.585 5.85 84.9 29.23 58.92 3.678 114.3 577.8 3.678
30 -243 -406 0.850 8.50 123 26.71 53.84 3.361 104.5 528.0 3.361
32 -241 -402 1.12 11.2 162 22.64 45.64 2.849 88.55 447.5 2.849
33.19 -240.0 -399.9 1.33 13.3 193 14.94 30.12 1.880 58.44 295.4 1.880
Gas at
equilibrium
pressure
13.96 -259.2 -434.5 0.00770 0.0770 1.12 0.06754 0.1362 8.50E-03 0.2642 1.335 8.50E-03
14 -259 -434 0.00789 0.0789 1.14 0.06902 0.1391 8.69E-03 0.2700 1.365 8.69E-03
16 -257 -431 0.0215 0.215 3.12 0.1676 0.3380 0.02110 0.6558 3.314 0.02110
18 -255 -427 0.0481 0.481 6.97 0.3413 0.6880 0.04295 1.335 6.747 0.04295
20 -253 -424 0.0932 0.932 13.5 0.6165 1.243 0.07759 2.412 12.19 0.07759
22 -251 -420 0.163 1.63 23.7 1.025 2.067 0.1290 4.010 20.27 0.1290
24 -249 -416 0.264 2.64 38.3 1.609 3.244 0.2025 6.294 31.81 0.2025
26 -247 -413 0.403 4.03 58.4 2.431 4.900 0.3059 9.508 48.06 0.3059
28 -245 -409 0.585 5.85 84.9 3.600 7.258 0.4531 14.08 71.18 0.4531
30 -243 -406 0.850 8.50 123 5.364 10.81 0.6751 20.98 106.1 0.6751
32 -241 -402 1.12 11.2 162 8.682 17.50 1.093 33.96 171.7 1.093
33.19 -240.0 -399.9 1.33 13.3 193 14.94 30.12 1.880
58.44
295.4 1.880
Liquid 20 -253 -424 0.1 1 14.5 35.29 71.14 4.441 138.0 698 4.44
20.23 -252.9 -423.3 0.1 1 14.5 35.16 70.88 4.425 137.5 695 4.42
Gas
20.23 -252.9 -423.3 0.1 1 14.5 0.6563 1.323 0.08259 2.567 13.0 0.0826
40 -233 -388 0.1 1 14.5 0.3054 0.6157 0.03844 1.195 6.04 0.0384
60 -213 -352 0.1 1 14.5 0.2013 0.4059 0.02534 0.7876 3.98 0.0253
80 -193 -316 0.1 1 14.5 0.1505 0.3035 0.01895 0.5889 2.98 0.0189
100 -173 -280 0.1 1 14.5 0.1203 0.2425 0.01514 0.4705 2.38 0.0151
120 -153 -244 0.1 1 14.5 0.1002 0.2020 0.01261 0.3919 1.98 0.0126
140 -133 -208 0.1 1 14.5 0.08586 0.1731 0.01081 0.3359 1.70 0.0108
160 -113 -172 0.1 1 14.5 0.07510 0.1514 0.009452 0.2938 1.48 0.00945
180 -93.2 -136 0.1 1 14.5 0.06677 0.1346 0.008403 0.2612 1.32 0.00840
200 -73.2 -99.7 0.1 1 14.5 0.06007 0.1211 0.007560 0.2350 1.19 0.00756
220 -53.2 -63.7 0.1 1 14.5 0.05461 0.1101 0.006873 0.2136 1.08 0.00687
240 -33.2 -27.7 0.1 1 14.5 0.05010 0.1010 0.006305 0.1960 0.990 0.00631
260 -13.2 8.3 0.1 1 14.5 0.04623 0.09319 0.005818 0.1808 0.914 0.00582
280 6.9 44.3 0.1 1 14.5 0.04293 0.08654 0.005403 0.1679 0.849 0.00540
300 26.9 80.3 0.1 1 14.5 0.04006 0.08077 0.005042 0.1567 0.792 0.00504
320 46.9 116 0.1 1 14.5 0.03756 0.07572 0.004727 0.1469 0.743 0.00473
340 66.9 152 0.1 1 14.5 0.03535 0.07127 0.004449 0.1383 0.699 0.00445
360 86.9 188 0.1 1 14.5 0.03339 0.06731 0.004202 0.1306 0.660 0.00420
400 127 260 0.1 1 14.5 0.03005 0.06058 0.003782 0.1175 0.594 0.00378
500 227 440 0.1 1 14.5 0.02404 0.04847 0.003026 0.09405 0.475 0.00303
600 327 620 0.1 1 14.5 0.02004 0.04040 0.002522 0.07839 0.396 0.00252
700 427 800 0.1 1 14.5 0.01718 0.03463 0.002162 0.06719 0.340 0.00216
800 527 980 0.1 1 14.5 0.01503 0.03030 0.001892 0.05879 0.297 0.00189
900 627 1160 0.1 1 14.5 0.01336 0.02693 0.001681 0.05225 0.264 0.00168
1000 727 1340 0.1 1 14.5 0.01202 0.02424 0.001513 0.04703 0.238 0.00151
Liquid 20 -253 -424 1 10 145 35.86 72.29 4.513 140.3 709 4.51
31.24 -241.9 -403.4 1 10 145 24.63 49.65 3.100 96.34 487 3.10
Gas
31.24 -241.9 -403.4 1 10 145 7.098 14.31 0.8933 27.77 140 0.893
40 -233 -388 1 10 145 3.606 7.270 0.4539 14.11 71.3 0.454
60 -213 -352 1 10 145 2.097 4.228 0.2639 8.204 41.5 0.264
80 -193 -316 1 10 145 1.524 3.072 0.1918 5.961 30.1 0.192
100 -173 -280 1 10 145 1.205 2.429 0.1516 4.713 23.8 0.152
120 -153 -244 1 10 145 0.9995 2.015 0.1258 3.910 19.8 0.126
140 -133 -208 1 10 145 0.8547 1.723 0.1076 3.343 16.9 0.108
160 -113 -172 1 10 145 0.7470 1.506 0.09402 2.922 14.8 0.0940
180 -93.2 -136 1 10 145 0.6637 1.338 0.08353 2.596 13.1 0.0835
200 -73.2 -99.7 1 10 145 0.5972 1.204 0.07516 2.336 11.8 0.0752
220 -53.2 -63.7 1 10 145 0.5432 1.095 0.06836 2.125 10.7 0.0684
240 -33.2 -27.7 1 10 145 0.4980 1.004 0.06268 1.948 9.85 0.0627
260 -13.2 8.3 1 10 145 0.4597 0.9267 0.05785 1.798 9.09 0.0579
280 6.9 44.3 1 10 145 0.4269 0.8607 0.05373 1.670 8.44 0.0537
300 26.9 80.3 1 10 145 0.3986 0.8035 0.05016 1.559 7.88 0.0502
320 46.9 116 1 10 145 0.3737 0.7534 0.04703 1.462 7.39 0.0470
340 66.9 152 1 10 145 0.3518 0.7093 0.04428 1.376 6.96 0.0443
360 86.9 188 1 10 145 0.3323 0.6700 0.04183 1.300 6.57 0.0418
400 127 260 1 10 145 0.2993 0.6033 0.03766 1.171 5.92 0.0377
500 227 440 1 10 145 0.2396 0.4830 0.03015 0.9372 4.74 0.0302
600 327 620 1 10 145 0.1998 0.4027 0.02514 0.7814 3.95 0.0251
700 427 800 1 10 145 0.1713 0.3454 0.02156 0.6702 3.39 0.0216
800 527 980 1 10 145 0.1500 0.3023 0.01887 0.5866 2.96 0.0189
900 627 1160 1 10 145 0.1333 0.2688 0.01678 0.5216 2.64 0.0168
1000 727 1340 1 10 145 0.1200 0.2420 0.01511 0.4696 2.37 0.0151
Liquid 25 -248 -415 5 50 725 35.66 71.89 4.488 139.5 705 4.49
Supercritical
phase
100 -173 -280 5 50 725 5.968 12.03 0.7511 23.35 118 5.98
175 -98.2 -145 5 50 725 3.313 6.679 0.4170 12.96 65.5 0.417
250 -23.2 -9.7 5 50 725 2.327 4.691 0.2928 9.102 46.0 0.293
325 51.9 125 5 50 725 1.799 3.627 0.2264 7.037 35.6 0.226
400 127 260 5 50 725 1.468 2.959 0.1848 5.742 29.0 0.185
Liquid 25 -248 -415 10 100 1450 37.93 76.47 4.774 148.4 750 4.77
Supercritical
phase
100 -173 -280 10 100 1450 11.42 23.02 1.437 44.66 226 1.44
175 -98.2 -145 10 100 1450 6.370 12.84 0.8017 24.92 126 0.802
250 -23.2 -9.7 10 100 1450 4.503 9.078 0.5667 17.61 89.0 0.567
325 51.9 125 10 100 1450 3.501 7.057 0.4406 13.69 69.2 0.441
400 127 260 10 100 1450 2.869 5.783 0.3610 11.22 56.7 0.361
Density units conversion:
Density converter
kilogram/cubic meter [kg/m3 ] = gram/liter [g/l], kilogram/liter [kg/l] = gram/cubic centimeter [g/cm3 ]= ton(metric)/cubic meter [t/m3 ], once/gallon(US liquid) [oz/gal(US liq)] pound/cubic inch [lb/in3 ], pound/cubic foot [lb/ft3 ], pound/gallon(UK) [lb/gal(UK)], pound/gallon(US liquid) [lb/gal(US liq)], slug/cubic foot [sl/ft3 ], ton(short)/cubic yard [ton(short)/yd3 ], ton(long)/cubic yard [yd3 ]
• 1 g/cm3 = 1 kg/l = 1000 kg/m3 = 62.428 lb/ft3 = 0.03613 lb/in3 = 1.9403 sl/ft3 = 10.0224 lb/gal(UK) = 8.3454 lb/gal(US liq) = 0.5780 oz/in3 = 0.7525 ton(long)/yr3
• 1 g/l = 1 kg/m3 = 0.001 kg/l = 0.000001 kg/cm3 = 0.001 g/cm3 = 0.99885 oz/ft3 = 0.0005780 oz/in3 = 0.16036 oz/gal(UK) = 0.1335 oz/gal(US liq) = 0.06243 lb/ft3 = 3.6127x10-5 lb/in3 = 1.6856 lb/yd3 = 0.010022 lb/gal(UK) = 0.0083454 lb/gal(US liq) = 0.0007525 ton(long)/yd3 = 0.0008428 ton(short)/yd3
• 1 kg/l = 1 g/cm3 = 1000 kg/m3 = 62.428 lb/ft3 = 0.03613 lb/in3 = 1.9403 sl/ft3 = 8.3454 lb/gal(US liq) = 0.5780 oz/in3 = 0.7525 ton(long)/yr3
• 1 kg/m3 = 1 g/l = 0.001 kg/l = 0.000001 kg/cm3 = 0.001 g/cm3 = 0.99885 oz/ft3 = 0.0005780 oz/in3 = 0.16036 oz/gal(UK) = 0.1335 oz/gal(US liq) = 0.06243 lb/ft3 = 3.6127x10-5 lb/in3 = 1.6856 lb/yd3 = 0.010022 lb/gal(UK) = 0.008345 lb/gal(US liq) = 0.0007525 ton(long)/yd3 = 0.0008428 ton(short)/yd3
• 1 lb/ft3 = 27 lb/yd3 = 0.009259 oz/in3 = 0.0005787 lb/in3 = 16.01845 kg/m3 = 0.01602 g/cm3 = 0.1605 lb/gal(UK) = 0.1349 lb/gal(US liq) = 2.5687 oz/gal(UK) = 2.1389 oz/gal(US liq) = 0.01205 ton(long)/yd3 = 0.0135 ton(short)/yd3
• 1 lb/gal(UK) = 0.8327 lb/gal(US liq) = 16 oz/gal(UK) = 13.323 oz/gal(US liq) = 168.179 lb/yd3 = 6.2288 lb/ft3 = 0.003605 lb/in3 = 0.05767 oz/in3 = 99.7764 kg/m3 = 0.09977 g/cm3 = 0.07508 ton(long)/yd3 = 0.08409 ton(short)/yd3
• 1 lb/gal(US liq) = 1.2009 lb/gal(UK) = 19.215 oz/gal(UK) = 16 oz/gal(US liq) = 201.97 lb/yd3 = 7.4805 lb/ft3 = 0.004329 lb/in3 = 0.06926 oz/in3 = 119.826 kg/m3 = 0.1198 g/cm3 = 0.09017 ton(long)/yd3 = 0.1010 ton(short)/yd3
• 1 lb/in3 = 1728 lb/ft3 = 46656 lb/yd3 = 16 oz/in3 = 27680 kg/m3 = 27.680 g/cm3 = 277.419 lb/gal(UK) = 231 lb/gal(US liq) =4438.7 oz/gal(UK) = 3696 oz/gal(US liq) = 20.8286 ton(long)/yd3 = 23.3280 ton(short)/yd3
• 1 oz/gal(UK) = 0.8327 oz/gal(US liq) = 6.2360 kg/m3 = 6.2288 oz/ft3 = 0.3893 lb/ft3 = 10.5112 lb/yd3
• 1 oz/gal(US liq) = 1.2009 oz/gal(UK) = 7.4892 kg/m3 = 7.4805 oz/ft3 = 0.4675 lb/ft3 = 12.6234 lb/yd3
• 1 sl/ft3 = 515.3788 kg/m3 = 514.7848 oz/ft3 = 0.2979 oz/in3 = 32.1741 lb/ft3 = 82.645 oz/gal(UK) = 68.817 oz/gal(US liq)
• 1 ton(long)/yd3 = 1.12 ton(short)/yd3 = 1328.94 kg/m3 = 0.7682 oz/in3 = 82.963 lb/ft3 = 2240 lb/yd3 = 2.5786 sl/ft3 = 13.319 lb/gal(UK) = 11.0905 lb/gal(US liq)
• 1 ton(short)/yd3 = 0.8929 ton(long)/yd3 = 1186.55 kg/m3 = 0.6859 oz/in3 = 74.074 lb/ft3 = 2000 lb/yd3 = 2.3023 sl/ft3 = 11.8921 lb/gal(UK) = 9.9023 lb/gal(US liq)
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# 6.4: The Equilibrium of Forces
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Figure 6-4 illustrates the forces on the layer of soil cut. The forces shown are valid in general. The forces acting on this layer are:
1. A normal force acting on the shear surface N1.
2. A shear force S1 as a result of internal friction N1·tan(φ).
3. A force W1 as a result of water under pressure in the shear zone.
4. A force normal to the blade N2.
5. A shear force S2 as a result of the soil/steel friction N2·tan(δ).
6. A force Was a result of water under pressure on the blade.
The normal force N1 and the shear force S1 can be combined to a resulting grain force K1.
$\ \mathrm{K}_{1}=\sqrt{\mathrm{N}_{1}^{2}+\mathrm{S}_{1}^{2}}\tag{6-8}$
The forces acting on a straight blade when cutting soil, can be distinguished as:
1. A force normal to the blade N2.
2. A shear force S2 as a result of the soil/steel friction N2·tan(δ).
3. A force W2 as a result of water under pressure on the blade.
These forces are shown in Figure 6-5. If the forces N2 and S2 are combined to a resulting force K2 and the adhesive force and the water under pressures are known, then the resulting force K2 is the unknown force on the blade. By taking the horizontal and vertical equilibrium of forces an expression for the force K2 on the blade can be derived.
$\ \mathrm{K}_{2}=\sqrt{\mathrm{N}_{2}^{2}+\mathrm{S}_{2}^{2}}\tag{6-9}$
Water saturated sand is also cohesionless, although in literature the phenomenon of water under pressures is sometimes referred to as apparent cohesion. It should be stated however that the water under pressures have nothing to do with cohesion or shear strength. The shear stresses still follow the rules of Coulomb friction. Due to dilatation, a volume increase of the pore volume caused by shear stresses, under pressures develop around the shear plane as described by Miedema (1987 September), resulting in a strong increase of the grain stresses. Because the permeability of the flow of water through the pores is very low, the stresses and thus the forces are dominated by the phenomenon of dilatancy and gravitation, inertia, adhesion and cohesion can be neglected.
The horizontal equilibrium of forces is:
$\ \mathrm{\sum F_{h}=K_{1} \cdot \sin (\beta+\varphi)-W_{1} \cdot \sin (\beta)+W_{2} \cdot \sin (\alpha)-K_{2} \cdot \sin (\alpha+\delta)=0}\tag{6-10}$
The vertical equilibrium of forces is:
$\ \mathrm{\sum F_{v}=-K_{1} \cdot \cos (\beta+\varphi)+W_{1} \cdot \cos (\beta)+W_{2} \cdot \cos (\alpha)-K_{2} \cdot \cos (\alpha+\delta)=0}\tag{6-11}$
The force Kon the shear plane is now:
$\ \mathrm{K}_{1}=\frac{\mathrm{W}_{2} \cdot \sin (\delta)+\mathrm{W}_{1} \cdot \sin (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)}\tag{6-12}$
The force K2 on the blade is now:
$\ \mathrm{K}_{2}=\frac{\mathrm{W}_{2} \cdot \sin (\alpha+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)}\tag{6-13}$
From equation (6-13) the forces on the blade can be derived. On the blade a force component in the direction of cutting velocity Fh and a force perpendicular to this direction Fv can be distinguished.
$\ \mathrm{F_{h}=-W_{2} \cdot \sin (\alpha)+K_{2} \cdot \sin (\alpha+\delta)}\tag{6-14}$
$\ \mathrm{F}_{v}=-\mathrm{W}_{2} \cdot \cos (\alpha)+\mathrm{K}_{2} \cdot \cos (\alpha+\delta)\tag{6-15}$
The normal force on the shear plane is now:
$\ \mathrm{N}_{1}=\frac{\mathrm{W}_{2} \cdot \sin (\delta)+\mathrm{W}_{1} \cdot \sin (\alpha+\beta+\delta)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\varphi)\tag{6-16}$
The normal force on the blade is now:
$\ \mathrm{N}_{2}=\frac{\mathrm{W}_{2} \cdot \sin (\alpha+\beta+\varphi)+\mathrm{W}_{1} \cdot \sin (\varphi)}{\sin (\alpha+\beta+\delta+\varphi)} \cdot \cos (\delta)\tag{6-17}$
Equations (6-16) and (6-17) show, that the normal forces on the shear plane and the blade are always positive. Positive means compressive stresses. In water saturated sand, always the shear type of cutting mechanism will occur. Figure 6-6 shows these forces on the layer cut.
This page titled 6.4: The Equilibrium of Forces is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Sape A. Miedema (TU Delft Open Textbooks) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://eng.libretexts.org/Bookshelves/Civil_Engineering/Book%3A_The_Delft_Sand_Clay_and_Rock_Cutting_Model_(Miedema)/06%3A_Saturated_Sand_Cutting/6.04%3A_The_Equilibrium_of_Forces", "fetch_time": 1660341173000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-33/segments/1659882571758.42/warc/CC-MAIN-20220812200804-20220812230804-00469.warc.gz", "warc_record_offset": 225045722, "warc_record_length": 26383, "token_count": 1645, "char_count": 5335, "score": 3.90625, "int_score": 4, "crawl": "CC-MAIN-2022-33", "snapshot_type": "latest", "language": "en", "language_score": 0.3396325409412384, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00003-of-00064.parquet"} |
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