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## Friday, 25 December 2009
### TETRAHEDRAL NUMBERS RECIPROCALS SUM
TETRAHEDRAL NUMBERS SERIES:
This post follows with the exercises on special numbers reciprocals related series, after the blog entries about Square Pyramidal Numbers and Polygonal Numbers . In fact, this example it is not very much interesting, but I wanted to write it before to deal with more difficult problems.
$latex \displaystyle T_{n}=\frac{n(n+1)(n+2)}{6}=\binom{n+2}{3}$
$latex \displaystyle S(n)=\sum_{k=1}^{n}{\frac{1}{T_{k}}=\sum_{k=1}^{n}{\frac{6}{k(k+1)(k+2)}$
If we split the main fraction into others:
$latex \displaystyle \frac{S(n)}{6}=\sum_{k=1}^{n}{\frac{1}{k(k+1)(k+2)}}=\sum_{k=1}^{n}{\left( \frac{A}{k}+\frac{B}{k+1}+\frac{C}{k+2} \right) }$
Solving the linear system of equations it gives:
$latex \displaystyle A=\frac{1}{2} \; ; B=-1 \; ; C=\frac{1}{2};$
This three series can be summed easily with the aid of the Harmonic Numbers:
$latex \displaystyle \sum_{k=1}^{n}{\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n}=H_n$
$latex \displaystyle \sum_{k=1}^{n}{\frac{1}{k+1}=\frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n}+\frac{1}{n+1}=H_n-1+\frac{1}{n+1}$
$latex \displaystyle \sum_{k=1}^{n}{\frac{1}{k+2}=\frac{1}{3}+\frac{1}{4}+ \cdots +\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}=H_n-1-\frac{1}{2} + \frac{1}{n+1}+\frac{1}{n+2}$
If we substitute everything in the expression for the reciprocals sum:
$latex \displaystyle \frac{S(n)}{6}=\frac{n}{n+1}-\frac{1}{2}-\frac{1}{4} +\frac{1}{2(n+1)}+\frac{1}{2(n+2)}$
In the previous step we can see what does exactly means to be a "telescoping series", the term $latex H_n$, has vanished and there is no need to handle Euler Mascheroni Gamma and Digamma Function:
$latex H_{n}=\gamma + \psi_{0}(n+1)$
Then the formula for the n-th partial sum is:
$latex \displaystyle S(n)=\frac{3n(3+n)}{2(1+n)(2+n)}$
And evaluating the limit we get:
$latex \displaystyle S(\infty)=\lim_{n \leftarrow \infty}{S(n)}=\frac{3}{2}$
References:
[1]-Tetrahedral Number at- Wikipedia
[2]-Weisstein, Eric W. "Tetrahedral Number." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/TetrahedralNumber.html
[3] A000292-Tetrahedral (or pyramidal) numbers: C(n+2,3) = n(n+1)(n+2)/6. The On-Line Encyclopedia of Integer Sequences!
## Thursday, 24 December 2009
### SPLITTING FTA FUNCTIONS (III)
PILTZ DIVISOR FUNCTIONS (2)
THEOREM-2:
$latex \tau_{j+k}(n)=\sum_{d|n}^{}{\tau_{j}(d)\cdot\tau_{k}\left(\frac{n}{d}\right)}$
Proof:
The Piltz Divisor functions are multiplicative, so it is only necessary to prove the case $latex p^{\alpha}$
In the previous post we saw how:
$latex \tau_{k}(p^\alpha)=\binom{\alpha+k-1}{k-1}$
And if we apply $latex p^\alpha$ to second member of the problem identity, then:
$latex f(p^\alpha)=\sum_{d|p^\alpha}^{}{\tau_{j}(d)\cdot\tau_{k}\left(\frac{p^\alpha}{d}\right)}=\sum_{i=0}^{\alpha}{\tau_{j}(p^i)\cdot\tau_{k}(p^{\alpha-i})=\sum_{i=0}^{\alpha}{\binom{i+j-1}{j-1}\binom{\alpha-i+k-1}{k-1}=$
$latex =\binom{\alpha+j+k-1}{j+k-1}=\tau_{j+k}(p^\alpha)$
This proof seems easy except for the binomial identity step:
$latex \sum_{i=0}^{\alpha}{\binom{i+j-1}{j-1}\binom{\alpha-i+k-1}{k-1}=\binom{\alpha+j+k-1}{j+k-1}$
After several unfruitful tries to prove it, due to my lack of mathematical skills, I resigned myself to look for information about this problem on the bibliography, this formula look very close to the one found on reference [1] or [2], but with an additional variable, and after many reviews, I was glad to find a combinatorial version of the proof into the book of Chuan Chong Chen and Khee-Meng Koh [3]
This proof solved the problem, but it let me very much unsatisfied, and I begun to rethink about this topic again.
This problem is about Number Theory and not Combinatorics, and I had to revise the first lesson about the properties of Dirichlet Product:
Dirichlet´s functional convolution is associative: we can put the brackets wherever we want, so:
$latex \tau_{j+k}(n)=(\underbrace{I_{0}(n)*...*I_{0}(n)}_{j}) *( \underbrace{I_{0}(n)*...*I_{0}(n)}_{k})=\tau_{j}(n)*\tau_{k}(n)$
This simple line proves this trivial property of the Piltz functions that I pedanticly considered as a Theorem, and it proves, as a tip, the binomial formula. The readers of this blog (if any) may forgive me.
But after all of this mess, I´ve learned many interesting things:
1) Number Theory counts with powerful mathematical tools than can be used for many unexpected purposes, just to mention the relationship between the Piltz functions and the Jacobi polynomials.
2) The properties of arithmetical functions can be used to get elegant proofs for binomial identities. (This is the opposite way that the one I took).
3) In my effort to deal with binomial identities, I discovered some formulas for determinants of matrices with binomial coefficients. (Well, there´s many articles about this topic, but I worked without previous knowledge of them). Anyhow, I haven´t found this formula somewhere but here.
References:
[1]-Matthew Hubbard and Tom Roby - Pascal's Triangle From Top To Bottom -Catalog #: 31000005
[2]-Ronald L. Graham, Donald E. Knuth, and Oren Patashnik (Reading, Massachusetts: Addison-Wesley, 1994 - Concrete Mathematics - Identity (5.26)
[3]-Chuan Chong Chen,Khee-Meng Koh - Principles and techniques in combinatorics page 88-Example 2.6.2-Shortest Routes in Rectangular Grid.
## Monday, 21 December 2009
### SPLITTING FTA FUNCTIONS (II)
PILTZ DIVISOR FUNCTIONS (1)
INTRO:
If we look for an example of "Functions that depend only on coefficients", our first idea should be the divisor function, $latex \tau_{2}(n)$ because it is multiplicative with:
$latex \tau_{2}(p^\alpha)=1+\alpha$
Here, they only appear the coefficients but not the primes.
With the help of recursive Dirichlet Convolution of the unit, $latex I_{0}(n)=1$, it is possible to construct a sequence of arithmetical functions only dependent on the coefficients of the prime factors of any number, known as Piltz Divisor Functions, $latex \tau_{k}(n)$, because they give the number of distinct solutions of the equation $latex x_{1}x_{2} \cdots x_{k}=n$, where $latex x_{1},x_{2},\cdots,x_{k}$ run indepently through the set of positive integers) or, if preferred, they give the number of ordered factorizations of $latex n$ as a product of $latex k$ terms. (References [3],[4] and [11])
DEFINITION:
$latex \displaystyle \tau_{1}(n)=I_{0}(n)=1$
$latex \tau_{k}(n)=\sum_{d|n}^{}{\tau_{k-1}(d)\cdot I_{0}(n/d)}=\sum_{d|n}^{}{\tau_{k-1}(d)}$
This recursion can also be notated in terms of Dirichlet Product as:
$latex (f*g)(n)=\sum_{d|n}{f(d)\cdot g(n/d)}$
$latex \tau_{k}(n)=\tau_{k-1}*I_{0}(n)$
NOTES ON NOTATION:
The divisor function can be found on the literature as: $latex d(n)$, $latex \sigma_{0}(n)$, $latex \tau(n)$, and in this post as $latex \tau_{2}(n)$.
The "$latex \sigma$´s", and "$latex \tau$´s" are two different series of arithmetical functions that share one element in common: The divisor function. With the help of this two notations, it is possible to remark what kind of series we are working with.
On the other hand the "$latex d$", it is a simple notation that can be used for another purposes, were the belonging to this series of functions, does not matters.
Unfortunately, this happens not only with the divisor function, the mathematical notation on arithmetical functions related to Dirichlet convolution (or product) varies from one book to another, and not only distinct functions are named the same, but all cases of "non-biyectivity" between notations and functions can be found.
Hereinafter we are going to use:
$latex I_{k}(n)=n^{k}$ ( like in Reference [3] but with $latex I_{0}(n)=1$ and $latex I_{1}(n)=n$)
The identity element for Dirichlet´s product (or unit function), using Kronecker´s delta notation, is:
$latex \displaystyle \delta_{1n}= \bigg\lfloor \frac{1}{n} \bigg\rfloor$ (Reference [2])
$latex \omega(n)$ means the number of
distinct prime factors of $latex n$
PROPERTY:
$latex \tau_{k}(n)$ is multiplicative because $latex \tau_{k-1}(n)$, and $latex \tau_{1}(n)$ are multiplicative.
This property can also be derived from the behavior of the Dirichlet Product, but we must note that although $latex \tau_{1}(n)=I_{0}(n)=1$ is a completely multiplicative function, its convolution: $latex \tau_{2}(n)$ is multiplicative, but it is not completely multiplicative.
THEOREM-1: [1] and [4]
$latex \displaystyle \tau_{k}(n)=\prod_{i=1}^{\omega(n)}{\prod_{j=1}^{k-1}\frac{\alpha_{i}+j}{j}}=\prod_{i=1}^{\omega(n)}{\binom{\alpha_{i}+k-1}{k-1}}; \; (k \ge 1)$
Proof:
$latex \displaystyle \tau_{k}(p^\alpha)=\sum_{d|p^\alpha}^{}{\tau_{k-1}(d)}=\tau_{k-1}(1)+\tau_{k-1}(p)+\tau_{k-1}(p^2)+ \cdots +\tau_{k-1}(p^\alpha)$
$latex \displaystyle \tau_{k}(p^\alpha)= \sum_{i=0}^{\alpha}{\tau_{k-1}(p^i)} =\sum_{i=0}^{\alpha}{\binom{i+k-2}{k-2}=\binom{\alpha+k-1}{k-1}$
Lemma:
$latex \displaystyle \sum_{i=0}^{\alpha}{\binom{i+k-2}{k-2}}=\binom{\alpha+k-1}{k-1}$
Proof:
From Parallel Summation Identity (References [6] and [8]):
$latex \displaystyle \sum_{k=0}^{n}{\binom{k+r}{k}}=\binom{n+r+1}{n}$
Substituing: $latex \displaystyle n\rightarrow{\alpha}$ and $latex \displaystyle k\rightarrow{i}$
$latex \displaystyle \sum_{i=0}^{\alpha}{\binom{i+r}{i}}=\binom{\alpha+r+1}{\alpha}$
$latex \displaystyle r\rightarrow{k-2}$ and with Pascal´s Symmetry Rule [7]:
$latex \displaystyle \sum_{i=0}^{\alpha}{\binom{i+k-2}{i}}= \sum_{i=0}^{\alpha}{\binom{i+k-2}{k-2}}= \binom{\alpha+k-1}{\alpha}=\binom{\alpha+k-1}{k-1}$
Corollary-1: Values of $latex \tau_{k}(s)$, being $latex s$ a squarefree number.
If $latex s$ is squarefree then all coefficients of its factorization are $latex \displaystyle \alpha_{i}(s)=1$, then:
$latex \displaystyle \tau_{k}(s)= \prod_{i=1}^{\omega(s)}{\binom{k}{k-1}=k^{\omega(s)}$
For a prime $latex p$, $latex \displaystyle \omega(p)=1$, and $latex \displaystyle \tau_{k}(p)=k$, and if $latex s=1$, then $latex \displaystyle \omega(1)=0$ and $latex \tau_{k}(1)=1$
Corollary-2:
$latex \displaystyle \tau_{k+1}(n^k)=\tau_{k}(n^k)\cdot \tau_2(n)$
Proof:
$latex \displaystyle \tau_{k+1}(n)=\tau_{k}(n)\cdot \prod_{i=1}^{\omega(n)}{\frac{\alpha_{i}+k}{k}}$
$latex \displaystyle \tau_{k+1}(n^k)=\tau_{k}(n^k)\cdot \prod_{i=1}^{\omega(n^k)}{\frac{\alpha_{i}+k}{k}}$
Like $latex \displaystyle \omega(n^k)=\omega(n)$, and $latex \displaystyle \alpha_{i}(n^k)=k\cdot\alpha_{i}(n)$:
$latex \displaystyle \tau_{k+1}(n^k)=\tau_{k}(n^k)\cdot \prod_{i=1}^{\omega(n)}{\frac{k \cdot \alpha_i+k}{k}}=\tau_{k}(n^k)\cdot \prod_{i=1}^{\omega(n)}{(\alpha_i+1)}=\tau_{k}(n^k)\cdot \tau_2(n)$
References:
[1]-p.167-Exercise 5.b - Leveque, William J. (1996) [1977]. Fundamentals of Number Theory. New York: Dover Publications. ISBN 9780486689067
[2]-T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, pages 29 and 38
[3]-J. Sándor: On the Arithmetical Functions $latex d_{k}(n)$ and $latex d_{k}^{*}(n)$, Portugaliæ Mathematica 53, No. 1 (1996)
[4]-I.Vinogradov, Fundamentos de la Teoría de los Números, Editorial RUBIÑOS, ISBN 84-2222-210-8, Segunda Edición,chapter II, Exercise 11, page 44
[5]-J. Sándor, B. Crstici, Handbook of Number Theory (Vol II), Kluwer Academic Publishers, Springer, 2004 ISBN 1402025467, 9781402025464
[6]-Ken.J.Ward, Ken Ward's Mathematics Pages, Parellel Summation-Formula 2.2.2
[7]-Matthew Hubbard and Tom Roby - Pascal's Triangle From Top To Bottom -Catalog #: 1000001
[8]-Matthew Hubbard and Tom Roby - Pascal's Triangle From Top To Bottom - Catalog #: 1100002
[9]-A000012-The simplest sequence of positive numbers: the all 1's sequence. The On-Line Encyclopedia of Integer Sequences!
[10]-A000005-d(n) (also called tau(n) or sigma_0(n)), the number of divisors of n. The On-Line Encyclopedia of Integer Sequences!
[11]-A007425-d_3(n), or tau_3(n), the number of ordered factorizations of n as n = rst. The On-Line Encyclopedia of Integer Sequences!
## Friday, 18 December 2009
### NOTES ON LEGENDRE'S FORMULA
MORE FACTS ABOUT PRIME FACTORIZATION OF FACTORIALS.
In the previous post, we introduced these functions, just as a small trick to calculate the limit we were looking for, but unlikely of what they seem to be, they are less artificial than expected.
Legendre´s formula for the exponent of p in the prime factorization of n!:
$latex \displaystyle\alpha(n,p)=\sum _{i=1}^{\lfloor log_p(n)\rfloor}{ \bigg\lfloor\frac{n}{p^i} \bigg\rfloor = \sum _{i=1}^{\infty} { \bigg\lfloor\frac{n}{p^i} \bigg\rfloor$
Integer Approximation for the Legendre's formula:
$latex \displaystyle \alpha^{*}(n,p)=\bigg\lfloor \frac{n}{p-1}\bigg\rfloor = \bigg\lfloor\sum _{i=1}^{\infty}{\frac{n}{p^i}}\bigg\rfloor$
The diference between one function and its approximation is the error function.
Error Function for the Legendre's formula:
$latex \displaystyle \epsilon(n,p)=|\alpha^{*}(n,p) - \alpha(n,p) |= \alpha^{*}(n,p) - \alpha(n,p)$
$latex \displaystyle \epsilon(n,p)= \bigg\lfloor\sum _{i=1}^{\infty}{\frac{n}{p^i}}\bigg\rfloor - \sum _{i=1}^{\infty} { \bigg\lfloor\frac{n}{p^i} \bigg\rfloor$
We can use $latex \lfloor x \rfloor = x - \left\{ x \right\}$ to get another beautiful expression for the error function:
$latex \displaystyle \epsilon(n,p)= \sum _{i=1}^{\infty} { \left\{ \frac{n}{p^i} \right\} - \left\{ \sum _{i=1}^{\infty}{\frac{n}{p^i}} \right\}$
This function shows fractal behavior:
Particular Values for $latex \epsilon(n,p)$:
$latex \epsilon(n,2)=A011371(n)=n-A000120(n)$, References [1] and [2]
$latex \epsilon(2^{n},2)=1$
$latex \epsilon(2^{n}+1,2)=2$
$latex \epsilon(p^{n},p)=0; \; (p > 2)$
$latex \epsilon(p^{n}-1,p)=n$
$latex \epsilon(p^{n}+1,p)=0; \; (p > 3)$
More facts about Legendre´s $latex \alpha(n,p)$
$latex \alpha(n,2)$ gives also the number of 1's in binary expansion of $latex n$ (or the sum of all its binary digits).
And if we extend the range of this formula, been $latex b$ any number not necessarily prime, then:
$latex \displaystyle \alpha(b^{n},b)=\frac{b^{n}-1}{b-1}=R_{n}^{(b)}$
It gives the base $latex b$ repunits, and so for base $latex 2$:
$latex \alpha(2^{n},2)=2^{n}-1=M_{n}$
It gives the Mersenne Numbers.
Amazingly, this uninteresting topic, at first sight, becomes a joint between: Repunits, Mersenne numbers, Factorials, primes, fractals, counting of digits...
Number Theory is it!
References:
[1]-A011371-n minus (number of 1's in binary expansion of n). Also highest power of 2 dividing n!. The On-Line Encyclopedia of Integer Sequences!
[2]-A000120-1's-counting sequence: number of 1's in binary expansion of n (or the binary weight of n). The On-Line Encyclopedia of Integer Sequences!
[3]-Cooper, Topher and Weisstein, Eric W. "Digit Sum." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/DigitSum.html
Archives:
## Saturday, 12 December 2009
### TRAILING ZEROS IN n!
BEHAVIOR OF THE PERCENTAJE OF TRAILING ZEROS IN THE DECIMAL EXPANSION OF n!
It was many and many a year ago, In a kingdom by the sea , I spent a very good time reading an spanish translation of Martin Gardner´s "Mathematical Magic Show" [1] , just because Annabel was not very much interested on me.
In this compilation from Scientific American, Gardner dedicated some pages to this topic in an article called "Factorial Oddities".
Gardner explained how, as $latex n$ increases, $latex n!$ is having more and more factors including the prime factor $latex 5$, that with other factors with any $latex 2$ it gives $latex 10$´s that accumulates in the decimal expansion of $latex n!$ creating a long tail of zeros [2] that fill the least significant digits of this kind of huge numbers.
As it is possible to calculate this number of trailing zeros without having to expand the whole factorial, I wondered (when I was sixteen) if this final zeros were giving very much information about the digits of the whole factorial number or not.
To answer this question, it is necessary to study the behaviour of $latex PTZ(n)$ the percentage between the trailing zeros and the total number of digits of $latex n!$ when $latex n$ tends to infinity.
Number of Digits of n!: [3]
$latex \displaystyle D_{10}(n!)=1+\lfloor log_{10}(n!) \rfloor$
Exponent of p in the prime factorization of n!: (Legendre´s formula)
$latex \displaystyle\alpha(n,p)=\sum _{i=1}^{\lfloor log_p(n)\rfloor}{ \bigg\lfloor\frac{n}{p^i} \bigg\rfloor$
Number of Trailing Zeros in n!. (See sequence [2])
$latex NTZ(n!)=\alpha(n,5)$
PTZ(n!) = Percentage of trailing zeros of the total digits in n! (%)
$latex \displaystyle PTZ(n!)=100*\frac{NTZ(n!)}{D_{10}(n!)}$
But if we want to test the limit of $latex PTZ(n)$ we need to work more handy bounds of this functions, notated with an added asterisk, and that they are going to hold for:
$latex \displaystyle D_{10}^{*}(n!) < D_{10}(n!)$ $latex \displaystyle NTZ^{*}(n!) \ge NTZ(n!)$ $latex \displaystyle PTZ^{*}(n!) > PTZ(n!)$
Approximation for the Number of Digits of n!:
We can sustitute the famous Stirling's approximation instead of $latex n!$ in the formula for $latex D_{10}(n!)$:
$latex \displaystyle n! \approx \sqrt{2\pi n} \left({\frac{n}{e}\right)}^n$
$latex \displaystyle D_{10}(n!)=\left\lfloor \frac{-n+\left(n+\frac{1}{2}\right) \log (n)+\frac{1}{2} \log (2 \pi )}{\log (10)}\right\rfloor +1 +\delta_{n,1}$
The last formula gives the exact value for the number of digits of $latex n!$, for $latex n>0$, because the error for the Stirling´s formula is $latex O\left(\frac{1}{n}\right)$
But for our purposes we must make some changes inside $latex D_{10}(n)$ to get a continuous lower bound:
$latex \displaystyle D_{10}^{*}(n!)=\frac{-n+\left(n+\frac{1}{2}\right) \log (n)+\frac{1}{2} \log (2 \pi )}{\log (10)} < D_{10}(n)$ Approximation for the Legendre's formula and for the Number of Trailing Zeros:
$latex \displaystyle\alpha(n,p)= \sum _{i=1}^{\lfloor log_p(n)\rfloor} { \bigg\lfloor\frac{n}{p^i} \bigg\rfloor = \sum _{i=1}^{\infty} { \bigg\lfloor\frac{n}{p^i} \bigg\rfloor \leq \bigg\lfloor\sum _{i=1}^{\infty}{\frac{n}{p^i}}\bigg\rfloor =\bigg\lfloor \frac{n}{p-1}\bigg\rfloor =\alpha^{*}(n,p)$
$latex NTZ^{*}(n!) =\frac{n}{4} \ge \bigg\lfloor \frac{n}{4}\bigg\rfloor = \alpha^{*}(n,5) \ge \alpha(n,5)=NTZ(n!)$
Final Result and Limit:
$latex \displaystyle PTZ(n!)\approx PTZ^{*}(n!)=100*\frac{NTZ^{*}(n!)}{D_{10}^{*}(n!)}$
$latex \displaystyle\lim_{n \to{+}\infty}{PTZ^{*}(n!)}=0$
$latex PTZ(n!)$ is always positive and it is upper bounded by a continuous function that tend to zero as $latex n$ tends to infinity.
So the number of trailing zeros of $latex n!$ is giving lesser information about the decimal digits of $latex n!$ when the more grows $latex n$
This result may not cause any surprise, but long time ago I had a lot of fun when I was able to prove and plot it.
This ideas does not finish here, but on the contrary there are many many things than can be derived from this introductory point, and that are going to be material for further development in this blog.
References:
[1]-Gardner, M. "Factorial Oddities." Ch. 4 in Mathematical Magic Show: More Puzzles, Games, Diversions, Illusions and Other Mathematical Sleight-of-Mind from Scientific American. New York: Vintage, pp. 50-65, 1978
[2]-A027868-Number of trailing zeros in n! The On-Line Encyclopedia of Integer Sequences!
[3]-A055642-Number of digits in decimal expansion of n. The On-Line Encyclopedia of Integer Sequences!
[5]-Trailing Zero @ Wikipedia
[6]-Stapel, Elizabeth. "Factorials and Trailing Zeroes." Purplemath. Available from
http://www.purplemath.com/modules/factzero.htm
[7]-A061010-Number of digits in (10^n)!. The On-Line Encyclopedia of Integer Sequences!
Archives:
## Saturday, 24 October 2009
### BINOMIAL MATRIX (III)
If $latex A_n$ is an square matrix with elements:
$latex \displaystyle a_{i,j}=\binom{i+j+k}{i}$
Then:
$latex \displaystyle |A_{n}|=\binom{n+k+1}{k+1}=\binom{n+k+1}{n}$
Proof:
The $latex \displaystyle A_{n}$ matrix can be decomposed as the product of a lower triangular matrix, $latex L_{n}$, and an upper triangular matrix, $latex U_{n}$:
$latex \displaystyle A_{n}=L_{n}*U_{n}$
Example:
$latex \displaystyle A_{3}=\left(\begin{array}{ccc} 2+k & 3+k & 4+k \\ \frac{1}{2} (2+k) (3+k) & \frac{1}{2} (3+k) (4+k) & \frac{1}{2} (4+k) (5+k) \\ \frac{1}{6} (2+k) (3+k) (4+k) & \frac{1}{6} (3+k) (4+k) (5+k) & \frac{1}{6} (4+k) (5+k) (6+k) \end{array}\right)$
$latex \displaystyle L_{3}= \left(\begin{array}{ccc} 1 & 0 & 0 \\ \frac{3+k}{2} & 1 & 0 \\ \frac{1}{6} (3+k) (4+k) & \frac{2 (4+k)}{3} & 1 \end{array} \right)$
$latex \displaystyle U_{3}=\left(\begin{array}{ccc} 2+k & 3+k & 4+k \\ 0 & \frac{3+k}{2} & 4+k \\ 0 & 0 & \frac{4+k}{3} \end{array} \right)$
After some trial and error puzzle, we can propose as decomposition:
$latex \displaystyle l_{i,j}=\frac{j}{i} \binom{i+k+1}{j+k+1}$
$latex \displaystyle u_{i,j}=\frac{j+k+1}{j}\binom{j}{i}$
If this decomposition is the correct one then the matrix product should be $latex a_{i,j}$:
$latex \displaystyle \sum_{r=1}^n{l_{i,r} \cdot u_{r,j}=\sum_{r=1}^{min(i,j)}{l_{i,r} \cdot u_{r,j}$
Where:
$latex \displaystyle l_{i,r}=\frac{r}{i} \binom{i+k+1}{r+k+1}$
$latex \displaystyle u_{r,j}=\frac{j+k+1}{j}\binom{j}{r}$
$latex l_{i,r}\cdot u_{r,j}$, can be simplified, changing the binomial coefficient to the gamma function, as:
$latex \displaystyle l_{i,r} \cdot u_{r,j}=\frac{j+k+1}{i}\binom{j-1}{r-1}\binom{i+k+1}{r+k+1}$
$latex \displaystyle \sum_{r=1}^{min(i,j)}{l_{i,r} \cdot u_{r,j}=\sum_{r=1}^{min(i,j)}{\frac{j+k+1}{i}\binom{j-1}{r-1}\binom{i+k+1}{r+k+1}} =$
$latex \displaystyle = \frac{j+k+1}{i} \cdot \sum_{r=1}^{min(i,j)}{\binom{j-1}{r-1}\binom{i+k+1}{r+k+1}} = \frac{j+k+1}{i} \cdot \binom{i+j+k}{j+k+1}=$
$latex \displaystyle=\frac{i}{i}\cdot \binom{i+j+k}{j+k}=\binom{i+j+k}{i}=a_{i,j}$
In the last part of the proof, involving binomial coefficients, we have used: (See ref [1] and [2])
1) $latex \displaystyle \sum_{r}^{}{\binom{l}{r+m} \binom{s}{r+n}}=\binom{l+s}{l-m+n}$
2) $latex \displaystyle r\binom{i}{r}=i \binom{i-1}{r-1}$
Now we can calculate easily $latex |A_n|$, because the LU-decomposition used is the so called Doolittle decomposition, where the matrix $latex L_{n}$ has all ones on its diagonal.
$latex \displaystyle |A_{n}|=\prod_{i=1}^{n}{l_{i,i} \cdot u_{i,i}}=\prod_{i=1}^{n}{u_{i,i}}=\prod_{i=1}^{n}{\frac{i+k+1}{i}}=\binom{n+k+1}{n}$
Archives:
References:
[1]-Ronald L. Graham, Donald E. Knuth, and Oren Patashnik (Reading, Massachusetts: Addison-Wesley, 1994 - Concrete Mathematics - Identity (5.32) in Table 169.
[2]-Matthew Hubbard and Tom Roby - Pascal's Triangle From Top To Bottom - the binomial coefficient website- Catalog #: 3100004.
### THE CLONE WARS
I´m tired of rewritting the same posts again and again due to the problems with the external $latex \LaTeX$ services that work with blogger, so I decided to make some changes in order to be the more self-sufficient as possible.
1) I´ve changed mathtex3.js, the java code that gives the blog $latex \LaTeX$ functionality (In particular some lines inside the Don't MOdify Under THis Line Unless You Know What You Are Doing !! piece of code). I hope this does not cause too much damage inside the Death Star.
2) Now, I´ve uploaded this modified java code, to my web page: (Psychedelic-mathtex.js). Now, this code, depends on me, but in the end, this java program invokes the external servers: http://www.forkosh.dreamhost.com/mathtex.cgi and http://mathcache.appspot.com/
3) With the modifications inside Psychedelic-mathtex.js, I can use the same post indistinctively in Blogger and in WordPress.
4) Now, this blog, has a clon copy in WordPress.com , under the original and unexpected name of Psychedelic Geometry
5) WordPress offers its own $latex \LaTeX$ service, so I expect some stability.
6) I´ve begun, also, another blog about technology, chemistry, and many other things I ignore, called: Psychedelic Thermodynamics
## Thursday, 24 September 2009
### BINOMIAL MATRIX (II)
If $latex A_n$ is an square matrix with elements:
$latex \displaystyle a_{i,j}=\binom{i+k}{j}$
Then:
$latex \displaystyle |A_{n}|=\binom{n+k}{n}$
Proof:
The $latex \displaystyle A_{n}$ matrix can be decomposed as the product of a lower triangular matrix, $latex \displaystyle L_{n}$, and an upper triangular matrix, $latex \displaystyle U_{n}$:
$latex \displaystyle A_{n}=L_{n}*U_{n}$
Example:
$latex \displaystyle A_{3}=\left(\begin{array}{ccc} 1+k & \frac{1}{2} k (1+k) & \frac{1}{6} (-1+k) k (1+k) \\ 2+k & \frac{1}{2} (1+k) (2+k) & \frac{1}{6} k (1+k) (2+k) \\ 3+k & \frac{1}{2} (2+k) (3+k) & \frac{1}{6} (1+k) (2+k) (3+k) \end{array} \right)$
$latex \displaystyle L_{3}=\left(\begin{array}{ccc} 1 & 0 & 0 \\ \frac{2+k}{1+k} & 1 & 0 \\ \frac{3+k}{1+k} & \frac{2 (3+k)}{2+k} & 1 \end{array} \right)$
$latex \displaystyle U_{3}=\left( \begin{array}{ccc} 1+k & \frac{1}{2} k (1+k) & \frac{1}{6} (-1+k) k (1+k) \\ 0 & \frac{2+k}{2} & \frac{1}{3} k (2+k) \\ 0 & 0 & \frac{3+k}{3}\end{array} \right)$
After some math "plumbing", we can propose as decomposition:
$latex \displaystyle l_{i,j}=\frac{i + k}{j + k} \binom{i - 1}{j - 1}$
$latex \displaystyle u_{i,j}=\binom{j - 1}{i - 1} \binom{i + k}{j} {\binom{k + i - 1}{i - 1}}^{-1}$
If this decomposition is correct then the matrix product should be $latex \displaystyle a_{i,j}$:
$latex \displaystyle \sum_{r=1}^n{l_{i,r} \cdot u_{r,j}}=\sum_{r=1}^{min(i,j)}{l_{i,r} \cdot u_{r,j}}$
Where:
$latex \displaystyle l_{i,r}=\frac{i + k}{r + k} \binom{i - 1}{r - 1}$
$latex \displaystyle u_{r,j}=\binom{j - 1}{r - 1} \binom{r + k}{j} {\binom{k + r - 1}{r - 1}}^{-1}$
$latex \displaystyle l_{i,r} \cdot u_{r,j}$, can be simplified, changing the binomial coefficient to the gamma function, as:
$latex \displaystyle l_{i,r} \cdot u_{r,j}=\frac{(i+k)\cdot r}{i \cdot j}\cdot\binom{i}{r}\binom{k}{j-r}$
$latex \displaystyle \sum_{r=1}^{min(i,j)}{l_{i,r} \cdot u_{r,j}}=\sum_{r=1}^{min(i,j)}{\frac{(i+k) r}{i \cdot j}\cdot\binom{i}{r}\binom{k}{j-r}}=\\ \frac{(i+k)}{i \cdot j} \cdot \sum_{r=1}^{min(i,j)}{r\binom{i}{r}\binom{k}{j-r}}$
This last expression can be simplified, with the help of:
$latex \displaystyle r\binom{i}{r}=i \binom{i-1}{r-1}$, the Vandermonde´s convolution (see reference [2]), and the absortion formula (see reference [3]), to:
$latex \displaystyle \sum_{r=1}^{min(i,j)}{l_{i,r} \cdot u_{r,j}}=\binom{i+k}{j}=a_{i,j}$
Now we can calculate easily $latex |A_n|$, because the LU-decomposition used is the so called Doolittle decomposition, where the matrix $latex L_{n}$ has all ones on its diagonal.
$latex \displaystyle |A_{n}|=\prod_{i=1}^{n}{l_{i,i}*u_{i,i}}=\prod_{i=1}^{n}{u_{i,i}}=\prod_{i=1}^{n}{\frac{i+k}{i}}=\binom{n+k}{k}=\binom{n+k}{n}$
Archives:
References:
[1]-John H. Mathews and Kurtis Fink, 2004 - Module for PA = LU Factorization with Pivoting
[2]-Matthew Hubbard and Tom Roby - Pascal's Triangle From Top To Bottom - the binomial coefficient website- Catalog #: 3100003.
[3]-Matthew Hubbard and Tom Roby - Pascal's Triangle From Top To Bottom - the binomial coefficient website- Catalog #: 2400002.
## Tuesday, 22 September 2009
### BINOMIAL MATRIX (I)
If we use yesterday's idea, with little variations, we can create new expressions for matrices with elements related to binomial coefficients, for instance:
Let be $latex U_n$ an square matrix with elements:
$latex \displaystyle u_{i,j}=\binom{j}{i}$
$latex U_n$ is an upper triangular matrix with all its diagonal elements equal to $latex 1$, and similar to Pascal's triangle.(That I prefer to name Tartaglia's Triangle)
Example:
$latex U_{10}= \left( \begin{array}{cccccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 0 & 1 & 3 & 6 & 10 & 15 & 21 & 28 & 36 & 45\\ 0 & 0 & 1 & 4 & 10 & 20 & 35 & 56 & 84 & 120 \\ 0 & 0 & 0 & 1 & 5 & 15 & 35 & 70 & 126 & 210 \\ 0 & 0 & 0 & 0 & 1 & 6 & 21 & 56 & 126 & 252 \\ 0 & 0 & 0 & 0 & 0 & 1 & 7 & 28 & 84 & 210 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 8 & 36 & 120 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 9 & 45 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 10 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{array} \right)$
And, of course:
$latex |U_{n}|=1$
If we multiply this matrix by its transposed one, then we get a symmetrical matrix with:
$latex |A_{n}|=|U_{n}^{T}*U_{n}|=|U_{n}|^2=1$
Example:
$latex A_{10}=U_{10}^{T}*U_{10}$
$latex \displaystyle A_{10}=\\ \\ \left( \begin{array}{cccccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 2 & 5 & 9 & 14 & 20 & 27 & 35 & 44 & 54 & 65 \\ 3 & 9 & 19 & 34 & 55 & 83 & 119 & 164 & 219 & 285 \\ 4 & 14 & 34 & 69 & 125 & 209 & 329 & 494 & 714 & 1000 \\ 5 & 20 & 55 & 125 & 251 & 461 & 791 & 1286 & 2001 & 3002 \\ 6 & 27 & 83 & 209 & 461 & 923 & 1715 & 3002 & 5004 & 8007 \\ 7 & 35 & 119 & 329 & 791 & 1715 & 3431 & 6434 & 11439 & 19447 \\ 8 & 44 & 164 & 494 & 1286 & 3002 & 6434 & 12869 & 24309 & 43757 \\ 9 & 54 & 219 & 714 & 2001 & 5004 & 11439 & 24309 & 48619 & 92377 \\ 10 & 65 & 285 & 1000 & 3002 & 8007 & 19447 & 43757 & 92377 & 184755 \end{array}\right)$
So we have constructed a new matrix with determinant equal to one:
$latex \displaystyle a_{i,j}=\sum_{r=1}^n{u_{i,r}^{*} \cdot u_{r,j}}=\sum_{r=1}^{min(i,j)}{u_{r,i} \cdot u_{r,j}}$
$latex \displaystyle a_{i,j}=\sum_{r=1}^{min(i,j)}{\binom{i}{r} \binom{j}{r}}=-1+\sum_{r=0}^{i}{\binom{i}{r} \binom{j}{r}}=\binom{i+j}{i}-1=\binom{i+j}{j}-1$
Note(i): For a proof on binomial identity see references [1] or [2]
OEIS Related Sequences:
Row/Column Sequence 1 A000027 2 A000096 3 A062748 4 A063258 5 A062988 6 A124089 7 A124090 8 A165618 9 A035927 10 -
References:
[1]-Ronald L. Graham, Donald E. Knuth, and Oren Patashnik (Reading, Massachusetts: Addison-Wesley, 1994 - Concrete Mathematics - Identity (5.32) in Table 169.
[2]-Matthew Hubbard and Tom Roby - Pascal's Triangle From Top To Bottom - the binomial coefficient website- Catalog #: 3100004.
## Monday, 21 September 2009
### BETA FUNCTION MATRIX DETERMINANT
If $latex A_n$ is an square matrix of order $latex n$, whose elements are defined as:
$latex \displaystyle a_{i,j}=\frac{1}{\beta(i,j)}$, where $latex \beta$ is the beta function. Then:
$latex |A_{n}|=n!$
Proof:
If we use Cholesky method, we can decompose this matrix as:
$latex A_{n}=U_{n}^{T}*U_n$
Where $latex U_n$ is an upper triangular matrix.
But instead of applying the algorithm to a generic case, we are going to propose a factorization, and after we will check that this decomposition generates the appropriate matrix. This mean to use software to speed up the proof, like:
$latex \displaystyle A_{5}=\left(\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 2 & 6 & 12 & 20 & 30 \\ 3 & 12 & 30 & 60 & 105 \\ 4 & 20 & 60 & 140 & 280 \\ 5 & 30 & 105 & 280 & 630 \end{array} \right)= U_{n}^{T}* \left( \begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 0 & \sqrt{2} & 3 \sqrt{2} & 6 \sqrt{2} & 10 \sqrt{2} \\ 0 & 0 & \sqrt{3} & 4 \sqrt{3} & 10 \sqrt{3} \\ 0 & 0 & 0 & 2 & 10 \\ 0 & 0 & 0 & 0 & \sqrt{5} \end{array} \right)$
The elements of $latex U_n$, seems to be:
$latex \displaystyle u_{i,j}=\sqrt{i}\cdot \binom{j}{i}$
$latex U_{n}^T$ is the transpose of $latex U_n$, so:
$latex \displaystyle u_{i,j}^{*}=u_{j,i}=\sqrt{j} \cdot \binom{i}{j}$
If we multiply both triangular matrices:
$latex \displaystyle a_{i,j}=\sum_{r=1}^n{u_{i,r}^{*} \cdot u_{r,j}} =\sum_{r=1}^n{r \cdot \binom{i}{r} \binom{j}{r}}=\sum_{r=1}^{min(i,j)}{r \cdot \binom{i}{r} \binom{j}{r}}$
This last binomial expression, can be added to a closed form, equal to the reciprocal of beta function, (See proof on reference [1])
$latex \displaystyle a_{i,j}=i \cdot \binom{i+j-1}{j-1}=\frac{1}{\beta(i,j)}$
Now, that we have found this decomposition, for $latex A_n$, then it is unique and $latex A_n$ is Hermitian and positive definite.
$latex \displaystyle |A_{n}|=|U_{n}^{T}*U_{n}|=|U_{n}|^{2}=\left(\prod_{i=1}^{n}{u_{i,i}}\right)^{2}=\prod_{i=1}^{n}{u_{i,i}^2}=\prod_{i=1}^{n}{\left(\sqrt{i}\cdot\binom{i}{i}\right)^2}=\prod_{i=1}^{n}{i}=n!$
Archives:
References:
[1]-Ronald L. Graham, Donald E. Knuth, and Oren Patashnik (Reading, Massachusetts: Addison-Wesley, 1994 - Concrete Mathematics - page 181 Problem 5
[2]-A000142-Factorial numbers: n! The On-Line Encyclopedia of Integer Sequences!
## Tuesday, 8 September 2009
### CURIOUS SERIES-002
This is the Dirichlet convolution of $latex \displaystyle \omega*1$:
$latex \displaystyle a(n)=\sum_{d|n}{\omega(d)}$
Where $latex \displaystyle \omega$ is the number of distinct prime factors function.
This function, $latex \displaystyle \omega(n)$ is an additive function:
$latex \displaystyle \omega(n)=\omega(d \cdot n/d ) \leq \omega(d)+\omega(n/d)$
Where the equal symbol holds iff $latex GCD(n/d,d)=1$
If we sum over all divisors:
$latex \displaystyle \sum_{d|n}{\omega(n)} \leq \sum_{d|n}{\bigg(\omega(d)+\omega(n/d)\bigg)}=\sum_{d|n}{\omega(d)}+\sum_{d|n}{\omega(n/d)}=2 \sum_{d|n}{\omega(d)}$
$latex \displaystyle \omega(n)\cdot \sum_{d|n}{1}=\omega(n)\cdot\tau_{2}(n) \leq 2 \sum_{d|n}{\omega(d)}=2\cdot a(n)$
If $latex s$ is a squarefree number and if $latex d|s$ then $latex GCD(d,d/s)=1$ and, the number of divisors , $latex \displaystyle \tau_{2}(s)=2^{\omega(s)}$, then:
$latex \displaystyle a(s)=\omega(s)\cdot 2^{\omega(s)-1}$
For every number $latex n>1$:
$latex \displaystyle a(n)=\sum_{d|n}{\omega(d)} \leq \sum_{d|n \; d\neq 1}{\omega(n)}= \omega(n) \cdot \sum_{d|n \; d\neq 1}{1}= \omega(n)\cdot(\tau_{2}(n)-1)$
$latex \displaystyle \omega(n)\cdot(\tau_{2}(n)-1) \geq a(n) \geq \bigg\lceil \frac{\omega(n)\cdot \tau_{2}(n)}{2} \bigg\rceil$
NOTE: Sequences in OEIS:
$latex a(n)=A062799(n)$
$latex \omega(n)=A001221(n)$
$latex \tau_{2}(n)=A000005(n)$
References:
[1]-A062799-Inverse Moebius transform of A001221, the number of distinct prime factors of n The On-Line Encyclopedia of Integer Sequences!
[2]-A001221-Number of distinct primes dividing n (also called omega(n)). The On-Line Encyclopedia of Integer Sequences!
[3]-A000005-d(n) (also called tau(n) or sigma_0(n)), the number of divisors of n.. The On-Line Encyclopedia of Integer Sequences!
## Friday, 28 August 2009
### SPLITTING FTA FUNCTIONS (I)
The Fundamental Theorem of Arithmetic (FTA) grants every natural number, $latex n>1$, a unique factorization of the form:
$latex \displaystyle n=p_1^{\alpha_1}p_2^{\alpha_2}...p_{\omega(n)}^{\alpha_{\omega(n)}} = \prod_{i=1}^{\omega(n)} p_i^{\alpha_i}$
Where $latex \omega(n)$ is the number of distinct prime factors of n.
The arithmetical functions can be evaluated once the factorization of $latex n$ is known, (although there are many of them that can be calculated without factorization)
In fact, the only way to "express some arithmetical property of $latex n$" is that the function, must be dependant on the primes, $latex p_i$ and (or) on the coefficients, $latex \alpha_i$
So the arithmetical functions can be classified, in a psychedelic and unorthodox way of course, in:
1) Functions that depend only on coefficients.
2) Functions that depend only on primes.
3) Functions that depend both on primes and coefficients.
This classification, mathematically speaking, seems to be useless, but it is only an alternative to the alphabetical order, when it comes to deal with this topic.
References:
[1]-D. Joyner, R. Kreminski, J. Turisco @ Applied Abstract Algebra The Fundamental Theorem of Arithmetic
[2]-Arithmetic Function @ Wikipedia Arithmetic Function
[3]-Prime Factor @ Wikipedia Prime Factor
## Thursday, 27 August 2009
### COROLLARY TO EULER´S COROLLARY
It is sure that, after a very short fraction of a second, the first time Euler saw (despite he was one-eyed and partially blind) Gauss´s Gamma Function Multiplication Formula:
$latex \displaystyle \prod _{k=0}^{n-1} \Gamma \left(\frac{k}{n}+z\right)=(2 \pi )^{\frac{n-1}{2}} n^{\frac{1}{2}-n z}\Gamma (n z)$
Euler tested the expresion with $latex \displaystyle z=\frac{1}{n}$ to get his corollary:
$latex \displaystyle \prod _{k=1}^{n-1} \Gamma \left(\frac{k}{n}\right)=\frac{(2 \pi )^{\frac{n-1}{2}}}{\sqrt{n}}$
Or maybe was Gauss who generalized, Legendre´s Gamma Duplication Formula with Euler´s ideas, I haven´t found anything about the real history.
Anyway, if we multiply Euler´s corollary by the Gamma Formula with $latex \displaystyle z=1$, and if we practice the "good habit" of multiplying things by $latex 1$:
$latex \displaystyle \Gamma \left(\frac{n}{n}\right)=\Gamma(1)=\Gamma \left(\frac{2n}{n}\right)=\Gamma(2)=1$
Then we get:
$latex \displaystyle \prod_{k=1}^{2n} \Gamma \left(\frac{k}{n}\right)=\frac{(2 \pi )^{n-1}}{n^n}\Gamma(n)$
References:
[1]-Xavier Gourdon and Pascal Sebah, Introduction to the Gamma Function
[2]-Gamma Function @ Wikipedia Gamma Function
## Sunday, 16 August 2009
### CURIOUS SERIES-001
There´s a very common finite series, that use to be, at the begining on every book:
$latex \displaystyle S_{n}(z)=\sum _{k=0}^n z^k =\frac{z^{n+1}-1}{z-1}$
Where $latex z$ can be real or complex.
There is a, very well known, particular case of this series where $latex z=2$:
$latex \displaystyle S_{n-1}(2)=\sum _{k=0}^{n-1} 2^{k} =2^{n}-1=M_n$
$latex \displaystyle M_n$ are the Mersenne Numbers, and due to this sum, is easy to see that the Mersenne numbers consist of all 1s in base-2 (they are base 2 repunits)
But this entry is about another particular case of this finite sum:
$latex \displaystyle S_{n}(\textbf{i})=\sum _{k=0}^{n} \textbf{i}^k$
Where: $latex \textbf{i}=\sqrt{-1}$, is the complex unit.
$latex \displaystyle S_{n}( \textbf{i} ) =\frac{1}{2}(1+\textbf{i}) \left(1-\textbf{i}^{n+1}\right)$
This sum shows periodical behaviour with a period of $latex 4$, and its values changes from one vertex to another in a square of side equal to $latex 1$, if we plot them in the complex plane:
$latex \displaystyle S_{n}( \textbf{i} )=\{1,1+\textbf{i},\textbf{i},0,1,1+\textbf{i},\textbf{i},...\}$
$latex \displaystyle S_{n}(\textbf{i})=1\;$ if $latex \;n\equiv 0\;mod\;4$
$latex \displaystyle S_{n}(\textbf{i})=1+\textbf{i}\;$ if $latex \;n\equiv 1\;mod\;4$
$latex \displaystyle S_{n}(\textbf{i})=\textbf{i}\;$ if $latex \;n\equiv 2\;mod\;4$
$latex \displaystyle S_{n}(\textbf{i})=0\;$ if $latex \;n\equiv 3\;mod\;4$
If we take a look at the real part of the complex number $latex S_{n}(\textbf{i})$:
$latex \displaystyle Re\bigg(\sum _{k=0}^{n} \textbf{i}^k\bigg)=\{1,1,0,0,1,1,0,0,...\}$
Then we had just found the sequence A133872 from OEIS, and then we can construct another expressions for this sequence, and also for the problem series:
$latex \displaystyle A133872(n)=Re\bigg(\sum _{k=0}^{n} \textbf{i}^k\bigg)$
$latex \displaystyle A133872(n)=\frac{1}{2}\bigg(\sum _{k=0}^{n} \textbf{i}^k + \sum _{k=0}^{n} \textbf{i}^{-k}\bigg)$
$latex \displaystyle A133872(n)=\frac{1}{2}+\frac{1}{2} \text{cos}\left(\frac{n \pi }{2}\right)+\frac{1}{2} \text{sin}\left(\frac{n \pi }{2}\right)$
Then, if we expand to trigonometrical functions $latex S_{n}( \textbf{i} )$:
$latex \displaystyle S_{n}( \textbf{i} ) =\left(\frac{1}{2}+\frac{1}{2} \text{cos}\left(\frac{n \pi }{2}\right)+\frac{1}{2} \text{sin}\left(\frac{n \pi }{2}\right)\right) + \textbf{i} \left( \frac{1}{2}-\frac{1}{2} \text{cos}\left(\frac{n \pi }{2}\right)+\frac{1}{2} \text{sin}\left(\frac{n \pi }{2}\right)\right)$
And finally using the information inside OEIS:
$latex \displaystyle S_{n}( \textbf{i} )= \text{mod}\left(\bigg\lfloor\frac{n+2}{2}\bigg\rfloor,2\right)+ \textbf{i}\cdot \text{mod}\left(\bigg\lfloor\frac{n+1}{2}\bigg\rfloor,2\right)\cdot \textbf{i}$
References:[1]-A133872-Period 4: repeat 1,1,0,0. The On-Line Encyclopedia of Integer Sequences!
## Sunday, 9 August 2009
### SUMMERTIME CHANGES
Now that the Fish are jumpin' and the cotton is high: the $latex \LaTeX$ code renderer this blog was using, is not working in the same way it was.
I´ve discovered today, that some older posts with tables and floor function symbols, are not been displayed properly. As far as I know, the server from http://yourequations.com/ has exceeded its files quota, and they have made some changes that are causing this kind of problems.
So I´ve added a new script that provides very good quality $latex \LaTeX$ rendering:
http://www.watchmath.com
with the added advantage that the code is easier to be included inside any page.
I´ve also updated my Mathematica´s version from 5.2 to 6.0 with some files I´ve found "somewhere" on the internet.
## Saturday, 18 July 2009
### INTEGRATING ROUNDING FUNCTIONS (IV)
FLOOR AND INTEGER PART PRODUCT DEFINITE INTEGRAL:
$latex \displaystyle I_4= \int_0^x \lfloor x \rfloor \left\{x\right\} dx$ $latex \displaystyle x^2=(\lfloor x \rfloor + \left\{x\right\})^2 = {\lfloor x \rfloor}^{2} +2 \lfloor x \rfloor \left\{x\right\} + \left\{x\right\}^2$
$latex \displaystyle \lfloor x \rfloor \left\{x\right\}=\frac{1}{2}(x^2 - \left\{x\right\}^2 - {\lfloor x \rfloor}^{2} )$
$latex \displaystyle I_4=\frac{1}{2} ( \frac{x^3}{3} - I_3 - I_1)$
$latex \displaystyle I_4=\frac{1}{2} ( \lfloor x \rfloor \left\{x\right\}^2 +\frac{{\lfloor x \rfloor}^{2}-\lfloor x \rfloor}{2})$
The same result can be derived just adding the areas under the curve.
## Saturday, 11 April 2009
### SQUARE PYRAMIDAL NUMBERS RECIPROCALS SUM
This post continues with the work on some special sets of integers related series:
The square pyramidal numbers expression can be found on [1] and it is:
$latex \displaystyle P(n)=\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$
The Square pyramidal numbers reciprocal finite serie is:
$latex \displaystyle S(n)=\sum_{k=1}^n{\frac{6}{k(k+1)(2k+1)}}$
Splitting the main fraction into others:
$latex \displaystyle \frac{S(n)}{6}=\sum_{k=1}^n{\frac{1}{k(k+1)(2k+1)}}=\sum_{k=1}^n{\frac{A}{k}+\frac{B}{k+1}+\frac{C}{2k+1}}$
Solving the equations gives:
$latex \displaystyle A=1 ;\; B=1;\; C=-4;$
Substituing the series (ii), and the expression for the Harmonic Numbers (i):
$latex \displaystyle \frac{S(n)}{6}= \sum_{k=1}^n {\frac{1}{k}}+\sum_{k=1}^n{\frac{1}{k+1}}-4\sum_{k=1}^n{\frac{1}{2k+1}}=2H_n-\frac{n}{n+1}-4\sum_{k=1}^n{\frac{1}{2k+1}}$
Taking into account the expressions: (i),(iv) and (vi), for the digamma function:
$latex \displaystyle \frac{S(n)}{6}= 2\gamma+2\psi_{0}(n+1)-\frac{n}{n+1}-4\sum_{k=1}^n{\frac{1}{2k+1}}$
$latex \displaystyle S(n)=6[ 2\gamma+2\psi_{0}(n+1) -\frac{n}{n+1}-2[\psi_{0}(n+\frac{3}{2})+\gamma+2\log{2}-2]]$
$latex \displaystyle S(n)=6[2\psi_{0}(n+1)-2\psi_{0}(n+\frac{3}{2})-\frac{n}{n+1} -4\log{2}+4]$
Calculating the limit, and using (vii):
$latex \displaystyle S(\infty)= \lim_{x \to{+}\infty}{S(n)}= 6(3-4\log{2})=18-24\log{2}\approx 1.3644676665...$
(see reference [4])
NOTES:
(i) Harmonic Numbers and Digamma Function at integer values:
$latex \displaystyle H_{n}=\sum_{k=1}^n {\frac{1}{k}}=\gamma+\psi_{0}(n+1)$
(ii) Changing series into Harmonic Numbers:
$latex \displaystyle \sum_{k=1}^n{\frac{1}{k+1}} =\sum_{k=1}^n{\frac{1}{k}}-1+\frac{1}{n+1}=H_{n}-\frac{n}{n+1}$
(iii) Changing one series into another:
$latex \displaystyle \sum_{k=1}^n{\frac{1}{2k-1}} -\sum_{k=1}^n{\frac{1}{2k+1}}=1-\frac{1}{2n+1}$
(iv) Digamma function at half-integer values:
$latex \displaystyle \psi_{0}(n+\frac{1}{2})= -\gamma-2\log{2}+2\sum_{k=1}^n{\frac{1}{2k-1}}$
(v) A Digamma function property:
$latex \displaystyle \psi_{0}(z+1)= \psi_{0}(z)+\frac{1}{z}$
(vi) Another expression for Digamma function at half-integer values:
Using (iii), (iv) and (v):
$latex \displaystyle \psi_{0}(n+\frac{3}{2}) = \psi_{0}(n+\frac{1}{2}) +\frac{1}{n+\frac{1}{2}}=\psi_{0}(n+\frac{1}{2})+\frac{2}{2n+1}=$
$latex = -\gamma-2\log{2} +2(1-\frac{1}{2n+1}+\sum_{k=1}^n{\frac{1}{2k+1}} )= -\gamma-2\log{2}+2+2\sum_{k=1}^n{\frac{1}{2k+1}}$
(vii) Limit for Digamma function at half-integer values:
$latex \displaystyle L_{1}(\infty)=\lim_{x \to{+}\infty}{L_{1}(n)}= \lim_{x \to{+}\infty}{[\psi_{0}(n+1)-\psi_{0}(n+\frac{3}{2})] }$
$latex \displaystyle L_{1}(n)=\psi_{0}(n+1)-\psi_{0}(n+\frac{3}{2})=2\log{2} + \sum_{k=1}^n {\frac{1}{k}}-2 \sum_{k=1}^n {\frac{1}{2k-1}}$
$latex \displaystyle L_{1}(n)=2\log{2}-\sum_{k=1}^n {\frac{1}{k(2k-1)}}$
The last serie limit can be derived from the Mercator-Mengoli infinite series for $latex \displaystyle \log{2}\;$. See [3].
This proof is interesting enough for another entry on the blog.
$latex \displaystyle \log{2}=\sum_{k=1}^\infty {\frac{1}{2k(2k-1)}}$
So:
$latex \displaystyle L_{1}(\infty)=0$
References:
[1]-Square Pyramidal Numbers @ Wikipedia Square Pyramidal Numbers
[2]-Weisstein, Eric W. "Digamma Function." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/DigammaFunction.html
[3]-Collection of formulae for log 2-Numbers Constants and Computation-Xavier Gourdon and Pascal Sebah.
[4]-A159354-Decimal expansion of 18-24*log(2) The On-Line Encyclopedia of Integer Sequences!
[5]-A000330-Square pyramidal numbers The On-Line Encyclopedia of Integer Sequences!
Archives:
## Tuesday, 24 March 2009
### INTEGRATING ROUNDING FUNCTIONS-(III)
SQUARE FLOOR FUNCTION DEFINITE INTEGRAL:
$latex \displaystyle I_3= \int_0^x \lfloor x \rfloor^2 dx = \int_0^{\lfloor x \rfloor} \lfloor x \rfloor^2 dx+ \int_{{\lfloor x \rfloor}}^ {x} \lfloor x \rfloor^2 dx$
$latex \displaystyle I_3=\sum_{k=1}^{ \lfloor x \rfloor-1}{k^2} + \left\{{x}\right\}\lfloor x \rfloor^2$
$latex \displaystyle I_3=P(\lfloor x \rfloor-1) + \left\{{x}\right\}\lfloor x \rfloor^2$
Where: $latex \displaystyle P(n)$ gives the n-th Square Pyramidal Number.
$latex \displaystyle P(n) =\frac{n(n+1)(2n+1)}{6}$
POWER FLOOR FUNCTION DEFINITE INTEGRAL:
$latex \displaystyle I_4= \int_0^x \lfloor x \rfloor^n dx = \sum_{k=1}^{ \lfloor x \rfloor-1}{k^n} + \left\{{x}\right\}\lfloor x \rfloor^n$
$latex \displaystyle S(n,m)=\sum_{k=1}^{m}{k^n} \;\;\;\;$ is the Faulhaber's formula.
If $latex \displaystyle n=1$ , the formula gives the Triangular Numbers.
And if $latex \displaystyle n=2$ , the formula gives the Square Pyramidal Numbers.
## Monday, 23 March 2009
### INTEGRATING ROUNDING FUNCTIONS-(II)
SQUARE FRACTIONAL PART DEFINITE INTEGRAL:
For any $latex \displaystyle x\geq 0$:
$latex \displaystyle I_1= \int_0^x \left\{{x}\right\}^2 dx = \int_0^{\lfloor x \rfloor} \left\{{x}\right\}^2 dx+ \int_{{\lfloor x \rfloor}}^ {x} \left\{{x}\right\}^2 dx$
The function $latex \displaystyle \left\{{x}\right\}^2$ has a periodical behaviour, so the limits can be reduced to the first period:
$latex \displaystyle I_1= \lfloor x \rfloor \int_0^{1} {\left\{{x}\right\}^2 dx}+ \int_0^{\left\{{x}\right\}}{ \left\{{x}\right\}^2 dx} = \lfloor x \rfloor \int_0^1 x^2 dx+ \int_0^{\left\{{x}\right\}} {x^2 dx}$
$latex \displaystyle I_1= \frac{ \lfloor x \rfloor + \left\{{x}\right\}^3}{3}$
$latex \displaystyle I_1= \int_0^x \left\{{x}\right\}^2 dx= \frac{x+ \left\{{x}\right\}^{3}-\left\{{x}\right\}}{3}$
POWER FRACTIONAL PART DEFINITE INTEGRAL:
The same formula can be extended to any given positive integer power:
$latex \displaystyle I_2= \int_0^x \left\{{x}\right\}^n dx= \frac{x+ \left\{{x}\right\}^{n+1}-\left\{{x}\right\}}{n+1}$
$latex \displaystyle I_2= \frac{ \lfloor x \rfloor + \left\{{x}\right\}^{n+1}}{n+1}$
## Sunday, 22 March 2009
### INTEGRATING ROUNDING FUNCTIONS-(I)
Integer Rounding Functions can be found in many Number Theory texts, but I wasn´t able to find something about its integrals.
The following expressions can be derived just from their plots, adding and subtracting areas. They hold if $latex \displaystyle \;x\geq 0$
The Triangular Numbers function is used to get shorter expressions.
$latex \displaystyle T(n)=\frac{n^{2}+n}{2}$
FLOOR FUNCTION DEFINITE INTEGRAL:
$latex \displaystyle \int_0^x \lfloor x \rfloor dx = \left\{{x}\right\}\lfloor x \rfloor +T(\lfloor x \rfloor-1)$
FRACTIONAL PART FUNCTION DEFINITE INTEGRAL:
$latex \displaystyle \int_0^x \left\{{x}\right\} dx = \frac{x}{2}- \left\{{x}\right\} +T( \left\{{x}\right\} )$
$latex \displaystyle \int_0^x \left\{{x}\right\} dx = \frac{x}{2} +T( \left\{{x}\right\}-1 )$
$latex \displaystyle \int_0^x \left\{{x}\right\} dx =\frac{x+ \left\{{x}\right\}^{2}-\left\{{x}\right\}}{2}$
CEILING FUNCTION DEFINITE INTEGRAL:
$latex \displaystyle \int_0^x \lceil x \rceil dx = \lceil x \rceil (x-\lceil x \rceil)+T(\lceil x \rceil)$
All these formulas can be changed using the relations between them:
$latex \displaystyle \left\{{x}\right\}=x- \lfloor x \rfloor$
This topic doesn't finish here it's going to be used on incoming posts.
Archives:
References:
[1]-Štefan Porubský:Retrieved 2009/3/22 from Interactive Information Portal for Algorithmic Mathematics, Institute of Computer Science of the Czech Academy of Sciences, Prague, Czech Republic http://www.cs.cas.cz/portal/AlgoMath/NumberTheory/ArithmeticFunctions/IntegerRoundingFunctions.htm
[2]-Greg Gamble: The University of Western Australia SCHOOL OF MATHEMATICS & STATISTICS The Floor or Integer Part function
## Thursday, 19 March 2009
### INVERSE POLYGONAL NUMBERS SERIES-Notes
The final result, in the preceeding post, can not be derived from a telescoping series [3], if $latex \displaystyle k$ is not integer (See comments at reference [1]).
$latex \displaystyle \sum_{n=1}^\infty \frac{1}{n(n+k)}=\frac{H_k}{k}$
This lack of generality, can be avoided, if we consider a more general definition for the Harmonic Numbers [4], extended to the complex plane, using the function:
$latex \displaystyle H_z=\gamma+\psi_0(z+1)$
Where $latex \displaystyle \psi_0 \;$ is the so called digamma function, and $latex \displaystyle \;\gamma\;$ is the Euler-Mascheroni constant.
If you take a look at the expresion (15), in the reference [2] : We can find that one asymptotic expansion for the digamma function is:
$latex \displaystyle \psi_0(k+1)=-\gamma+\sum_{n=1}^\infty{\frac{k}{n(n+k)}$
This is why the Polygonal Numbers Series sum is working:
$latex \displaystyle H_k=\gamma-\gamma+ \sum_{n=1}^\infty{\frac{k}{n(n+k)}$
$latex \displaystyle \frac{H_k}{k}=\sum_{n=1}^\infty{\frac{1}{n(n+k)}=\frac{\gamma+\psi_0(k+1)}{k}$
And the polygonal numbers infinite sum, can be expressed (if $latex \displaystyle \;s\neq4\;$) as:
$latex \displaystyle S_{\infty}(s)=\frac{2}{4-s}*(\gamma+\psi_{0}\left(\frac{2}{s-2}\right))$
This expresion works for all $latex \displaystyle s>2$, as well as for all nonreal $latex \displaystyle s$, It also works for all $latex \displaystyle s<2$, except if $latex \displaystyle s<2$, and $latex \displaystyle s$ is $latex \displaystyle \;\;0, 1, 4/3, 6/4, 8/5, 10/6, ... \;$, because $latex \displaystyle \;\psi_0\;$ is not defined for negative integers (See reference) [1]
References:
[1]-Charles R Greathouse IV - Comments @ My Math Forum Inverse Polygonal Series
[2]-Weisstein, Eric W. "Digamma Function." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/DigammaFunction.html
[3]-Telescoping Series @ Wikipedia Telescoping Series
[4]-Sondow, Jonathan and Weisstein, Eric W. "Harmonic Number." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/HarmonicNumber.html
[5]-Photo Martin Gardner, Mathematical Games, Scientific American, 211(5):126-133, taken from http://bit-player.org/2007/hung-over
## Tuesday, 10 March 2009
### INVERSE POLYGONAL NUMBERS SERIES
Let there be $latex \displaystyle P(n,s)$ the n-th Polygonal Number [4] of s sides, then:
$latex \displaystyle P(n,s)=\frac{ (s-2)*n^{2}-(s-4)*n }{2}$
For s=3, we get the formula for Triangular Numbers
$latex \displaystyle P(n,3)=\frac{n^{2}+n}{2}=T_n$
and with s=4 then we get the Squares:
$latex \displaystyle P(n,4)=n^{2}$
Polygonal numbers hold for the next identity:
$latex \displaystyle P(n,s+1)=P(n,s)+ P(n-1,3)=P(n,s)+ T_{n-1}$
The sum of the inverse of the first k polygonal numbers with side s, is:
$latex \displaystyle S_{k}(s)=\sum_{n=1}^k{} \frac{1}{P(n,s)}$
And its infinite series sum:
$latex \displaystyle S_{\infty}(s)=\sum_{n=1}^{\infty{}} \frac{1}{P(n,s)}$
Than this infinite series is convergent, can be proved, using some convergence test:
Raabe's convergence Test:
$latex \displaystyle \rho=\displaystyle\lim_{n \to{+}\infty}{n*\left(\frac{P(n+1,s)}{P(n,s)}-1\right) }=2>1$
A series with a lower number of sides, upper bounds, series with higher sides: So the convergence in triangular numbers, implies the convergence of the remaining polygonal numbers series:
If $latex \displaystyle s_1>s_2 \longrightarrow{} \frac{1}{P(n,s_1)}<\frac{1}{P(n,s_2)} \longrightarrow{} S_{\infty}(s_1)$
The inverse series with Triangular numbers is a telescoping series [3]:
$latex \displaystyle S_{\infty}(3)=2\cdot\sum_{n=1}^{\infty{}} \frac{1}{n\cdot(n+1)}= 2$
The squares sum is the so called Basel Problem, [1] [2], related with the Riemann Zeta Function:
$latex \displaystyle S_{\infty}(4)=\sum_{n=1}^{\infty{}} \frac{1}{n^2}=\zeta(2)=\frac{\pi^2}{6}$
If $latex \displaystyle s\neq4$, then:
The series is a more general case of a telescoping series [3], related with the Harmonic Numbers.
$latex \displaystyle S_{\infty}(s)=2\cdot \sum_{n=1}^{\infty{}} \frac{1}{(s-2)\cdot n^{2}-(s-4)*n}=\frac{2}{s-2}\cdot\sum_{n=1}^{\infty{}}\frac{1}{n\cdot(n+\frac{4-s}{s-2})}$
$latex \displaystyle \sum_{n=1}^{\infty{}} \frac{1}{n*(n+k)}=\frac{H_k}{k}$
$latex \displaystyle S_{\infty}(s)= \frac{2}{4-s}\cdot H_{\frac{4-s}{s-2}}$
Example: In the hexagonal numbers case:
$latex \displaystyle S_{\infty}(6)= -H_{-\frac{1}{2}} \approx 1.38629$
Archives:
References:
[1]-Estimating Basel Problem@ MAA Online How Euler did it, Ed.Sandifer
[2]-Basel Problem @ Wikipedia Basel Problem
[3]-Telescoping Series @ Wikipedia Telescoping Series
[4]-Polygonal Numbers @ Wikipedia Polygonal Number
## Tuesday, 17 February 2009
### MERSENNE NUMBERS TREASURE MAP
The Mersenne generating function splits the integer set in some subsets:
$latex \displaystyle \mathbb{N}\longrightarrow\mathbb{N}$
$latex \displaystyle f(n)\longrightarrow{2^{n}-1=M_n}$
1-INTEGERS PARTITION SET
$latex \displaystyle Composites =\{4,6,8,9,10,12,...\}; \;[2]$
$latex \displaystyle Primes =\{2,3,5,7,11,13,...\}; \;[3]$
$latex \displaystyle Square Free =\{1,2,3,5,6,7,10,...\}; \;[4]$
$latex \displaystyle Integers =\{0,1,2,3,4,5,6,...\}=\mathbb{N}$
$latex \displaystyle \{0,1\} \cup Primes \cup Composites=Integers$
$latex \displaystyle Primes \subset Square Free$
$latex \displaystyle (Square Free-Primes-\{1\}) \subset Composites$
$latex \displaystyle Primes \cap Composites =\emptyset \\$
2-RANGE PARTITION SET
$latex \displaystyle f(0)=0$
$latex \displaystyle f(1)=1$
$latex \displaystyle f(Primes)\cap f(Composites) = \emptyset$
$latex \displaystyle f(Primes) \subset (Square Free \; \cup$ Composite factors of unknown Wieferich Primes $latex )$
$latex \displaystyle f(Primes) \cap Primes = \textrm{Prime Mersenne Numbers}$
$latex \displaystyle f(Composites)\subset Composites$
$latex \displaystyle f(Composites)\cap Primes=\emptyset$
$latex \displaystyle f(Square Free)\subset (Composites \cup \{ 1 \} )$
If $latex \displaystyle n=a*b$ is a composite number, then $latex \displaystyle M_{n}=M_{a*b}$ is also composite, because:
$latex \displaystyle M_{a*b}=2^{a*b}-1=(2^{a}-1)\cdot (1+2^a+2^{2a}+\dots+2^{(b-1)a})=$
$latex \displaystyle M_{a}*\sum_{i=1}^b{2^{(b-i)\cdot a}}$
And also if $latex \displaystyle d|n$ , and if $latex \displaystyle M_d$ , is not squarefree, then $latex M_n$, can not be squarefree [8].
The only known Wieferich primes are 1093 and 3511, but they can not be prime factors of a Mersenne prime, see [6] and [7].
Note: See $latex \displaystyle \frac{M_{n^2}}{M_n}=\sum_{i=1}^n{2^{(n-i)\cdot n}}$ on link [5]
Archives:
References:
[1]- N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences.
A000225: 2^n - 1. (Sometimes called Mersenne numbers, although that name is usually reserved for A001348.)
[2]- N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences.
A002808: The composite numbers: numbers n of the form x*y for x > 1 and y > 1.
[3]- N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences.
A000040: The prime numbers.
[4]- N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences.
A005117: Squarefree numbers: numbers that are not divisible by a square greater than 1.
[5]- Leroy Quet Apr 19 2007, The On-Line Encyclopedia of Integer Sequences.
A128889: a(n) = (2^(n^2) -1) /(2^n -1).
[6]- Labos E., The On-Line Encyclopedia of Integer Sequences.
A049094: 2^n - 1 is divisible by a square >
[7]-Wieferich primes and Mersenne primes Miroslav Kures, Wieferich@Home - search for Wieferich prime.
[8]-Pacific J. Math. Volume 22, Number 3 (1967), 563-564 Henry G. Bray and Leroy J. Warren.
## Monday, 16 February 2009
### FERMAT AND MERSENNE NUMBERS CONJECTURE-(3)
(2)-CONSTRUCTING SOME FUNCTION ZEROS (ii)
(2.5) POWERS OF A PRIME * ODD SQUAREFREE:
$latex \displaystyle f(N)=f(p_{1}^{k}*M)=f(p_{1}^{k}*p_2*...*p_n)=0\;\rightarrow (k+1)|p_{1}^{k-1}* \phi\left(\frac{N}{p_{1}^{k-1}}\right)$, then the zeros can fall into two cases:
(2.5.1) Case: $latex \displaystyle (k+1)| \phi\left(\frac{N}{p_{1}^{k-1}}\right)$
Then if $latex \displaystyle d_i| \phi\left(\frac{N}{p_{1}^{k-1}}\right)$, $latex \displaystyle f(p_{1}^{d_{i}-1}*M)=0$, and we can built $latex \displaystyle \sigma_{0} \left(\frac{N}{p_{1}^{k-1}}\right)$, zeros,
one for every divisor of the product applied to every distinct prime factor
$latex \displaystyle \prod_{i=1}^{i=n}{ (p_{i}-1)}$ of $latex \displaystyle N$
Note that, in the particular case:
$latex \displaystyle (k+1)=(p_{i}-1) \rightarrow{k=p_{i}-2}\rightarrow{f(p_{1}^{p_{i}-2}*M)=0}$.
And $latex \displaystyle f(p_{1}^{p_{1}-1}*M)=Mod(p_{1}^{p_{1}-2}*\phi\left(\frac{N}{p_{1}^{k-1}}\right),p_{1})=0$.
(2.5.2) Case: $latex \displaystyle (k+1)|p_{1}^{k-1}$
$latex \displaystyle f(p_{1}^{p_{1}^{n}-1})=0$
## Saturday, 14 February 2009
### FERMAT AND MERSENNE NUMBERS CONJECTURE-(2)
(2)-CONSTRUCTING SOME FUNCTION ZEROS
$latex f(n)=Mod( \phi(n),\sigma_0(n))$
(2.1) PRIMES:
$latex \displaystyle f(p)=Mod(p-1,2)=0$, holds for every odd prime $latex p \in \mathbb{P} -\{2\}$.
$latex f(2)=1$
(2.2) PRODUCT OF DISTINCT PRIMES NOT 2:
$latex \displaystyle f(p_1*p_2*...*p_n)=0$, because $latex \displaystyle \sigma_0(p_i)|\phi(p_i) \rightarrow {2 |(p_i-1)}$, always holds if $latex \displaystyle p_{i} \neq 2$
With $latex \displaystyle k$, distinct primes, none of them equal two, it is possible to combine them in $latex \displaystyle 2^n$, products to find $latex \displaystyle 2^n$ zeros.
This set of zeros can be described as odd squarefree numbers [1].
(2.3) POWERS OF 2:
$latex \displaystyle f(2^k)=0\;\rightarrow (k+1)|2^{k-1}$, so $latex \displaystyle k+1=2^n$ must be a power of 2:
$latex \displaystyle k=2^n-1=M_n$
A power of 2, is a function zero, iff the exponent is a Mersenne number.
$latex \displaystyle f(2^{M_n})=f(2^{2^{n}-1})=0$
(2.4) POWERS OF A PRIME:
$latex \displaystyle f(p^k)=0\;\rightarrow (k+1)|(p-1)*p^{k-1}$, then the zeros can fall into two cases:
(2.4.1) Case: $latex \displaystyle (k+1)|(p-1)$
Then if $latex \displaystyle d|(p-1)$, $latex \displaystyle f(p^{d-1})=0$, and we can built $latex \displaystyle \sigma_{0}(p-1)$, zeros, one for every divisor of $latex \displaystyle (p-1)$.
Note that, in the particular case:
$latex \displaystyle (k+1)=(p-1) \rightarrow{k=p-2}\rightarrow{f(p^{p-2})=0}$.
And $latex \displaystyle f(p^{p-1})=Mod((p-1)*p^{p-2},p)=0$.
(2.4.2) Case: $latex \displaystyle (k+1)|p^{k-1}$
This is more general than (2.3):
$latex \displaystyle f(p^{p^{n}-1})=0$
Archives:
References:
[1]-CRCGreathouse at My Math Forum/Number Theory: Mersenne and Fermat Numbers congruence
## Sunday, 8 February 2009
### FERMAT AND MERSENNE NUMBERS CONJECTURE-(1)
(1)-INITIAL IDEAS:
Playing, like always, with my computer, I´ve been plotting this function, that includes the Euler totient function, and the Divisor function.
$latex \displaystyle f(n)=Mod( \phi(n),\sigma_0(n))\: , \; n\in\mathbb{N_{*}^{+}}$
The first 25 values of $latex \displaystyle f(n)$, (not in OEIS) are:
$latex \displaystyle \{0, 1, 0, 2, 0, 2, 0, 0, 0, 0, 0, 4, 0, 2, 0, 3, 0, 0, 0, 2, 0, 2, 0, 0, 2,...\}$
Applying $latex \displaystyle f(x)$ to the Fermat Numbers, $latex \displaystyle F_n=2^{2^{n}}+1$, and to the Mersenne Numbers, $latex \displaystyle M_n=2^n-1$, we can conjecture the following congruences:
(1) $latex \displaystyle \phi(F_n) \equiv 0\; (mod \; \sigma_0(F_n))$
(2) $latex \displaystyle \phi(F_n-2) \equiv 0\; (mod \; \sigma_0(F_n-2))$
(3) $latex \displaystyle \phi(M_n) \equiv 0\; (mod \; \sigma_0(M_n))$
(4) $latex \displaystyle \phi(M_n+2) \equiv 0\; (mod \; \sigma_0(M_n+2))$
¿How do they can be proved? [1]
Archives:
References:
[1]-My Math Forum/Number Theory: Mersenne and Fermat Numbers congruence
## Friday, 6 February 2009
### SUMS INSIDE POWER SET
Let's consider the set of integers less or equal than a given one:
$latex \displaystyle A=\{1,2,3,...,n\}\:\:n\in\mathbb{N}$
If we add all the elements in this set, we have:
$latex \displaystyle S=\sum_{i\in A}{i}=\sum_{i=1}^n{i}=\frac{n^2+n}{2}$
If we look, at the figure, we can see how the sum ,is equal to the area below the "ladder" plotted:
$latex \displaystyle S=S_{(i)}+S_{(ii)}$
Where $latex S_{(i)}$ is the area formed for $latex n$ small triangles, with a $latex \displaystyle \frac{1}{2}$ area, each one; And $latex S_{(ii)}$ is the area of an isosceles triangle with an equal base and height of $latex n$.
$latex \displaystyle S_{(i)}=\frac{1}{2}\cdot n;\quad S_{(ii)}=\frac{1}{2}\cdot n \cdot n;$
And if the aim, of this blog, where to be rigurous instead of, to give simple ideas, an induction proof, should fit here, perfectly [1].
To extend this result, to the sum of all the elements, in the subsets included, in the Power Set of integers less or equal to a given one:
$latex \displaystyle P(A) = \{X:X\subseteq A=\{1,2,3,...,n\}\}$
There are $latex 2^n$ subsets in $latex \displaystyle P(A)$, (see [3] and [4]):
$latex \displaystyle |P(A)| = 2^{|A|} =2^n$
If $latex \displaystyle i\in A$, then this element is in half of the subsets of $latex \displaystyle P(A)$ (observe the relation between the binary digits and the power set [3]): $latex \displaystyle 2^{n-1}$
To get the final result, it is only necessary to multiply both expresions [2]:
$latex \displaystyle h(n)=\sum_{i\in X\subseteq P(A)}{i}= \frac{n^2+n}{2} \cdot 2^{n-1}=(n^2+n) \cdot 2^{n-2}$
NOTE:
$latex \displaystyle h(n)=A001788(n)$
$latex \displaystyle\lim_{x \to{+}\infty}{\frac{h(n+1)}{h(n)}}\displaystyle\lim_{x \to{+}\infty}{2+\frac{4}{x}}=2$
Archives:
References:
[1]-A000217 The On-Line Encyclopedia of Integer Sequences!
[2]-A001788 The On-Line Encyclopedia of Integer Sequences!
[3]-Powerset Construction Algorithm Shriphani Palakodety.
[4]-Course Notes 8: Size of the Power Set. Chris Nowlin.
## Monday, 2 February 2009
### n! ARITHMETIC FUNCTIONS
Theorem:
If $latex p_j$ is prime, then $latex \displaystyle \alpha_j(n)=\sum_{i=1}^\infty \bigg\lfloor\frac{n}{p_j^i}\bigg\rfloor$ is the exponent of $latex p_j$ appearing in the prime factorization of $latex n!$
Proof: (see [1] page 104, and [3])
But the infinity summation is applied to a infinity of zeros, because when $latex \displaystyle \frac{n}{p_j^i}<1$, the terms vanish.
So the bigger exponent, $latex m_j$, that makes the terms not null is:
$latex \displaystyle \\n{<}p_j^{m_j}; \rightarrow \frac{\log{n}}{\log{p_j}}{<}m_j; \rightarrow m_j=\bigg\lfloor\frac{\log{n}}{\log{p_j}}\bigg\rfloor; (m_j\in\mathbb{N})$
$latex \displaystyle \alpha_{j}(n)=\sum_{i=1}^{\big\lfloor\frac{\log{n}}{\log{p_j}}\big\rfloor}{\bigg\lfloor\frac{n}{p_j^i}\bigg\rfloor};$
Now we can calculate any arithmetic function, like, e.g., the divisor sigma:
$latex \displaystyle \sigma_{0}(n!)=\prod_{j=1}^{\pi(n)} (\alpha_j(n)+1)$
References:
[1]-Number Theory George E. Andrews - Courier Dover Publications, 1994 - ISBN 0486682528, 9780486682525
[2]-ASYMPTOTIC ORDER OF THE SQUARE-FREE PART OF N! Kevin A. Broughan - Department of Mathematics, University of Waikato, Hamilton, New Zealand
[3]-FUNDAMENTOS DE LA TEORÍA DE LOS NÚMEROS - I. Vinogradov - Editorial MIR
## Sunday, 1 February 2009
### EUCLINACCI - [1]
Maybe the roughest lesson, someone can ever receive, is this:
$latex 1+1=2$
The implications related to that simple line, can fill a whole library: One of them, is the, so called, Fibonacci sequence:
$latex \{0,1, 1, 2, 3, 5, 8, 13, 21, 34, 55,...\}$
Defined as:
$latex \displaystyle f_n = \left\{ \begin{array}{lcr} f_0=0;\\ f_1=1;\\ f_n=f_{n-1}+f_{n-2}; \left n \ge 2\\ \end{array}$
Tabulating the ratio between two consecutive terms of the sequence, appear on insight, two properties:
1) This ratio has a finite limit.
2) Two consecutive Fibonacci numbers are
relatively prime.
$latex \begin{array}{rrc} \hline {n} & f_n & f_{n}/f_{n-1} \\ \hline 0 & 0 & -\\ 1 & 1 & -\\ 2 & 1 & 1.0000 \\ 3 & 2 & 2.0000 \\ 4 & 3 & 1.5000 \\ 5 & 5 & 1.6667 \\ 6 & 8 & 1.6000 \\ 7 & 13 & 1.6250 \\ 8 & 21 & 1.6154 \\ 9 & 34 & 1.6190 \\ 10 & 55 & 1.6176 \\ 11 & 89 & 1.6182 \\ 12 & 144 & 1.6180 \\ 13 & 233 & 1.6181 \\ 14 & 377 & 1.6180 \\ 15 & 610 & 1.6180 \\ 16 & 987 & 1.6180 \\ \hline \end{array}$
To prove the first property, we have:
$latex \displaystyle I = \lim_{x \to +\infty} \frac{f_n}{f_{n-1}} = \lim_{x \to +\infty} \frac{f_{n-1}+f_{n-2}}{f_{n-1}} = \\ 1 + \lim_{x \to +\infty} \frac{f_{n-2}}{f_{n-1}} = 1 + \lim_{x \to +\infty} \frac{f_{n-1}}{f_n}; I=1+\frac{1}{I} \Rightarrow I=1+\frac{1}{I}$
With positive solution, equal to the golden ratio.
$latex \displaystyle I=\frac{1+\sqrt{5}}{2}=\Phi$
The proof, of this second property, used to be by induction or contradiction [2], but this can also be proved using the oldest procedure designed to calculate de greatest common divisor, GCD, of two integers, known as The Euclidean Algorithm
The Euclidean method constructs a decreasing sequence of integers who share the same GCD.
$latex (a,b)=(r_0,r_1)=(r_2,r_3)=...=(r_{n-1},r_n)=r_n$
$latex \left \{ \begin{array}{l} a=r_0; \\ b=r_1; \\ r_{i+1}=r_{i-1}-r_i*\bigg \lfloor\frac{r_{i-1}}{r_i}\bigg\rfloor;\\ \end{array}$
if $latex r_{i+1}=0$ then $latex (a,b)=r_i$
Where the function $latex \displaystyle \lfloor x \rfloor$ is the Floor Function, see [3].
Proof:
$latex \displaystyle (f_{n},f_{n-1})=(r_0,r_1)$
$latex \displaystyle r_2=r_0-r_1*\bigg\lfloor\frac{r_0}{r_1}\bigg\rfloor$
$latex \displaystyle r_2=f_n-f_{n-1}*\bigg\lfloor\frac{f_n}{f_{n-1}}\bigg\rfloor=f_n-f_{n-1}*\bigg\lfloor\frac{f_{n-1}+f_{n-2}}{f_{n-1}} \bigg\rfloor$
$latex \displaystyle r_2=f_n-f_{n-1}*\bigg\lfloor 1+\frac{f_{n-2}}{f_{n-1}}\bigg\rfloor =f_n-f_{n-1}*\bigg(1+\bigg\lfloor\frac{f_{n-2}}{f_{n-1}}\bigg\rfloor\bigg)$
$latex \displaystyle f_{n-2}\le f_{n-1} \implies \bigg\lfloor\frac{f_{n-2}}{f_{n-1}}\bigg\rfloor=0$
$latex \displaystyle r_2=f_n-f_{n-1}=f_{n-2}$
$latex \displaystyle(f_{n},f_{n-1})=(r_0,r_1)=(r_1,r_2)=(f_{n-1},f_{n-2})=...=(2,1)=1$
The Euclidean Algorithm reproduces all Fibonacci´s sequence and proves that two consecutive terms are relatively prime.
References:
[1]-Fundamentals of Number Theory William J. LeVeque - Courier Dover Publications, 1996 - ISBN 0486689069, 9780486689067
[2]-Consecutive Fibonacci Numbers Relatively Prime - The Math Forum@Drexel
[3]Weisstein, Eric W. "Floor Function." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/FloorFunction.html
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# How many even numbers less than 500 can be formed using the digits 1,2,3,4,5? Each digit can only be used once.
Dec 19, 2017
$28$
#### Explanation:
The number must end with $2$ or $4$, yielding two choices. Then the remaining digits are $1$, $3$, $5$ and an even digit less than $5$.
• Three digit numbers: Having chosen $2$ or $4$ for the last digit, there are $3$ admissible possibilities for the first digit, namely $1$, $3$ or whichever of $2$ or $4$ is left. Finally, any of the $3$ remaining digits can be used for the middle digit. So there are $2 \times 3 \times 3 = 18$ possible $3$ digit numbers.
• Two digit numbers: Having chosen $2$ or $4$ for the last digit, there are $4$ possibilities for the first digit, so a total of $2 \times 4 = 8$ possibly $2$ digit numbers.
• One digit numbers: $2$ or $4$, which makes $2$ possibilities.
So there are a total of $18 + 8 + 2 = 28$ possible numbers satisfying the criteria.
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# Trigonometric Ratios of Multiple and Substitute Angles
## Trigonometric Ratios of Multiple and Substitute Angles
1. Sin2A = 2sinA cosA
2. Cos2A = cos²A – sin²A
3. Cos2A = 2cos²A – 1 (or) 1 + cos2A = cos²A
4. Cos2A = 1 – 2sin²A (or) 1 – cos2A = 2cos²A
5. $$\,\tan 2A=\frac{2\tan A}{1-{{\tan }^{2}}A}$$
6. $$\sin 2A=\frac{2\tan A}{1+{{\tan }^{2}}A}$$
7. $$\,\cos 2A=\frac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}$$
The above relation are true for all values of angle A. replacing A by A/2 in the above relation, we obtain the following formulae.
1. sin A = 2 sin A/2 cos A/2
2. cos A = cos² A/2 – sin² A/2
3. cos A = 2 cos² A/2 – 1 or 1 + cos A = 2 cos² A/2
4. cos A = 1 – 2 sin² A/2 or 1 – cos A = 2 sin² A/2
5. $$\tan A=\frac{2\tan \frac{A}{2}}{1-{{\tan }^{2}}\frac{A}{2}}$$
6. $$\sin A=\frac{2\tan \frac{A}{2}}{1+{{\tan }^{2}}\frac{A}{2}}$$
7. $$\cos A=\frac{1-{{\tan }^{2\grave{\ }}}\frac{A}{2}}{1+{{\tan }^{2}}\frac{A}{2}}$$
Example: Find the $$\sqrt{2+\sqrt{2+\sqrt{2+2\cos \theta }}}$$
Solution: We have,
=$$\sqrt{2+\sqrt{2+\sqrt{2\left( 1+\cos \theta \right)}}}$$
= $$\sqrt{2+\sqrt{2+\sqrt{4{{\cos }^{2}}4\theta }}}$$
= $$\sqrt{2+\sqrt{2\left( 1+\cos 4\theta \right)}}$$
= $$\sqrt{2+\sqrt{4{{\cos }^{2}}2\theta }}$$
= $$\sqrt{2+2\cos 2\theta }$$
= $$\sqrt{2\left( 1+\cos 2\theta \right)}$$
Example: Prove that cos²A + cos² (A + 120) + cos² (A – 120) = 3/2
Solution: ½ [2cos²A + cos² (A + 120) + cos² (A – 120)]
= ½ [1 + cos2A + (1 + cos2 (A + 120)) + (1 + cos2 (A – 120))]
= ½ [3 + cos2A + cos2 (A + 120) + cos2 (A – 120))]
= ½ [3 + cos2A + 2cos2A cos240]
= ½ [3 + cos2A + 2 cos 2A (- ½)]
= 3/2
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# Let a, b, c, d, e be natural numbers in an arithmetic progression such that $a+b+c+d+e$ is the cube of an integer and b + c + d is square of an integer. The least possible value of the number of digits of c is?(a) 2(b) 3(c) 4(d) 5
Last updated date: 21st Jul 2024
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Hint: We know that in an arithmetic series, the numbers are in such sequence that the difference between two consecutive terms is constant throughout the series.
Let a, b, c, d, e be natural numbers in an arithmetic progression. We know that arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between two consecutive terms is constant.
Let us assume that the common difference between them is D and the first term of series is A.
So according to our assumption, the numbers can be written as,
a = A – 2D
b = A – D
c = A
d = A + D
e = A + 2D
We know that it is given in the question that $a+b+c+d+e$ is the cube of an integer. So, let us suppose that the integer is p.
So, we can write that,
$a+b+c+d=e={{p}^{3}}....\left( i \right)$
Also, it is given in the question that b + c + d is the square of an integer. So, let us suppose that the integer is q.
So, we can write that,
$b+c+d={{q}^{2}}....\left( ii \right)$
Now we can substitute the values of A, B, C and D in equation (i) as shown below,
$A-2D+A-D+A+A+D+A+2D={{p}^{3}}$
On simplifying the above equation, we can observe that the variable D will get cancelled out and we will be left with the variable A in the left-hand side of the equation.
$5A={{p}^{3}}....\left( iii \right)$
Now substituting the value of B, C and D in equation (ii), we will get
$A-D+A+A+D={{q}^{2}}$
Again, on simplifying the above equation, we can observe that the variable D will get cancelled out and we will be left with variable A in the left-hand side of the equation.
$3A={{p}^{2}}....\left( iv \right)$
Now we have to divide equation (iv) by equation (iii). We will get,
$\dfrac{5A}{3A}=\dfrac{{{p}^{3}}}{{{q}^{2}}}$
We can observe that variable A is cancelled out and we will be left with
$\dfrac{5}{3}=\dfrac{{{p}^{3}}}{{{q}^{2}}}$
$\dfrac{{{p}^{3}}}{5}=\dfrac{{{q}^{2}}}{3}$
It is said in the question that we have to find the least possible value of the number of digits of c.
So by hit and trial method, the least possibility is
\begin{align} & p=5\times 3 \\ & q=5\times 3\times 3 \\ \end{align}
As it will satisfy the equation, we have $p=15\text{ and }q=45$.
Now we know that
$5A={{p}^{3}}$
So putting the value of p = 15 in it, we will get
\begin{align} & 5A={{15}^{3}} \\ & 5A=3375 \\ & A=675 \\ \end{align}
Since we have the value of c = A, we get that c = 675.
It consists of 3 numbers of digits, so the answer to the question is 3. Hence, we get option b as the correct answer.
Note: We can approach this question with a different method. We can take elements of the A.P series as $a,a+d,a+2d...$ instead of $a=A-2D,\text{ }b=A-D,c=A,d=A+D,e=A+2D$. But, this method will increase the number of unknown variables in the equation and make it difficult to solve the question.
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# How do you simplify 2-(30/5)+9?
May 18, 2017
$5$
#### Explanation:
First, handle the fraction inside the parenthesis.
$\frac{30}{5} = 6$
This gives us
$2 - \left(6\right) + 9$
Next, handle the subtraction on the left
$- 4 + 9 = 5$
May 18, 2017
See the Explanation
#### Explanation:
$2 - \frac{\frac{30}{5}}{\frac{5}{5}} + 9$
$2 - 6 + 9$
$- 4 + 9$
$5$
May 18, 2017
$2 - \left(\frac{30}{5}\right) + 9$
$2 - \left(6\right) + 9$
$2 - 6 + 9$
$5$
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# Point Slope Slope Intercept And Standard Form
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
Point Slope Slope Intercept And Standard Form – There are many forms employed to depict a linear equation, one of the most commonly used is the slope intercept form. You may use the formula for the slope-intercept to find a line equation assuming you have the straight line’s slope , and the y-intercept, which is the y-coordinate of the point at the y-axis crosses the line. Learn more about this specific line equation form below.
## What Is The Slope Intercept Form?
There are three main forms of linear equations: standard slope-intercept, the point-slope, and the standard. Though they provide similar results when used, you can extract the information line produced more quickly with the slope-intercept form. It is a form that, as the name suggests, this form makes use of an inclined line where you can determine the “steepness” of the line reflects its value.
The formula can be used to find the slope of a straight line. It is also known as the y-intercept, also known as x-intercept in which case you can use a variety of available formulas. The equation for this line in this specific formula is y = mx + b. The straight line’s slope is represented through “m”, while its y-intercept is represented via “b”. Every point on the straight line is represented as an (x, y). Note that in the y = mx + b equation formula, the “x” and the “y” have to remain as variables.
## An Example of Applied Slope Intercept Form in Problems
The real-world in the real world, the slope-intercept form is frequently used to depict how an object or problem evolves over an elapsed time. The value provided by the vertical axis demonstrates how the equation deals with the magnitude of changes in what is represented via the horizontal axis (typically times).
A basic example of this formula’s utilization is to determine how much population growth occurs in a particular area as the years pass by. Based on the assumption that the area’s population grows annually by a specific fixed amount, the value of the horizontal axis will increase one point at a time as each year passes, and the amount of vertically oriented axis will rise in proportion to the population growth by the set amount.
It is also possible to note the starting point of a problem. The starting value occurs at the y-value of the y-intercept. The Y-intercept is the place where x is zero. In the case of a problem above the starting point would be at the time the population reading begins or when time tracking starts, as well as the associated changes.
The y-intercept, then, is the place that the population begins to be recorded for research. Let’s suppose that the researcher starts with the calculation or measure in the year 1995. This year will serve as considered to be the “base” year, and the x 0 points would be in 1995. This means that the population of 1995 represents the “y”-intercept.
Linear equations that employ straight-line formulas are nearly always solved this way. The beginning value is represented by the y-intercept, and the change rate is represented through the slope. The main issue with this form usually lies in the interpretation of horizontal variables particularly when the variable is accorded to an exact year (or any type or unit). The most important thing to do is to ensure that you know the variables’ meanings in detail.
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# A radius circular cylinder bring having diameter 12cm and height 15cm is full ice-cream.
+1 vote
170 views
edited
A radius circular cylinder bring having diameter 12cm and height 15cm is full ice-cream. The ice-cream is to be filled in cones of height 12cm and diameter 6cm having a hemisphere shape on top find the number of such cones which can be filled with icecream?
+1 vote
by (7.9k points)
selected
Given radius of cylinder (r1) = 12/2 = 6cm
Given radius of hemisphere (r2) = 6/2 = 3cm
Given height of cylinder (h) = 15cm
Height of cones (l) = 12cm
Volume of each cone = volume of cone + volume of hemisphere
Let number of cones be ‘n’
n(Volume of each cone) = volume of cylinder
Given volume of a hollow cylinder = 99cm3
Volume of a hollow cylinder
Equating (1) and (2) equations we get
Substituting r2 value in (1)
r1= 2cm
by (3.8k points)
Let r and R be the inner and outer radius of the cylindrical metallic pipe respectively.
h be the height of the metallic pipe = 14 cm
Difference between the Curved surface area of the outer cylinder and Curved surface area of the inner cylinder = 2πRh - 2πrh.
Given that difference between the outside and inside curved surface area of cylinder is 44 cm2 .
⇒ 2πh( R - r) = 44
⇒ 44 / 7 x 14 ( R - r) = 44
⇒ R - r = 1 / 2 = 0.5 ----------(1)
Given the pipe is made up of 99 cubic cm of metal so that
Volume of cylindrical metallic pipe = πR2h - πr2h.
⇒ 22/7 x 14 (R2 - r2) = 99 cm3 .
⇒ 44 x (R2 - r2) = 99
⇒ (R2 - r2) = 9 / 4 = 2.25
⇒ ( R - r)(R + r) = 2.25
= (0.5)x(R + r) = 2.25
R + r = 2.25 / 0.5 = 4.5
R + r = 4.5 ------------ (2)
Adding (1) and (2) we get
2R = 4.5 + 0.5 = 5
∴ R = 2.5 cm and r = 2 cm
∴ Outer side radius R = 2.5 cm and inner side radius r = 2 cm.
by (7.9k points)
by (3.8k points)
you have changed the question......
and the previous question for which i have given the solution is
A cylindrical metallic pipe is 14cm long.The difference between the outside and inside surfaces is 44cm square.If the pipe is made up of 99 cubic cm of metal.Find the outer and inner radii of the pipe.
by (3.8k points)
A radius circular cylinder bring having diameter 12cm and height 15cm is full ice-cream. The ice-cream is to be filled in cones of height 12cm and diameter 6cm having a hemisphere shape on top find the number of such cones which can be filled with icecream?
its solution is.....
Given:
For right circular cylinder
Diameter = 12 cm
Radius(R1) = 12/2= 6 cm & height (h1) = 15 cm
Volume of Cylindrical ice-cream container= πr1²h1= 22/7 × 6× 6× 15= 11880/7 cm³
Volume of Cylindrical ice-cream container=11880/7 cm³
For cone,
Diameter = 6 cm
Radius(r2) =6/2 = 3 cm & height (h2) = 12 cm
Volume of cone full of ice-cream= volume of cone + volume of hemisphere
= ⅓ πr2²h2 + ⅔ πr2³= ⅓ π ( r2²h2 + 2r2³)
= ⅓ × 22/7 (3²× 12 + 2× 3³)
= ⅓ × 22/7 ( 9 ×12 + 2 × 27)
= 22/21 ( 108 +54)
= 22/21(162)
= (22×54)/7
= 1188/7 cm³
Let n be the number of cones full of ice cream.
Volume of Cylindrical ice-cream container =n × Volume of one cone full with ice cream
11880/7 = n × 1188/7
11880 = n × 1188
n = 11880/1188= 10
n = 10
Hence, the required Number of cones = 10
by (3.8k points)
don't be oversmart bachche
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# What Is 9/49 as a Decimal + Solution With Free Steps
The fraction 9/49 as a decimal is equal to 0.183.
The Long Division method is the fundamental method of converting a fraction into decimal form. Some divisions give zero remainders after some decimal values and others do not end. 9/49 also has unending decimals but we will take 3 values after the decimal point.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 9/49.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 9
Divisor = 49
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 9 $\div$ 49
This is when we go through the Long Division solution to our problem. The following figure shows the solution for fraction 9/49.
Figure 1
## 9/49 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 9 and 49, we can see how 9 is Smaller than 49, and to solve this division, we require that 9 be Bigger than 49.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 9, which after getting multiplied by 10 becomes 90.
We take this 90 and divide it by 49; this can be done as follows:
90 $\div$ 49 $\approx$ 1
Where:
49 x 1 = 49
This will lead to the generation of a Remainder equal to 90 – 49 = 41. Now this means we have to repeat the process by Converting the 41 into 410 and solving for that:
410 $\div$ 49 $\approx$ 8
Where:
49 x 8 = 392
This, therefore, produces another Remainder which is equal to 410 – 392 = 18. Now this means we have to repeat the process by Converting the 18 into 180 and solving for that:
180 $\div$ 49 $\approx$ 3
Where:
49 x 3 = 147
Finally, we have a Quotient generated after combining the three pieces of it as 0.183, with a Remainder equal to 33.
Images/mathematical drawings are created with GeoGebra.
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# 5.08 Perimeter of composite shapes
Lesson
### Composing a shape
When a shape is made up, or composed, of separate shapes, it's called a composite shape. The lines where they join aren't usually visible, but we can still make out the separate shapes. We can use what we already know, including how to find the perimeter of a polygon, to calculate the perimeter of our composite shape.
When finding the perimeter of composite shapes there are two main approaches.
The first approach is finding the length of all the sides and adding them together like we would for an irregular shape. We can do this by using the lengths we are given to find any missing lengths.
The other approach is less obvious and relies on some visualization. We can see in the image below that the composite shape actually has the same perimeter as a rectangle.
So the perimeter of this composite shape can be calculated as:
Perimeter $=$= $2\times\left(8+13\right)$2×(8+13) $=$= $2\times21$2×21 $=$= $42$42
Careful!
When using this method it is important to keep track of any sides that do not get moved.
An example of a shape that we need to be careful with is:
Notice that we moved the indented edge to complete the rectangle but we still need to count the two edges that weren't moved.
We can calculate the perimeter of this shape as:
Perimeter $=$= $2\times\left(5+11\right)+2+2$2×(5+11)+2+2 $=$= $2\times16+4$2×16+4 $=$= $32+4$32+4 $=$= $36$36
With our knowledge of the perimeter of simple shapes like rectangles and squares we can often find creative ways to work out the perimeter of more complicated composite shapes.
#### Practice questions
##### Question 1
Consider the following figure.
1. Find the length $x$x.
2. Find the length $y$y.
3. Calculate the perimeter of the figure.
##### Question 2
Find the perimeter of the shape.
##### Question 3
Find the perimeter of the following figure. Use the $\pi$π button on your calculator, rounding your final answer to one decimal place.
### Outcomes
#### 8.1
Compare and order real numbers
#### 8.10
Solve area and perimeter problems, including practical problems, involving composite plane figures.
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# Texas Go Math Grade 4 Unit 1 Answer Key Number and Operations: Place Value, Fraction Concepts, and Operations
Refer to our Texas Go Math Grade 4 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 4 Unit 1 Answer Key Number and Operations: Place Value, Fraction Concepts, and Operations.
## Texas Go Math Grade 4 Unit 1 Answer Key Number and Operations: Place Value, Fraction Concepts, and Operations
Show What You Know
Read and Write Numbers Within 1,000
Write the number in different ways.
Question 1.
The given figure is:
From the given figure,
We can observe that
There are a total of 27 blocks
Now,
We know that,
“Expanded form” means writing the number so you show the value of each digit.
“Word form” means writing the number with words instead of numerals.
“Standard form” is a way of writing down very large or very small numbers easily
Hence, from the above,
We can conclude that
The representation of 27 in the following forms is:
Standard form: 27
Word form: Twenty-Seven
Expanded form: 20 + 7
Question 2.
The given figure is:
From the given figure,
We can observe that
There are a total of 43 blocks
Now,
We know that,
“Expanded form” means writing the number so you show the value of each digit.
“Word form” means writing the number with words instead of numerals.
“Standard form” is a way of writing down very large or very small numbers easily
Hence, from the above,
We can conclude that
The representation of 43 in the following forms is:
Standard form: 43
Word form: Forty-Three
Expanded form: 40 + 3
Fractions of a Whole
Write a fraction that names the shaded part.
Question 3.
The given model is:
Now,
From the above model,
We can observe that
The total number of parts are: 8
The number of shaded parts is: 1
Hence, from the above,
We can conclude that
The representation of the fraction that names the shaded part is: $$\frac{1}{8}$$
4th Grade Standard Form Expanded Form Unit Form Question 4.
The given model is:
Now,
From the above model,
We can observe that
The total number of parts are: 3
The number of shaded parts is: 2
Hence, from the above,
We can conclude that
The representation of the fraction that names the shaded part is: $$\frac{2}{3}$$
Question 5.
The given model is:
Now,
From the above model,
We can observe that
The total number of parts are: 4
The number of shaded parts is: 3
Hence, from the above,
We can conclude that
The representation of the fraction that names the shaded part is: $$\frac{3}{4}$$
Question 6.
The given model is:
Now,
From the above model,
We can observe that
The total number of parts are: 6
The number of shaded parts is: 2
Hence, from the above,
We can conclude that
The representation of the fraction that names the shaded part is: $$\frac{2}{6}$$
Question 7.
The given model is:
Now,
From the above model,
We can observe that
The total number of parts are: 12
The number of shaded parts is: 6
Hence, from the above,
We can conclude that
The representation of the fraction that names the shaded part is: $$\frac{6}{12}$$
Unit Form and Standard Form 4th Grade Question 8.
The given model is:
Now,
From the above model,
We can observe that
The total number of parts are: 10
The number of shaded parts is: 2
Hence, from the above,
We can conclude that
The representation of the fraction that names the shaded part is: $$\frac{2}{10}$$
Fractions on a Number Line
Complete the number line. Draw a point to show the fraction.
Question 9.
$$\frac{4}{5}$$
The given fraction is: $$\frac{4}{5}$$
Now,
The representation of the complete number line is:
Hence, from the above,
We can conclude that
The representation of the given fraction on the given number line is:
Question 10.
$$\frac{3}{8}$$
The given fraction is: $$\frac{3}{8}$$
Now,
The representation of the complete number line is:
Hence, from the above,
We can conclude that
The representation of the given fraction on the given number line is:
Vocabulary Builder
Visualize It
Complete the flow map by using the words with a ✓.
Numbers and Operations
Understand Vocabulary
Complete the sentences by using preview words.
Question 1.
A fraction is in ____________ if the numerator and denominator have only 1 as a common factor.
We know that,
A fraction is in “Simplest form” if the numerator and denominator have only 1 as a common factor.
Hence, from the above,
We can conclude that
The term that is suitable for the given statement is: Simplest form
Standard Form Go Math 4th Grade Question 2.
A number close to an exact amount is an _____________ .
We know that,
A number close to an exact amount is an “Estimate” (or) “Approximate number”
Hence, from the above,
We can conclude that
The term that is suitable for the given statement is: Estimate (or) Approximate number
Question 3.
________________ is a way to write a number by showing the value of each digit.
We know that,
“Expanded form” is a way to write a number by showing the value of each digit.
Hence, from the above,
We can conclude that
The term that is suitable for the given statement is: Expanded form
Question 4.
A ________________ is a known size or amount that helps you understand a different size or amount.
We know that,
A “Benchmark” is a known size or amount that helps you understand a different size or amount.
Hence, from the above,
We can conclude that
The term that is suitable for the given statement is: Benchmark
Question 5.
A ________________ is one of ten equal parts.
We know that,
A “Tenth” is one of ten equal parts.
Hence, from the above,
We can conclude that
The term that is suitable for the given statement is: Tenth
Question 6.
A ________________ is one of one hundred equal parts.
We know that,
A “Hundredth” is one of one hundred equal parts.
Hence, from the above,
We can conclude that
The term that is suitable for the given statement is: Hundredth
Reading and Writing Math
Vocabulary
Look at the number in the place-value chart. Use this number to complete the word web.
The given number is: 8,827,610,731
Now,
The given place value chart for the given number is:
Now,
We know that,
“Expanded form” means writing the number so you show the value of each digit.
“Word form” means writing the number with words instead of numerals.
“Standard form” is a way of writing down very large or very small numbers easily
Hence, from the above,
We can conclude that
The complete word web for the given place value chart is:
Expanded form:
8,000,000,000 + 800,000,000 + 20,000,000 + 7,000,000 + 600,000 + 10,000 + 0 + 700 + 30 + 1
Word form:
Eight billion Eight hundred twenty-seven million Six hundred ten thousand seven hundred thirty-one
Standard form:
8,827,610,731
Writing How can you write the 7536,301398 in word form?
The given number is: 7536,301398
Now,
We know that,
“Word form” means writing the number with words instead of numerals.
Hence, from the above,
We can conclude that
The representation of the word form for the given number is:
Seven billion Five hundred thirty-six million three hundred one thousand three hundred ninety-eight
Reading Look for this book in your library. Digging for Bird Dinosaurs: An Expedition to Madagascar, by Nic Bishop.
How to Play
Question 1.
Player 1 draws a card from the stack. He or she makes the fraction with the fraction strips and places them under the 1 whole strip.
4th Grade Go Math Standard Form Question 2.
Player 2 draws a card from the stack. He or she makes the fraction with the fraction strips and places them under Player 1’s strips.
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# 7.1: Sums of Discrete Random Variables
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In this chapter we turn to the important question of determining the distribution of a sum of independent random variables in terms of the distributions of the individual constituents. In this section we consider only sums of discrete random variables, reserving the case of continuous random variables for the next section.
We consider here only random variables whose values are integers. Their distribution functions are then defined on these integers. We shall find it convenient to assume here that these distribution functions are defined for all integers, by defining them to be 0 where they are not otherwise defined.
## Convolutions
Suppose $$X$$ and $$Y$$ are two independent discrete random variables with distribution functions $$m_{1}(x)$$ and $$m_{2}(x)$$. Let $$Z=X+Y$$. We would like to determine the distribution function $$m_{3}(x)$$ of $$Z$$. To do this, it is enough to determine the probability that $$Z$$ takes on the value $$z$$, where $$z$$ is an arbitrary integer. Suppose that $$X=k$$, where $$k$$ is some integer. Then $$Z=z$$ if and only if $$Y=z-k$$. So the event $$Z=z$$ is the union of the pairwise disjoint events
$(X=k) \text { and }(Y=z-k),$
where $$k$$ runs over the integers. Since these events are pairwise disjoint, we have
$P(Z=z)=\sum_{k=-\infty}^{\infty} P(X=k) \cdot P(Y=z-k)$
Thus, we have found the distribution function of the random variable $$Z$$. This leads to the following definition.
##### Definition: Term
Let $$X$$ and $$Y$$ be two independent integer-valued random variables, with distribution functions $$m_{1}(x)$$ and $$m_{2}(x)$$ respectively. Then the convolution of $$m_{1}(x)$$ and $$m_{2}(x)$$ is the distribution function $$m_{3}=m_{1} * m_{2}$$ given by
$m_{3}(j)=\sum_{k} m_{1}(k) \cdot m_{2}(j-k)$
for $$j=\ldots,-2,-1,0,1,2, \ldots$$. The function $$m_{3}(x)$$ is the distribution function of the random variable $$Z=X+Y$$.
It is easy to see that the convolution operation is commutative, and it is straightforward to show that it is also associative.
Now let $$S_{n}=X_{1}+X_{2}+\cdots+X_{n}$$ be the sum of $$n$$ independent random variables of an independent trials process with common distribution function $$m$$ defined on the integers. Then the distribution function of $$S_{1}$$ is $$m$$. We can write
$S_{n}=S_{n-1}+X_{n} .$
Thus, since we know the distribution function of $$X_{n}$$ is $$m$$, we can find the distribution function of $$S_{n}$$ by induction.
##### Example $$\PageIndex{1}$$
A die is rolled twice. Let $$X_{1}$$ and $$X_{2}$$ be the outcomes, and let $$S_{2}=X_{1}+X_{2}$$ be the sum of these outcomes. Then $$X_{1}$$ and $$X_{2}$$ have the common distribution function:
$m=\left(\begin{array}{cccccc} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 / 6 & 1 / 6 & 1 / 6 & 1 / 6 & 1 / 6 & 1 / 6 \end{array}\right) .$
The distribution function of $$S_{2}$$ is then the convolution of this distribution with itself. Thus,
\begin{aligned} P\left(S_{2}=2\right) & =m(1) m(1) \\ & =\frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}, \\ P\left(S_{2}=3\right) & =m(1) m(2)+m(2) m(1) \\ & =\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}=\frac{2}{36}, \\ P\left(S_{2}=4\right) & =m(1) m(3)+m(2) m(2)+m(3) m(1) \\ & =\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}=\frac{3}{36} . \end{aligned}
Continuing in this way we would find $$P\left(S_{2}=5\right)=4 / 36, P\left(S_{2}=6\right)=5 / 36$$, $$P\left(S_{2}=7\right)=6 / 36, P\left(S_{2}=8\right)=5 / 36, P\left(S_{2}=9\right)=4 / 36, P\left(S_{2}=10\right)=3 / 36$$, $$P\left(S_{2}=11\right)=2 / 36$$, and $$P\left(S_{2}=12\right)=1 / 36$$.
The distribution for $$S_{3}$$ would then be the convolution of the distribution for $$S_{2}$$ with the distribution for $$X_{3}$$. Thus
$P\left(S_{3}=3\right)=P\left(S_{2}=2\right) P\left(X_{3}=1\right)$
\begin{aligned} & =\frac{1}{36} \cdot \frac{1}{6}=\frac{1}{216}, \\ P\left(S_{3}=4\right) & =P\left(S_{2}=3\right) P\left(X_{3}=1\right)+P\left(S_{2}=2\right) P\left(X_{3}=2\right) \\ & =\frac{2}{36} \cdot \frac{1}{6}+\frac{1}{36} \cdot \frac{1}{6}=\frac{3}{216}, \end{aligned}
and so forth.
This is clearly a tedious job, and a program should be written to carry out this calculation. To do this we first write a program to form the convolution of two densities $$p$$ and $$q$$ and return the density $$r$$. We can then write a program to find the density for the sum $$S_{n}$$ of $$n$$ independent random variables with a common density $$p$$, at least in the case that the random variables have a finite number of possible values.
Running this program for the example of rolling a die $$n$$ times for $$n=10,20,30$$ results in the distributions shown in Figure 7.1. We see that, as in the case of Bernoulli trials, the distributions become bell-shaped. We shall discuss in Chapter 9 a very general theorem called the Central Limit Theorem that will explain this phenomenon.
##### Example $$\PageIndex{1}$$
A well-known method for evaluating a bridge hand is: an ace is assigned a value of 4 , a king 3 , a queen 2 , and a jack 1 . All other cards are assigned a value of 0 . The point count of the hand is then the sum of the values of the cards in the hand. (It is actually more complicated than this, taking into account voids in suits, and so forth, but we consider here this simplified form of the point count.) If a card is dealt at random to a player, then the point count for this card has distribution
$p_{X}=\left(\begin{array}{ccccc} 0 & 1 & 2 & 3 & 4 \\ 36 / 52 & 4 / 52 & 4 / 52 & 4 / 52 & 4 / 52 \end{array}\right)$
Let us regard the total hand of 13 cards as 13 independent trials with this common distribution. (Again this is not quite correct because we assume here that we are always choosing a card from a full deck.) Then the distribution for the point count $$C$$ for the hand can be found from the program NFoldConvolution by using the distribution for a single card and choosing $$n=13$$. A player with a point count of 13 or more is said to have an opening bid. The probability of having an opening bid is then
$P(C \geq 13)$
Since we have the distribution of $$C$$, it is easy to compute this probability. Doing this we find that
$P(C \geq 13)=.2845$
so that about one in four hands should be an opening bid according to this simplified model. A more realistic discussion of this problem can be found in Epstein, The Theory of Gambling and Statistical Logic. $${ }^{1}$$
$${ }^{1}$$ R. A. Epstein, The Theory of Gambling and Statistical Logic, rev. ed. (New York: Academic Press, 1977).
For certain special distributions it is possible to find an expression for the distribution that results from convoluting the distribution with itself $$n$$ times.
The convolution of two binomial distributions, one with parameters $$m$$ and $$p$$ and the other with parameters $$n$$ and $$p$$, is a binomial distribution with parameters $$(m+n)$$ and $$p$$. This fact follows easily from a consideration of the experiment which consists of first tossing a coin $$m$$ times, and then tossing it $$n$$ more times.
The convolution of $$k$$ geometric distributions with common parameter $$p$$ is a negative binomial distribution with parameters $$p$$ and $$k$$. This can be seen by considering the experiment which consists of tossing a coin until the $$k$$ th head appears.
## Exercises
Exercise $$\PageIndex{1}$$: A die is rolled three times. Find the probability that the sum of the outcomes is
(a) greater than 9.
(b) an odd number.
Exercise $$\PageIndex{2}$$: The price of a stock on a given trading day changes according to the distribution
$p_{X}=\left(\begin{array}{cccc} -1 & 0 & 1 & 2 \\ 1 / 4 & 1 / 2 & 1 / 8 & 1 / 8 \end{array}\right) .$
Find the distribution for the change in stock price after two (independent) trading days.
Exercise $$\PageIndex{3}$$: Let $$X_{1}$$ and $$X_{2}$$ be independent random variables with common distribution
$p_{X}=\left(\begin{array}{ccc} 0 & 1 & 2 \\ 1 / 8 & 3 / 8 & 1 / 2 \end{array}\right)$
Find the distribution of the sum $$X_{1}+X_{2}$$.
Exercise $$\PageIndex{4}$$: In one play of a certain game you win an amount $$X$$ with distribution
$p_{X}=\left(\begin{array}{ccc} 1 & 2 & 3 \\ 1 / 4 & 1 / 4 & 1 / 2 \end{array}\right)$
Using the program NFoldConvolution find the distribution for your total winnings after ten (independent) plays. Plot this distribution.
Exercise $$\PageIndex{5}$$: Consider the following two experiments: the first has outcome $$X$$ taking on the values 0,1 , and 2 with equal probabilities; the second results in an (independent) outcome $$Y$$ taking on the value 3 with probability $$1 / 4$$ and 4 with probability $$3 / 4$$. Find the distribution of
(a) $$Y+X$$.
(b) $$Y-X$$.
Exercise $$\PageIndex{6}$$: People arrive at a queue according to the following scheme: During each minute of time either 0 or 1 person arrives. The probability that 1 person arrives is $$p$$ and that no person arrives is $$q=1-p$$. Let $$C_{r}$$ be the number of customers arriving in the first $$r$$ minutes. Consider a Bernoulli trials process with a success if a person arrives in a unit time and failure if no person arrives in a unit time. Let $$T_{r}$$ be the number of failures before the $$r$$ th success.
(a) What is the distribution for $$T_{r}$$ ?
(b) What is the distribution for $$C_{r}$$ ?
(c) Find the mean and variance for the number of customers arriving in the first $$r$$ minutes.
Exercise $$\PageIndex{7}$$: (a) A die is rolled three times with outcomes $$X_{1}, X_{2}$$, and $$X_{3}$$. Let $$Y_{3}$$ be the maximum of the values obtained. Show that
$P\left(Y_{3} \leq j\right)=P\left(X_{1} \leq j\right)^{3} .$
Use this to find the distribution of $$Y_{3}$$. Does $$Y_{3}$$ have a bell-shaped distribution?
(b) Now let $$Y_{n}$$ be the maximum value when $$n$$ dice are rolled. Find the distribution of $$Y_{n}$$. Is this distribution bell-shaped for large values of $$n$$ ?
Exercise $$\PageIndex{8}$$: A baseball player is to play in the World Series. Based upon his season play, you estimate that if he comes to bat four times in a game the number of hits he will get has a distribution
$p_{X}=\left(\begin{array}{ccccc} 0 & 1 & 2 & 3 & 4 \\ .4 & .2 & .2 & .1 & .1 \end{array}\right)$
Assume that the player comes to bat four times in each game of the series.
(a) Let $$X$$ denote the number of hits that he gets in a series. Using the program NFoldConvolution, find the distribution of $$X$$ for each of the possible series lengths: four-game, five-game, six-game, seven-game.
(b) Using one of the distribution found in part (a), find the probability that his batting average exceeds .400 in a four-game series. (The batting average is the number of hits divided by the number of times at bat.)
(c) Given the distribution $$p_{X}$$, what is his long-term batting average?
Exercise $$\PageIndex{9}$$: Prove that you cannot load two dice in such a way that the probabilities for any sum from 2 to 12 are the same. (Be sure to consider the case where one or more sides turn up with probability zero.)
Exercise $$\PageIndex{10}$$:$$10\left(\right.$$ Lévy $$\left.^{2}\right)$$ Assume that $$n$$ is an integer, not prime. Show that you can find two distributions $$a$$ and $$b$$ on the nonnegative integers such that the convolution of
$${ }^{2}$$ See M. Krasner and B. Ranulae, "Sur une Proprieté des Polynomes de la Division du Circle"; and the following note by J. Hadamard, in C. R. Acad. Sci., vol. 204 (1937), pp. 397-399. $$a$$ and $$b$$ is the equiprobable distribution on the set $$0,1,2, \ldots, n-1$$. If $$n$$ is prime this is not possible, but the proof is not so easy. (Assume that neither $$a$$ nor $$b$$ is concentrated at 0 .)
Exercise $$\PageIndex{11}$$: Assume that you are playing craps with dice that are loaded in the following way: faces two, three, four, and five all come up with the same probability $$(1 / 6)+r$$. Faces one and six come up with probability $$(1 / 6)-2 r$$, with $$0<$$ $$r<.02$$. Write a computer program to find the probability of winning at craps with these dice, and using your program find which values of $$r$$ make craps a favorable game for the player with these dice.
This page titled 7.1: Sums of Discrete Random Variables is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Charles M. Grinstead & J. Laurie Snell (American Mathematical Society) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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# Algebra Expressions with Fraction Bars
## Evaluate the numerator, evaluate the denominator, then simplify
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Practice Algebra Expressions with Fraction Bars
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Algebra Expressions with Fraction Bars
What if you knew that the volume of a sphere could be represented by the formula V=4πr33\begin{align*}V=\frac{4 \pi r^3}{3}\end{align*}, where r\begin{align*}r\end{align*} is the radius, and that the radius of a sphere is 4 feet? How would the fraction bar in the formula affect the way that you found the sphere's volume? Which operation do you think the fraction bar represents? After completing this Concept, you'll be able to correctly interpret the fraction bar when finding the sphere's volume.
### Guidance
Fraction bars count as grouping symbols for PEMDAS, and should be treated as a set of parentheses. All numerators and all denominators can be treated as if they have invisible parentheses. When real parentheses are also present, remember that the innermost grouping symbols should be evaluated first. If, for example, parentheses appear in a numerator, they would take precedence over the fraction bar. If the parentheses appear outside of the fraction, then the fraction bar takes precedence.
#### Example A
Use the order of operations to simplify the following expression:
z+341
when
z=2
Solution: Begin by substituting the appropriate value for the variable:
(2+3)41=541\begin{align*}\frac{(2+3)}{4} -1 = \frac{5}{4} -1\end{align*}.
Rewriting 1 as a fraction, the expression becomes:
5444=14
#### Example B
Use the order of operations to simplify the following expression:
\begin{align*}\left (\frac{a+2}{b+4} - 1 \right ) + b\end{align*} when \begin{align*}a = 3\end{align*} and \begin{align*}b = 1\end{align*}
Solution: Begin by substituting the appropriate value for the variable:
\begin{align*}\frac{(3+2)}{(1+4)} = \frac{5}{5} = 1\end{align*}
\begin{align*}(1 - 1) + b\end{align*} Substituting 1 for b, the expression becomes \begin{align*} 0 + 1 = 1\end{align*}
#### Example C
Use the order of operations to simplify the following expression:
\begin{align*}2 \times \left ( \frac{w + (x - 2z)}{(y + 2)^2} - 1 \right )\end{align*} when \begin{align*}w = 11, \ x = 3, \ y = 1\end{align*} and \begin{align*}z = -2\end{align*}
Solution: Begin by substituting the appropriate values for the variables:
\begin{align*}2 \left ( \frac{[11+(3-2(-2))]}{[(1+2)^2)]} - 1 \right ) = 2 \left ( \frac{(11+7)}{3^2} -1 \right ) = 2 \left (\frac{18}{9} - 1 \right )\end{align*}
Continue simplifying:
\begin{align*}2\left ( \frac{18}{9} - \frac{9}{9} \right ) = 2 \left ( \frac{9}{9} \right ) = 2(1)= 2\end{align*}
-->
### Guided Practice
Use the order of operations to evaluate the following expression when
and
:
\begin{align*}\left (\frac{x-1}{y-1} \right )^2 + \frac{2x}{y^2}\end{align*}
Solution:
Begin by substituting in 6 for \begin{align*}x\end{align*} and 1 for \begin{align*}y\end{align*}.
\begin{align*}\left (\frac{6-1}{2-1} \right )^2 + \frac{2(6)}{2^2}\end{align*}
First, we work with what is inside the parentheses. There, we have a fraction so we have to simplify the fraction first, simplifying the numerator and then the denominator before dividing. We can simply the other fraction at the same time.
\begin{align*}\left (\frac{5}{1} \right )^2 + \frac{12}{4}\end{align*}
Next we simplify the fractions, and finish with exponents and then addition:
\begin{align*} (5 )^2 + 3=25+3=28.\end{align*}
### Explore More
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Order of Operations (14:23)
Use the order of operations to simplify the following expressions.
1. \begin{align*}\frac{2 \cdot (3 + (2 - 1))}{4 - (6 + 2)} - (3 - 5)\end{align*}
2. \begin{align*}\frac{(2+3)^2}{3-8} - \frac{3\cdot(10-4)}{7-4}\end{align*}
Evaluate the following expressions involving variables.
1. \begin{align*}\frac{jk}{j + k}\end{align*} when \begin{align*}j = 6\end{align*} and \begin{align*}k = 12\end{align*}
2. \begin{align*}\frac{4x}{9x^2 - 3x + 1}\end{align*} when \begin{align*}x = 2\end{align*}
3. \begin{align*}\frac{z^2}{x + y} + \frac{x^2}{x - y}\end{align*} when \begin{align*}x = 1, \ y = -2\end{align*}, and \begin{align*}z = 4\end{align*}
4. \begin{align*}\frac{4xyz}{y^2 - x^2}\end{align*} when \begin{align*}x = 3, \ y = 2\end{align*}, and \begin{align*}z = 5\end{align*}
5. \begin{align*}\frac{x^2 - z^2}{xz - 2x(z - x)}\end{align*} when \begin{align*}x = -1\end{align*} and \begin{align*}z = 3\end{align*}
The formula to find the volume of a square pyramid is \begin{align*}V=\frac{s^2 (h)}{3}\end{align*}. Evaluate the volume for the given values.
1. \begin{align*}s=4\ inches,h=18\ inches\end{align*}
2. \begin{align*}s=10\ feet,h=50\ feet\end{align*}
3. \begin{align*}h=7\ meters,s=12\ meters\end{align*}
4. \begin{align*}h=27\ feet,s=13\ feet\end{align*}
5. \begin{align*}s=16\ cm,h=90\ cm\end{align*}
### Vocabulary Language: English Spanish
Fraction Bar
Fraction Bar
A fraction bar is a line used to divide the numerator and the denominator of a fraction. The fraction bar means division.
fraction
fraction
A fraction is a part of a whole. A fraction is written mathematically as one value on top of another, separated by a fraction bar. It is also called a rational number.
Order of Operations
Order of Operations
The order of operations specifies the order in which to perform each of multiple operations in an expression or equation. The order of operations is: P - parentheses, E - exponents, M/D - multiplication and division in order from left to right, A/S - addition and subtraction in order from left to right.
Parentheses
Parentheses
Parentheses "(" and ")" are used in algebraic expressions as grouping symbols.
PEMDAS
PEMDAS
PEMDAS (Please Excuse My Daring Aunt Sally) is a mnemonic device used to help remember the order of operations: Parentheses, Exponents, Multiplication/Division, Addition/Subtraction.
Volume
Volume
Volume is the amount of space inside the bounds of a three-dimensional object.
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# When two dice are rolled what is the probability of getting a sum less than 11?
Contents
## When two dice are rolled what is the probability you roll a sum less than 4?
To get a number less than 4 on a die, 3/6 is the probability.
## When two dice are rolled find the probability of getting a sum less than 10?
so I wrote out all the possabilities of combinations of 10 or higher. That totals 8 combination out of 36 that could be ten or higher, so 8/36= 2/9. since I wanted less than ten 1-(2/9) = 7/9 probability of getting less than 10.
## What is the probability of throwing less than 11 with two dice?
Probability of rolling a certain number or less for two 6-sided dice.
Two (6-sided) dice roll probability table.
Roll a… Probability
9 4/36 (11.111%)
10 3/36 (8.333%)
11 2/36 (5.556%)
12 1/36 (2.778%)
## What is the probability of getting a sum of 4 or less?
There are six faces for each of two dice, giving 36 possible outcomes. If the two dice are fair, each of 36 outcomes is equally likely. Three outcomes sum to 4: (1+3), (2+2) and (3+1). Probability of getting a sum of 4 on one toss of two dice is 3/36, or 1/12.
IT IS INTERESTING: What does folding do in blackjack?
## What is the probability of getting a sum of 7 when rolling two dice?
For each of the possible outcomes add the numbers on the two dice and count how many times this sum is 7. If you do so you will find that the sum is 7 for 6 of the possible outcomes. Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7 is 6/36 = 1/6.
## What is the probability of rolling a sum of 8?
Probabilities for the two dice
Total Number of combinations Probability
7 6 16.67%
8 5 13.89%
9 4 11.11%
10 3 8.33%
## What is the probability that the sum of two dice is 13?
To find: The sum of the two numbers appearing on the top is 13. Hence there is no possibility to get sum of the two number is 13. Hence, there is no possibility to get sum of the two numbers appearing on the top is 13.
## What is the probability of getting a sum of 13 when rolling a pair of dice?
Probability of a sum of 13: 21/216 = 9.7%
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# 3.1 What Are Functions?
Functions are what we use to describe things we want to talk about mathematically. I find, though, that I get a bit tongue tied when I try to define them.
The simplest definition is: a function is a bunch of ordered pairs of things (in our case the things will be numbers, but they can be otherwise), with the property that the first members of the pairs are all different from one another.
Thus, here is an example of a function:
$[\{1, 1\}, \{2, 1\}, \{3, 2\}]$
This function consists of three pairs, whose first members are $$1, 2$$ and $$3$$.
It is customary to give functions names, like $$f, g$$ or $$h$$, and if we call this function $$f$$, we generally use the following notation to describe it:
$f(1) = 1, f(2) = 1, f(3) = 2$
The first members of the pairs are called arguments and the whole set of them is called the domain of the function. Thus the arguments of $$f$$ here are $$1, 2$$ and $$3$$, and the set consisting of these three numbers is its domain.
The second members of the pairs are called the values of the functions, and the set of these is called the range of the function.
The standard terminology for describing this function f is:
The value of $$f$$ at argument $$1$$ is $$1$$, its value at argument $$2$$ is $$1$$, and its value at argument $$3$$ is $$2$$, which we write as $$f(1) = 1, f(2) = 1, f(3) = 2$$.
We generally think of a function as a set of assignments of values (second members of our pairs) to arguments (their first members).
The condition that the first members of the pairs are all different is the condition that each argument in the domain of $$f$$ is assigned a unique value in its range by any function.
Exercise 3.1 Consider the function $$g$$, defined by the pairs $$(1, 1), (2, 5), (3, 1)$$ and $$(4, 2)$$. What is its domain? What is the value of $$g$$ at argument $$3$$? What is $$g(4)$$?
If you stick a thermometer in your mouth, you can measure your temperature, at some particular time. You can define a function $$T$$ or temperature, which assigns the temperature you measure to the time at which you remove the thermometer from your mouth. This is a typical function. Its arguments are times of measurement and its values are temperatures.
Of course your mouth has a temperature even when you don't measure it, and it has one at every instant of time and there are an infinite number of such instants.
This means that if you want to describe a function $$T$$ whose value at any time t is the temperatures in your mouth at that time, you cannot really list all its pairs. There are an infinite number of possible arguments $$t$$ and it would take you forever to list them.
Instead, we employ a trick to describe a function $$f$$: we generally provide a rule which allows you, the reader, to choose any argument you like in $$f$$'s domain, and, by using the rule, to compute the value of your function at that argument. This rule is often called a formula for the function. The symbol $$x$$ is often used to denote the argument you will select, and the formula tells you how to compute the function at that argument.
The simplest function of all, sometimes called the identity function, is the one that assigns as value the argument itself. If we denote this function as $$f$$, it obeys
$f(x) = x$
for $$x$$ in whatever domain we choose for it. In other words, both members of its pairs are the same wherever you choose to define it.
We can get more complicated functions by giving more complicated rules, (These rules are often called formulae as we have noted already). Thus we can define functions by giving any of the following formulae among an infinity of possibilities:
$3x, x^2, x^2-1, \frac{3}{x}, x^3, \frac{x}{x^2 + 1}, 3x + 5, x^2 + 7x - 1$
These represent, respectively, $$3$$ times $$x$$, $$x$$ squared, $$x$$ squared minus $$1$$, $$3$$ divided by $$x$$, $$x$$ cubed, $$x$$ divided by the sum of the square of $$x$$ and $$1$$, and so on.
We can construct functions by applying the operations of addition, subtraction, multiplication and division to copies of $$x$$ and numbers in any way we see fit to do so.
There are two very nice features of functions that we construct in this way, and the first applies to all functions.
We can draw a picture of a function, called its graph, on a piece of graph paper, or on a spreadsheet, chart, or with a graphing calculator. We can do it by taking argument-value pairs of the function and describing each by a point in the plane, with $$x$$ coordinate given by the argument and y coordinate given by the value for its pair.
Of course it is impossible to plot all the pairs of a function that has an infinite domain, but we can get a pretty good idea of what its graph looks like by taking perhaps a hundred evenly spaced points in any interval of interest to us. This sounds like an impossibly tedious thing to do and it used to be so, but now it is not. On a spreadsheet, the main job is to enter the function once (with its argument given by the address of some other location). That and some copying is all you have to do, and with practice it can be done in $$30$$ seconds for a very wide variety of functions.
The second nice feature is that we can enter any function formed by adding, subtracting, multiplying, dividing and performing still another operation, on the contents of some address very easily on a spreadsheet or graphing calculator. Not only that, these devices have some other built in functions that we can use as well.
The two of these facts mean that we can actually look at any function formed by adding subtracting multiplying or dividing copies of the identity function $$x$$ and other built in functions, and any number we want, and see how they behave, with very limited effort.
We will soon see that we can use the same procedure used for graphing functions to graph their derivatives (we have not defined these yet) as well, but that is getting ahead of the story. You should realize though that we can compute derivatives for most functions numerically with only a small amount of effort as well.
Exercise 3.2 What is the value of the function $$x^2 + 3$$ at $$x = 5$$? At argument $$10$$?
Would you please give some examples?
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There are different types of lines (or segments or rays) or pairs of lines (or segments or rays). You can identify single lines based on the direction they’re pointing (horizontal, vertical, or neither), and you can categorize pairs of lines based on whether they intersect or not, the angle they make with each other, and, whether they are coplanar or non-coplanar.
Defining horizontal and vertical lines
You probably already know what the terms horizontal and vertical mean, but here’s a refresher along with a handy-dandy figure.
Horizontal lines, segments, or rays: Horizontal lines, segments, and rays go straight across, left and right, not up or down at all — you know, like the horizon.
Vertical lines, segments, or rays: Lines or parts of a line that go straight up and down are vertical. (Rocket science this is not.)
Defining pairs of lines
There are five terms that describe pairs of lines. The first four are about coplanar lines — you use these a lot. The fifth term describes non-coplanar lines. This term comes up only in 3-D problems.
Coplanar lines
Coplanar lines are lines in the same plane. Below are four terms that describe different ways coplanar lines may interact:
Parallel lines, segments, or rays: Lines that run in the same direction and never cross (like two railroad tracks) are called parallel. Segments and rays are parallel if the lines that contain them are parallel. See the above figure.
Intersecting lines, segments, or rays: Lines, segments, or rays that cross or touch are intersecting. The point where they cross or touch is called the point of intersection.
Perpendicular lines, segments, or rays: Lines, segments, or rays that intersect at right angles (90° angles) are perpendicular.
In the above figure, the little boxes in the corners of the angles indicate right angles.
Oblique lines, segments, or rays: Lines, segments, or rays that intersect at any angle other than 90° are called oblique. The above figure shows oblique lines and rays on the right.
Because lines extend forever, a pair of coplanar lines must be either parallel or intersecting. (However, this is not true for coplanar segments and rays. Segments and rays can be nonparallel and at the same time non-intersecting, because their endpoints allow them to stop short of crossing.)
Non-coplanar lines
Non-coplanar lines are simply lines that aren’t in the same plane.
Skew lines, segments, or rays: Lines that don’t lie in the same plane are called skew lines — skew simply means non-coplanar. Or you can say that skew lines are lines that are neither parallel nor intersecting. The above figure shows an example. (You probably won’t ever hear anyone refer to skew segments or rays, but there’s no reason they can’t be skew. They’re skew if they’re non-coplanar.)
Here’s a good way to get a handle on skew lines. Take two pencils or pens, one in each hand. Hold them a few inches apart, both of them pointing away from you. Now, keep one where it is and point the other one up at the ceiling. That’s it. You’re holding skew lines.
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# How do you find the perimeter of a Midsegment?
## How do you find the perimeter of a Midsegment?
To find the perimeter, we’ll just add all the outside lengths together.
What is the length of the Midsegment AB?
The length of the midsegment is the sum of the two bases divided by 2. Remember that the bases of a trapezoid are the two parallel sides.
What is the measure of angle a Midsegment?
Also recall that same side interior angles are supplementary, meaning that the measure of one angle is 180° – (the measure of the other). The angle on the same side of the midsegment as the third side is a same side interior angle with the base angle of the triangle.
### What is the formula for midpoint of a line?
The formula for midpoint = (x1 + x2)/2, (y1 + y2)/2. Substituting in the two x coordinates and two y coordinates from the endpoints, we get (–1 + 3)/2. (4 + 6)/2 or (1, 5) as the midpoint.
What makes a midsegment of a trapezoid special?
Explanation:A midsegment of a trapezoid is a segment that connects the midpoints of the two non-parallel sides of a trapezoid. This segment has two special properties. It is parallel to the bases of the trapezoid. The length of the midsegment is the average of the lengths of the two bases. Remember that the bases of a trapezoid are parallel.
How are the midsegments of a triangle conjectured?
Midsegment Conjectures Explanation: A midsegmentof a triangle is a segment connecting the midpoints of two sides of a triangle. This segment has two special properties. It is always parallel to the third side, and the length of the midsegment is half the length of the third side. The precise statement of the conjecture is:
#### How do you adjust the length of a trapezoid?
Adjust the trapezoid above by dragging any vertex and convince yourself this is so. The length of the median is the average length of the bases, or using the formula: If one of the bases is zero length, the result is a triangle. and is identical to the triangle midsegment case. The median line will then be half the remaining base.
Which is the midpoint of a trapezoid ABCD?
The midsegment of a trapezoid is the segment connecting the midpoints of the two non-parallel sides. In trapezoid ABCD below, segment PQ is the midsegment. The length of the midsegment of trapezoid is half the sum of the lengths of the two parallel sides.
11/03/2021
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# Complex Number
Complex numbers
Complex numbers are numbers that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit, defined as i2 = -1.
The real part of a complex number a + bi is the real number a, and the imaginary part is the real number b multiplied by i.
For example, 2+5i is a complex number, where 2 is a real number (Re) and 5i is an imaginary number (Im).
The set of all complex numbers is denoted by the symbol C
Note:-1)Euler was the first mathematician to introduce the symbol i (iota) for the square root of -1 with property. i2=-1 He also called this symbol as the imaginary unit.
2) Iota (i) is neither 0, nor greater than 0, nor less than 0.
3) The square root of a negative real number is called an imaginary unit.
4) For any positive real number a, we have √-a=i√a
5)The property √a√b = √ab is valid only if at least one of a and b is non-negative. If
a and b are both negative then √a√b = -√ab
Integral powers of iota (i) :
Since i= $\sqrt{-1}$1 hence we have i2=-1,i3=I and i4=1 .
In general we have the following results
i4k+1 = i.i4k+2 = -1 i4k+3 = -i.i4k = 1
Where k can have an integral value (positive or negative).
Similarly, we can find for the negative power of i, which are as follows;
i-1 =-i, i-2=-1,i-3=i and i-4=1
Important Tips
→The sum of four consecutive powers of i is always zero i.e ik+ik+1+ik+2+ik=3=0
(i+1)2=2i , (1-i)2=-2i
$\frac{1+i}{1-i}=i,\frac{1-i}{i+i}=-i$1+i1i =i,1ii+i =i
Real Part of a Complex Number
If x and y are two real numbers, then a number of the form z=x+iy is called a complex
number. Here ‘x’ is called the real part of z. The real
part of z is denoted by Re(z)
Imaginary Part of a Complex Number
If x and y are two real numbers, then a number of the form z=x+iy. Here ‘y’ is called the Imaginary part of z. The Imaginary
part of z is denoted by Im(z)
If z = 5 – 4i, then Re(z) = 5 and Im(z) = – 4
Note :
• A complex number z is purely real if its imaginary part is zero i.e., Im(z) = 0 and
purely imaginary if its real part is zero i.e., Re(z) = 0.
• I can be denoted by the ordered pair (0,1).
• The complex number (a, b) can also be split as (a, 0) + (0, 1) (b, 0)
Equality of Two Complex Numbers.
Two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 are said to be equal if and only if their
real parts and imaginary parts are separately equal.
i.e., z1= z2 Þ x1 + iy1 = x2 + iy2 Û x1 = x2 and y1= y2
Thus , one complex equation is equivalent to two real equations.
Note :
• A complex number z=x+iy.=0 iff . x=0, y=0
• The complex number do not possess the property of order i.e., (a + ib) < (or) > (c + id) is
not defined. For example, the statement 9 + 6i > 3 + 2i makes no sense.
Representation of Complex Number.
A complex number can be represented in the following from:
(1) Geometrical representation (Cartesian representation): The complex number z = a + ib =(a, b)
is represented by a point P whose coordinates are referred to rectangular axes XOX ¢ and YOY ¢ which are called real and imaginary axis respectively. Thus a complex number z is represented by a point in a plane, and corresponding to every point in this plane there exists a complex number such a plane is called argand plane or argand diagram or complex plane or gaussian plane.
Note :
• Distance of any complex number from the origin is called the modules of complex number and is denoted by |z|, i.e |z|= $\sqrt{a^2+b^2}$a2+b2
• Angle of any complex number with positive direction of x– axis is called amplitude or argument of z. i.e., amp (z)= arg(z)= tan -1(b/a)
(2)Trigonometrical (Polar) representation : In D OPM, let OP = r , then a= rcosq and b=. r sinq
Hence z can be expressed as z = r(cosq + i sinq )
where r = |z| and q = principal value of argument of z.
For general values of the argument z = r[cos(2np +q )+ i sin(2np +q )]
(3) Vector representation :
If P is the point (a, b) on the argand plane corresponding to the complex number z = a + ib .
Then OP̅=aî+bĵ, \| OP̅|=√a²+b² and arg z = direction of the vector OP̅= tan-1(b/a)
Therefore, complex number z can also be represented by OP̅
(4) Eulerian representation (Exponential form) :
Since we have eiq =cosq + i sinq and thus z can be expressed as z = r eiq , where r = |z| and q = arg(z).
Logarithm of complex number
Geometry of complex number
Square Root of a Complex Number
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# Multiplying 2-3 Digit Numbers Without Regrouping Worksheets
This multiplication worksheet concentrates on training college students how you can psychologically grow complete amounts. Individuals can use customized grids to fit precisely one particular issue. The worksheets also deal withfractions and decimals, and exponents. You can even find multiplication worksheets having a distributed property. These worksheets are a should-have for your personal mathematics school. They may be employed in class to figure out how to psychologically multiply entire numbers and line them up. Multiplying 2-3 Digit Numbers Without Regrouping Worksheets.
## Multiplication of whole phone numbers
You should consider purchasing a multiplication of whole numbers worksheet if you want to improve your child’s math skills. These worksheets will help you master this basic concept. You are able to choose to use one particular digit multipliers or two-digit and a few-digit multipliers. Capabilities of 10 can also be an excellent alternative. These worksheets will enable you to process extended practice and multiplication reading through the figures. They are also a great way to support your youngster comprehend the value of knowing the different kinds of complete figures.
## Multiplication of fractions
Possessing multiplication of fractions on a worksheet will help teachers prepare and get ready lessons successfully. Utilizing fractions worksheets allows educators to quickly determine students’ understanding of fractions. Individuals can be questioned to finish the worksheet within a specific time and then symbol their answers to see where they require more instruction. Individuals may benefit from phrase problems that associate maths to real-life circumstances. Some fractions worksheets consist of instances of contrasting and comparing numbers.
## Multiplication of decimals
Once you increase two decimal phone numbers, make sure to team them vertically. If you want to multiply a decimal number with a whole number, the product must contain the same number of decimal places as the multiplicant. For instance, 01 x (11.2) by 2 will be similar to 01 x 2.33 x 11.2 except if the merchandise has decimal spots of less than two. Then, the product is rounded for the nearest entire amount.
## Multiplication of exponents
A math concepts worksheet for Multiplication of exponents can help you practice multiplying and dividing amounts with exponents. This worksheet will also offer conditions that requires students to multiply two different exponents. By selecting the “All Positive” version, you will be able to view other versions of the worksheet. In addition to, you can even enter specific instructions in the worksheet on its own. When you’re concluded, you can click “Generate” and also the worksheet will be acquired.
## Division of exponents
The essential guideline for section of exponents when multiplying figures would be to subtract the exponent from the denominator from the exponent inside the numerator. However, if the bases of the two numbers are not the same, you can simply divide the numbers using the same rule. For instance, \$23 divided by 4 will equivalent 27. However, this method is not always accurate. This technique can result in misunderstandings when multiplying amounts that happen to be too big or too small.
## Linear capabilities
You’ve probably noticed that the cost was \$320 x 10 days if you’ve ever rented a car. So the total rent would be \$470. A linear function of this kind has got the form f(by), in which ‘x’ is the amount of times the automobile was hired. In addition, it offers the form f(x) = ax b, in which ‘b’ and ‘a’ are real phone numbers.
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## Simplifying Expressions Using the Distributive Property
### Learning Outcomes
• Simplify expressions using the distributive property
## Simplify Expressions Using the Distributive Property
Suppose three friends are going to the movies. They each need $9.25$; that is, $9$ dollars and $1$ quarter. How much money do they need all together? You can think about the dollars separately from the quarters.
They need $3$ times $9$, so $27$, and $3$ times $1$ quarter, so $75$ cents. In total, they need $27.75$.
If you think about doing the math in this way, you are using the Distributive Property.
### Distributive Property
If $a,b,c$ are real numbers, then
$a\left(b+c\right)=ab+ac$
Back to our friends at the movies, we could show the math steps we take to find the total amount of money they need like this:
$3(9.25)\\3(9\quad+\quad0.25)\\3(9)\quad+\quad3(0.25)\\27\quad+\quad0.75\\27.75$
In algebra, we use the Distributive Property to remove parentheses as we simplify expressions. For example, if we are asked to simplify the expression $3\left(x+4\right)$, the order of operations says to work in the parentheses first. But we cannot add $x$ and $4$, since they are not like terms (Go here for a review on adding like terms). So we use the Distributive Property, as shown in the next example.
### example
Simplify: $3\left(x+4\right)$
Solution:
$3\left(x+4\right)$ Distribute. $3\cdot x+3\cdot 4$ Multiply. $3x+12$
Some students find it helpful to draw in arrows to remind them how to use the Distributive Property. Then the first step in the previous example would look like this:
$3\cdot x+3\cdot 4$
Now you try.
### try it
In our next example, there is a coefficient on the variable y. When you use the distributive property, you multiply the two numbers together, just like simplifying any product. You will also see another example where the expression in parentheses is subtraction, rather than addition. You will need to be careful to change the sign of your product.
### example
Simplify: $6\left(5y+1\right)$
Simplify: $2\left(x - 3\right)$
Now you try.
### try it
The distributive property can be used to simplify expressions that look slightly different from $a\left(b+c\right)$. Here are two other forms.
### different Forms of the Distributive Property
If $a,b,c$ are real numbers, then
$a\left(b+c\right)=ab+ac$
Other forms
$a\left(b-c\right)=ab-ac$
$\left(b+c\right)a=ba+ca$
In the following video we show more examples of using the distributive property.
## Using the Distributive Property With Fractions and Decimals
Do you remember how to multiply a fraction by a whole number? We’ll need to do that in the next two examples. The distributive property comes in all shapes and sizes, and can include fractions or decimals as well.
### example
Simplify: $\Large\frac{3}{4}\normalsize\left(n+12\right)$
Simplify: $8\Large\left(\frac{3}{8}\normalsize x+\Large\frac{1}{4}\right)$.
Now you try.
### try it
Using the Distributive Property as shown in the next example will be very useful when we solve money applications later.
### example
Simplify: $100\left(0.3+0.25q\right)$
Now you try.
## Distributing a Variable
In the next example we’ll multiply by a variable. We’ll need to do this in a later chapter.
### example
Simplify: $m\left(n - 4\right)$
Now you try.
## The Backwards Form of the Distributive Property
The next example will use the ‘backwards’ form of the Distributive Property, $\left(b+c\right)a=ba+ca$.
### example
Simplify: $\left(x+8\right)p$
## Distributing a Negative Term
When you distribute a negative number, you need to be extra careful to get the signs correct.
### example
Simplify: $-2\left(4y+1\right)$
Simplify: $-11\left(4 - 3a\right)$
### try it
In the next example, we will show how to use the Distributive Property to find the opposite of an expression. Remember, $-a=-1\cdot a$.
### example
Simplify: $-\left(y+5\right)$
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# 2013 AIME II Problems/Problem 10
## Problem 10
Given a circle of radius $\sqrt{13}$, let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$. A line passing through the point $A$ intersects the circle at points $K$ and $L$. The maximum possible area for $\triangle BKL$ can be written in the form $\frac{a - b\sqrt{c}}{d}$, where $a$, $b$, $c$, and $d$ are positive integers, $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$.
## Solution 1
Now we put the figure in the Cartesian plane, let the center of the circle $O (0,0)$, then $B (\sqrt{13},0)$, and $A(4+\sqrt{13},0)$
The equation for Circle O is $x^2+y^2=13$, and let the slope of the line$AKL$ be $k$, then the equation for line$AKL$ is $y=k(x-4-\sqrt{13})$
Then we get $(k^2+1)x^2-2k^2(4+\sqrt{13})x+k^2\cdot (4+\sqrt{13})^2-13=0$, according to Vieta's Formulas, we get
$x_1+x_2=\frac{2k^2(4+\sqrt{13})}{k^2+1}$, and $x_1x_2=\frac{(4+\sqrt{13})^2\cdot k^2-13}{k^2+1}$
So, $LK=\sqrt{1+k^2}\cdot \sqrt{(x_1+x_2)^2-4x_1x_2}$
Also, the distance between $B$ and $LK$ is $\frac{k\times \sqrt{13}-(4+\sqrt{13})\cdot k}{\sqrt{1+k^2}}=\frac{-4k}{\sqrt{1+k^2}}$
So the ares $S=0.5ah=\frac{-4k\sqrt{(16-8\sqrt{13})k^2-13}}{k^2+1}$
Then the maximum value of $S$ is $\frac{104-26\sqrt{13}}{3}$
So the answer is $104+26+13+3=\boxed{146}$.
## Solution 2
Draw $OC$ perpendicular to $KL$ at $C$. Draw $BD$ perpendicular to $KL$ at $D$.
$$\frac{\triangle OKL}{\triangle BKL} = \frac{OC}{BD} = \frac{AO}{AB} = \frac{4+\sqrt{13}}{4}$$
Therefore, to maximize area of $\triangle BKL$, we need to maximize area of $\triangle OKL$.
$$\triangle OKL = \frac12 r^2 \sin{\angle KOL}$$
So when area of $\triangle OKL$ is maximized, $\angle KOL = \frac{\pi}{2}$.
Eventually, we get $$\triangle BKL= (\frac12 \cdot \sqrt{13}^2)\cdot(\frac{4}{4+\sqrt{13}})=\frac{104-26\sqrt{13}}{3}$$
So the answer is $104+26+13+3=\boxed{146}$.
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Formula: (30/100) × 200 = 60 copy
Leave: Feedback
## 30 percent of 200?
How to calculator the 30% of the number 200?. We will explain how to do that. and you can see the values in the charts below. And you can calculate more percentages of any values other than (30% and 200), You can do this with the calculator above.
30%
60
70%
140
+200
30%
60
70%
140
+200
### Solution for What is 30% of 200?:
Firstly, to calculate (30% of 200) we use the next formula.
Formula: ( X / 100 ) × Y copy
• X refers to the percentage in our case, it's 30%.
• 100 It's 100% (One hundred percent).
• Y refers to the number, it's 200.
#### Calculation steps
Step 1 30 / 100 × 200
We divide 30 (thirty ) by 100 (one hundred) then multiply the result by 200 (two hundred ).
Step 2 = 0.3 × 200
30 divided by 100 equals 0.3 ( point three). The next step, we multiply the 0.3 by 200 (two hundred ).
Step 3 = 60 (sixty )
0.3 multiplied by 200 equals 60.
The result30% of 200 is 60. In words: thirty percent of two hundred is sixty .
copy
## 200 percent of 30?
### Solution for What is 200 percent of 30?:
#### Calculation steps
Step 1 200 / 100 × 30
We divide 200 (two hundred ) by 100 (one hundred) then multiply the result by 30 (thirty ).
Step 2 = 2 × 30
200 divided by 100 equals 2 (two ). The next step, we multiply the 2 by 30 (thirty ).
Step 3 = 60 (sixty )
2 multiplied by 30 equals 60.
The resultTherefore, 200% of 30 is 60. In words: two hundred percent of thirty is sixty .
Related Calculations
### 30% of 200 similar calculation
30 percent of 200? = 60 30 percent of 201? = 60.3 30 percent of 202? = 60.6 30 percent of 203? = 60.9 30 percent of 204? = 61.2 30 percent of 205? = 61.5 30 percent of 206? = 61.8 30 percent of 207? = 62.1 30 percent of 208? = 62.4 30 percent of 209? = 62.7
30 percent of 210? = 63 30 percent of 220? = 66 30 percent of 230? = 69 30 percent of 240? = 72 30 percent of 250? = 75 30 percent of 260? = 78 30 percent of 270? = 81 30 percent of 280? = 84 30 percent of 290? = 87 30 percent of 300? = 90
30 percent of 400? = 120 30 percent of 500? = 150 30 percent of 600? = 180 30 percent of 700? = 210 30 percent of 800? = 240 30 percent of 900? = 270 30 percent of 1000? = 300 30 percent of 1100? = 330 30 percent of 1200? = 360 30 percent of 1300? = 390
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How do you find the slope and intercept to graph y = 5x - 9?
Nov 10, 2015
Looking at the $m$ and $c$ values of the equation.
Explanation:
General Form of a Straight Line Equation
The general form of an equation for a straight line is $y = m x + c$ In your case $m = 5$ and $c = - 9$
$m$ tells us the gradient or 'slope'. In your case this is 5, which means for every one coordinate you move along on the $x$ axis, your line will move up 5 coordinates in the $y$ axis.
The $y$ Intercept
As for the $y$ intercept, this is given by the value of $c$. In your case this is -9. Since we know that $x = 0$ on the $y$ axis then we know -9 is reffering to the $y$ coordinate. This means the $y$ intercept is at the coordinate $\left(0 , - 9\right)$
The $x$ Intercept
If you want to figure out the $x$ intercept as well then follow the method below.
On the $x$ axis we know that $y = 0$ we can put this into our equation and rearrange to find the value for $x$ at this point.
$0 = 5 x - 9$
$9 = 5 x$ (Add 9 to both sides)
$\frac{9}{5} = x$ (Divide by 5 on both sides)
We know know $x = \frac{9}{5}$ or $x = 1.8$
So the $x$ intercept is at the point $\left(1.8 , 0\right)$
This should be enough information to draw the graph!
Nov 10, 2015
See the explanation.
Explanation:
$y = 5 x - 9$ is in slope-intercept form for a linear equation, $y = m x + b$, where $m$ is the slope and $b$ is the y-intercept.
By definition, $5$ is the slope or $\left(\frac{5}{1}\right)$, and the y-intercept is $- 9$.
The y-intercept is the value of $y$ when $x = 0$.
$y = 5 \left(0\right) - 9 = - 9$
The point is $\left(0 , - 9\right)$
The x-intercept is the value of $x$ when $y = 0$.
$0 = 5 x - 9$
Add $9$ to both sides.
$9 = 5 x$
Divide both sides by $5$.
$\frac{9}{5} = x$
Switch sides.
$x = \frac{9}{5} = 1.8$
The point is $\left(\frac{9}{5} , 0\right)$
You can plot the x and y-intercepts and draw a straight line through the two points.
graph{y=5x-9 [-14.49, 17.53, -11.41, 4.61]}
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# Math Expressions Grade 1 Unit 5 Lesson 4 Answer Key Small Group Practice with Teen Problems
## Math Expressions Common Core Grade 1 Unit 5 Lesson 4 Answer Key Key Small Group Practice with Teen Problems
Math Expressions Grade 1 Unit 5 Lesson 4 Homework
Show your work. Use drawings, numbers, or words.
Solve the Story problem.
Small Group Practice with Teen Problems Grade 1 Unit 5 Lesson 4 Answer Key Question 1.
Hakim draws 8 stars. Lisa draws 7 stars. How many stars do they draw altogether?
Explanation:
Hakim draws 8 stars. Lisa draws 7 stars
15 stars do they draw altogether.
Unit 5 Lesson 4 Practice Problems Answer Key Grade 1 Question 2.
There are 13 markers in a box. Jorge takes some out. Now there are 8 in the box. How many markers does Jorge take out of the box?
Explanation:
There are 13 markers in a box.
Jorge takes some out. Now there are 8 in the box.
5 markers Jorge take out of the box
Unit 5 Lesson 4 Small Group Practice with Teen Problems Question 3.
Karla plants 7 flowers. Then she plants 5 more flowers. How many flowers does Karla plant?
Explanation:
Karla plants 7 flowers.
Then she plants 5 more flowers
Karla plant 12 flowers in all
Subtract.
Unit 5 Lesson 4 Practice Problems Answer Key Grade 1 Question 4.
Explanation:
The difference of 16 and 7 is 9
16 – 7 = 9
Unit 5 Lesson 4 Small Group Practice with Teen Problems Question 5.
Explanation:
The difference of 13 and 5 is 8
13 – 5 =8
Explanation:
The difference of 15 and 8 is 7
15 – 8 = 7
Small Group Practice with Teen Problems Answers Grade 1 Unit 5 Question 7.
Explanation:
The difference of 14 and 9 is 5
Math Expressions Grade 1 Unit 5 Lesson 4 Remembering
Find the total number of toys.
Math Expressions Grade 1 Unit 5 Lesson 4 Answer Key Question 1.
7 train cars in the box
Explanation:
The box has 7 trains inside
3 trains outside so, total there are 10 toy trains
7 + 3 = 10
Question 2.
5 bears in the box
Explanation:
The box contains 5 bears inside
3 bears outside so, total they are 7 bears
5 + 2 = 7
Question 3.
6 + 2 = ___
Explanation :
The sum of 6 and 2 is 8
6 + 2 = 8
Question 4.
5 + 1 = ___
Explanation :
The sum of 5and 1 is 6
5 + 1 = 6
Question 5.
7 + 3 = ___
Explanation :
The sum of 7 and 3 is 10
7 + 3 = 10
Question 6.
5 + 5 = ___
Explanation :
The sum of 5 and 5 is 10
5 + 5 = 10
Question 7.
1 + 6 = ___
Explanation :
The sum of 1 and 6 is 7
1 + 6 = 7
Question 8.
4 + 5 = ___
Explanation :
The sum of 4 and 5 is 9
4 + 5 = 9
Question 9.
Stella picks 8 red flowers and 9 yellow flowers. How many flowers does she pick?
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# 1.4: Using Fractions, Decimals and Percents to Describe Charts
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Learning Outcomes
1. Interpret bar charts using fractions, decimals and percents
2. Interpret pie charts using fractions, decimals and percents
Charts, such as bar charts and pie charts are visual ways of presenting data. You can think of each slice of the pie or each bar as a part of the whole. The numerical versions of this are a list of fractions, decimals and percents. By the end of this section we will be able to look at one of these charts and produce the corresponding fractions, decimals, and percents.
Bar charts occur frequently and it is definitely required to understand how to read them and interpret them in statistics. Often we want to convert the information of a bar chart to information shown numerically. We need fractions and/or percents to do this.
Example $$\PageIndex{1}$$
The above bar chart shows the demographics of California in 2019 where the numbers represent millions of people. Here are some questions that might come up in a statistics class.
1. What fraction of Californians was Hispanic in 2019?
2. What proportion of all Californians was White in 2019? Write your answer as a decimal number rounded to four decimal places.
3. What percent of Californians who were neither Hispanic nor White in 2019? Round your answer to the nearest percent.
Solution
1. To find the fraction of California that was Hispanic in 2019, the numerator will be the total number of Hispanics and the denominator will be the total number of people in California in 2019. The height of the bar that represents Hispanics is 15. Therefore the numerator is 15. To find the total number of people in California, we add up the heights of the three bars:
$15+13+10\:=\:38 \nonumber$
Now we can just write down the fraction:
$\frac{15}{38} \nonumber$To find the proportion of Californians who were White in 2019, we start in the same way. The numerator will be the number of Whites: 13. The denominator will be the total number of Californians which we already computed as 38. Therefore the fraction of Californians who were White is:
$\frac{13}{38} \nonumber$ To convert this to a decimal, we use a calculator to get:
$\frac{13}{38}\approx0.342105 \nonumber$Next round to four decimal places. Since the digit to the right of the fourth decimal place is $$0\:<5$$, we round down to:
$0.3421 \nonumber$
2. To find the percent of Californians who were neither Hispanic nor White in 2019, we first find the fraction who were neither. The numerator will be the number of "Other" which is: 10. The denominator will be the total which is 38. Thus the fraction is: $\frac{10}{38} \nonumber$ Next, use a calculator to divide these numbers to get: $\frac{10}{38}\approx0.263158 \nonumber$ To convert this to a percent we multiply by 100% by moving the decimal two places to the right: $0.263158\:\times100\%\:=\:26.3158\% \nonumber$ Finally we round to the nearest whole number. Noting that $$3 < 5$$, we round down to get: 26%
Exercise
The bar chart below shows the grade distribution for a math class.
2. Find the proportion of grades below a "C". Write your answer as a decimal number rounded to the nearest hundredth.
Another important chart that is used to display the components of a whole is a pie chart. With a pie chart, it is very easy to determine the percent of each item.
Example $$\PageIndex{2}$$
The pie chart below shows the makeup of milk. Write the proportion of fat contained in milk as a decimal.
Solution
We see that 31% of milk is fat. To convert a percent to a decimal, we just move the decimal over two places to the left. Thus, 31% becomes 0.31.
Example $$\PageIndex{3}$$
The pie chart above shows the number of pets of each type that had to be euthanized by the humane society due to incurable illnesses.
1. What fraction of the euthanized pets were dogs?
2. What percent of the euthanized pets were cats? Round to the nearest whole number percent.
Solution
1. We take the number of dogs over the total. There were 334 euthanized dogs. To find the total we add: $737+37+334\:=\:1108 \nonumber$
Therefore, the fraction of euthanized dogs is $\frac{334}{1108} \nonumber$
2. To find the percent of euthanized cats, we first find the fraction. There were 737 cats over a total of 1108 pets. The fraction is $\frac{737}{1108} \nonumber$
Next use a calculator to get the decimal number: 0.66516. Now multiply by 100% by moving the decimal place two digits to the right to get: 66.516%. Finally, we need to round to the nearest whole number percent. Since $$5\ge\:5$$, we round up. Thus the percent of euthanized cats is 67%.
This page titled 1.4: Using Fractions, Decimals and Percents to Describe Charts is shared under a CC BY license and was authored, remixed, and/or curated by Larry Green.
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Giáo trình
# Introductory Statistics
Mathematics and Statistics
## Test of a Single Variance
Tác giả: OpenStaxCollege
A test of a single variance assumes that the underlying distribution is normal. The null and alternative hypotheses are stated in terms of the population variance (or population standard deviation). The test statistic is:
$\frac{\left(n-1\right){s}^{2}}{{\sigma }^{2}}$
where:
• n = the total number of data
• s2 = sample variance
• σ2 = population variance
You may think of s as the random variable in this test. The number of degrees of freedom is df = n - 1. A test of a single variance may be right-tailed, left-tailed, or two-tailed. [link] will show you how to set up the null and alternative hypotheses. The null and alternative hypotheses contain statements about the population variance.
Math instructors are not only interested in how their students do on exams, on average, but how the exam scores vary. To many instructors, the variance (or standard deviation) may be more important than the average.
Suppose a math instructor believes that the standard deviation for his final exam is five points. One of his best students thinks otherwise. The student claims that the standard deviation is more than five points. If the student were to conduct a hypothesis test, what would the null and alternative hypotheses be?
Even though we are given the population standard deviation, we can set up the test using the population variance as follows.
• H0: σ2 = 52
• Ha: σ2 > 52
With individual lines at its various windows, a post office finds that the standard deviation for normally distributed waiting times for customers on Friday afternoon is 7.2 minutes. The post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes.
With a significance level of 5%, test the claim that a single line causes lower variation among waiting times (shorter waiting times) for customers.
Since the claim is that a single line causes less variation, this is a test of a single variance. The parameter is the population variance, σ2, or the population standard deviation, σ.
Random Variable: The sample standard deviation, s, is the random variable. Let s = standard deviation for the waiting times.
• H0: σ2 = 7.22
• Ha: σ2 < 7.22
The word "less" tells you this is a left-tailed test.
Distribution for the test: ${\chi }_{24}^{2}$, where:
• n = the number of customers sampled
• df = n – 1 = 25 – 1 = 24
Calculate the test statistic:
where n = 25, s = 3.5, and σ = 7.2.
Graph:
Probability statement: p-value = P ( χ2 < 5.67) = 0.000042
Compare α and the p-value:
• α = 0.05
• p-value = 0.000042
• α > p-value
• Make a decision: Since α > p-value, reject H0. This means that you reject σ2 = 7.22. In other words, you do not think the variation in waiting times is 7.2 minutes; you think the variation in waiting times is less.
Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that a single line causes a lower variation among the waiting times or with a single line, the customer waiting times vary less than 7.2 minutes.
# References
“AppleInsider Price Guides.” Apple Insider, 2013. Available online at http://appleinsider.com/mac_price_guide (accessed May 14, 2013).
Data from the World Bank, June 5, 2012.
# Chapter Review
To test variability, use the chi-square test of a single variance. The test may be left-, right-, or two-tailed, and its hypotheses are always expressed in terms of the variance (or standard deviation).
# Formula Review
${\chi }^{2}=$ $\frac{\left(n-1\right)\cdot {s}^{2}}{{\sigma }^{2}}$ Test of a single variance statistic where:
n: sample size
s: sample standard deviation
σ: population standard deviation
df = n – 1 Degrees of freedom
Test of a Single Variance
• Use the test to determine variation.
• The degrees of freedom is the number of samples – 1.
• The test statistic is $\frac{\left(n–1\right)\cdot {s}^{2}}{{\sigma }^{2}}$, where n = the total number of data, s2 = sample variance, and σ2 = population variance.
• The test may be left-, right-, or two-tailed.
Use the following information to answer the next three exercises: An archer’s standard deviation for his hits is six (data is measured in distance from the center of the target). An observer claims the standard deviation is less.
What type of test should be used?
a test of a single variance
State the null and alternative hypotheses.
Is this a right-tailed, left-tailed, or two-tailed test?
a left-tailed test
Use the following information to answer the next three exercises: The standard deviation of heights for students in a school is 0.81. A random sample of 50 students is taken, and the standard deviation of heights of the sample is 0.96. A researcher in charge of the study believes the standard deviation of heights for the school is greater than 0.81.
What type of test should be used?
State the null and alternative hypotheses.
H0: σ2 = 0.812;
Ha: σ2 > 0.812
df = ________
Use the following information to answer the next four exercises: The average waiting time in a doctor’s office varies. The standard deviation of waiting times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the variance of waiting times is greater than originally thought.
What type of test should be used?
a test of a single variance
What is the test statistic?
What is the p-value?
0.0542
What can you conclude at the 5% significance level?
# Homework
Use the following information to answer the next twelve exercises: Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes.
Is the traveler disputing the claim about the average or about the variance?
A sample standard deviation of 15 minutes is the same as a sample variance of __________ minutes.
225
Is this a right-tailed, left-tailed, or two-tailed test?
H0: __________
H0: σ2 ≤ 150
df = ________
chi-square test statistic = ________
36
p-value = ________
Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade the p-value.
Check student’s solution.
Let α = 0.05
Decision: ________
Conclusion (write out in a complete sentence.): ________
How did you know to test the variance instead of the mean?
The claim is that the variance is no more than 150 minutes.
If an additional test were done on the claim of the average delay, which distribution would you use?
If an additional test were done on the claim of the average delay, but 45 flights were surveyed, which distribution would you use?
a Student's t- or normal distribution
For each word problem, use a solution sheet to solve the hypothesis test problem. Go to [link] for the chi-square solution sheet. Round expected frequency to two decimal places.
A plant manager is concerned her equipment may need recalibrating. It seems that the actual weight of the 15 oz. cereal boxes it fills has been fluctuating. The standard deviation should be at most 0.5 oz. In order to determine if the machine needs to be recalibrated, 84 randomly selected boxes of cereal from the next day’s production were weighed. The standard deviation of the 84 boxes was 0.54. Does the machine need to be recalibrated?
Consumers may be interested in whether the cost of a particular calculator varies from store to store. Based on surveying 43 stores, which yielded a sample mean of \$84 and a sample standard deviation of \$12, test the claim that the standard deviation is greater than \$15.
1. H0: σ = 15
2. Ha: σ > 15
3. df = 42
4. chi-square with df = 42
5. test statistic = 26.88
6. p-value = 0.9663
7. Check student’s solution.
1. Alpha = 0.05
2. Decision: Do not reject null hypothesis.
3. Reason for decision: p-value > alpha
4. Conclusion: There is insufficient evidence to conclude that the standard deviation is greater than 15.
Isabella, an accomplished Bay to Breakers runner, claims that the standard deviation for her time to run the 7.5 mile race is at most three minutes. To test her claim, Rupinder looks up five of her race times. They are 55 minutes, 61 minutes, 58 minutes, 63 minutes, and 57 minutes.
Airline companies are interested in the consistency of the number of babies on each flight, so that they have adequate safety equipment. They are also interested in the variation of the number of babies. Suppose that an airline executive believes the average number of babies on flights is six with a variance of nine at most. The airline conducts a survey. The results of the 18 flights surveyed give a sample average of 6.4 with a sample standard deviation of 3.9. Conduct a hypothesis test of the airline executive’s belief.
1. H0: σ ≤ 3
2. Ha: σ > 3
3. df = 17
4. chi-square distribution with df = 17
5. test statistic = 28.73
6. p-value = 0.0371
7. Check student’s solution.
1. Alpha: 0.05
2. Decision: Reject the null hypothesis.
3. Reason for decision: p-value < alpha
4. Conclusion: There is sufficient evidence to conclude that the standard deviation is greater than three.
The number of births per woman in China is 1.6 down from 5.91 in 1966. This fertility rate has been attributed to the law passed in 1979 restricting births to one per woman. Suppose that a group of students studied whether or not the standard deviation of births per woman was greater than 0.75. They asked 50 women across China the number of births they had had. The results are shown in [link]. Does the students’ survey indicate that the standard deviation is greater than 0.75?
# of births Frequency 0 5 1 30 2 10 3 5
According to an avid aquarist, the average number of fish in a 20-gallon tank is 10, with a standard deviation of two. His friend, also an aquarist, does not believe that the standard deviation is two. She counts the number of fish in 15 other 20-gallon tanks. Based on the results that follow, do you think that the standard deviation is different from two? Data: 11; 10; 9; 10; 10; 11; 11; 10; 12; 9; 7; 9; 11; 10; 11
1. H0: σ = 2
2. Ha: σ ≠ 2
3. df = 14
4. chi-square distiribution with df = 14
5. chi-square test statistic = 5.2094
6. p-value = 0.0346
7. Check student’s solution.
1. Alpha = 0.05
2. Decision: Reject the null hypothesis
3. Reason for decision: p-value < alpha
4. Conclusion: There is sufficient evidence to conclude that the standard deviation is different than 2.
The manager of "Frenchies" is concerned that patrons are not consistently receiving the same amount of French fries with each order. The chef claims that the standard deviation for a ten-ounce order of fries is at most 1.5 oz., but the manager thinks that it may be higher. He randomly weighs 49 orders of fries, which yields a mean of 11 oz. and a standard deviation of two oz.
You want to buy a specific computer. A sales representative of the manufacturer claims that retail stores sell this computer at an average price of \$1,249 with a very narrow standard deviation of \$25. You find a website that has a price comparison for the same computer at a series of stores as follows: \$1,299; \$1,229.99; \$1,193.08; \$1,279; \$1,224.95; \$1,229.99; \$1,269.95; \$1,249. Can you argue that pricing has a larger standard deviation than claimed by the manufacturer? Use the 5% significance level. As a potential buyer, what would be the practical conclusion from your analysis?
The sample standard deviation is \$34.29.
H0 : σ2 = 252
Ha : σ2 > 252
df = n – 1 = 7.
test statistic: ;
p-value:
Alpha: 0.05
Decision: Do not reject the null hypothesis.
Reason for decision: p-value > alpha
Conclusion: At the 5% level, there is insufficient evidence to conclude that the variance is more than 625.
A company packages apples by weight. One of the weight grades is Class A apples. Class A apples have a mean weight of 150 g, and there is a maximum allowed weight tolerance of 5% above or below the mean for apples in the same consumer package. A batch of apples is selected to be included in a Class A apple package. Given the following apple weights of the batch, does the fruit comply with the Class A grade weight tolerance requirements. Conduct an appropriate hypothesis test.
(a) at the 5% significance level
(b) at the 1% significance level
Weights in selected apple batch (in grams): 158; 167; 149; 169; 164; 139; 154; 150; 157; 171; 152; 161; 141; 166; 172;
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# learn how to graph tangent features with transformations – Lisbdnet.com
Graph a Tangent Transformation within the Kind: y=atan(bx+c)+d See additionally what number of bayous in houston
After which if we transfer proper one-fourth of the interval. We’re up at this level right here that has a y-Extra After which if we transfer proper one-fourth of the interval. We’re up at this level right here that has a y-coordinate of 1. If we return to the purpose of the origin. And transfer left one-fourth of the interval.
Graph a Tangent Operate
1. Discover the vertical asymptotes so you’ll find the area. These steps use x as an alternative of theta as a result of the graph is on the x–y aircraft. …
2. Decide values for the vary. …
3. Calculate the graph’s x-intercepts. …
4. Determine what’s taking place to the graph between the intercepts and the asymptotes.
## What’s the relation between SEC and tan?
How do folks bear in mind these items?
Verbal description Mathematical relationship
secant The secant is the reciprocal of the cosine. sec ( A ) = 1 cos ( A ) sec(A)=dfrac{1}{cos(A)} sec(A)=cos(A)1
cotangent The cotangent is the reciprocal of the tangent. cot ( A ) = 1 tan ( A ) cot(A)=dfrac{1}{tan(A)} cot(A)=tan(A)1
Including or subtracting a quantity from the angle variable of the tangent operate has the identical impact as with the sine and cosine — it strikes the curve to the left or proper. The graph of y = tan (x + 1) shifts one unit to the left, together with the asymptotes.
## How do you discover the section shift of a tangent operate?
If we take a look at a trigonometrical operate written within the kind:
1. y=atan(bx+c)+d.
2. Interval = πb ( That is the traditional interval of the operate divided by b )
3. Part shift = −cb.
4. y=tan(x+60)
5. interval =πc on this case we’re utilizing levels so:
6. interval =1801=180∘
7. Part shift =−cb=−601=60∘
8. Vertical shift =d=0 ( no vertical shift )
See additionally what number of kingdoms do scientists use to categorise earth’s organisms
As we all know, tan is the ratio of sin and cos, similar to tan θ = sin θ/cos θ. Thus, we are able to get the values of tan ratio for the precise angles.
## How do you draw a tangent?
Level to Tangents on a Circle
1. Draw a line connecting the purpose to the middle of the circle.
2. Assemble the perpendicular bisector of that line.
3. Place the compass on the midpoint, modify its size to succeed in the tip level, and draw an arc throughout the circle.
4. The place the arc crosses the circle would be the tangent factors.
Click on on the “Chart Instruments: Format” tab after which click on the “Trendline” button on the suitable. Choose “Linear” to create a straight line that approximates the progress of the curved line. It would intersect the curve at some factors.
## What are the important thing factors of a tangent graph?
Essential Traits of the Graph of the Tangent Operate
• The interval of the tangent operate is π : …
• The area of the tangent operate excludes π2+kπ π 2 + ok π for all integers ok ; …
• The vary of the tangent operate is the set of all actual numbers;
cot(x) = 1/tan(x) , so cotangent is principally the reciprocal of a tangent, or, in different phrases, the multiplicative inverse.
## What’s reciprocal transformation?
a metamorphosis of uncooked knowledge that entails (a) changing the unique knowledge models with their reciprocals and (b) analyzing the modified knowledge. In contrast to different transformations, a reciprocal transformation modifications the order of the unique knowledge. … Additionally referred to as inverse transformation.
## How do you discover asymptotes of a tan graph?
For any y=tan(x) y = tan ( x ) , vertical asymptotes happen at x=π2+nπ x = π 2 + n π , the place n is an integer. Use the fundamental interval for y=tan(x) y = tan ( x ) , (−π2,π2) ( – π 2 , π 2 ) , to search out the vertical asymptotes for y=tan(x) y = tan ( x ) .
## The place are the asymptotes of a tangent graph?
The asymptotes for the graph of the tangent operate are vertical traces that happen often, every of them π, or 180 levels, aside. They separate every bit of the tangent curve, or every full cycle from the subsequent. is as a result of these factors are the place the cosine operate is the same as 0.
## What’s the by-product of Secx TANX?
Calculus Examples
By the Sum Rule, the by-product of sec(x)+tan(x) sec ( x ) + tan ( x ) with respect to x is ddx[sec(x)]+ddx[tan(x)] d d x [ sec ( x ) ] + d d x [ tan ( x ) ] .
## Is tan 2x 1 Secx a Pythagorean identification?
That is readily derived straight from the definition of the fundamental trigonometric features sin and cos and Pythagoras’s Theorem. Confirming that the result’s an identification. … Sure, sec2−1=tan2x is an identification.
## Which is the same as Sinθ?
sin(2π + θ) = sinθ, cos(2π + θ) = cosθ, tan(2π + θ) = tanθ, cot(2π + θ) = cotθ.
## How do you change trigonometric features?
The section shift of a trigonometric operate is calculated utilizing the components section shift=CB section shift = C B When C is constructive, the graph will seem to shift to the suitable. When C is detrimental, the graph will shift to the left. Including a worth D to a trig operate will translate its graph vertically.
## How do you discover the section shift on a graph?
To seek out the section shift from a graph, it’s essential to:
1. Decide whether or not it’s a shifted sine or cosine.
2. Have a look at the graph to the suitable of the vertical axis.
3. Discover the primary: …
4. Calculate the space from the vertical line to that time.
5. If the operate was a sine, subtract π/2 from that distance.
## Graphing Sine and Cosine Trig Features With Transformations, Part Shifts, Interval – Area & Vary
graphing tangent transformations worksheetgraph of trigonometric features pdfgraphing trig features calculatorhorizontal shift tangent graphgraphing trig features examples
graphing trig features apply with solutions
See extra articles in class: FAQ
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# Trig identities
Here you will learn about trigonometric identities, including recognizing and working with key trigonometric identities, as well as applying algebraic skills to simplify the identities.
Students first learn how to work with trigonometric identities in Algebra II and expand that knowledge through Precalculus.
## What are trigonometric identities?
Trigonometric identities are mathematical equations that involve trigonometric functions, like sine, cosine, and tangent and they are true for all values of the variables involved.
An identity is an equation that is always true.
For example, when expanding the algebraic expression 2(x+1) using the distributive property you get 2 x+2. These two functions of x are equivalent so you can write the statement: 2(x+1) \equiv 2 x+2.
The same is true for trigonometric ratios and expressions.
Let’s take a look at several equivalent expressions that are trigonometric identities.
Using the right triangle:
Where O= Opposite side, A= Adjacent side, and H= Hypotenuse for the right-angled triangle with respect to angle \theta (theta). The following relationships are considered trigonometric identities because they are equivalent, equal or identical.
• \sin(\theta)=\cfrac{O}{H}
• \cos(\theta)=\cfrac{A}{H}
• \tan(\theta)=\cfrac{O}{A}
Where O= Opposite side, A= Adjacent side, and H= Hypotenuse for the right-angled triangle with respect to angle \theta (theta).
Let’s take a look at another trigonometric relationship that is equivalent, equal, or identical, hence making it a trigonometric identity.
\tan (\theta)=\cfrac{\sin (\theta)}{\cos (\theta)}
This relationship can be proven true using the same right triangle. Substitute \sin \theta with \cfrac{O}{H} and \cos \theta with \cfrac{A}{H}.
\tan (\theta)=\cfrac{\sin \theta}{\cos \theta}=\cfrac{\cfrac{O}{H}}{\cfrac{A}{H}}=\cfrac{O}{A}
This proves that \tan (\theta)=\cfrac{O}{A} which also equals \cfrac{\sin \theta}{\cos \theta} meaning that \tan (\theta)=\cfrac{\sin \theta}{\cos \theta} is always a true relationship so it is a trigonometric identity.
Let’s take a look at another trigonometric relationship that is equivalent, equal, or identical, hence making it a trigonometric identity.
\sin ^2(\theta)+\cos ^2(\theta)=1.
This relationship can be proven true using a right triangle. Notice how values are assigned to each of the sides. Using the numerical values might be easier to see the relationship.
In this case, \sin \theta=\cfrac{3}{5} and \cos \theta=\cfrac{4}{5} so, let’s substitute those values in for \sin \theta and \cos \theta.
\begin{aligned} \sin ^2(\theta)+\cos ^2(\theta)&=1 \\\\ \left(\cfrac{3}{5}\right)^2+\left(\cfrac{4}{5}\right)^2&=1 \\\\ \cfrac{9}{25}+\cfrac{16}{25}&=1 \\\\ \cfrac{25}{25}&=1 \\\\ 1&=1 \end{aligned}
This proves that \sin ^2(\theta)+\cos ^2(\theta)=1 is a true relationship meaning that it is equivalent, equal, or identical so it is a trigonometric identity.
\sin ^2(\theta)+\cos ^2(\theta)=1 is known as a Pythagorean theorem identity because you use the Pythagorean theorem to prove it true.
Both, \tan (\theta)=\cfrac{\sin \theta}{\cos \theta} and \sin ^2(\theta)+\cos ^2(\theta)=1 are identities that you can use to prove other identities.
## Common Core State Standards
How does this apply to high school math?
• High School Functions – Trigonometric Functions (HSF-TF.C.8)
Prove the Pythagorean identity \sin ^2(\theta)+\cos ^2(\theta)=1 and use it to find \sin(\theta), \, \cos(\theta), or \tan(\theta) given \sin(\theta), \, \cos(\theta), or \tan(\theta) and the quadrant of the angle.
## How to prove trigonometric identities
In order to prove trigonometric identities:
1. Choose one side of the identity to simplify.
2. Substitute values using the trigonometric ratios or another trigonometric identity.
3. Simplify the identity.
## Trigonometric identities examples
### Example 1: the relationship between sin, cos and tan
Prove that the trigonometric identity is true:
\cfrac{\sin \theta)}{\cos(\theta)}=\tan(\theta)
1. Choose one side of the identity to simplify.
In this case, use the left hand side of the identity to simplify.
2Substitute values using the trigonometric ratios or another trigonometric identity.
Using the relationships of the sides of a right triangle, you know that:
\sin (\theta)=\cfrac{O}{H}, \, \cos (\theta)=\cfrac{A}{H}, and \tan (\theta)=\cfrac{O}{A}
The equation can be rewritten to be:
\cfrac{\cfrac{O}{H}}{\cfrac{A}{H}}=\tan \theta
3Simplify the identity.
Use the rules of division of fractions:
\begin{aligned}\cfrac{\cfrac{O}{H}}{\cfrac{A}{H}}&=\tan \theta \\\\ \cfrac{O}{H} \div \cfrac{A}{H}&=\tan \theta \\\\ \cfrac{O}{H} \times \cfrac{H}{A}&=\tan \theta \\\\ \cfrac{O}{A}&=\tan \theta \\\\ \tan \theta&=\tan \theta \end{aligned}
Both sides of the equation are identical, meaning that the identity is proven to be true.
### Example 2: the relationship between sin, cos and tan
Verify the identity:
\sin \theta \cot \theta \sec \theta=1
Choose one side of the identity to simplify.
Substitute values using the trigonometric ratios or another trigonometric identity.
Simplify the identity.
### Example 3: the relationship between sin, cos and tan
Verify the identity:
\cot \theta=\cfrac{\cos \theta}{\sin \theta}
Choose one side of the identity to simplify.
Substitute values using the trigonometric ratios or another trigonometric identity.
Simplify the identity.
### Example 4: the relationship between sin, cos and tan
Verify the identity:
\cfrac{\cot \theta \sec \theta}{\csc \theta}=1
Choose one side of the identity to simplify.
Substitute values using the trigonometric ratios or another trigonometric identity.
Simplify the identity.
### Example 5: the relationship between sin, cos and tan
Verify the identity:
\sin x \sec x=\tan x
Choose one side of the identity to simplify.
Substitute values using the trigonometric ratios or another trigonometric identity.
Simplify the identity.
### Example 6: using Pythagoras’ theorem to prove a trigonometric identity
Verify the identity:
\sin ^2(\theta)+\cos ^2(\theta)=1
Choose one side of the identity to simplify.
Substitute values using the trigonometric ratios or another trigonometric identity.
Simplify the identity.
### Example 7: proving identities using exact trig values
Verify the identity \cos(60)=1-\sin(30) using the given right triangle.
Choose one side of the identity to simplify.
Substitute values using the trigonometric ratios or another trigonometric identity.
Simplify the identity.
### Teaching tips for trigonometric identities
• Have students investigate trig identities using inquiry based platforms such as Desmos.
• Instead of having students practice problems with worksheets, have them engage in collaborative learning activities such as jigsaw tasks, scavenger hunts, and/or gallery walks.
• Engage students with team game-playing to review skills with platforms such as Blooket and Quizziz.
### Easy mistakes to make
• Treating an identity like an equation
You do not solve an identity by balancing both sides of the equal sign like an equation. An identity is a statement where two expressions or functions are mathematically identical. For example, 3x+6\equiv{3}(x+2) because when it is factored, the expression becomes 3(x+2).
• Mixing up the right triangle trigonometric ratios
It is key to remember the three trig ratios that are written as the mnemonic SOHCAHTOA. This way you will not confuse the relationships between the sides and the angles.
\begin{aligned}\sin \theta & =\cfrac{O}{H} \\\\ \cos \theta & =\cfrac{A}{H} \\\\ \tan \theta & =\cfrac{O}{A} \\\\ \csc \theta & =\cfrac{H}{O} \text { (reciprocal of the sine) } \\\\ \sec \theta & =\cfrac{H}{A} \text { (reciprocal of the cosine) } \\\\ \cot \theta & =\cfrac{A}{O} \text { (reciprocal of the tangent) } \end{aligned}
### Practice trigonometric identities questions
1. Which of the following trigonometric identities is mathematically correct?
\sin(\theta)\times\cos(\theta)=\tan(\theta)
\sin(\theta)+\cos(\theta)=\tan(\theta)
\sin(\theta)\div\cos(\theta)=\tan(\theta)
\sin(\theta)-\cos(\theta)=\tan(\theta)
From SOHCAHTOA, you know that \sin(\theta)=\cfrac{O}{H} and \cos(\theta)=\cfrac{A}{H},
This means that:
\sin(\theta)\div\cos(\theta)=\cfrac{O}{H} \div \cfrac{A}{H}=\cfrac{O}{H}\times \cfrac{H}{A}=\cfrac{OH}{HA}=\cfrac{O}{A}=\tan(\theta)
2. Which of the following trigonometric identities is mathematically correct?
\sin \theta \csc \theta=\cot \theta
\cos \theta \csc \theta=\cot \theta
\cos \theta \csc \theta=\tan \theta
\cos \theta \sin \theta=\cot \theta
Using the trigonometric ratios from SOHCAHTOA, you know that \cos \theta=\cfrac{A}{H} and \csc \theta=\cfrac{H}{O}
Rewrite the trigonometric identity to be:
\left(\cfrac{A}{H}\right)\left(\cfrac{H}{O}\right)=\cot \theta
Simplify the left hand side of the identity,
\left(\cfrac{A}{O}\right)=\cot \theta
From SOHCAHTOA, the cotangent is the reciprocal of the tangent. So, if \tan \theta=\cfrac{O}{A} then it is true that \cot \theta=\cfrac{A}{o}
So,
\begin{aligned}& \left(\cfrac{A}{O}\right)=\cot \theta \\\\ & \cot \theta=\cot \theta \end{aligned}
You can also use the identity that \csc \theta=\cfrac{1}{\sin \theta} because the sine is the reciprocal of the cosecant which means you can rewrite the identity to be:
\begin{aligned} \cos \theta \times \cfrac{1}{\sin \theta}&=\cot \theta \\\\ \cfrac{\cos \theta}{\sin \theta}&=\cot \theta \end{aligned}
3. Which of the following trigonometric functions is identical to \cfrac{\tan(\theta)}{\sin(\theta)}?
\cos(\theta)
\sec(\theta)
\cos^{2}(\theta)
\cfrac{1}{\cos(\theta)}
Using the right triangle ratios from SOHCAHTOA, you know that
\sin(\theta)=\cfrac{O}{H}, \, \tan(\theta)=\cfrac{O}{A}, and \cos(\theta)=\cfrac{A}{H},
\tan(\theta)\div\sin(\theta) is equivalent to
\begin{aligned}=&\cfrac{O}{A}\div\cfrac{O}{H} \\\\ =&\cfrac{O}{A}\times\cfrac{H}{O} \\\\ =&\cfrac{OH}{AO} \\\\ =&\cfrac{H}{A} \\\\ =&\cfrac{1}{\cos(\theta)} \end{aligned}
\begin{aligned}& \cfrac{\tan \theta}{\sin \theta}=\cfrac{1}{\cos \theta} \\\\ & \cfrac{1}{\cos \theta}=\cfrac{1}{\cos \theta} \end{aligned}
4. Which solution is identical to 2\cos(45)\times\sin(45)?
\sqrt{2}
1
2\sqrt{2}
2
Using the right triangle and the ratios from SOHCAHTOA,
\begin{aligned}&\begin{aligned}& \cos \theta=\cfrac{A}{H} \\\\ & \cos 45=\cfrac{1}{\sqrt{2}} \end{aligned} \\\\ &\begin{aligned}& \sin \theta=\cfrac{O}{H} \\\\ & \sin 45=\cfrac{1}{\sqrt{2}} \end{aligned} \end{aligned}
\begin{aligned}& \cfrac{\tan \theta}{\sin \theta}=\cfrac{1}{\cos \theta} \\\\ & \cfrac{1}{\cos \theta}=\cfrac{1}{\cos \theta} \end{aligned}
Rewrite the identity to be:
\begin{aligned}& 2\left(\cfrac{1}{\sqrt{2}}\right) \times 2\left(\cfrac{1}{\sqrt{2}}\right)=4\left(\cfrac{1}{\sqrt{4}}\right) \\\\ & 4\left(\cfrac{1}{\sqrt{4}}\right)=4 \times \cfrac{1}{2}=\cfrac{4}{2}=2\end{aligned}
This verifies that
2=2
5. Select the expression that is identical to \cfrac{\cos \theta \tan \theta}{\sin \theta}.
1
\tan^{2}(\theta)
\cfrac{1}{\tan(\theta)}
\cfrac{\cos^{2}(\theta)}{\sin^{2}(\theta)}
Using the ratios from SOHCAHTOA, you know that:
\begin{aligned}\cos \theta & =\cfrac{A}{H} \\\\ \tan \theta & =\cfrac{A}{O} \\\\ \sin \theta & =\cfrac{O}{H} \end{aligned}
Rewrite the identity to be:
\left(\cfrac{A}{H} \times \cfrac{O}{A}\right) \div \cfrac{O}{H}
Using rules for multiplying and dividing fractions,
\cfrac{O}{H} \times \cfrac{H}{O}=\cfrac{O H}{O H}=1
Another way to solve it would be to use the identity that
\tan \theta=\cfrac{\sin \theta}{\cos \theta}
Rewrite the identity to be:
\begin{aligned}& \cfrac{\cos \theta \times \cfrac{\sin \theta}{\cos \theta}}{\sin \theta} \\\\ & \cfrac{\sin \theta}{\sin \theta}=1 \end{aligned}
6. Select the expression that is identical to \cfrac{\sin(\theta)}{\sqrt{1-\sin^{2}(\theta)}}?
\sin(\theta)
\cos(\theta)
\tan(\theta)
\cos^{2}(\theta)
Using the Pythagorean identity, \sin ^2 \theta+\cos ^2 \theta=1
The identity can be rearranged by solving for \cos \theta\text{:}
\begin{aligned}& \cos ^2 \theta=1-\sin ^2 \theta \\\\ & \cos \theta=\sqrt{1-\sin ^2 \theta} \end{aligned}
Rewrite the identity to be:
\cfrac{\sin \theta}{\cos \theta}
You know from SOHCAHTOA that
\begin{aligned}& \sin \theta=\cfrac{O}{H} \\\\ & \cos \theta=\cfrac{A}{H} \end{aligned}
So, \cfrac{\cfrac{O}{H}}{\cfrac{A}{H}}=\cfrac{O}{H} \div \cfrac{A}{H}=\cfrac{O}{H} \times \cfrac{H}{A}=\cfrac{O}{A}
\cfrac{O}{A}=\tan \theta
But, \cfrac{\sin \theta}{\cos \theta}=\tan \theta is an identity that can be used to verify other identities.
## Trigonometric identities FAQs
Do you always have to use the ratios from SOHCAHTOA?
No, you can use other relationships to help verify identities such as secant being the reciprocal of cosine, cotangent being the reciprocal of the tangent, etc.
Are there other identities other than reciprocal identities and Pythagorean trigonometric identities?
Yes, as you continue to learn about trig identities, there are other fundamental trigonometric identities that you will learn how to verify and use, cofunction identities, sum identities, difference identities, product identities, half-angle identities, and double angle identities. They are also referred to as double-angle formulas, half-angle formulas, difference formulas, and sum formulas.
Yes, the unit circle is helpful when verifying identities because it can be used as a base for proving identities. The unit circle has a radius of 1 and special right triangles can be formed within it where the radius of 1 is the hypotenuse of the triangles.
How do you solve trigonometric equations?
In order to solve trigonometric equations, you may have to apply fundamental trigonometric identities and the unit circle to find the radian measure of the angles in the equation.
Are the sine and cosine complements of each other?
Yes, the sine and cosine are complementary.
## The next lessons are
• Circle math
• Sectors, arcs and segments
• Circle theorems
## Still stuck?
At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts.
Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence.
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# Video: Multiply Monomials by Polynomials
Lucy Murray
Through a series of increasingly complicated examples, we learn how to use the distributive property to multiply monomials (expressions that have just one term, such as 3) by polynomials (expressions that have more than one term, such as 2𝑥 − 7).
08:25
### Video Transcript
Multiplying monomials by polynomials. So we can see that this is a monomial as we have three. That’s a monomial, meaning basically one term. And we’re multiplying it by a polynomial inside the parentheses which means basically more than one term. Well what we’re gonna do for this is apply the distributive property, where 𝑎 multiplied by all of 𝑏 plus 𝑐 is equal to 𝑎𝑏 plus 𝑎𝑐. So what we mean by that is we gonna take three or take 𝑎 and multiply it by the first term. So in this case will do three multiplied by two 𝑥. And then we’ll add to that three multiplied by the second term. So three multiplied by negative seven. And now we need to simplify it. So we know that three multiplied by two is six and then we can put an 𝑥 after it, cause there is an 𝑥 there as well. So six 𝑥 and then three multiplied by negative seven is negative twenty-one. So nice and simply, all we did there is we took a monomial and multiply it by the first term in the parentheses and then we added that to the multiplication of our monomial by the second term of the parentheses.
Let’s have a look at a different example. So we can see here with this next example, we’ve got a couple of things that are different. First of all, a monomial on the outside of the parentheses is not just a constant. It’s got a variable and also is negative. The other thing we can see is inside the parentheses we have three terms. Well it doesn’t really matter to us, because we’re going to use exactly the same method to expand these parentheses. We’re gonna use the distributive property. So first thing first good place to start, we’ll take the monomial and we’ll multiply it by our first term. And then we’ll add on to that monomial negative five 𝑥 multiplied by our second term. Be careful! Our second term don’t forget the negative in front of the two, because it’s always the sign in front of it that’s attached. So we’ve got negative five 𝑥 multiplied by negative two 𝑥. And then we need to multiply the last term, so we’ll have negative five 𝑥 multiplied by negative eight.
So then let’s try and simplify. We’ve got negative five. We multiply it by three. We get negative fifteen and then we know that 𝑥 multiplied by 𝑥 squared is the same as 𝑥 multiplied by 𝑥 multiplied by 𝑥. So we’ve got negative fifteen 𝑥 to the power of three or 𝑥 cubed. Then look at the signs for the next one. So we’ve got two negatives multiplied together. So that will cancel out. So we’ll have a positive. Now we’ve got five multiplied by two which is ten, and then 𝑥 multiplied by 𝑥 which is 𝑥 squared. So we’re adding on ten 𝑥 squared.
Now again let’s look at the signs. First, we’ve got a negative multiplied by a negative which gives us a positive. Then we’ve got five multiplied by eight which is forty. Then we’ve got an 𝑥, so add on forty 𝑥. So there we have it. We have completely multiplied out this set of parentheses. We’ve multiplied this monomial by this polynomial. For our next one, let’s have a look when it’s not just 𝑥 but also 𝑦.
So now we have five 𝑥 squared all multiplied by three 𝑦 plus two 𝑥 plus 𝑥𝑦. So again, we can see that we’ve got some 𝑥s and some 𝑦s, but we could have any variables. We’re gonna do exactly the same thing either way. We’re gonna take our first term and we’re gonna multiply it by a monomial. So we’ve got five 𝑥 squared multiplied by three 𝑦. And then we’re gonna add on to that monomial by the second term, so five 𝑥 squared multiplied by two 𝑥. And then five 𝑥 squared multiplied by 𝑥𝑦.
And we’re just gonna take it one term at a time. So we’ll do the numbers first. We’ve got five multiplied by three. We know that’s fifteen. And then 𝑥 squared multiplied by 𝑦, so it’s fifteen 𝑥 squared 𝑦. Then for our next one, it’s five multiplied by two which we know is ten. And then looking at the 𝑥s, we’ll have 𝑥 squared which is 𝑥 times 𝑥, then we’re timesing that by another 𝑥. So that gives us ten 𝑥 cubed or ten 𝑥 to the power of three. Then for the last one we’ve just got five, because it’s five times one. And then looking at the 𝑥 powers, we’ve got five 𝑥 to the power of three and then 𝑦. So there we have it. We’ve done that one as well. All we’ve done is we take the monomial and multiplied it by each term individually. Let’s have a look at our final example.
So before we do anything for this question, I want you to just have a look at it and think what’s gonna be the very first step to multiply out this monomial. And now I hope that none of you have thought two plus three, because it’s not two plus three then multiplied by the rest. It’s two plus three multiplied by everything. So the two is just by itself to add at the end. So what we’re gonna do and what you should always do in these things is first of all write the thing by itself straight down, so you don’t get tempted to do anything with it.
And now we’ve got a nice and simple multiplication to get on with and then collect the light terms after. So we’ve got three multiplied by the first term: so three multiplied by 𝑥 squared then three multiplied by two 𝑥 and then three multiplied by negative seven. So taking it one term at a time, we’d just write two again straight down, not to do anything with it. And we’ll add three multiplied by 𝑥 squared which is three 𝑥 squared. Three multiplied by two 𝑥, well three times two is six, so we’ve got six 𝑥. And then three multiplied by negative seven, well three multiplied by seven is twenty-one. Check the negative in front of it, because it’s a positive and a negative gives us a negative. So we’ve got negative twenty-one.
In this case, we’re not actually finished yet and that is because we have got some like terms that we need to collect. So what we’re gonna do is we’ve got negative twenty-one, add two. So we’ll have three 𝑥 squared plus six 𝑥 minus nineteen. And there we have it. We finished it. So be careful that whenever you’ve got something plus a monomial multiplied by a polynomial, it’s there to trick you. Make sure that you do not add it first, or in the case where it was, say two 𝑥 plus three all multiplied by 𝑥 squared plus two 𝑥 minus seven, that you don’t even worse try and make it a binomial multiplied by that polynomial. Because that is a much more challenging thing to work out and you’ll end up wasting loads of time doing something that’s not going to get you any marks. So be careful. Pay attention to parentheses to make sure that you know exactly what you’re doing.
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The square source of the number 138 is the turning back of squaring the number 11.7473 or increasing the number 11.7473 to the second power (11.74732). Come undo squaring, we take the square root.
You are watching: What is the square root of 138
Square root of 138 = 11.7473
## Is 138 a Perfect Square Root?
No. The square root of 138 is not an integer, for this reason √138 isn"t a perfect square.
Previous perfect square source is: 121
Next perfect square root is: 144
## How carry out You simplify the Square root of 138 in Radical Form?
The main point of simplification (to the simplest radical kind of 138) is as follows: acquiring the number 138 within the radical authorize √ as low as possible.
138 is currently simplified (have no pair prime factors).
## Is the Square source of 138 rational or Irrational?
Since 138 isn"t a perfect square (it"s square source will have an infinite number of decimals), it is one irrational number.
## The Babylonian (or Heron’s) method (Step-By-Step)
StepSequencing
1
In step 1, we have to make our first guess about the value of the square root of 138. To carry out this, division the number 138 by 2.
See more: What Are The Major Types Of Buying Situations In The Business Market?
As a result of splitting 138/2, we acquire the very first guess: 69
2
Next, we must divide 138 by the an outcome of the previous action (69).138/69 = 2
Calculate the arithmetic mean of this value (2) and the an outcome of action 1 (69).(69 + 2)/2 = 35.5 (new guess)
Calculate the error by individually the previous worth from the new guess.|35.5 - 69| = 33.533.5 > 0.001
Repeat this action again together the margin the error is greater than than 0.001
3
Next, we have to divide 138 by the an outcome of the previous action (35.5).138/35.5 = 3.8873
Calculate the arithmetic median of this value (3.8873) and the result of step 2 (35.5).(35.5 + 3.8873)/2 = 19.6937 (new guess)
Calculate the error by individually the previous worth from the brand-new guess.|19.6937 - 35.5| = 15.806315.8063 > 0.001
Repeat this action again as the margin the error is higher than than 0.001
4
Next, we have to divide 138 by the result of the previous step (19.6937).138/19.6937 = 7.0073
Calculate the arithmetic average of this value (7.0073) and also the an outcome of action 3 (19.6937).(19.6937 + 7.0073)/2 = 13.3505 (new guess)
Calculate the error by individually the previous value from the new guess.|13.3505 - 19.6937| = 6.34326.3432 > 0.001
Repeat this step again as the margin the error is higher than 보다 0.001
5
Next, we must divide 138 by the an outcome of the previous step (13.3505).138/13.3505 = 10.3367
Calculate the arithmetic typical of this value (10.3367) and also the an outcome of action 4 (13.3505).(13.3505 + 10.3367)/2 = 11.8436 (new guess)
Calculate the error by individually the previous value from the new guess.|11.8436 - 13.3505| = 1.50691.5069 > 0.001
Repeat this step again together the margin the error is better than than 0.001
6
Next, we should divide 138 by the an outcome of the previous action (11.8436).138/11.8436 = 11.6519
Calculate the arithmetic mean of this worth (11.6519) and also the result of step 5 (11.8436).(11.8436 + 11.6519)/2 = 11.7478 (new guess)
Repeat this action again together the margin of error is greater than 보다 0.001
7
Next, we should divide 138 through the result of the previous action (11.7478).138/11.7478 = 11.7469
Calculate the arithmetic mean of this worth (11.7469) and the an outcome of action 6 (11.7478).(11.7478 + 11.7469)/2 = 11.7474 (new guess)
Calculate the error by subtracting the previous value from the new guess.|11.7474 - 11.7478| = 0.00040.0004
Result✅ We found the result: 11.7474 In this case, that took united state seven procedures to discover the result.
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# Prime and Composite Numbers Prime Factorization
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## Transcription
1 Prime and Composite Numbers Prime Factorization Reteaching Math Course, Lesson A prime number is a whole number greater than that has exactly two factors, the number itself and. Examples: Factors of are and. Factors of are and. A composite number has more than two factors. Examples: 4 Factors of 4 are,, and 4. 6 Factors of 6 are,,, and 6. Prime factorization expresses a composite number as a product of its prime factors. Examples: 4 = 6 = 9 = = 40 To factor a number using a factor tree: Example:. Write any two factors of the given number Continue this process until each factor is a 6 7 prime number.. Circle the prime factors. 4. Write the prime factors in order. 40 = 7 Notice that some factors will occur more than once. To factor a number using division by primes:. Write the given number in a division box.. Begin dividing by a prime number that is a factor.. Divide the answer by a prime number that is a factor. 4. Repeat this process until the quotient is.. The divisors are the prime factors of the given number. Write the prime factors in order.. List the composite numbers from to 0? Write the prime factorization of the following numbers Write the prime factorization of 64 and 64, using exponents. Example: 7 ) 7 ) ) 0 ) 0 ) = 7 Saxon Math Course Harcourt Achieve Inc. and Stephen Hake. All rights reserved.
2 Problems About a Fraction of a Group Reteaching Math Course, Lesson To solve problems about a fraction of a group use diagrams. Example: Debbie scored of her team s 6 points. How many points did she score?. Draw a rectangle. This stands for the total. 6 points. Divide the rectangle into the same number of parts as the denominator. The denominator of is, so divide the rectangle into equal parts. 6 points. Divide the total by the denominator. 6 points 6 = points Write that answer in each part. points points 4. Bracket the parts into fractions. 6 points Debbie scored 4 points. points points points Debbie scored Team scored Draw a diagram of each statement. Then answer the questions that follow. Fifty people saw the movie. Two fifths were children.. How many children saw the movie?. How many adults saw the movie? 0 people 0 people 0 people 0 people 0 people 0 people children adults Twenty-five percent of the books were mysteries.. What fraction of the books were not mysteries? 4. How many books were mysteries? books books books books books 4 4 mystery books non-mystery books 4 Harcourt Achieve Inc. and Stephen Hake. All rights reserved. Saxon Math Course
3 Subtracting Mixed Numbers with Regrouping Reteaching Math Course, Lesson To subtract mixed numbers that require regrouping:. Borrow from the whole number and rename as a fraction.. Combine the top fraction with the renamed.. Then subtract. 4. Reduce if possible. Example: Simplify % % 4. 6 % 6 % % 7 4 % Saxon Math Course Harcourt Achieve Inc. and Stephen Hake. All rights reserved.
4 Reducing Fractions, Part Reteaching 4 Math Course, Lesson 4 Reduce large numbers by prime factorization. Cancel matching factors. Example: 4 0 = = Cancel means reduce before multiplying. 8 Pair any numerator 8 with any denominator. 4 = 4 Cancel. (Reduce the pairs.) Multiply the reduced terms. Another way to find the greatest common factor (GCF) of two numbers:. Write the prime factorization of each number.. Circle the common factors.. Multiply these common factors to find the GCF. Example: Find the GCF of 4 and = 64 = GCF = = 8 Reduce Simplify 4 and Harcourt Achieve Inc. and Stephen Hake. All rights reserved. Saxon Math Course
5 Dividing Fractions Reteaching Math Course, Lesson The reciprocal of any whole number or fraction is the reverse of terms. The product of a term and its reciprocal is. Examples: The reciprocal of is. The reciprocal of is, or. The reciprocal of 6 is 6. To divide whole numbers sometimes multiply by the reciprocal of the divisor. Example: 4 = 4 = 8 4 = 8 To divide fractions, multiply by the reciprocal of the divisor. Example: = =. Copy the first number (fraction).. Change to.. Write the reciprocal of the second fraction. 4. Cancel (reduce pairs) if possible.. Multiply. = Simplify Saxon Math Course Harcourt Achieve Inc. and Stephen Hake. All rights reserved. 7
6 Multiplying and Dividing Mixed Numbers Reteaching 6 Math Course, Lesson 6 To multiply and divide mixed numbers:. Change mixed numbers to improper ( top heavy ) fractions.. Then multiply or divide.. Simplify (reduce and/or convert) as necessary. Examples: Multiply Divide Change mixed numbers to improper fractions. = 6 Multiply. 0 0 = 4 Change mixed numbers to improper fractions. Multiply by reciprocal of the divisor. = 4 6 Simplify. = Simplify. Simplify 6: Harcourt Achieve Inc. and Stephen Hake. All rights reserved. Saxon Math Course
7 Multiples Least Common Multiple Equivalent Division Problems Reteaching 7 Math Course, Lesson 7 Use times table in Reference Guide to find the least common multiple (LCM) of small numbers. Example: Find the LCM of 6 and 8. Look down the 6 s and the 8 s columns. Find the first number that is the same in both columns. LCM = 4 Use prime factorization to find the LCM of larger numbers. Example: Find the LCM of 4 and 60.. Factor each number into prime factors. 4 = 60 =. Write each prime factor the greatest number of times it was used to form either number.. Multiply these factors to find the LCM. LCM = = 40 Use equivalent division to make complicated problems easier. Cancel matching zeros. Multiply or divide the dividend and divisor by the same number to form a new problem that is easier to calculate. Example: 0 Multiply both numbers by to form an equivalent problem. 0 = 44 0 = 44 0 Find the least common multiple (LCM) of each pair or group of numbers.. 4 and 0. and., 7, and 0 4., 6, and 8. 4 and and 6 Saxon Math Course Harcourt Achieve Inc. and Stephen Hake. All rights reserved. 9
8 Two-Step Word Problems Average, Part Reteaching 8 Math Course, Lesson 8 Two-step word problems can be written as two-step computation problems. Example: 0 (6 + ) Parentheses can make the problem easier. Example: Julie went to the store with \$0. She bought 8 cans of dog food for 67 per can. How much money did she have left? \$0 (0.67 8). Find out how much she spent = \$.6. Then find out how much money she had left. \$0 \$.6 = \$4.64 Calculating an average is often a two-step process.. Add the items.. Divide by the number of items. The answer must be between the smallest and the largest numbers. Another name for average is mean. Example: There were people in the first row, 7 in the second row, and 0 in the third row. What was the average number of people in each of the rows? people 7 people 0 people per row + 0 people rows ) 0 people 0 people The average of two numbers is the number halfway between the given numbers.. Hilda s scores on five games were the following: 8, 89, 94, 99, and 00. What was her average score?. Myrna bought 7 pounds of meat for a barbecue. She paid \$.49 per pound and gave the clerk a \$0 bill. How much change should she receive?. Jim drove 7 miles in hours 0 minutes. How many miles per hour did Jim drive? 4. What is the average (mean) of 0 and 40? 0 Harcourt Achieve Inc. and Stephen Hake. All rights reserved. Saxon Math Course
9 Rounding Whole Numbers Rounding Mixed Numbers Estimating Answers Reteaching 9 Math Course, Lesson 9 Before working a problem, estimate by rounding the numbers first. To round whole numbers:. Circle the place value you are rounding to.. Underline the digit to its right.. Ask Is the underlined number or more? Yes Add to the circled number. No Circled number stays the same. 4. Replace the underlined number (and any numbers after it) with zero. Examples: To round mixed numbers, compare each fraction to. If the fraction is equal to or greater than If the fraction is less than Examples: Round, to the nearest thousand., round up to the next whole number., round down. 7 ( because > ) 6 ( because 8 < ). Round 7,90 to the nearest hundred.. Estimate the quotient when 7,80 is divided by Estimate the sum of 4 and 0. Saxon Math Course Harcourt Achieve Inc. and Stephen Hake. All rights reserved.
10 Reteaching 0 Math Course, Lesson 0 Common Denominators Adding and Subtracting Fractions with Different Denominators To add or subtract fractions with different denominators:. Find the least common multiple (LCM) of the denominators. If the denominators are small numbers, use the times table. Look down the columns of each denominator. The first number that is the same in both columns is the LCM. If the denominators are large numbers, use prime factorization. Factor each denominator into prime factors. Write each prime factor the greatest number of times it was used to form either number. Multiply these factors to find the LCM.. Rename the fractions using the LCM as the new common denominator.. Add or subtract. 4. Simplify. Example: = = = 4. Rewrite 4 7 and so they have common denominators. Then find the sum and simplify. Simplify Compare: Harcourt Achieve Inc. and Stephen Hake. All rights reserved. Saxon Math Course
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# For what values of x is f(x)=(2x-2)(x-4)(3x-3) concave or convex?
Feb 13, 2018
$\setminus$
\qquad \quad "concave up:" \qquad \quad \quad \quad \ \ "on the interval" \quad ( 2, \infty );
$\setminus q \quad \setminus \quad \text{concave down:" \qquad \quad \ "on the interval} \setminus \quad \left(- \setminus \infty , 2\right) .$
#### Explanation:
$\setminus$
$\text{To answer the question, we want to find where the second}$
$\text{derivative of the function is positive, and where it is negative.}$
$\text{We are given:}$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f \left(x\right) \setminus = \setminus \left(2 x - 2\right) \left(x - 4\right) \left(3 x - 3\right) .$
$\text{Taking a second look at the function, we see we can rewrite it as:}$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f \left(x\right) \setminus = \setminus 2 \left(x - 1\right) \left(x - 4\right) \setminus \cdot 3 \left(x - 1\right)$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f \left(x\right) \setminus = \setminus 6 \left(x - 4\right) {\left(x - 1\right)}^{2.}$
$\text{This makes it easier to differentiate, starting with the}$
$\text{Product Rule:}$
$\setminus q \quad \setminus \quad f ' \left(x\right) \setminus = \setminus 6 \setminus \cdot \left(\left(x - 4\right) \left[{\left(x - 1\right)}^{2}\right] ' + \left[\left(x - 4\right)\right] ' {\left(x - 1\right)}^{2}\right)$
$\setminus q \quad \setminus \quad f ' \left(x\right) \setminus = \setminus 6 \setminus \cdot \left(\left(x - 4\right) \left[2 \left(x - 1\right) \setminus \cdot 1\right] + \left[1\right] {\left(x - 1\right)}^{2}\right)$
$\setminus q \quad \setminus \quad f ' \left(x\right) \setminus = \setminus 6 \setminus \cdot \left(2 \left(x - 4\right) \left(x - 1\right) + {\left(x - 1\right)}^{2}\right)$
$\text{Factor out" \ ( x - 1 ) \ "from the inside:}$
$\setminus q \quad \setminus \quad f ' \left(x\right) \setminus = \setminus 6 \setminus \cdot \left(x - 1\right) \left[2 \left(x - 4\right) + {\left(x - 1\right)}^{1}\right]$
$\setminus q \quad \setminus \quad f ' \left(x\right) \setminus = \setminus 6 \setminus \cdot \left(x - 1\right) \left[2 x - 8 + x - 1\right]$
$\setminus q \quad \setminus \quad f ' \left(x\right) \setminus = \setminus 6 \setminus \cdot \left(x - 1\right) \left[3 x - 9\right] \setminus = \setminus 6 \setminus \cdot \left(x - 1\right) \setminus \cdot 3 \setminus \cdot \left(x - 3\right)$
$\setminus q \quad \setminus \quad f ' \left(x\right) \setminus = \setminus 18 \left(x - 1\right) \left(x - 3\right) .$
$\text{Now calculate" \ f''(x)":}$
$\setminus q \quad \setminus \quad f ' ' \left(x\right) \setminus = \setminus \left[18 \left(x - 1\right) \left(x - 3\right)\right] '$
$\setminus q \quad \setminus \quad f ' ' \left(x\right) \setminus = \setminus 18 \setminus \cdot \left[\left(x - 1\right) \left(x - 3\right)\right] '$
$\setminus q \quad \setminus \quad f ' ' \left(x\right) \setminus = \setminus 18 \setminus \cdot \left[\left(x - 1\right) \left(x - 3\right) ' + \left(x - 1\right) ' \left(x - 3\right)\right]$
$\setminus q \quad \setminus \quad f ' ' \left(x\right) \setminus = \setminus 18 \setminus \cdot \left[\left(x - 1\right) \setminus \cdot 1 + 1 \setminus \cdot \left(x - 3\right)\right]$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus 18 \setminus \cdot \left[x - 1 + x - 3\right] \setminus = \setminus 18 \setminus \cdot \left(2 x - 4\right)$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus 18 \setminus \cdot 2 \setminus \cdot \left(x - 2\right) \setminus = \setminus 36 \setminus \cdot \left(x - 2\right)$
$\setminus$
$\therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus f ' ' \left(x\right) \setminus = \setminus 36 \setminus \cdot \left(x - 2\right) .$
$\setminus$
$\text{Now for concavity, we find where" \ f''(x) \ "is positive, and where}$
$\text{it's negative.}$
\qquad \qquad \qquad \qquad \quad "concave up:" \qquad \quad \ \quad \ 36 \cdot ( x - 2 ) > 0;
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{concave down:} \setminus q \quad \setminus 36 \setminus \cdot \left(x - 2\right) < 0.$
$\text{There are several ways to solve such inequalities. With the}$
$\text{expression as simple as it is, probably the easiest method is to}$
$\text{do it directly.}$
$\text{As 36 is positive, we can divide through both sides of the}$
$\text{inequalities by 36, leaving us with:}$
\qquad \qquad \qquad \qquad \quad "concave up:" \qquad \quad \quad \ \ x - 2 > 0;
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{concave down:} \setminus q \quad \setminus x - 2 < 0.$
$\text{So:}$
\qquad \qquad \qquad \quad "concave up:" \qquad \quad \quad \ \ x > 2, \quad "or the interval" \quad ( 2, \infty );
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{concave down:" \qquad \ x < 2, \quad "or the interval} \setminus \quad \left(- \setminus \infty , 2\right) .$
$\text{This our answer.}$
$\setminus$
$\text{Summarizing:}$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f \left(x\right) \setminus = \setminus \left(2 x - 2\right) \left(x - 4\right) \left(3 x - 3\right) :$
\qquad \qquad \quad "is concave up:" \qquad \quad \quad \quad \ \ "on the interval" \quad ( 2, \infty );
$\setminus q \quad \setminus q \quad \setminus \quad \text{is concave down:" \qquad \quad \ "on the interval} \setminus \quad \left(- \setminus \infty , 2\right) .$
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# How do you find a solution?
## How do you find a solution?
To solve for , you must first combine the ‘s on the right side of the equation. This will give you . Then, subtract and from both sides of the equation to get . Finally, divide both sides by to get the solution .
## How do you tell if there are infinitely many solutions?
The equation simplifies to the point that it no longer contains a variable, but expresses a true equation, e.g. 0=0 . For example: 2x+2=2(x+1) simplifies in this way. The equation has an identifiable solution and is periodic in nature. For example: tan2x+tanx−5=0 has infinitely many solutions since tanx has period π .
## Does the initial value problem have a unique solution?
That is, the theorem guarantees that the given initial value problem will always have (existence of) exactly one (uniqueness) solution, on any interval containing t0 as long as both p(t) and g(t) are continuous on the same interval.
## How do you know if a solution is unique?
Condition for Unique Solution to Linear Equations A system of linear equations ax + by + c = 0 and dx + ey + g = 0 will have a unique solution if the two lines represented by the equations ax + by + c = 0 and dx + ey + g = 0 intersect at a point. i.e., if the two lines are neither parallel nor coincident.
## What is the unique solution?
By the term unique solution, one mean to say that only one specific solution set exists for a given equation. So, if we have two equations, then unique solution will mean that there is one and only point at which the two equations intersect.
## What is the condition of no solution?
If there are infinitely many solutions of the given pair of linear equations, the equations are called dependent (consistent). If the lines are parallel, there is no solution for the pair of linear equations. If there is no solution of the given pair of linear equations, the equations are called inconsistent.
## How do you find no solution?
To find the missing number, compare both sides of the equation. If the variable terms are the same and the constant terms are different, then the equation has no solutions.
## How many solutions do 2 parallel lines have?
Each shows two lines that make up a system of equations. If the graphs of the equations intersect, then there is one solution that is true for both equations. If the graphs of the equations do not intersect (for example, if they are parallel), then there are no solutions that are true for both equations.
symbol Ø
## How do you write a solution?
When you write your solution you should:
1. Give each important definition or equation its own line.
2. Don’t bury too much algebra in a paragraph.
3. Label equations or formulas or lemmas or cases you will use later very clearly.
4. Remember that there’s always more paper.
## How do you tell if an equation has no solution?
The constants are the numbers alone with no variables. If the coefficients are the same on both sides then the sides will not equal, therefore no solutions will occur. Use distributive property on the right side first.
## How do you tell the number of solutions an equation has?
If solving an equation yields a statement that is true for a single value for the variable, like x = 3, then the equation has one solution. If solving an equation yields a statement that is always true, like 3 = 3, then the equation has infinitely many solutions.
## How do you know if an equation has 2 solutions?
The discriminant is the part under the square root in the quadratic formula, b²-4ac. If it is more than 0, the equation has two real solutions. If it’s less than 0, there are no solutions. If it’s equal to 0, there is one solution.
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# Solve by factoring: 3x^2 + 5x - 2 = 0. Solve by factoring. 1. x^2 + 5x + 6 = 0 2. 2x^2 - x - 3 = 0 3. 3x^2 + 5x - 2 = 0
1. x^2 + 5x + 6 = 0
Begin with this set up:
(x + __)(x + __) = 0
The missing numbers must have a product of 6 and a sum of 5.
2*3=6 2+3=5
Therefore...
(x + 2)(x + 3) = 0
Now take each binomial and set it equal to 0. Solve for x.
x + 2 = 0 x = -2
x + 3 = 0 x = -3
Solution set: {-2, -3}
2. 2x^2 - x - 3 = 0
When there are two subtraction signs in the trinomial, the set up is:
(2x + __)(x - __) = 0 or (2x - __)(x + __) = 0
Now we need two numbers whose product is 3. Since 3 is prime, the only options are 1 and 3. Try substituting 1 and 3 into the set ups above, use FOIL to see which set up works.
(2x + 1)(x - 3) = 2x^2 - 6x + 1x - 3 = 2x^2 - 5x - 3 NO
(2x + 3)(x - 1) = 2x^2 - 2x + 3x - 3 = 2x^2 + 1x - 3 NO
(2x - 1)(x + 3) = 2x^2 + 6x - 1x - 3 = 2x^2 + 5x - 3 NO
(2x - 3)(x + 1) = 2x^2 + 2x - 3x - 3 = 2x^2 - 1x - 3 YES
So now we know the trinomial can be factored as...
(2x - 3)(x + 1) = 0
Again, set each binomial equal to 0 and solve for x.
2x - 3 = 0 x = 1.5
x + 1 = 0 x = -1
Solution set: {1.5, -1}
3. 3x^2 + 5x - 2 = 0
This one is solved similarly to #2. Here is the set up:
(3x + __)(x - __) = 0 or (3x - __)(x + __) = 0
Again, since 2 is prime, the only options for the blanks are 1 and 2. Try each combination, use FOIL to see which one works.
(3x + 1)(x - 2) = 3x^2 - 6x + 1x - 2 = 3x^2 - 5x - 2 NO
(3x + 2)(x - 1) = 3x^2 - 3x + 2x - 2 = 3x^2 + 1x - 2 NO
(3x - 1)(x + 2) = 3x^2 + 6x - 1x - 2 = 3x^2 + 5x - 2 YES
Therefore...
(3x - 1)(x + 2) = 0
Set each binomial equal to 0 and solve for x.
3x - 1 = 0 x = 1/3
x + 2 = 0 x = -2
Solution set: {1/3, -2}
Remember, you can always check these answers by graphing the equations. The x-intercepts should equal the solution set.
Approved by eNotes Editorial Team
You are allowed to ask only one question at a time. I am solving one of the problems, you can apply the same method for the others.
To solve 3x^2 + 5x - 2 = 0, we express 5 in terms of two numbers that add up to 5 and the product of the numbers is 3*-2 = -6. The numbers that satisfy this are 6 and -1.
3x^2 + 5x - 2 = 0
=> 3x^2 + 6x - x - 2 = 0
=> 3x(x + 2) - 1(x + 2) = 0
=> (3x - 1)(x + 2) = 0
3x - 1 = 0
=> x = 1/3
x + 2 = 0
=> x = -2
The solution of 3x^2 + 5x - 2 = 0 is x = 1/3 and x = -2
Approved by eNotes Editorial Team
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There’s a renowned story that Gauss, mathematician extraordinaire, had actually a lazy teacher. The so-called educator wanted to store the children busy for this reason he could take a nap; that asked the course to add the numbers 1 come 100.
You are watching: Sum of all numbers from 1 to 1000
Gauss approached v his answer: 5050. So soon? The teacher suspected a cheat, yet no. Manual enhancement was for suckers, and also Gauss found a formula to sidestep the problem:
Let’s re-publishing a couple of explanations of this an outcome and really recognize it intuitively. For these examples we’ll add 1 to 10, and then see how it uses for 1 come 100 (or 1 to any number).
## Technique 1: Pair Numbers
Pairing numbers is a common approach to this problem. Rather of composing all the numbers in a solitary column, stop wrap the number around, choose this:
1 2 3 4 510 9 8 7 6An amazing pattern emerges: the sum of each tower is 11. Together the height row increases, the bottom row decreases, for this reason the sum continues to be the same.
Because 1 is paired with 10 (our n), we deserve to say the each column has actually (n+1). And also how countless pairs perform we have? Well, we have actually 2 equal rows, us must have n/2 pairs.
which is the formula above.
## Wait — what around an odd number of items?
Ah, ns glad you lugged it up. What if we are adding up the numbers 1 come 9? us don’t have an even number of items to pair up. Plenty of explanations will certainly just offer the explanation above and leaving it in ~ that. Ns won’t.
Let’s include the number 1 come 9, but instead of beginning from 1, let’s count from 0 instead:
0 1 2 3 49 8 7 6 5By counting native 0, we acquire an “extra item” (10 in total) therefore we can have one even variety of rows. However, our formula will certainly look a little bit different.
Notice that each column has actually a amount of n (not n+1, like before), since 0 and 9 space grouped. And instead the having exactly n items in 2 rows (for n/2 bag total), we have actually n + 1 items in 2 rows (for (n + 1)/2 pairs total). If friend plug this numbers in you get:
which is the very same formula together before. It constantly bugged me that the same formula worked for both odd and also even number – won’t you gain a fraction? Yep, you get the same formula, but for various reasons.
## Technique 2: Use two Rows
The above method works, however you handle odd and also even numbers differently. Isn’t there a much better way? Yes.
Instead that looping the number around, let’s compose them in two rows:
1 2 3 4 5 6 7 8 9 1010 9 8 7 6 5 4 3 2 1Notice the we have actually 10 pairs, and also each pair adds approximately 10+1.
The total of every the numbers over is
But we only want the amount of one row, no both. For this reason we division the formula over by 2 and get:
Now this is cool (as cool together rows the numbers can be). It works for an odd or even number of items the same!
## Technique 3: make a Rectangle
I newly stumbled upon another explanation, a fresh approach to the old pairing explanation. Various explanations work better for different people, and I tend to prefer this one better.
Instead of composing out numbers, pretend we have actually beans. We desire to include 1 bean to 2 bean to 3 beans… every the means up to 5 beans.
xx xx x xx x x xx x x x xSure, we can go to 10 or 100 beans, but with 5 you acquire the idea. Just how do us count the variety of beans in ours pyramid?
Well, the amount is clearly 1 + 2 + 3 + 4 + 5. Yet let’s look at it a various way. Stop say we winter our pyramid (I’ll usage “o” because that the mirrored beans), and also then topple the over:
x o x o o o o ox x o o x x o o o ox x x o o o => x x x o o ox x x x o o o o x x x x o ox x x x x o o o o o x x x x x oCool, huh? In case you’re wondering even if it is it “really” lines up, the does. Take it a look in ~ the bottom heat of the continuous pyramid, with 5′x (and 1 o). The following row that the pyramid has 1 less x (4 total) and also 1 much more o (2 total) to fill the gap. Just like the pairing, one next is increasing, and the other is decreasing.
Now for the explanation: How countless beans do we have total? Well, that’s simply the area that the rectangle.
We have actually n rows (we didn’t change the number of rows in the pyramid), and also our repertoire is (n + 1) devices wide, because 1 “o” is combine up through all the “x”s.
Notice the this time, us don’t care around n being odd or also – the total area formula works out just fine. If n is odd, we’ll have actually an even variety of items (n+1) in every row.
But the course, we don’t want the full area (the number of x’s and also o’s), we just want the number of x’s. Since we doubled the x’s to gain the o’s, the x’s through themselves room just half of the complete area:
And we’re ago to our original formula. Again, the number of x’s in the pyramid = 1 + 2 + 3 + 4 + 5, or the amount from 1 come n.
## Technique 4: median it out
We all understand that
average = sum / variety of items
which we deserve to rewrite to
sum = average * variety of items
So let’s number out the sum. If we have actually 100 number (1…100), then we clearly have 100 items. That was easy.
To gain the average, notification that the numbers are all equally distributed. Because that every huge number, yes sir a little number ~ above the other end. Let’s look in ~ a tiny set:
1 2 3The average is 2. 2 is currently in the middle, and 1 and also 3 “cancel out” for this reason their typical is 2.
For an even number of items
1 2 3 4the mean is in between 2 and 3 – that 2.5. Also though we have a fractional average, this is yes — due to the fact that we have actually an even number of items, when we multiply the mean by the count that ugly portion will disappear.
Notice in both cases, 1 is ~ above one side of the average and N is equally far away top top the other. So, we can say the median of the entire set is actually simply the median of 1 and n: (1 + n)/2.
Putting this into our formula
And voila! We have actually a fourth way of thinking around our formula.
## So why is this useful?
Three reasons:
1) adding up numbers quickly can be beneficial for estimation. An alert that the formula increases to this:
Let’s to speak you want to include the numbers from 1 to 1000: suppose you acquire 1 extr visitor come your site each day – exactly how many full visitors will you have actually after 1000 days? due to the fact that thousand squared = 1 million, we acquire million / 2 + 1000/2 = 500,500.
2) This concept of adding numbers 1 to N mirrors up in various other places, prefer figuring out the probability for the date of birth paradox. Having actually a firm grasp of this formula will help your expertise in numerous areas.
3) most importantly, this example shows there are countless ways to understand a formula. Possibly you like the pairing method, probably you prefer the rectangle technique, or maybe there’s another explanation that functions for you. Don’t offer up once you don’t know — try to find one more explanation that works. Happy math.
By the way, over there are more details about the background of this story and the an approach Gauss may have used.
## Variations
Instead that 1 come n, how around 5 come n?
Start through the continual formula (1 + 2 + 3 + … + n = n * (n + 1) / 2) and subtract off the component you don’t desire (1 + 2 + 3 + 4 = 4 * (4 + 1) / 2 = 10).
Sum because that 5 + 6 + 7 + 8 + … n = – 10And because that any beginning number a:
Sum indigenous a to n = – <(a - 1) * a / 2>We desire to remove every number from 1 approximately a – 1.
See more: How Much Is Magic Johnson Autograph Worth, Magic Johnson
How around even numbers, favor 2 + 4 + 6 + 8 + … + n?
Just dual the continual formula. To include evens native 2 to 50, discover 1 + 2 + 3 + 4 … + 25 and dual it:
Sum the 2 + 4 + 6 + … + n = 2 * (1 + 2 + 3 + … + n/2) = 2 * n/2 * (n/2 + 1) / 2 = n/2 * (n/2 + 1)So, to gain the evens native 2 come 50 you’d execute 25 * (25 + 1) = 650
How about odd numbers, like 1 + 3 + 5 + 7 + … + n?
That’s the exact same as the also formula, other than each number is 1 less than its counterpart (we have 1 instead of 2, 3 rather of 4, and also so on). We gain the following biggest even number (n + 1) and take turn off the extra (n + 1)/2 “-1″ items:
Sum of 1 + 3 + 5 + 7 + … + n = <(n + 1)/2 * ((n + 1)/2 + 1)> – <(n + 1) / 2>To include 1 + 3 + 5 + … 13, acquire the next biggest also (n + 1 = 14) and do
<14/2 * (14/2 + 1)> – 7 = 7 * 8 – 7 = 56 – 7 = 49Combinations: evens and offset
Let’s to speak you want the evens native 50 + 52 + 54 + 56 + … 100. Find all the evens
2 + 4 + 6 + … + 100 = 50 * 51and subtract turn off the people you nothing want
2 + 4 + 6 + … 48 = 24 * 25So, the sum from 50 + 52 + … 100 = (50 * 51) – (24 * 25) = 1950
Phew! expect this helps.
Ruby nerds: friend can inspect this using
(50..100).select x.inject(:+)1950Javascript geeks, carry out this:
<...Array(51).keys()>.map(x => x + 50).filter(x => x % 2 == 0).reduce((x, y) => x + y)1950// Note: There room 51 number from 50-100, inclusive. Fencepost!
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### Dr. Wilson
5. In problem 2, assume that there is no point inside the circle where three of the line meet. How many points of intersection are there inside the circle?
The picture from problem 2,
will not work in this problem, because there are several point where more than two lines intersect, and this will throw off our theoretical count. We will need to adjust the points somewhat so that there is no point inside the circle where more than two lines intersect.
We see in the figure above that it is possible to arrange 8 points on a circle so that there is no point inside the circle where more than two of the lines meet. We also see that there are quite a few points of intersection inside the circle for these lines. While it is possible to count them, it would probably be better to solve enough of the simpler problems to be able to find a pattern which would work in general and enable us to solve this problem.
If there are 1, 2, or 3 points on the circle
there are no points of intersection inside the circle. The first time we actually get a point of intersection inside the circle is with 4 points.
It gets a little more interesting with 5
There are 5 points of intersection.
With 6
there are 15 points of intersection. With 6 points, we need to take a little bit of care to make sure that there is no point where more than two lines meet.
At this point, the figures are getting more and more complicated, so let's see if we can deduce the pattern at this point.
We can find these numbers in Pascal's triangle
These are the number of 4 element subsets of a given set. The next number in the sequence is 35 which should be the number of points of intersection inside the circle if there are 7 points on the circle.
you can count that there are 35 points of intersection inside, and in our problem, there are 70 points inside the circle.
The reason that this procedure works is because if you have a point inside a circle where 2 lines intersect,
each line is determined by 2 points on the circle, so if you have two lines, they will be determined by 4 points on the circle. This sets up a one to one correspondence between points inside the circle and 4 element subsets of the set of points on the circle.
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Groupwork
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# How do you differentiate f(x)=(x^2+x)(x^3-1) using the product rule?
May 26, 2016
$d f \left(x\right) = \left(5 {x}^{4} + 4 {x}^{3} - 2 x - 1\right) \cdot d x$
#### Explanation:
$f \left(x\right) = \left({x}^{2} + x\right) \left({x}^{3} - 1\right)$
$d f \left(x\right) = \left({\left({x}^{2} + x\right)}^{'} \cdot \left({x}^{3} - 1\right) + {\left({x}^{3} - 1\right)}^{'} \cdot \left({x}^{2} + x\right)\right) d x$
$d f \left(x\right) = \left[\left(2 x + 1\right) \left({x}^{3} - 1\right) + 3 {x}^{2} \left({x}^{2} + x\right)\right] \cdot d x$
$d f \left(x\right) = \left[2 {x}^{4} - 2 x + {x}^{3} - 1 + 3 {x}^{4} + 3 {x}^{3}\right] \cdot d x$
$d f \left(x\right) = \left(5 {x}^{4} + 4 {x}^{3} - 2 x - 1\right) \cdot d x$
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# Lesson 17
Squares and Cubes
Let’s investigate perfect squares and perfect cubes.
### 17.1: Perfect Squares
1. The number 9 is a perfect square. Find four numbers that are perfect squares and two numbers that are not perfect squares.
2. A square has side length 7 in. What is its area?
3. The area of a square is 64 sq cm. What is its side length?
### 17.2: Building with 32 Cubes
Use the 32 snap cubes in the applet’s hidden stack to build the largest single cube you can. Each small cube has side length of 1 unit.
1. How many snap cubes did you use?
2. What is the side length of the cube you built?
3. What is the area of each face of the built cube? Show your reasoning.
4. What is the volume of the built cube? Show your reasoning.
This applet has a total of 64 snap cubes. Build the largest single cube you can.
1. How many snap cubes did you use?
2. What is the edge length of the new cube you built?
3. What is the area of each face of this built cube? Show your reasoning.
4. What is the volume of this built cube? Show your reasoning.
### 17.3: Perfect Cubes
1. The number 27 is a perfect cube. Find four other numbers that are perfect cubes and two numbers that are not perfect cubes.
2. A cube has side length 4 cm. What is its volume?
3. A cube has side length 10 inches. What is its volume?
4. A cube has side length $$s$$ units. What is its volume?
### 17.4: Introducing Exponents
Make sure to include correct units of measure as part of each answer.
1. A square has side length 10 cm. Use an exponent to express its area.
2. The area of a square is $$7^2$$ sq in. What is its side length?
3. The area of a square is 81 m2. Use an exponent to express this area.
4. A cube has edge length 5 in. Use an exponent to express its volume.
5. The volume of a cube is $$6^3$$ cm3. What is its edge length?
6. A cube has edge length $$s$$ units. Use an exponent to write an expression for its volume.
The number 15,625 is both a perfect square and a perfect cube. It is a perfect square because it equals $$125^2$$. It is also a perfect cube because it equals $$25^3$$. Find another number that is both a perfect square and a perfect cube. How many of these can you find?
### Summary
When we multiply two of the same numbers together, such as $$5\boldcdot 5$$, we say we are squaring the number. We can write it like this: $$\displaystyle 5^2$$
Because $$5\boldcdot 5 = 25$$, we write $$5^2 = 25$$ and we say, “5 squared is 25.”
When we multiply three of the same numbers together, such as $$4\boldcdot 4 \boldcdot 4$$, we say we are cubing the number. We can write it like this: $$\displaystyle 4^3$$
Because $$4\boldcdot 4\boldcdot 4 = 64$$, we write $$4^3 = 64$$ and we say, “4 cubed is 64.”
We also use this notation for square and cubic units.
• A square with side length 5 inches has area 25 in2.
• A cube with edge length 4 cm has volume 64 cm3.
To read 25 in2, we say “25 square inches,” just like before.
The area of a square with side length 7 kilometers is $$7^2$$ km2. The volume of a cube with edge length 2 millimeters is $$2^3$$ mm3.
In general, the area of a square with side length $$s$$ is $$s^2$$, and the volume of a cube with edge length $$s$$ is $$s^3$$.
### Glossary Entries
• cubed
We use the word cubed to mean “to the third power.” This is because a cube with side length $$s$$ has a volume of $$s \boldcdot s \boldcdot s$$, or $$s^3$$.
• exponent
In expressions like $$5^3$$ and $$8^2$$, the 3 and the 2 are called exponents. They tell you how many factors to multiply. For example, $$5^3$$ = $$5 \boldcdot 5 \boldcdot 5$$, and $$8^2 = 8 \boldcdot 8$$.
• squared
We use the word squared to mean “to the second power.” This is because a square with side length $$s$$ has an area of $$s \boldcdot s$$, or $$s^2$$.
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It is taken for granted that the reader knows how to perform the common operations of addition, subtraction, etc., where only whole numbers are used; but, when there are mixed or fractional numbers, a little refreshment of the memory may be desirable to some; hence, a little space is devoted to this elementary branch of arithmetic.
## Common Fractions
The numerator of a fraction is the number above the bar; and the denominator is the number beneath it; thus, in the fraction 3/4, 3 is the numerator and 4 is the denominator. Two or more fractions having the same denominator are said to have a common denominator. By "reducing fractions to a common denominator" is meant finding such a denominator as will contain each of the given denominators without a remainder, and multiplying each numerator by the number of times its denominator is contained in the common denominator. Thus, the fractions 1/4, 7/8, and ft have, as a common denominator, 16; then, 1/4=4/16; 7/8=14/16; 9/16=9/16.
By "reducing a fraction to its lowest terms" is meant dividing both numerator and denominator by the greatest number that each will contain without a remainder; for example, in 14/16, the greatest number that will thus divide 14 and 16 is 2; so that, (14/2)/(16/2)=7/8, which is 14/16 reduced to the lowest terms.
A mixed number is one consisting of a whole number and a fraction, as 7 3/8.
An improper fraction is one in which the numerator is equal to, or greater than, the denominator, as 17/8. This is reduced to a mixed number by dividing 17 by 8, giving 2 1/8. If the numerator is less than the denominator, the fraction is termed proper. A mixed number is reduced to a fraction by multiplying the whole number by the denominator, adding the numerator, and placing the sum over the denominator; thus 1 7/8=[(1x8)+7] / 8=15/8.
To add fractions or mixed numbers. If fractions only, reduce them to a common denominator, add partial results, and reduce sum to a whole or mixed number. If mixed numbers are to be added, add the sum of the fractions to that of the whole numbers; thus, 1 7/8+2 1/4=(1 + 2)+(7/8+2/8)= 4 1/8.
To subtract two fractions or mixed numbers. If they are fractions only, reduce them to a common denominator, take less from greater, and reduce result; as, 7/8 in. - ft in. = (14-9)/16
=5/16 in. If they are mixed numbers, subtract fractions and whole numbers separately, placing remainders beside one another; thus, 3 7/8 in. - 2 1/4 in.= (3 - 2)+ (7/8-2/8)= 1 5/8 in. With fractions like the following, proceed as indicated: 3 7/16 in.-1 13/16 in.= (2+16/16 +7/16)-1 13/16=2 23/16 - l 13/16 = l 10/16 = 1 5/8in.; 7in.- 4 3/4in. = (6 + 4/4)- 4 3/4 = 2 1/4in.
To multiply fractions. Multiply the numerators together, and likewise the denominators, and divide the former by the latter; thus, 1/2 in. x3/4 in x5/8 in.=(1x3x5)/(2x4x8)=15/64 cu. in. If mixed numbers are to be multiplied, reduce them to frac-tions, and proceed as above shown; thus, l 1/2 in. x 3 1/4 in. = 3/2x13/4 =39/8 = 4 7/8 sq. in.
To divide fractions. Invert the divisor (i. e., exchange places of numerator and denominator) and multiply the dividend by it, reducing the result, if necessary; thus, (7/8)/(3/4) =(7/8)x(4/3)=28/24=7/6=1 1/6. If there are mixed numbers, reduce them to fractions, and then divide as just shown; thus, (l 5/8)/(3 1/4)=(13/8)/(13/4),or(13/8)x(4/13) =52/104 =1/2.
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# Thread: Probability of having one of two unknown bags
1. ## Probability of having one of two unknown bags
Bag #1
900 red eggs
100 blue eggs
Bag #2
100 red eggs
900 blue eggs
You are given one of these two bags (50/50 chance) randomly.
You perform a set of trials where you take out one egg randomly from your bag, note the color, then put it back in, and repeat.
You drew 8 red eggs and 2 blue eggs from your bag during this trial. What is the probabilty (%) that you have Bag #1?
I thought it was 50% because you could have drawn 8 red and 2 blue from bag #2 as well, since drawing eggs and putting them back does not change the bag's contents...
2. Suppose that $\displaystyle A$ is the event of drawing eight red and two blue eggs.
First we must calculate $\displaystyle P(A)$.
$\displaystyle P(A)=P(A\cap B_1)+ P(A\cap B_2)= P(A| B_1)P(B_1)+ P(A| B_2)P(B_2)$.
This problem is asking you to find $\displaystyle P(B_1|A)$.
$\displaystyle P(B_1|A)=\frac{P(B_1\cap A)}{P(A)}$.
Can you finish?
3. Hello, daigo!
$\displaystyle \text{Bag \#1:}\;\; \begin{array}{c}\text{900 red eggs} \\ \text{100 blue eggs} \end{array} \qquad\text{Bag \#2:}\;\;\begin{array}{c}\text{ 100 red eggs} \\ \text{900 blue eggs} \end{array}$
You are given one of these two bags randomly (50/50 chance).
You perform a set of trials where you take out one egg randomly from your bag,
note the color, then put it back in, and repeat.
You drew 8 red eggs and 2 blue eggs from your bag during this trial.
What is the probabilty (%) that you have Bag #1?
This is a Conditional Probability problem:
. . Given that 8R, 2B are drawn,
. . what is the probability that they were drawn from Bag #1?
Bayes' Theorem: .$\displaystyle P\bigg(\text{Bag 1 }|\text{ 8R,2B}\bigg) \;=\;\frac{P\bigg(\text{[Bag 1] }\wedge\text{[8R,2B]}\bigg)}{P\bigg(\text{8R, 2B}\bigg)}$ .[1]
Numerator: .$\displaystyle P(\text{Bag \#1}) \,=\,\frac{1}{2}$
. . From Bag #1: .$\displaystyle P(R) = \frac{9}{10},\;\;\;P(B) = \frac{1}{10}$
. . Then: .$\displaystyle P(\text{8R,2B}) \:=\:{10\choose8}\left(\frac{9}{10}\right)^8\left( \frac{1}{10}\right)^2 \;=\;(45)\,\frac{9^8}{10^{10}}$
. . Hence: .$\displaystyle P\bigg(\text{Bag 1 }\wedge \text{ 8R,2B}\bigg) \;=\;\frac{45}{2}\cdot\frac{9^8}{10^{10}}$ .[2]
Denominator: .$\displaystyle P(\text{Bag 2}) \,=\,\frac{1}{2}$
. . From Bag #2: .$\displaystyle P(R) =\frac{1}{10},\;\;P(B) =\frac{9}{10}$
. . Then: .$\displaystyle P(\text{8R,2B}) \:=\:{10\choose8}\left(\frac{1}{10}\right)^8\left( \frac{9}{10}\right)^2 \;=\;(45)\cdot \frac{9^2}{10^{10}}$
. . Hence: .$\displaystyle P(\text{Bag 2 }\wedge\text{ 8R,2B}) \:=\: \frac{45}{2}\cdot\frac{9^2}{10^{10}}$ .(A)
And we have: .$\displaystyle P\bigg(\text{8R,2B}\bigg) \;=\;\frac{45}{2}\cdot\frac{9^8}{10^{10}} + \frac{45}{2}\cdot\frac{9^2}{10^{10}} \;=\;\frac{45}{2}\cdot\frac{9^2}{10^{10}}\cdot\lef t(9^6 + 1\right)$ .[3]
Substitute [2] and [3] into [1]:
. . $\displaystyle P(\text{Bag 1 }|\text{ 8R,2B}) \;=\;\frac{\dfrac{45}{2}\cdot\dfrac{9^8}{10^{10}} } {\dfrac{45}{2}\cdot\dfrac{9^2}{10^{10}}\cdot(9^6 + 1)}\;=\;\frac{9^6}{9^6 + 1}$
Therefore: .$\displaystyle P(\text{Bag 1 }|\text{ 8R,2B}) \;=\;0.999998118$
If we draw 8 Reds and 2 Blues from a bag,
. . the probability that they came from Bag #1 is virtually 100%.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
The answer is reasonable when we consider (A).
$\displaystyle P(\text{Bag 2 }\wedge\text{ 8R,2B}) \;=\;\frac{45}{2}\cdot\frac{9^2}{10^{10}} \;=\;0.000000182$
So if we have Bag #2, it is virtually impossible to draw 8 Reds and 2 Blues.
4. Thanks. I was shown another method where I do 100 - ([1^8 * 100] / [9^8]) but I did not understand why that was the answer. Your solution clarified it better.
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Paul's Online Notes
Home / Calculus I / Applications of Integrals / Area Between Curves
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Section 6.2 : Area Between Curves
5. Determine the area of the region bounded by $$x = 3 + {y^2}$$, $$x = 2 - {y^2}$$, $$y = 1$$ and $$y = - 2$$.
Show All Steps Hide All Steps
Hint : It’s generally best to sketch the bounded region that we want to find the area of before starting the actual problem. Having the sketch of the graph will usually help with determining the right/left functions and the limits for the integral.
Start Solution
Let’s start off with getting a sketch of the region we want to find the area of.
We are assuming that, at this point, you are capable of graphing most of the basic functions that we’re dealing with in these problems and so we won’t be showing any of the graphing work here.
Here is a sketch of the bounded region we want to find the area of.
Show Step 2
It should be clear from the graph that the right function is $$x = 3 + {y^2}$$ and the left function is $$x = 2 - {y^2}$$.
Next, we were given limits on $$y$$ in the problem statement and we can see that the two curves do not intersect in that range. Note that this is something that we can’t always guarantee and so we need the graph to verify if the curves intersect or not. We should never just assume that because limits on $$y$$ were given in the problem statement that the curves will not intersect anywhere between the given limits.
So, because the curves do not intersect we will be able to find the area with a single integral using the limits : $$- 2 \le y \le 1$$.
Show Step 3
At this point there isn’t much to do other than step up the integral and evaluate it.
We are assuming that you are comfortable with basic integration techniques so we’ll not be including any discussion of the actual integration process here and we will be skipping some of the intermediate steps.
The area is,
$A = \int_{{ - 2}}^{1}{{3 + {y^2} - \left( {2 - {y^2}} \right)\,dy}} = \int_{{ - 2}}^{1}{{1 + 2{y^2}\,dy}} = \left. {\left( {y + \frac{2}{3}{y^3}} \right)} \right|_{ - 2}^1 = \require{bbox} \bbox[2pt,border:1px solid black]{9}$
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# Inscribed angle
(Redirected from Inscribed angle theorem)
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In geometry, an inscribed angle is formed when two secant lines of a circle (or, in a degenerate case, when one secant line and one tangent line of that circle) intersect on the circle.
Typically, it is easiest to think of an inscribed angle as being defined by two chords of the circle sharing an endpoint.
The basic properties of inscribed angles are discussed in Book 3, Propositions 20–22 of Euclid's Elements. These are the inscribed angle is half the central angle, inscribed angles on the same arc of a chord are equal and the sum of the two distinct inscribed angles of a chord is 180°.
## Property
An inscribed angle is said to intersect an arc on the circle. The arc is the portion of the circle that is in the interior of the angle. The measure of the intercepted arc (equal to its central angle) is exactly twice the measure of the inscribed angle.
This single property has a number of consequences within the circle. For example, it allows one to prove that when two chords intersect in a circle, the products of the lengths of their pieces are equal. It also allows one to prove that the opposite angles of a cyclic quadrilateral are supplementary.
## Proof
To understand this proof, it is useful to draw a diagram.
### Inscribed angles where one chord is a diameter
Let O be the center of a circle. Choose two points on the circle, and call them V and A. Draw line VO and extended past O so that it intersects the circle at point B which is diametrically opposite the point V. Draw an angle whose vertex is point V and whose sides pass through points A and B.
Draw line OA. Angle BOA is a central angle; call it θ. Lines OV and OA are both radii of the circle, so they have equal lengths. Therefore triangle VOA is isosceles, so angle BVA (the inscribed angle) and angle VAO are equal; let each of them be denoted as ψ.
Angles BOA and AOV are supplementary. They add up to 180°, since line VB passing through O is a straight line. Therefore angle AOV measures 180° − θ.
It is known that the three angles of a triangle add up to 180°, and the three angles of triangle VOA are:
180° − θ
ψ
ψ.
Therefore
$2 \psi + 180^\circ - \theta = 180^\circ.$
Subtract 180° from both sides,
$2 \psi = \theta, \,$
where θ is the central angle subtending arc AB and ψ is the inscribed angle subtending arc AB.
### Inscribed angles with the center of the circle in their interior
Given a circle whose center is point O, choose three points V, C, and D on the circle. Draw lines VC and VD: angle DVC is an inscribed angle. Now draw line VO and extend it past point O so that it intersects the circle at point E. Angle DVC subtends arc DC on the circle.
Suppose this arc includes point E within it. Point E is diametrically opposite to point V. Angles DVE and EVC are also inscribed angles, but both of these angles have one side which passes through the center of the circle, therefore the theorem from the above Part 1 can be applied to them.
Therefore
$\angle DVC = \angle DVE + \angle EVC. \,$
then let
$\psi_0 = \angle DVC,$
$\psi_1 = \angle DVE,$
$\psi_2 = \angle EVC,$
so that
$\psi_0 = \psi_1 + \psi_2. \qquad \qquad (1)$
Draw lines OC and OD. Angle DOC is a central angle, but so are angles DOE and EOC, and
$\angle DOC = \angle DOE + \angle EOC.$
Let
$\theta_0 = \angle DOC,$
$\theta_1 = \angle DOE,$
$\theta_2 = \angle EOC,$
so that
$\theta_0 = \theta_1 + \theta_2. \qquad \qquad (2)$
From Part One we know that $\theta_1 = 2 \psi_1$ and that $\theta_2 = 2 \psi_2$. Combining these results with equation (2) yields
$\theta_0 = 2 \psi_1 + 2 \psi_2 \,$
therefore, by equation (1),
$\theta_0 = 2 \psi_0. \,$
### Inscribed angles with the center of the circle in their exterior
[The previous case can be extended to cover the case where the measure of the inscribed angle is the difference between two inscribed angles as discussed in the first part of this proof.]
Given a circle whose center is point O, choose three points V, C, and D on the circle. Draw lines VC and VD: angle DVC is an inscribed angle. Now draw line VO and extend it past point O so that it intersects the circle at point E. Angle DVC subtends arc DC on the circle.
Suppose this arc does not include point E within it. Point E is diametrically opposite to point V. Angles DVE and EVC are also inscribed angles, but both of these angles have one side which passes through the center of the circle, therefore the theorem from the above Part 1 can be applied to them.
Therefore
$\angle DVC = \angle EVC - \angle DVE$.
then let
$\psi_0 = \angle DVC,$
$\psi_1 = \angle DVE,$
$\psi_2 = \angle EVC,$
so that
$\psi_0 = \psi_2 - \psi_1. \qquad \qquad (3)$
Draw lines OC and OD. Angle DOC is a central angle, but so are angles DOE and EOC, and
$\angle DOC = \angle EOC - \angle DOE.$
Let
$\theta_0 = \angle DOC,$
$\theta_1 = \angle DOE,$
$\theta_2 = \angle EOC,$
so that
$\theta_0 = \theta_2 - \theta_1. \qquad \qquad (4)$
From Part One we know that $\theta_1 = 2 \psi_1$ and that $\theta_2 = 2 \psi_2$. Combining these results with equation (4) yields
$\theta_0 = 2 \psi_2 - 2 \psi_1$
therefore, by equation (3),
$\theta_0 = 2 \psi_0.$
## Theorem
The inscribed angle θ is half of the central angle 2θ that subtends the same arc on the circle (magenta). Thus, the angle θ does not change as its vertex is moved around on the circle (green, blue and gold angles).
Given the two points A and B, the set of points M in the plane for which the angle AMB is equal to α is an arc of a circle. The measure of the angle AOB, which O is the center of the circle, is 2α.
The inscribed angle theorem states that an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle. Therefore, the angle does not change as its vertex is moved to different positions on the circle.
The inscribed angle theorem is used in many proofs of elementary Euclidean geometry of the plane. A special case of the theorem is Thales' theorem, which states that the angle subtended by a diameter is always 90°, i.e., a right angle. As a consequence of the theorem, opposite angles of cyclic quadrilaterals sum to 180°; conversely, any quadrilateral for which this is true can be inscribed in a circle. As another example, the inscribed angle theorem is the basis for several theorems related to the power of a point with respect to a circle.
### Proof
In the simplest case, one leg of the inscribed angle is a diameter of the circle, i.e., passes through the center of the circle. Since that leg is a straight line, the supplement of the central angle equals 180° − 2θ. Drawing a segment from the center of the circle to the other point of intersection of the inscribed angle produces an isosceles triangle, made from two radii of the circle and the second leg of the inscribed angle. Since two of the angles in an isosceles triangle are equal and since the angles in a triangle must add up to 180°, it follows that the inscribed angle equals θ, half of the central angle.
This result may be extended to an arbitrarily inscribed angle by drawing a diameter from the vertex of the angle. This converts the general problem into two sub-cases in which a diameter is a leg of each angle. The arbitrary angle equals half of the sum of the two central angles that share the diameter as a leg. Adding the two subangles again yields the result that the inscribed angle is half of the central angle.
Note that the central angle for the golden inscribed angle is 360° − 2θ. Therefore, the half of it (and thus the measure of the golden inscribed angle) is 180° − θ.
The set of all points (locus) for which a line segment can be seen at angle measured θ contains two arcs (one of each side of the line segment with central angle 2θ). In the special case of 90°, there is exactly one circle with center the middle of the line segment.
### Corollaries
By a similar argument, the angle between a chord and the tangent line at one of its intersection points equals half of the central angle subtended by the chord. See also Tangent lines to circles.
## References
• Ogilvy, C. S. (1990). Excursions in Geometry. Dover. pp. 17–23. ISBN 0-486-26530-7.
• Gellert W, Küstner H, Hellwich M, Kästner H (1977). The VNR Concise Encyclopedia of Mathematics. New York: Van Nostrand Reinhold. p. 172. ISBN 0-442-22646-2.
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## Fractions are Hard!
### 3.5 Percentages
In the first century B.C.E., Roman Emperor Augustus levied for the first time a tax of one part per hundred on the proceeds of all goods sold at markets and auctions in ancient Rome. From the Latin phrase per centum meaning “by the hundred” came the term percent.
A percent is just a fraction with denominator one hundred (the number of pies per 100 boys). For example, $$\dfrac{1}{2}$$ written as a fraction with denominator one hundred is
$$\dfrac{1}{2}=\dfrac{1\times 50}{2\times 50}=\dfrac{50}{100}$$.
We say that $$\dfrac{1}{2}$$ equals 50 percent, and write
$$\dfrac{1}{2}=50\%$$.
Levying a tax of 1 dollar for every 100 dollars results in $$\dfrac{1}{100}$$, one one-hundredth, of all cash exchanged being given to the government. This is tax rate of $$1\%$$.
The symbol $$\%$$ developed in Italy during the 1500s. Clerks started shortening per cento to P 00, which then eventually became $$%$$.
Working backwards is actually easier. For example, 80% is a fraction with denominator 100. We have
$$80\%=\dfrac{80}{100}$$
which is equivalent to the fraction $$\dfrac{4}{5}$$.
Some more examples:
$$\dfrac{3}{10}=\dfrac{30}{100}=100\%$$
$$2 = \dfrac{200}{100}=200\%$$
$$0.632=\dfrac{0.632}{1}=\dfrac{0.632\times 100}{1 \times 100}=\dfrac{63.2}{100}=63.2\%$$
Question: Nervous Nelly wants a rule for understanding percentages. Someone once told her that to write a number $$x$$ as a percentage, just multiply that number by 100 and write a $$\%$$ sign after the result. For example: $$\dfrac{1}{2}\times 100 = 50$$ and indeed $$\dfrac{1}{2}$$ is $$50\%$$. $$0.632\times 100 = 63.2$$ and indeed $$0.632$$ is $$63.2\%$$. $$\dfrac{3}{7}\times 100 = \dfrac{300}{7}=42\dfrac{6}{7}$$ and indeed $$\dfrac{3}{27}$$ is $$42\dfrac{6}{7}\%$$. One day, maybe in the distant future, Nervous Nelly will ask why this rule works. What would you say to her? How would you explain why this rule is true? (And did she even need to memorize this as rule in the first place?)
And backwards
$$53\% = \dfrac{53}{100}$$
$$25\% = \dfrac{25}{100}=\dfrac{1}{4}$$
$$150\% = \dfrac{150}{100}=1\dfrac{1}{2}$$
$$1000\% = \dfrac{1000}{100}=10$$
Question: The price of a Nice-and-Spicy Soup Cup yesterday was $1.50. It’s price went up to$1.80 overnight. What percentage increase is that?
Question: The ancient Romans also spoke of parts per thousand. Today this is called, if it is ever used, per millage with the symbol o/oo. For example, 467 o/oo equals $$\dfrac{467}{1000}$$, and one tenth, which is $$\dfrac{1}{10} = \dfrac{100}{1000}$$, equals $$100$$ o/oo. Convert each of these per millages into fractions a) 40 o/oo b) 500 o/oo c) 6000 o/oo d) 2 o/oo Convert each of the following fractions into per millages: a) $$\dfrac{1}{100}$$ b) $$\dfrac{2}{5}$$ c) $$30\%$$ d) $$1$$ e) $$0$$
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# Math Expressions Grade 4 Student Activity Book Unit 2 Lesson 5 Answer Key Estimate Products
This handy Math Expressions Grade 4 Student Activity Book Answer Key Unit 2 Lesson 5 Estimate Products provides detailed solutions for the textbook questions.
## Math Expressions Grade 4 Student Activity Book Unit 2 Lesson 5 Estimate Products Answer Key
Estimate Products
It is easier to estimate the product of a two-digit number and a one-digit number when you think about the two multiples of ten close to the two-digit number. This is shown in the drawings below.
Question 1.
In each drawing, find the rectangles that represent 4 × 70 and 4 × 60. These rectangles “frame” the rectangles for
4 × 68 and 4 × 63. Find the values of 4 × 70 and 4 × 60.
4 × 70 = ____ 4 × 60 = _____
280, 240,
Explanation:
The value of 4 X 70 from the rectangles frame is 4 X 70 = 280 and 4 X 60 = 240.
Question 2.
Look at the rectangle that represents 4 × 68. Is 4 × 68 closer to 4 × 60 or to 4 × 70? So is 4 × 68 closer to 240 or 280?
4 × 70 = ____ 4 × 60 = _____
Closer to 4 X 70 = 280, 280, 240,
Explanation:
Looking at the rectangle that represents 4 X 68. 4 X 68 = 272 is closer to 4 × 70 =280 as 4 X 70 = 280 is much closer than 4 X 60 = 240.
Question 3.
Look at the rectangle that represents 4 × 63. Is 4 × 63 closer to 4 × 60 or to 4 × 70? Is 4 × 63 closer to 240 or 280?
Closer to 4 X 60 = 240,
Explanation:
Looking at the rectangle that represents 4 X 63 = 252 is closer to 4 X 60 = 240 than to 4 X 70 = 280.
Question 4.
Explain how to use rounding to estimate the product of a one-digit number and a two-digit number.
In order to estimate products, we round the given factors to the required place value. Estimating products help us to check the reasonableness of an answer.
To estimate the product, we first round off the multiplier and the multiplicand to the nearest tens, hundreds and then multiply the rounded numbers.
Explanation:
In order to estimate products, we round the given factors to the required place value. Estimating products helps us to check the reasonableness of an answer.
To estimate the product, we first round off the multiplier and the multiplicand to the nearest tens, and hundreds and then multiply the rounded numbers. Estimate the products of 331 and 267 by rounding to the nearest hundred: 331 ⟶ 300, 267 ⟶ 300, 300 × 300 = 90000, The estimated product is 90000.Practice Estimation
Discuss how rounding and estimation could help solve these problems.
Question 5.
Keesha’s school has 185 fourth-grade students. The library has 28 tables with 6 chairs at each table. Can all of the fourth-graders sit in the library at one time? How do you know?
No,
Explanation:
Given Keesha’s school has 185 fourth-grade students. The library has 28 tables with 6 chairs at each table. So to check all of the fourth- graders sit in the library at one time is 28 X 6 = 168 as 168 < 185 so no.
Question 6.
Ameena is printing the class newsletter. There are 8 pages in the newsletter, and she needs 74 copies. Each package of paper contains 90 sheets. How many packages of paper does she need to print the newsletter?
7 packages,
Explanation:
Given Ameena is printing the class newsletter. There are 8 pages in the newsletter, and she needs 74 copies. Each package of paper contains 90 sheets. So many packages of paper do she need to print the newsletter are 74 X 8 = 592 pages so 592/90 = 6.577 is approximately equal to 7 packages.
Estimate each product. Then solve to check your estimate.
Question 7.
3 × 52
Estimate:150,
156, 150 is near 156,
Explanation:
Estimate: 3 X 52 is 32 to 30 so 3 X 50 = 150,
3 X 52 = 156. 150 is near 156.
Question 8.
7 × 48
Estimate:350,
336, 350 is near to 336,
Explanation:
Estimate: 7 X 48 is 48 to 50 so 7 X 50 = 350,
7 X 48 = 336. 350 is near to 336.
Question 9.
9 × 27
Estimate: 270,
243, 270 is near to 243,
Explanation:
Estimate: 9 X 27 is 27 to 30 so 9 X 30 = 270,
9 X 27 = 243. 270 is near to 243.
Question 10.
8 × 34
Estimate: 240,
272, 240 is near to 272,
Explanation:
Estimate: 8 X 34 is 34 to 30 so 8 X 30 = 240,
8 X 34 = 272. 240 is near to 272.
Question 11.
8 × 35
Estimate: 320,
280, 320 is near to 280,
Explanation:
Estimate: 8 X 35 is 35 to 40 so 8 X 40 = 320,
8 X 35 = 280. 320 is near to 280.
Question 12.
5 × 22
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# Triangle A has an area of 4 and two sides of lengths 8 and 4 . Triangle B is similar to triangle A and has a side with a length of 13 . What are the maximum and minimum possible areas of triangle B?
Mar 31, 2017
$\text{Max} = \frac{169}{40} \left(5 + \sqrt{15}\right) \approx 37.488$
$\text{Min} = \frac{169}{40} \left(5 - \sqrt{15}\right) \approx 4.762$
#### Explanation:
Let the vertices of triangle $A$ be labelled $P$, $Q$, $R$, with $P Q = 8$ and $Q R = 4$.
Using Heron's Formula,
$\text{Area} = \sqrt{S \left(S - P Q\right) \left(S - Q R\right) \left(S - P R\right)}$, where
$S = \frac{P Q + Q R + P R}{2}$ is the half-perimeter,
we have
$S = \frac{8 + 4 + P R}{2} = \frac{12 + P R}{2}$
Thus,
$\sqrt{S \left(S - P Q\right) \left(S - Q R\right) \left(S - P R\right)}$
$= \sqrt{\left(\frac{12 + P Q}{2}\right) \left(\frac{12 + P Q}{2} - 8\right) \left(\frac{12 + P Q}{2} - 4\right) \left(\frac{12 + P Q}{2} - P Q\right)}$
$= \frac{\sqrt{\left(12 + P Q\right) \left(P Q - 4\right) \left(4 + P Q\right) \left(12 - P Q\right)}}{4}$
$= \text{Area} = 4$
Solve for $C$.
$\sqrt{\left(144 - P {Q}^{2}\right) \left(P {Q}^{2} - 16\right)} = 16$
$\left(P {Q}^{2} - 144\right) \left(P {Q}^{2} - 16\right) = - 256$
$P {Q}^{4} - 160 P {Q}^{2} + 2304 = - 256$
${\left(P {Q}^{2}\right)}^{2} - 160 P {Q}^{2} + 2560 = 0$
Complete the square.
${\left({\left(P {Q}^{2}\right)}^{2} - 80\right)}^{2} + 2560 = {80}^{2}$
${\left({\left(P {Q}^{2}\right)}^{2} - 80\right)}^{2} = 3840$
$P {Q}^{2} = 80 + 16 \sqrt{15}$ or $P {Q}^{2} = 80 - 16 \sqrt{15}$
$P Q = 4 \sqrt{5 + \sqrt{15}} \approx 11.915$ or
$P Q = 4 \sqrt{5 - \sqrt{15}} \approx 4.246$
This shows that there are 2 possible kinds of triangle that satisfy the conditions given.
In the case of max area for triangle be, we want the side with length 13 to be similar to the side PQ for the triangle with $P Q = 4 \sqrt{5 - \sqrt{15}} \approx 4.246$.
Therefore, the linear scale ratio is
$\frac{13}{4 \sqrt{5 - \sqrt{15}}} \approx 3.061$
The area is therefore enlarged to a factor that is the square of the linear scale ratio. Therefore, The max area triangle B can have is
$4 \times {\left(\frac{13}{4 \sqrt{5 - \sqrt{15}}}\right)}^{2} = \frac{169}{40} \left(5 + \sqrt{15}\right) \approx 37.488$
Similarly, in the case of min area for triangle be, we want the side with length 13 to be similar to the side PQ for the triangle with $P Q = 4 \sqrt{5 + \sqrt{15}} \approx 11.915$.
Therefore, the linear scale ratio is
$\frac{13}{4 \sqrt{5 + \sqrt{15}}} \approx 1.091$
The area is therefore enlarged to a factor that is the square of the linear scale ratio. Therefore, The min area triangle B can have is
$4 \times {\left(\frac{13}{4 \sqrt{5 + \sqrt{15}}}\right)}^{2} = \frac{169}{40} \left(5 - \sqrt{15}\right) \approx 4.762$
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## Finding and Graphing Points for Linear Relationships
At this level, students will begin to see the relationship between equations and straight-line graphs on a coordinate grid.
Materials: poster paper or a transparency and overhead projector for demonstration; straightedge; one copy of a coordinate grid, a straightedge, and lined paper for each student
Preparation: Draw a coordinate grid on poster paper or a transparency. Label the x- and y- axes from 0 through 10. Make copies of the grid for students.
Prerequisite Skills and Concepts: Students should know about ordered pairs and locating points on a grid. They should also be able to recognize and interpret an equation.
• Write the equation x + 5 = y on the board.
• Ask: How could you say this equation in words?
Students should say that the equation means "a number plus five equals another number", or some comparable statement.
• Draw a table with four columns and five rows on the board. Have students draw their own table. Label the first column x, the second column x + 5, and the third column y. Leave the fourth column blank for now. Write 1 in the first column below x.
• Ask: What happens to the equation if we replace x with 1? Elicit from students the equation 1 + 5 = 6. Write 1 + 5 in the second column below x + 5. Then write 6 in the third column below y.
• Continue to replace x with 2, 3, then 4. Have students complete the first three columns of their tables on their own. Ask for a volunteer to complete the table on the board.
• Say: Let's write ordered pairs using the values of x and y. Label the fourth column of your table Ordered Pairs. Remind students that when they locate points on a grid, they first move right on the x-axis, then up on the y-axis. Therefore, the first number in an ordered pair is a value for x, and the second number is a value for y. These numbers are called the x- and y-coordinates.
• Ask: What is the first number we used for x? (1) What is the first number we calculated for y? (6) So, what is the first ordered pair? (1,6)
Have students complete their tables. When they are finished, record the ordered pairs in the table on the board.
• Say: Now we're going to graph the equation x + 5 = y on a grid. (Point to the grid you made.) This grid is called a coordinate grid. Let's take a closer look at the different parts of the grid.
• Point to the horizontal line on the grid.
• Say: This line is called the x-axis.
• Point to the vertical line on the grid.
• Ask: What do you think this line is called?
Students should make the connection to the y-axis.
• Say: Now, let's locate the ordered pairs on the grid. Who can find (1,6)?
Have a volunteer describe the location of the ordered pair. Mark the location on the coordinate grid at the front of the class. Then have students locate the rest of the ordered pairs on their own grids.
• Say: Let's connect all of the points. What figure did we make?
Have students use a straight edge to connect the points. Show students how extending both ends of the line slightly, and drawing arrows, shows that the line goes on in both directions. Students should identify the figure as a straight line.
• Have students repeat this activity with the equation x – 2 = y. Use the numbers 5, 6, 7, 8, and 9 for x.
Wrap-Up and Assessment Hints
These skills will need lots of practice. Reinforce the need for students to work carefully, so their graph is accurate. When you assess students' progress, keep the number of exercises small enough that they have time to complete each step without rushing.
Finding and Graphing Points
for Linear Relationships
Finding the Length of a Line
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# Multiplying Rational Numbers Worksheet Answer Key Lesson 2-4
A Realistic Amounts Worksheet will help your kids become more informed about the methods powering this percentage of integers. In this worksheet, pupils can solve 12 distinct troubles relevant to rational expressions. They will likely discover ways to flourish several phone numbers, group them in sets, and figure out their products. They will also practice simplifying realistic expression. Once they have enhanced these principles, this worksheet will certainly be a important device for advancing their studies. Multiplying Rational Numbers Worksheet Answer Key Lesson 2-4.
## Reasonable Phone numbers really are a ratio of integers
There are 2 types of phone numbers: rational and irrational. Realistic numbers are understood to be total figures, whereas irrational phone numbers will not repeat, and get an unlimited number of digits. Irrational phone numbers are non-no, non-terminating decimals, and square beginnings that are not best squares. They are often used in math applications, even though these types of numbers are not used often in everyday life.
To define a rational quantity, you need to realize what a reasonable variety is. An integer is really a total number, plus a realistic quantity is actually a proportion of two integers. The rate of two integers will be the amount on the top separated through the quantity on the bottom. For example, if two integers are two and five, this would be an integer. There are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They could be made into a portion
A reasonable number features a numerator and denominator which are not no. Because of this they could be conveyed being a small percentage. In addition to their integer numerators and denominators, reasonable numbers can also have a adverse importance. The unfavorable value ought to be located left of along with its total importance is its distance from absolutely no. To streamline this case in point, we shall point out that .0333333 is a fraction that could be written as a 1/3.
Together with unfavorable integers, a reasonable amount can even be made right into a portion. For example, /18,572 can be a realistic amount, while -1/ is not really. Any small percentage comprised of integers is reasonable, given that the denominator will not consist of a and can be written as an integer. Similarly, a decimal that leads to a level is yet another rational number.
## They are sensation
In spite of their label, reasonable phone numbers don’t make significantly perception. In math, they can be single organizations with a special size around the quantity collection. Which means that when we matter one thing, we are able to order the shape by its proportion to its authentic amount. This contains correct regardless if there are unlimited reasonable amounts in between two particular numbers. If they are ordered, in other words, numbers should make sense only. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
If we want to know the length of a string of pearls, we can use a rational number, in real life. To get the duration of a pearl, as an example, we could add up its width. A single pearl is twenty kilos, which is a realistic quantity. Furthermore, a pound’s excess weight equates to ten kgs. Therefore, we should certainly divide a lb by twenty, without concern yourself with the duration of one particular pearl.
## They may be indicated being a decimal
If you’ve ever tried to convert a number to its decimal form, you’ve most likely seen a problem that involves a repeated fraction. A decimal number may be written being a multiple of two integers, so 4 times five is the same as 8. A similar issue involves the recurring portion 2/1, and each side should be divided up by 99 to find the appropriate response. But how do you create the conversion process? Here are some good examples.
A realistic quantity may also be printed in various forms, which include fractions plus a decimal. One way to represent a logical quantity inside a decimal would be to break down it into its fractional comparable. You will find three ways to break down a logical variety, and every one of these methods yields its decimal equivalent. One of these approaches is to separate it into its fractional comparable, and that’s what’s called a terminating decimal.
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Students can Download Maths Chapter 8 Linear Equations in One Variable Ex 8.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
## Karnataka Board Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.1
1. Solve the following:
Question i.
x + 3 = 11
x + 3 = 11
x = 11 – 3
x = 8
Question ii.
y – 9 = 21
y – 9 = 21
y = 21 + 9
y = 30
Question iii.
10 = z + 3
10 = z + 3 10-3 =z
7 = z or z = 7
Question iv.
$$\frac{3}{11}+x=\frac{9}{11}$$
Question v.
10x = 30
10x = 30
x = $$\frac { 30 }{ 10 }$$
x = 3
Question vi.
$$\frac{s}{7}=4$$
$$\frac{s}{7}=4$$
S = 4 × 7
S = 28
Question vii.
$$\frac{3 x}{6}=10$$
$$\frac{3 x}{6}=10$$
3x = 10 × 6
3x = 60
x = $$\frac { 60 }{ 3 }$$
x = 20
Question viii.
$$1.6=\frac{x}{1.5}$$
$$1.6=\frac{x}{1.5}$$
1.6 × 1.5 = x
2.40 = x
x = 2.4
Question ix.
8x – 8 = 48
8x – 8 = 48
8x = 48 + 8
8x = 56
x = $$\frac { 56 }{ 8 }$$
x = 7
Question x.
$$\frac{x}{3}+1=\frac{7}{15}$$
Question xi.
$$\frac{x}{5}=12$$
$$\frac{x}{5}=12$$
x = 12 × 5
x = 60
Question xii.
$$\frac{3 x}{5}=15$$
Question xiii.
3(x + 6) = 24
3(x + 6) = 24
3x+ 18 = 24
3x = 24 – 18
3x = 6
x = $$\frac { 6 }{ 3 }$$ = 2
Question xiv.
$$\frac{x}{4}-8=1$$
Question xv.
3(x+2) – 2(x—1) = 7
3(x + 2) – 2(x – 1) = 7
3x + 6 – 2x + 2 = 7
x + 8 = 7 .
x = 7 – 8
x = -1
2. Solve the equations :
Question i.
5x = 3x + 24
5x = 3x + 24
2x = 24
x = $$\frac { 24 }{ 2 }$$
x = 12
Question ii.
8t + 5 = 2t – 31
8t + 5 = 2t – 31
6t = -36
t = $$\frac { -36 }{ 6 }$$
t = -6
Question iii.
7x -10 = 4x + 11
7x – 10 = 4x + 11
7x – 4x = 11 + 10
3x = 21
x = $$\frac { 21 }{ 3}$$
x = 7
Question iv.
4z + 3 = 6 + 2z
4z + 3 = 6 + 2z
4z – 2z = 6 – 3
2z = 3
Question v.
2x – 1 = 14 – x
2x – 1 = 14 – x
2x + x = 14 + 1
3x = 15
x = $$\frac { 15 }{ 3 }$$
x = 5
Question vi.
6x + 1 = 3(x -1) + 7
6x + 1 = 3(x —1) + 7
6x + l =3x-3 + 7
6x + l = 3x + 4
6x – 3x = 4 – 1
3x = 3
x = $$\frac { 3 }{ 3 }$$
x = 1
Question vii.
$$\frac{2 x}{5}-\frac{3}{2}=\frac{x}{2}+1$$
Question viii.
$$\frac{x-3}{5}-2=\frac{2 x}{5}$$
Question ix.
3(x + 1) = 12 + 4(x – 1)
3(x + 1)= 12 + 4(x – 1)
3x + 3 = 12 + 4x – 4
3x + 3 = 4x + 8
3x – 4x = 8 – 3
-x = 5
x = -5
Question x.
2x – 5 = 3(x – 5)
2x – 5 = 3(x – 5)
2x – 5 = 3x – 15
2x – 3x = -15 + 5
-x = -10
x= 10
Question xi.
6(1 – 4x) + 7(2 + 5x) = 53
6( 1 – 4x) + 7(2 + 5x) = 53
6 – 24x+ 14 + 35x = 53
35x – 24x + 6 + 14 = 53
11x+ 20 = 53
11x = 53 – 20
x = $$\frac { 33 }{ 11 }$$
x = 3
Question xii.
3(x + 6) + 2(x + 3) = 64
3(x + 6) + 2(x + 3) = 64
3x + 18 + 2x + 6 = 64
3x + 2x + 18 + 6 = 64
5x + 24 = 64
5x = 64 – 24
5x = 40
x = $$\frac { 40 }{ 5 }$$
∴ x = 8
Question xiii.
$$\frac{2 m}{3}+8=\frac{m}{2}-1$$
$$\frac{3}{4}(x-1)=(x-3)$$
$$\frac{3}{4}(x-1)=(x-3)$$
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# Factors - Atreya Roy
• Author: Atreya Roy is pursuing his BTech From Kalyani Government Engineering College, Bengal.
Factor of a number is a number smaller or equal to it which divides the number wholly. Hence factor of a number cannot exceed the number itself. Factors are always expressed as the product of primes present in that number.
Example : Find the factors of 30
Solution : 30 = 2 * 3 * 5
In simple terms we can say that the factors of 30 = 1,2,3,5,6,10,15,30 : Total 8 factors.
Let us look into a formula we can keep in mind so that we can calculate the number of factors fast.
If the Number is represented in the form N = (a p)*(b q)*(c r)*(d s )…. Where a,b,c,d.. are the primes present in the number and p,q,r,s… are their respective exponents.
So the number of factors in the number = (p+1)*(q+1)*(r+1)*(s+1)……. And so on.
Example : Find the factors of 30
Solution : 30 = 2^1 * 3^1 *5^1
P=1
Q=1
R=1
Hence the number of factors of 30 = (1+1)(1+1)(1+1) = 2*2*2 = 8 factors.
Factors are of two types :
1. Even factors : the factors which are divisible by 2
2. Odd factors : the factors which are not divisible by 2
Example :
Find the number of even factors in the number 3600
Solution :
Break 3600 into its prime factors. 3600 = 100*36 = 2^4*3^2*5^2
Hence the total number of factors of 3600 = (4+1)*(2+1)*(2*1) = 45
Out of these how many are even. For being even factors the factors must have atleast one 2 in them. Hence out of the four 2s present. Take out one. We are left with :
2^3*3^2*5^2
Hence, with these factors if we multiply the 2 we took out, we will get the total even factors.
Total even factors = (3+1)*(2+1)*(2+1) = 4*3*3 = 36
From this we can also find out the odd factors.
Total Factors = Even Factors + Odd factors.
Hence if the total number of factors = 45 and even factors are 36 then 45-36 = 9 odd factors are present in the number.
Note : To find the number of odd factors present in a number we can also calculate them by removing all the 2s present in the number.
Problem :
We are given a number 64800. What are the total number of :
1. Prime Factors
2. Composite Factors
3. Odd factors
4. Even Factors
Breaking 64800 into its factors, we get : 2^5 * 3^4 * 5^2 .
1. Prime factors : 2,3,5 : hence 3 factors
2. Total factors = (5+1)*(4+1)*(2+1) = 6*5*3 = 90
Hence composite factors = total factors – prime factors-1 (since 1 is neither prime nor composite) = 90-3-1
3. Odd Factors = factors in the number : 3^4 * 5^2 = (4+1)*(2+1 ) = 15
4. Even Factors = total factors – odd factors = 90-15 = 75
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# Trigonometry for Dopes - The Touchy Tangent
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The tangent is the last of the three principal trigonometric functions. It derives its name from the Latin tangere for “to touch. " This derivation is relevant to the actual mechanics of the function and the manner in which the tangent works to give us some important measurements in life. You see, the tangent allows us to compute the maximum and minimum values of a function, and this application has significant weight in the real world.
The tangent of a given angle in a right triangle tells us the quotient of the side opposite this angle to the side adjacent to it. In the oft taught SOHCAHTOA, the TOA part stands for tangent = opposite/adjacent. This is the mnemonic that most students are taught when they are introduced to the basic trigonometric functions, of which sine, cosine, and tangent form the core.
Like the sine and cosine, the tangent is also a periodic function. Unlike the sine and cosine, however, the tangent is not defined for certain values along the x-axis, and these values occur at the points which are the odd multiples of pi/2. By undefined is meant that the tangent grows increasingly positive or negative at these values. Mathematicians say that the tangent “grows without bound" here, or that the tangent “approaches infinity" at these values.
The tangent is related to a very important concept in algebra: the slope. If you recall, the slope of a line is defined by the rise over the run, or the change in the y values over the change in the x values. What the slope is measuring is nothing more than the inclination of the given line. Thus a higher value for the slope means that the line is steeper than a line with a smaller slope. If you draw a right triangle in the coordinate plane, with one of the sides parallel to the x-axis, and the hypotenuse with a positive slope (rising from left to right), then the tangent of the acute angle formed by the side parallel to the x-axis and the hypotenuse, is the opposite side over the adjacent side or the change in the y values over the change in the x values. This is precisely the slope of the hypotenuse.
The tangent line to a curve or surface is a line that passes through the curve in only one point, unlike a secant line which passes through the curve or surface in two points. The tangent line thus gently “touches" the surface of the curve and does not cut it. Now where this becomes very important is in the calculus, where the tangent line to a curve is found by calculating the derivative and evaluating at a given point. You see the tangent line can tell us where a given curve reaches both its highest and lowest values. How is this so?
Well if you think about it, picture a curve drawn in the coordinate plane. More specifically, picture a curvy line going from left to right with a number of “hills" and “valleys. " The hills represent where the curve reaches a local high point, and the valleys are the points where the curve reaches a local low point. If the tangent is a line which crosses the curve in one point, all we need to do, to find these highs and lows, is see where the tangent line is horizontal. This is where the curve tops or bottoms out. This is another way of saying we seek to find where the slope of the tangent line is zero, since horizontal lines have zero slopes.
Thus the “touchy" tangent finds itself involved in a very important application in mathematics: that of giving us the maximum and minimum values of a function. Since functions model real life phenomena, this would seem an extremely important thing. For instance, if a function modeled the profits of a major corporation, then knowing the maximum and minimum values would tell us where the profit was highest and where lowest. Having this information just might allow us to tweak the parameters of the profit model to make more or less money. Now wouldn't you think companies might want to know this kind of stuff?
Yes, trigonometry finds itself enmeshed within our world. The three trigonometric functions do more than tell us the ratio of sides of a right triangle, they help tell us about life itself. Keep this in mind next time you run into the sine, cosine, or tangent.
See more at Innovative Math Ebooks and see these cool math poems Curious Math Poems E-Book
Joe is a prolific writer of self-help and educational material and an award-winning former teacher of both college and high school mathematics. Joe is the creator of the Wiz Kid series of math ebooks, Arithmetic Magic, the little classic on the ABC's of arithmetic, the original collection of poetry, Poems for the Mathematically Insecure, and the short but highly effective fraction troubleshooter Fractions for the Faint of Heart. The diverse genre of his writings (novel, short story, essay, script, and poetry)-particularly in regard to its educational flavor- continues to captivate readers and to earn him recognition.
Joe propagates his teaching philosophy through his articles and books and is dedicated to helping educate children living in impoverished countries. Toward this end, he donates a portion of the proceeds from the sale of every ebook. For more information go to http://www.mathbyjoe.com
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Calculus for Dopes - It Ain' That Hard - The Limit Part I
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## AravindG 2 years ago Note this is not a question! This is my tutorial on "MATH TRICKS" .hope you all like these .Also provide your feedback !
1. mathmate
What is it, mentally calculate 92*96?
2. AravindG
Find Your Age and Change Tell a friend that with this trick you can tell him his age, and how much change (less than \$1.00) he is carrying. The friend starts with his age and follows these steps: 1. Multiply his age by 4. 2. Add 10. 3. Multiply by 25. 4. Subtract the number of days in a non-leap year. 5. Add in the change. 6. Add 115. The first two digits are the person's age, and the second two digits are the change in his pocket. (As with the Birthday trick, you can get the number before the last stem, and do that one in your head.)
3. AravindG
When Is Your Birthday? Ask someone to follow these steps: 1. Multiply the number of the month by 5. 2. Add 7. 3. Multiply by 4. 4. Add 13. 5. Multiply by 5. 6. Add the day of the month. 7. Subtract 205. The last two digits are the day of the month, and the first digit(s) is the month number. (To make this seem more amazing, have them tell you the result before they subtract 205, and you do that part in your head.)
4. AravindG
How Much Change? Tell a friend you can tell how much change he/she has in his/her pocket if he/she will tell you the answer to these steps: 1. Multiply the amount by 2. 2. Add 3 to the product. 3. Multiply the sum by 5. 4. Subtract 6. The last digit is 9, and the rest are her change!
5. craigmeg1
wow! thats a lot of math! lol. but great tricks
6. robtobey
Find Your Age and Change $(((4a+10)25)-365)+c+115\text{ //}\text{ Expand}$$100 a+c$
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# Equation for slope of exponential function of arbitrary base
Say I have some general exponential function defined as such;
$$f(x,a,b) = a\left(1-b^{-x}\right)$$
Given any arbitrary value for $$a$$, how can I solve for $$b$$ such that $$\frac{d}{dx}f(0,a,b) = 1$$? Is there a general equation for this?
• Note that $\frac{d}{dx}\,f(x,a,b)=a\ln(b)\,b^{-x}.$ Can you go from there? – Adrian Keister Jul 12 at 13:19
• I've managed to simplify the equation down to $b^{b^{-x}}=e^{\frac{1}{a}}$, but I can't make it any further on my own. Do you have any advice? – Maurdekye Jul 12 at 13:54
You are solving for when $$\frac{d}{dx}\,f(0,a,b)=1.$$ Technically, you probably mean that you're plugging in $$x=0$$ after taking the derivative, and not before. The right way to write that expression is this: solve $$\left(\frac{d}{dx}\,f(x,a,b)\right)_{x=0}=1$$ for $$b,$$ given $$a$$. So, moving forward, we have: \begin{align*} \frac{d}{dx}\,f(x,a,b)&=a\ln(b)\,b^{-x},\quad\text{so solve} \\ a\ln(b)&=1\\ \ln(b)&=\frac1a \\ b&=e^{1/a}. \end{align*}
• You've left out the $b^{-x}$ in one step of your equation. – Maurdekye Jul 12 at 14:22
• You specifically said to solve $f'(0,a,b)=0,$ where you've plugged in $x=0.$ When you do that, you get $b^{-0}=1.$ – Adrian Keister Jul 12 at 14:23
• Hmm, you're right. This appears to be the correct answer. – Maurdekye Jul 12 at 14:25
Since the derivative of $$a\left(1 - b^{-x}\right) = a\ln(b)b^{-x}$$, the solution would be to solve the following equation:
$$a\ln(b)b^{-x} = 1$$
However, I hit a wall and got stuck trying to solve this function symbolically, so I gave in and punched the equation into WolframAlpha. This is the answer I was given back;
$$b=\sqrt[x]{-\frac{aW\left(-\frac{x}{a}\right)}{x}}$$
I don't doubt this is a correct answer, but what I don't understand is the use of the Lambert W-Function. Is this really necessary in the solution, or is WolframAlpha complicating it more than need be? In any case, it's use is inconvenient, as I don't have an easy way to approximate the function in my use case. As well, WolframAlpha wasn't able to tell me the steps it took to reach this point, so if someone could provide that for me, it'd be a great help.
• The Lambert $W$ function is there because you still have $x$ in your expression. It's the inverse of the function $f(x)=xe^x,$ and is useful at times. – Adrian Keister Jul 12 at 14:28
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## 17Calculus - Polar Coordinates
##### 17Calculus
Polar coordinates is one of those topics that can be taught in many different courses. Some students come across the topic in physics for the first time. Sometimes, it's in precalculus or trig. No matter what course you are in right now, you will find everything you need here on polar coordinates. If you've already learned this topic, take some time to review the material on these pages and watch a few videos anyway, so that it is fresh in your mind.
Basic Idea of Polar Coordinates
You should already be familiar with graphing in rectangular coordinates (sometimes called cartesian coordinates). We can use trigonometry to describe the same point(s) or graph another way, as shown in plot 1.
From basic trig, you know that a point in the plane can be described as $$(x,y)$$ or as $$( r \cos(\theta), r \sin(\theta) )$$. Comparing these two forms gives you the equations
These equations are used to convert between polar coordinates and rectangular coordinates.
Remember from trig that angles can be described in an infinite number of ways, since $$\theta = \theta + 2\pi$$ and $$\theta = \theta - 2\pi$$. It is always best to use the smallest possible angle in the interval $$(-\pi, \pi]$$ or $$[0, 2\pi)$$ or whatever is required by the context.
plot 2
One of the biggest differences you will find between trig and polar coordinates is that in trig, r in the above equation is usually $$1$$. Trig focuses on the unit circle (when $$r=1$$). However, in polar coordinates we generalize the equations so that r is usually not $$1$$.
The positive x-axis is called the polar axis, labeled L in plot 2 and the point O is called the pole. All angles are measured from the polar axis with positive angles in a counter-clockwise direction.
Polar coordinates are just parametric equations where the parameter is the angle $$\theta$$ and r is a function of $$\theta$$. It will help you to understand polar coordinates if you have a good understanding of parametrics. Go to the parametrics section for more information.
Note: In plot 2, the angles are measured in degrees. However, in calculus we almost always specify angles in radians. To convert between radians and degrees, remember that $$2\pi = 360^o$$.
Okay, it's time to watch some videos. This first video is really good to give you an overview of polar coordinates.
### Krista King Math - Overview of Polar Coordinates [7min-24secs]
video by Krista King Math
Here is another good video introduction to polar coordinates. He uses graphs and examples very effectively in this video.
### PatrickJMT - Introduction to Polar Coordinates [10min-34secs]
video by PatrickJMT
Okay, now you know the basics of polar coordinates and you can work most problems you come across. However, if you really want to understand polar coordinates, then this video clip is good to watch. It gives a more in-depth discussion with some very good examples, some unique, which will help you a lot.
### MIT OCW - Understand Polar Coordinates [20min-35secs]
The first half of this video discusses parametric equations but you don't need that material to understand the second half on polar coordinates.
video by MIT OCW
Graphing In Polar Coordinates
### Krista King Math - How do you sketch a polar curve (cardioid)? [6min-32secs]
video by Krista King Math
Next
With the basics of polar coordinates under your belt, now it is time to work directly with the points and equations to convert between rectangular and polar coordinates. You will find discussion, videos and practice problems on the next page.
### polar coordinates 17calculus youtube playlist
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Investigation #1 by Jake Klerlein
I would like to discuss the graphs of the sinusoidal equation y = a sin (bx + c). This type of graph in my opinion is one of the first types of graphs that high school students learn about which is truly different from the basic linear, quadratic and other polynomial equation graphs that high schoolers discuss. Also, I find these types of cyclic graphs interesting. Not only to these graphs have applications in the real world, fields such as electronics, but they create a high "OOOOO, AHHHH" factor for students who have no experience with secondary school mathematics. Putting a sine graph on the overhead calculator in my 8th grade math classes created a buzz of interest. "How do you do that one? What is that? Why does it keep going up and down like that?", were some of the types of questions generated simply by showing my students one example. We then chose to experiment with different values to see what would change. This desire from a student to learn more is essential to them being able to make the most of their education. For these reasons it is worthwhile to discuss the equations of the form y = a sin (bx + c).
To begin let's look at the graph of y = 1 sin x, the simplest form of the above equation. We would like to be able to show our students that all other forms of the equation can be graphed if we know this base form's graph. The other forms of this equation will simply change the amplitude of the graph, its period, or will create a phase shift. By changing the values of a, b, and c one at a time we can see what affect they each have on the graph.
Here it is important that we and our students recognize the cyclic nature of this graph. This can be seen by starting at any point on the graph and tracing along the x-axis. We may begin at x = 0 and trace the graph just beyond x = 6. Here we see one period of this figure. Another period is noticed when we begin just left of -3 on the x-axis and trace just past 3. This cycle always repeats itself if you choose a starting point and trace the graph for a particular distance. Of course, we know this distance to be 2 * pi, but as teachers we would like to engage our students in a discussion so that they may determine the period of this graph for themselves. Also, we see the peaks and valleys of the graph and will change a constant to see how those change. We will also see what changes make the graph slide along the x-axis or move up and down the y-axis. Finally, we will experiment to see if we can make the period of this graph change.
It is important to change only one variable at a time, so that we may view the affect that is created. Let's begin by varying a.
Here we can see that as "a" gets larger the graph reaches higher maximums and lower minimums. Thus, we know that the value "a" controls the amplitude of the graph. Let's see what happens for a < 1.
By viewing this set of graphs we may lead our students in a discussion that will enlighten them to the fact that when 0 < a <1, the amplitude lessens. Also, the students should come to the conclusion that if a < 0, then the graph reflects on the x-axis and maintains the same amplitude as the graph with the -a multiplied by the sine function.
Now we investigate what changing values of b will do to our graph.
Certainly this picture is quite confusing. Here it is important that we help our students to see the differences in these graphs. The main concept to lead discussion towards is the fact that the graph repeats itself faster as b increases. Thus, the period is shortened bases on the choice of b and by a factor of 1/b. When b = 2, the period of the graph is half that of when b = 1. Likewise, b = 4 has a fourth the period of b = 1 and half that of b = 2.
What happens if 0 < b < 1?
From this graph we can see that if 0 < b < 1 that the period will get longer by the factor 1/b.
What will happen if b < 0.
By viewing the purple and green graphs we see instantly that making b < 0 will reflect the graph about the x-axis, while preserving the period and amplitude. This reflection property can also be seen if we compare our red graph here to the red graph in the diagram above (jump to that graph), as well as looking at the blue curve here and the red curve two graphs above (jump to that graph)
Let's now look at graphs where the value of c varies.
With some careful inspection we can see that changing values of c simply slides the graph left or right along the x-axis. If c is positive then the graph moves the value of c to the left. Likewise if c is negative then the graph slides that distance to the right.
Now that we have seen how the values of a, b, and c affect the graph of the sine function, we should be able to predict how the following sine graphs will appear. Try to determine what each graph will look like and then click the equation to view the graph.
The natural question one would now ask is what will happen to the graph of the sine function if a number is added or subtracted outside the parentheses. Pick your favorite graphing calculator and have fun investigating.
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# Cutting a Cube/Cuboid
## Number of smaller cubes
(Finding the number of smaller cubes when a bigger cube or cuboid is cut)
When we cut a rod once, then it divides into how many parts?
When we cut a rod once, then it gets divided into two equal parts.
Similarly, if we cut a rod twice, then it divides into three equal parts. And so on…
Can we generalize?
If number of cuts = (n -1), then number of parts in which a rod or a line gets divided into = n
A cube can also be cut as a rod/stick/line.
If a cube of X cm long sides is cut into equal smaller cubes of sides Y cm, then n = X/Y
(n is the number of smaller cubes on each edge.)
If these cuts are made along all the three dimensions, i.e. along the length, breadth and height, then we will get smaller cubes of same size.
Total number of such smaller cubes = $n^3$
##### Q. If the edge of a cube is 10 cm and it has to be cut into smaller cubes each having edge of 2 cm, then:
1. Each edge of the cube has to be cut how many times?
2. Whole cube will has to be cut how many times?
3. What is the number of smaller cubes thus obtained?
Explanation:
Number of smaller cubes on each edge, i.e. n = X/Y = 10/2 = 5, i.e. each edge has to be divided into 5 parts. (here X = 10 and Y = 2).
And to divide into five parts, each edge will have to be cut 5-1 = 4 times Hence, the number of times the cube has to be cut = 4×3 = 12
Number of smaller cubes = $n^3$ = $5^3$ = 125 (Basically, it’s length × breadth × height)
What if it’s a cuboid?
Can we still count the number of smaller cubes?
When a cuboid is cut into smaller cubes of equal volume, then:
Total number of Cubes = $\frac{Vol. \hspace{1ex} of \hspace{1ex} Cuboid}{Vol. \hspace{1ex} of \hspace{1ex} smaller \hspace{1ex} cube}$ = $\frac{Length × Breadth × Height \hspace{1ex} of \hspace{1ex} cuboid}{(Side \hspace{1ex} of \hspace{1ex} smaller \hspace{1ex} cubes)^3}$
Let’s consider an example.
##### Q. If a cuboid with length = 10 cm, breadth = 8 cm and height = 6 cm is cut into smaller cubes of edge 2 cm each, then find the number of smaller cubes.
(a) 72 (b) 42 (c) 48 (d) 60
Explanations :
Explanation 1:
Total number of Cubes = $\frac{Length × Breadth × Height \hspace{1ex} of \hspace{1ex} cuboid}{(Side \hspace{1ex} of \hspace{1ex} smaller \hspace{1ex} cubes)^3}$ = $\frac{10 × 8 × 6}{2^3}$ = 60
Explanation 2:
Total number of Cubes = No. of cubes along length × No. of cubes along breadth × No. of cubes along height = (10/2) × (8/2) × (6/2) = 5 × 4 × 3 = 60
##### Q. How many cubes are there in this cuboid?
Explanation:
Number of cubes = Number of cubes along the Height × Number of cubes along the Length × Number of cubes along the Width = 5 × 3 × 3 = 45
## Types of smaller cubes
After cutting a cube, following four types of smaller cubes are obtained:
(we are taking an example of a cube where there are 3 cubes on each of its edges, i.e. n = 3)
### Corner cubes
Number of corner cubes = 8 (always)
If a cube is painted on all faces/sides and the cuts are made, then these smaller corner cubes will have three faces/sides painted (as they have three exposed faces).
No smaller cube can have more than three painted faces under normal circumstances.
### Cubes at the middle of each edge
Number of middle cubes = 12 (n - 2)
OR
Number of middle cubes = Number of middle cubes on each edge × 12 (as there are 12 edges in a cube) In this case, Number of middle cubes = 12 (n - 2) = 12 (3 - 2) = 12
If a cube is painted on all faces/sides and the cuts are made, then these middle smaller cubes will have two faces/sides painted (as they have two exposed faces).
### Cubes at the centre of each face
Number of such cubes = 6 $(n – 2)^2$
OR
Number of such cubes = Number of middle cubes in each face × 6 (as there are 6 faces in a cube) In this case, such cubes = 6 $(3 - 2)^2$ = 6
If a cube is painted on all faces/sides and the cuts are made, then these smaller cubes will have one face/side painted (as they have one exposed face).
### Inner cubes that are hidden
Number of inner cubes = $(n - 2)^3$ In this case, such cubes = $(n - 2)^3$ = $(3 - 2)^3$ = 1
If a cube is painted on all faces/sides and the cuts are made, then these inner smaller cubes will have no face/side painted (as they do not have any face exposed).
Total number of smaller cubes = Corner cubes + Middle of edge cubes + Centre of face cubes + Inner cubes
or $n^3$ = 8 + 12 (n - 2) + 6 $(n – 2)^2$ + $(n - 2)^3$
In our example:
$3^3$ = 8 + 12 + 6 + 1
or 27 = 27
Hence, we have accounted for all kinds of smaller cubes.
##### Q. If a larger cube of 8 cm edge is cut into smaller cubes of equal volumes, each having edge of 2 cm, then:
Number of smaller cubes = ?
Number of corner cubes = ?
Number of cubes at the centre of edges = ?
Number of cubes at the centre of faces = ?
Number of inner cubes (N) = ?
Explanation:
n = 8/2 = 4
Number of smaller cubes = $n^3$ = $4^3$ = 64
Number of corner cubes = 8 (always)
Number of cubes at the centre of edges = 12 (n - 2) = 12 (4 - 2) = 24
Number of cubes at the centre of faces = 6 $(n – 2)^2$ = 6 $(4 – 2)^2$ = 24
Number of inner cubes = $(n - 2)^3$ = $(4 - 2)^3$ = 8
Note: If we check, total number of cubes = 8 + 24 + 24 + 8 = 64
##### Q. A cube of side 4 cm is painted blue on all of its surfaces and then divided into various smaller cubes of side 1 cm each. Find the number of cubes having none, one, two and three faces painted.
Explanation:
n = 4/1 = 4
Smaller cubes with three faces painted = Number of corner cubes = 8 (always)
Smaller cubes with two faces painted = Number of cubes at the centre of edges = 12 (n - 2) = 12 (4 - 2) = 24
Smaller cubes with one face painted = Number of cubes at the centre of faces = 6 $(n – 2)^2$ = 6 $(4 – 2)^2$ = 24
Smaller cubes with no face painted = Number of inner cubes = $(n - 2)^3$ = $(4 - 2)^3$ = 8
For those who are more into mugging up the numbers, here’s a list of the number of four types of smallers cubes/blocks (got after cutting bigger cubes of various sizes):
Note: We will solve questions wherein the larger cube is painted with more than one colours and then cut, in a separate module. Here we covered the basics of cutting and painting a cube.
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# Regression analysis
The regression analysis is a very interesting topic that many times the utility that we see is null, some people like you do see a great utility to regression analysis since statistics is a very useful tool for modeling equations. And in the regression analysis we will analyze an experiment, so remember that the ultimate goal of an experiment is to predict the behavior of a certain phenomenon, which can be represented in a regression graph.
When performing an experiment, we obtain pairs of data (X, Y) that are placed in a Cartesian plane and thus these points form a scatter diagram.
The points can approach a curve in such a way that it adjusts to the behavior of the phenomena. In addition, this curve can be of many types, such as parabola, exponential or geometric.
## Simple linear regression
This type of regression uses the independent variable (x) and a dependent variable (y) for a population.
Suppose that the relation between both is a straight line, therefore, the line can be written in the following way:
Y = \beta_{0} + \beta_{1}x
An experiment consists of random results, that is, we can not predict exactly what value will be measured from the independent variable. To which it refers that since we are using samples obtained by an experiment, the data refer to estimates, resulting in the following graph:
\hat{y} =\hat{\beta}_{0} + \hat{\beta}_{1}x
## How to calculate the regression coefficients
\hat{y} is the regression line, but that \hat{y} is matched to other things, which are \hat{\beta}_{0} plus \hat{\beta}_{1} x . To find the values of \hat{\beta}_0 and \hat{\beta}_{1}, we will give you at once the formulas for you to calculate those values called regression coefficients. Calm down, the formulas are a little tedious where they come from, but they are not so tedious to solve:
\hat{\beta}_{1} = \cfrac{n\sum xy - \left[ \left( \sum x\right)\left(\sum y \right)\right]}{n\sum x^{2} - \left( \sum x \right)^{2}}
\hat{\beta}_{0} = \bar{y} - \hat{\beta}_{1}\bar{x}
## Some considerations of the regression line
• Do not use the least squares line when the data is not linear.
• Estimators are not the same as true values.
• That there is a relationship does not indicate that there is causality between the two.
## Example of regression analysis
Let’s go with the statement:
In this regression line example that we will see, we measured the inertial weight (in tons) and the fuel savings (in miles/gallon) for a sample of seven diesel trucks, predicts how different are the truck’s mileage if they have a difference of 5 tons. The following table presents the results:
\begin{array}{|c|c|}
\hline
\text{Weight }(x) & \text{Mileage } (y) \\
\hline
8\text{.}00 & 7\text{.}69 \\
24\text{.}50 & 4\text{.}97 \\
27\text{.}00 & 4\text{.}56 \\
14\text{.}50 & 6\text{.}49 \\
28\text{.}50 & 4\text{.}34 \\
12\text{.}75 & 6\text{.}24 \\
21\text{.}25 & 4\text{.}45 \\
\hline
\end{array}
Very well, what can be observed is that the weight is the independent variable x and the mileage is the dependent variable y. Now, to make the calculation of the regression line, we need the elements of the regression coefficients, so we need \sum x, \sum y, \sum xy, \sum x^{2} and \left(\sum x\right)^{2}
Let’s first determine the sum of x and y:
\begin{array}{| c | c |}
\hline
\text{Weight } (x) & \text{Mileage } (y) \\
\hline
8\text{.}00 & 7\text{.}69 \\
24\text{.}50 & 4\text{.}97 \\
27\text{.}00 & 4\text{.}56 \\
14\text{.}50 & 6\text{.}49 \\
28\text{.}50 & 4\text{.}34 \\
12\text{.}75 & 6\text{.}24 \\
21\text{.}25 & 4\text{.}45 \\
\hline
\sum x = 136\text{.}5 & \sum y = 38.74\\
\hline
\end{array}
Now let’s multiply x by y and we will calculate the sum:
\begin{array}{| c | c | c |}
\hline
\text{Weight }(x) & \text{Mileage } (y) & xy \\
\hline
8\text{.}00 & 7\text{.}69 & 61\text{.}52\\
24\text{.}50 & 4\text{.}97 & 121\text{.}765\\
27\text{.}00 & 4\text{.}56 & 123\text{.}12 \\
14\text{.}50 & 6\text{.}49 & 94\text{.}105 \\
28\text{.}50 & 4\text{.}34 & 123\text{.}69 \\
12\text{.}75 & 6\text{.}24 & 79\text{.}56 \\
21\text{.}25 & 4\text{.}45 & 94\text{.}5625 \\
\hline
& & \sum xy = 698\text{.}3225 \\
\hline
\end{array}
Then we will calculate the sum of x squared:
\begin{array}{| c | c | c |}
\hline
\text{Weight }(x) & \text{Mileage }(y) & x^{2} \\
\hline
8\text{.}00 & 7\text{.}69 & 64 \\
24\text{.}50 & 4\text{.}97 & 600\text{.}25 \\
27\text{.}00 & 4\text{.}56 & 729 \\
14\text{.}50 & 6\text{.}49 & 210\text{.}25 \\
28\text{.}50 & 4\text{.}34 & 812\text{.}25 \\
12\text{.}75 & 6\text{.}24 & 162\text{.}5625 \\
21\text{.}25 & 4\text{.}45 & 451\text{.}5625 \\
\hline
& & \sum x^{2} = 3029\text{.}875 \\
\hline
\end{array}
Next we are going to calculate the sum of y squared:
\begin{array}{| c | c | c |}
\hline
\text{Weight }(x) & \text{Mileage }(y) & y^{2} \\
\hline
8\text{.}00 & 7\text{.}69 & 59\text{.}1361 \\
24\text{.}50 & 4\text{.}97 & 24\text{.}7009 \\
27\text{.}00 & 4\text{.}56 & 20\text{.}7936 \\
14\text{.}50 & 6\text{.}49 & 42\text{.}1201 \\
28\text{.}50 & 4\text{.}34 & 18\text{.}8356 \\
12\text{.}75 & 6\text{.}24 & 38\text{.}9376 \\
21\text{.}25 & 4\text{.}45 & 19\text{.}8025 \\
\hline
& & \sum y^{2} = 224\text{.}3264 \\
\hline
\end{array}
We need to calculate the average of x and the average of y, let’s calculate it:
\bar{y} = \cfrac{\sum y}{n} = \cfrac{38.74}{7} = 5.5342
\bar{x} = \cfrac{\sum x}{n} = \cfrac{136.5}{7} = 19.5
Finally, taking the equations of the regression coefficients, substitute all the values in the formulas:
\hat{\beta}_{1} = \cfrac{7(698.3225)-[(136.5)(38.74)]}{7(3029.875) - 18632.25} = - 0.1551
\hat{\beta}_{0} = \bar{y} - \hat{\beta}_{1}\hat{x} = 5.5342 - ( - 0.1551)(19.5) = 8.5593
Our regression line will be as follows:
\hat{y} = 8.5593 - 0.1551x
### Predict how different the mileage of two 5-ton difference trucks is
To do this part of the exercise, two values of x were taken with a difference of 5 tons and substituted in the equation of the line that was found, so values of x = 9 and x = 14 were taken:
x = 9
\hat{y} = 8.5593 - 0.1551(9) = 7.1631
x = 14
\hat{y} = 8.5593 - 0.1551(14) = 6.3875
Then you have to make the difference of the two values obtained:
7.1631 - 6.3875 = 0.7756
So it is predicted that your mileage has 0.7756 difference if the trucks have 5 tons difference.
Thank you for being at this moment with us:)
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How many degrees does the hour hand of a clock turn in 5 minutes?
Nov 22, 2015
$\frac{360}{144} = {2.5}^{\circ}$
Explanation:
Each hour is $\frac{1}{12}$ of a complete circle for the hour hand (on a standard $12$ hour clock).
$5$ minutes is $\frac{1}{12}$ of an hour.
So in $5$ minutes, the hour hand moves $\frac{1}{12} \cdot \frac{1}{12} = \frac{1}{144}$ of a complete circle.
A complete circle is ${360}^{\circ}$, so this is $\frac{360}{144} = {2.5}^{\circ}$
Jul 20, 2017
2.5° in $5$ minutes
Explanation:
What fraction does $5$ minutes represent of a full revolution of the hour hand?
The hour hand rotates 360° in $12$ hours.
This is $12 \times 60 = 720$ minutes
Therefore $5$ minutes is
5/720 xx 360° = 2.5°
The same result is obtained if we determine through how many degrees the hour hand turns in $1$ hour:
360 div 12 = 30° in one hour
$5$ minutes is $\frac{1}{12}$ of an hour.
1/12 xx 30° = 2.5°
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# Application of Arithmetic Sequence and Series
Example:
Tickets for a certain show were printed bearing numbers from $1$ to $100$. The odd number tickets were sold by receiving cents equal to thrice the number on the ticket while the even number tickets were issued by receiving cents equal to twice the number on the ticket. How much was received by the issuing agency?
Solution:
Let ${S_1}$ and ${S_2}$ be the amounts received for odd number and even number tickets respectively, then
${S_1} = 3\left[ {1 + 3 + 5 + \cdots + 99} \right]$ and ${S_2} = 2\left[ {2 + 4 + 6 + \cdots + 100} \right]$
Thus, ${S_1} + {S_2} = 3 \times \frac{{50}}{2}\left( {1 + 99} \right) + 2 \times \frac{{50}}{2}\left( {2 + 100} \right)$
$\because$ there are $50$ terms in each series
${S_1} + {S_2} = 7500 + 5100 = 12600$
Hence the total amount received by the issuing agency $= 12600$ cents $= \ 126$
Example:
A clock strikes once when its hour hand is at $1$, twice when it is at $2$ and so on. How many times does the clock strike in six hours?
Solution:
Since the clock strikes once when its hour hand is at $1$, twice when it is at $2$ and so on, so the sequence of strikes from $1$ hour to $6$ hours is $1,2,3,4,5,6$.
Here ${a_1} = 1$, $d = 2 - 1 = 1$, $n = 6$ so
${S_n} = \frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right]$
$= \frac{6}{2}\left[ {2\left( 1 \right) + \left( {6 - 1} \right)\left( 1 \right)} \right] = 3\left( {2 + 5} \right) = 3\left( 7 \right) = 21$
This shows that the clock strikes $21$ times in $6$ hours.
Example:
A factory owner repays his loan of $\ 2088000$ by $\ 20000$ in the first monthly installment and then increases the payment by $\ 1000$ in every installment. How many installments will it take for him to clear his loan?
Solution:
By the given conditions, we have
${a_1} = 20000$, $d = 1000$, ${S_n} = 2088000$
${S_n} = \frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right]$
$2088000 = \frac{n}{2}\left[ {2\left( {20000} \right) + \left( {n - 1} \right)\left( {1000} \right)} \right]$
$\Rightarrow 2088 = \frac{n}{2}\left[ {40 + n - 1} \right]$
$\Rightarrow 4176 = {n^2} + 39n$
$\Rightarrow {n^2} + 39n - 4176 = 0$
$\Rightarrow n = \frac{{ - 39 \pm \sqrt {{{\left( {39} \right)}^2} - 4\left( 1 \right)\left( { - 4176} \right)} }}{{2\left( 1 \right)}} = \frac{{ - 39 \pm \sqrt {1521 + 16704} }}{2}$
$= \frac{{ - 39 \pm \sqrt {18225} }}{2} = \frac{{ - 39 \pm 135}}{2}$
$= \frac{{ - 39 + 135}}{2},\frac{{ - 39 - 135}}{2}$
$\Rightarrow n = 48, - 87$
Now $n$, being the number of installments, cannot be negative, so $n = 48$. This shows that the factory owner will clear his loan in $48$ monthly installments.
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Mathematics NCERT Grade 9, Chapter 13: Surface Areas And Volumes- The chapter begins with recalling plane and solid figures. The chapter explains that how the relationship between length and breadth of the figure can give us area and inclusion of the concept of heights can provide us the volume of any particular given object.
The first section is about the surface area of a cuboid and a cube. This includes lateral surface area of a cuboid as well.
Exercise 13.1 is about the same consisting of lot of word problems.
The next topic is- Surface area of a right circular cylinder.
• The topic is also accounted for Curved Surface Area and Total Surface Area of Right Circular Cylinder.
Exercise 13.2 covers the questions based on the surface area of a cylinder.
Moving on to the next solid shape which is right circular cone is explained followed by the formulas for finding the Curved Surface Area and Total surface area of Right Circular Cone. Exercise 13.3 contains 8 questions based on the surface area of a cone.
The next shape is sphere and hemisphere.
• A sphere is a three-dimensional figure(solid figure), which is made up of all points in the space, which lie at a constant distance called the radius, from a fixed point called the centre of the sphere.
• A solid sphere when sliced exactly ‘through the middle’ with a plane that passes through its centre gets divided into two equal parts called the hemisphere.
Curved Surface Area and Total Surface Area of sphere and Hemisphere are discussed in the chapter.
Later, in this chapter volume will be discussed.
• If an object is solid, then the space occupied by such an object is measured and is termed the volume of the object.
Firstly, the Volume of the cuboid and Cube are discussed, then the volume of the cylinder is explained followed by the volume of a right circular cone.
Lastly, we will discuss the Volume of a Sphere and Hemisphere.
The chapter is filled with solved examples and exercises. It requires knowing the correct formulas and its application. Thus, practising the questions is the best way to master this chapter.
In the end summary of the chapter is discussed.
#### Question 1:
A plastic box 1.5 m long, 1.25 m wide and 65 cm deep, is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine:
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1 m2 costs Rs 20.
It is given that, length (l) of box = 1.5 m
Breadth (b) of box = 1.25 m
Depth (h) of box = 0.65 m
(i) Box is to be open at top.
Area of sheet required
= 2lh + 2bh + lb
= [2 × 1.5 × 0.65 + 2 × 1.25 × 0.65 + 1.5 × 1.25] m2
= (1.95 + 1.625 + 1.875) m2 = 5.45 m2
(ii) Cost of sheet per m2 area = Rs 20
Cost of sheet of 5.45 m2 area = Rs (5.45 × 20)
= Rs 109
##### Video Solution for surface areas and volumes (Page: 213 , Q.No.: 1)
NCERT Solution for Class 9 maths - surface areas and volumes 213 , Question 1
#### Question 2:
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 per m2.
It is given that
Length (l) of room = 5 m
Breadth (b) of room = 4 m
Height (h) of room = 3 m
It can be observed that four walls and the ceiling of the room are to be white-washed. The floor of the room is not to be white-washed.
Area to be white-washed = Area of walls + Area of ceiling of room
= 2lh + 2bh + lb
= [2 × 5 × 3 + 2 × 4 × 3 + 5 × 4] m2
= (30 + 24 + 20) m2
= 74 m2
Cost of white-washing per m2 area = Rs 7.50
Cost of white-washing 74 m2 area = Rs (74 × 7.50)
= Rs 555
##### Video Solution for surface areas and volumes (Page: 213 , Q.No.: 2)
NCERT Solution for Class 9 maths - surface areas and volumes 213 , Question 2
#### Question 3:
The floor of a rectangular hall has a perimeter 250 m. If the cost of panting the four walls at the rate of Rs.10 per m2 is Rs.15000, find the height of the hall.
[Hint: Area of the four walls = Lateral surface area.]
Let length, breadth, and height of the rectangular hall be l m, b m, and h m respectively.
Area of four walls = 2lh + 2bh
= 2(l + b) h
Perimeter of the floor of hall = 2(l + b)
= 250 m
∴ Area of four walls = 2(l + b) h = 250h m2
Cost of painting per m2 area = Rs 10
Cost of painting 250h m2 area = Rs (250h × 10) = Rs 2500h
However, it is given that the cost of paining the walls is Rs 15000.
∴ 15000 = 2500h
h = 6
Therefore, the height of the hall is 6 m.
##### Video Solution for surface areas and volumes (Page: 213 , Q.No.: 3)
NCERT Solution for Class 9 maths - surface areas and volumes 213 , Question 3
#### Question 4:
The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?
Total surface area of one brick = 2(lb + bh + lh)
= [2(22.5 ×10 + 10 × 7.5 + 22.5 × 7.5)] cm2
= 2(225 + 75 + 168.75) cm2
= (2 × 468.75) cm2
= 937.5 cm2
Let n bricks can be painted out by the paint of the container.
Area of n bricks = (n ×937.5) cm2 = 937.5n cm2
Area that can be painted by the paint of the container = 9.375 m2 = 93750 cm2
∴ 93750 = 937.5n
n = 100
Therefore, 100 bricks can be painted out by the paint of the container.
##### Video Solution for surface areas and volumes (Page: 213 , Q.No.: 4)
NCERT Solution for Class 9 maths - surface areas and volumes 213 , Question 4
#### Question 5:
A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?
(i) Edge of cube = 10 cm
Length (l) of box = 12.5 cm
Breadth (b) of box = 10 cm
Height (h) of box = 8 cm
Lateral surface area of cubical box = 4(edge)2
= 4(10 cm)2
= 400 cm2
Lateral surface area of cuboidal box = 2[lh + bh]
= [2(12.5 × 8 + 10 × 8)] cm2
= (2 × 180) cm2
= 360 cm2
Clearly, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box.
Lateral surface area of cubical box − Lateral surface area of cuboidal box = 400 cm2 − 360 cm2 = 40 cm2
Therefore, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box by 40 cm2.
(ii) Total surface area of cubical box = 6(edge)2 = 6(10 cm)2 = 600 cm2
Total surface area of cuboidal box
= 2[lh + bh + lb]
= [2(12.5 × 8 + 10 × 8 + 12.5 × 10] cm2
= 610 cm2
Clearly, the total surface area of the cubical box is smaller than that of the cuboidal box.
Total surface area of cuboidal box − Total surface area of cubical box = 610 cm2 − 600 cm2 = 10 cm2
Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 cm2.
##### Video Solution for surface areas and volumes (Page: 213 , Q.No.: 5)
NCERT Solution for Class 9 maths - surface areas and volumes 213 , Question 5
#### Question 6:
A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12 edges?
(i) Length (l) of green house = 30 cm
Breadth (b) of green house = 25 cm
Height (h) of green house = 25 cm
Total surface area of green house
= 2[lb + lh + bh]
= [2(30 × 25 + 30 × 25 + 25 × 25)] cm2
= [2(750 + 750 + 625)] cm2
= (2 × 2125) cm2
= 4250 cm2
Therefore, the area of glass is 4250 cm2.
(ii)
It can be observed that tape is required along side AB, BC, CD, DA, EF, FG, GH, HE, AH, BE, DG, and CF.
Total length of tape = 4(l + b + h)
= [4(30 + 25 + 25)] cm
= 320 cm
Therefore, 320 cm tape is required for all the 12 edges.
##### Video Solution for surface areas and volumes (Page: 213 , Q.No.: 6)
NCERT Solution for Class 9 maths - surface areas and volumes 213 , Question 6
#### Question 7:
Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.
Length (l1) of bigger box = 25 cm
Breadth (b1) of bigger box = 20 cm
Height (h1) of bigger box = 5 cm
Total surface area of bigger box = 2(lb + lh + bh)
= [2(25 × 20 + 25 × 5 + 20 × 5)] cm2
= [2(500 + 125 + 100)] cm2
= 1450 cm2
Extra area required for overlapping
= 72.5 cm2
While considering all overlaps, total surface area of 1 bigger box
= (1450 + 72.5) cm2 =1522.5 cm2
Area of cardboard sheet required for 250 such bigger boxes
= (1522.5 × 250) cm2 = 380625 cm2
Similarly, total surface area of smaller box = [2(15 ×12 + 15 × 5 + 12 × 5] cm2
= [2(180 + 75 + 60)] cm2
= (2 × 315) cm2
= 630 cm2
Therefore, extra area required for overlappingcm2
Total surface area of 1 smaller box while considering all overlaps
= (630 + 31.5) cm2 = 661.5 cm2
Area of cardboard sheet required for 250 smaller boxes = (250 × 661.5) cm2
= 165375 cm2
Total cardboard sheet required = (380625 + 165375) cm2
= 546000 cm2
Cost of 1000 cm2 cardboard sheet = Rs 4
Cost of 546000 cm2 cardboard sheet
Therefore, the cost of cardboard sheet required for 250 such boxes of each kind will be Rs 2184.
##### Video Solution for surface areas and volumes (Page: 213 , Q.No.: 7)
NCERT Solution for Class 9 maths - surface areas and volumes 213 , Question 7
#### Question 8:
Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m?
Length (l) of shelter = 4 m
Breadth (b) of shelter = 3 m
Height (h) of shelter = 2.5 m
Tarpaulin will be required for the top and four wall sides of the shelter.
Area of Tarpaulin required = 2(lh + bh) + l b
= [2(4 × 2.5 + 3 × 2.5) + 4 × 3] m2
= [2(10 + 7.5) + 12] m2
= 47 m2
Therefore, 47 m2 tarpaulin will be required.
##### Video Solution for surface areas and volumes (Page: 213 , Q.No.: 8)
NCERT Solution for Class 9 maths - surface areas and volumes 213 , Question 8
#### Question 1:
The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. Assume π =
Height (h) of cylinder = 14 cm
Let the diameter of the cylinder be d.
Curved surface area of cylinder = 88 cm2
⇒ 2πrh = 88 cm2 (r is the radius of the base of the cylinder)
⇒ πdh = 88 cm2 (d = 2r)
d = 2 cm
Therefore, the diameter of the base of the cylinder is 2 cm.
##### Video Solution for surface areas and volumes (Page: 216 , Q.No.: 1)
NCERT Solution for Class 9 maths - surface areas and volumes 216 , Question 1
#### Question 2:
It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same?
Height (h) of cylindrical tank = 1 m
Base radius (r) of cylindrical tank
Therefore, it will require 7.48 m2 area of sheet.
##### Video Solution for surface areas and volumes (Page: 216 , Q.No.: 2)
NCERT Solution for Class 9 maths - surface areas and volumes 216 , Question 2
#### Question 3:
A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm.
(i) Inner curved surface area,
(ii) Outer curved surface area,
(iii) Total surface area.
Inner radius of cylindrical pipe
Outer radius of cylindrical pipe
Height (h) of cylindrical pipe = Length of cylindrical pipe = 77 cm
(i) CSA of inner surface of pipe
(ii) CSA of outer surface of pipe
(iii) Total surface area of pipe = CSA of inner surface + CSA of outer surface + Area of both circular ends of pipe
Therefore, the total surface area of the cylindrical pipe is 2038.08 cm2.
##### Video Solution for surface areas and volumes (Page: 216 , Q.No.: 3)
NCERT Solution for Class 9 maths - surface areas and volumes 216 , Question 3
#### Question 4:
The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2?
It can be observed that a roller is cylindrical.
Height (h) of cylindrical roller = Length of roller = 120 cm
Radius (r) of the circular end of roller =
CSA of roller = 2πrh
Area of field = 500 × CSA of roller
= (500 × 31680) cm2
= 15840000 cm2
= 1584 m2
##### Video Solution for surface areas and volumes (Page: 217 , Q.No.: 4)
NCERT Solution for Class 9 maths - surface areas and volumes 217 , Question 4
#### Question 5:
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs.12.50 per m2.
Height (h) cylindrical pillar = 3.5 m
Radius (r) of the circular end of pillar =
= 0.25 m
CSA of pillar = 2πrh
Cost of painting 1 m2 area = Rs 12.50
Cost of painting 5.5 m2 area = Rs (5.5 × 12.50)
= Rs 68.75
Therefore, the cost of painting the CSA of the pillar is Rs 68.75.
##### Video Solution for surface areas and volumes (Page: 217 , Q.No.: 5)
NCERT Solution for Class 9 maths - surface areas and volumes 217 , Question 5
#### Question 6:
Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.
Let the height of the circular cylinder be h.
Radius (r) of the base of cylinder = 0.7 m
CSA of cylinder = 4.4 m2
rh = 4.4 m2
h = 1 m
Therefore, the height of the cylinder is 1 m.
##### Video Solution for surface areas and volumes (Page: 217 , Q.No.: 6)
NCERT Solution for Class 9 maths - surface areas and volumes 217 , Question 6
#### Question 7:
The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) Its inner curved surface area,
(ii) The cost of plastering this curved surface at the rate of Rs 40 per m2.
Inner radius (r) of circular well
Depth (h) of circular well = 10 m
Inner curved surface area = 2πrh
= (44 × 0.25 × 10) m2
= 110 m2
Therefore, the inner curved surface area of the circular well is 110 m2.
Cost of plastering 1 m2 area = Rs 40
Cost of plastering 110 m2 area = Rs (110 × 40)
= Rs 4400
Therefore, the cost of plastering the CSA of this well is Rs 4400.
##### Video Solution for surface areas and volumes (Page: 217 , Q.No.: 7)
NCERT Solution for Class 9 maths - surface areas and volumes 217 , Question 7
#### Question 8:
In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Height (h) of cylindrical pipe = Length of cylindrical pipe = 28 m
Radius (r) of circular end of pipe = = 2.5 cm = 0.025 m
CSA of cylindrical pipe = 2πrh
= 4.4 m2
The area of the radiating surface of the system is 4.4 m2.
##### Video Solution for surface areas and volumes (Page: 217 , Q.No.: 8)
NCERT Solution for Class 9 maths - surface areas and volumes 217 , Question 8
#### Question 9:
Find
(i) The lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) How much steel was actually used, if of the steel actually used was wasted in making the tank.
Height (h) of cylindrical tank = 4.5 m
Radius (r) of the circular end of cylindrical tank =
(i) Lateral or curved surface area of tank = 2πrh
= (44 × 0.3 × 4.5) m2
= 59.4 m2
Therefore, CSA of tank is 59.4 m2.
(ii) Total surface area of tank = 2πr (r + h)
= (44 × 0.3 × 6.6) m2
= 87.12 m2
Let A m2 steel sheet be actually used in making the tank.
Therefore, 95.04 m2 steel was used in actual while making such a tank.
##### Video Solution for surface areas and volumes (Page: 217 , Q.No.: 9)
NCERT Solution for Class 9 maths - surface areas and volumes 217 , Question 9
#### Question 10:
In the given figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.
Height (h) of the frame of lampshade = (2.5 + 30 + 2.5) cm = 35 cm
Radius (r) of the circular end of the frame of lampshade =
Cloth required for covering the lampshade = rh
= 2200 cm2
Hence, for covering the lampshade, 2200 cm2 cloth will be required.
##### Video Solution for surface areas and volumes (Page: 217 , Q.No.: 10)
NCERT Solution for Class 9 maths - surface areas and volumes 217 , Question 10
#### Question 11:
The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?
Radius (r) of the circular end of cylindrical penholder = 3 cm
Height (h) of penholder = 10.5 cm
Surface area of 1 penholder = CSA of penholder + Area of base of penholder
= 2πrh + πr2
Area of cardboard sheet used by 1 competitor
Area of cardboard sheet used by 35 competitors
= = 7920 cm2
Therefore, 7920 cm2 cardboard sheet will be bought.
##### Video Solution for surface areas and volumes (Page: 217 , Q.No.: 11)
NCERT Solution for Class 9 maths - surface areas and volumes 217 , Question 11
#### Question 1:
Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Radius (r) of the base of cone == 5.25 cm
Slant height (l) of cone = 10 cm
CSA of cone = πrl
Therefore, the curved surface area of the cone is 165 cm2.
##### Video Solution for surface areas and volumes (Page: 221 , Q.No.: 1)
NCERT Solution for Class 9 maths - surface areas and volumes 221 , Question 1
#### Question 2:
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Radius (r) of the base of cone == 12 m
Slant height (l) of cone = 21 m
Total surface area of cone = πr(r + l)
##### Video Solution for surface areas and volumes (Page: 221 , Q.No.: 2)
NCERT Solution for Class 9 maths - surface areas and volumes 221 , Question 2
#### Question 3:
Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find
(i) radius of the base and (ii) total surface area of the cone.
(i) Slant height (l) of cone = 14 cm
Let the radius of the circular end of the cone be r.
We know, CSA of cone = πrl
Therefore, the radius of the circular end of the cone is 7 cm.
(ii) Total surface area of cone = CSA of cone + Area of base
= πrl + πr2
Therefore, the total surface area of the cone is 462 cm2.
##### Video Solution for surface areas and volumes (Page: 221 , Q.No.: 3)
NCERT Solution for Class 9 maths - surface areas and volumes 221 , Question 3
#### Question 4:
A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent
(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs 70.
(i) Let ABC be a conical tent.
Height (h) of conical tent = 10 m
Radius (r) of conical tent = 24 m
Let the slant height of the tent be l.
In ΔABO,
AB2 = AO2 + BO2
l2 = h2 + r2
= (10 m)2 + (24 m)2
= 676 m2
l = 26 m
Therefore, the slant height of the tent is 26 m.
(ii) CSA of tent = πrl
Cost of 1 m2 canvas = Rs 70
Cost of canvas =
= Rs 137280
Therefore, the cost of the canvas required to make such a tent is
Rs 137280.
##### Video Solution for surface areas and volumes (Page: 221 , Q.No.: 4)
NCERT Solution for Class 9 maths - surface areas and volumes 221 , Question 4
#### Question 5:
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. [Use π = 3.14]
Height (h) of conical tent = 8 m
Radius (r) of base of tent = 6 m
Slant height (l) of tent =
CSA of conical tent = πrl
= (3.14 × 6 × 10) m2
= 188.4 m2
Let the length of tarpaulin sheet required be l.
As 20 cm will be wasted, therefore, the effective length will be (l − 0.2 m).
Breadth of tarpaulin = 3 m
Area of sheet = CSA of tent
[(l − 0.2 m) × 3] m = 188.4 m2
l − 0.2 m = 62.8 m
l = 63 m
Therefore, the length of the required tarpaulin sheet will be 63 m.
##### Video Solution for surface areas and volumes (Page: 221 , Q.No.: 5)
NCERT Solution for Class 9 maths - surface areas and volumes 221 , Question 5
#### Question 6:
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs 210 per 100 m2.
Slant height (l) of conical tomb = 25 m
Base radius (r) of tomb = 7 m
CSA of conical tomb = πrl
= 550 m2
Cost of white-washing 100 m2 area = Rs 210
Cost of white-washing 550 m2 area =
= Rs 1155
Therefore, it will cost Rs 1155 while white-washing such a conical tomb.
##### Video Solution for surface areas and volumes (Page: 221 , Q.No.: 6)
NCERT Solution for Class 9 maths - surface areas and volumes 221 , Question 6
#### Question 7:
A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Radius (r) of conical cap = 7 cm
Height (h) of conical cap = 24 cm
Slant height (l) of conical cap =
CSA of 1 conical cap = πrl
CSA of 10 such conical caps = (10 × 550) cm2 = 5500 cm2
Therefore, 5500 cm2 sheet will be required.
##### Video Solution for surface areas and volumes (Page: 221 , Q.No.: 7)
NCERT Solution for Class 9 maths - surface areas and volumes 221 , Question 7
#### Question 8:
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per m2, what will be the cost of painting all these cones? (Use π = 3.14 and take= 1.02).
Radius (r) of cone = = 0.2 m
Height (h) of cone = 1 m
Slant height (l) of cone =
CSA of each cone = πrl
= (3.14 × 0.2 × 1.02) m2 = 0.64056 m2
CSA of 50 such cones = (50 × 0.64056) m2
= 32.028 m2
Cost of painting 1 m2 area = Rs 12
Cost of painting 32.028 m2 area = Rs (32.028 × 12)
= Rs 384.336
= Rs 384.34 (approximately)
Therefore, it will cost Rs 384.34 in painting 50 such hollow cones.
##### Video Solution for surface areas and volumes (Page: 221 , Q.No.: 8)
NCERT Solution for Class 9 maths - surface areas and volumes 221 , Question 8
#### Question 1:
Find the surface area of a sphere of radius:
(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm
(i) Radius (r) of sphere = 10.5 cm
Surface area of sphere = 4πr2
Therefore, the surface area of a sphere having radius 10.5cm is 1386 cm2.
(ii) Radius(r) of sphere = 5.6 cm
Surface area of sphere = 4πr2
Therefore, the surface area of a sphere having radius 5.6 cm is 394.24 cm2.
(iii) Radius (r) of sphere = 14 cm
Surface area of sphere = 4πr2
Therefore, the surface area of a sphere having radius 14 cm is 2464 cm2.
##### Video Solution for surface areas and volumes (Page: 225 , Q.No.: 1)
NCERT Solution for Class 9 maths - surface areas and volumes 225 , Question 1
#### Question 2:
Find the surface area of a sphere of diameter:
(i) 14 cm (ii) 21 cm (iii) 3.5 m
(i) Radius (r) of sphere =
Surface area of sphere = 4πr2
Therefore, the surface area of a sphere having diameter 14 cm is 616 cm2.
(ii) Radius (r) of sphere =
Surface area of sphere = 4πr2
Therefore, the surface area of a sphere having diameter 21 cm is 1386 cm2.
(iii) Radius (r) of sphere = m
Surface area of sphere = 4πr2
Therefore, the surface area of the sphere having diameter 3.5 m is 38.5 m2.
##### Video Solution for surface areas and volumes (Page: 225 , Q.No.: 2)
NCERT Solution for Class 9 maths - surface areas and volumes 225 , Question 2
#### Question 3:
Find the total surface area of a hemisphere of radius 10 cm. [Use π = 3.14]
Radius (r) of hemisphere = 10 cm
Total surface area of hemisphere = CSA of hemisphere + Area of circular end of hemisphere
Therefore, the total surface area of such a hemisphere is 942 cm2.
##### Video Solution for surface areas and volumes (Page: 225 , Q.No.: 3)
NCERT Solution for Class 9 maths - surface areas and volumes 225 , Question 3
#### Question 4:
The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Radius (r1) of spherical balloon = 7 cm
Radius (r2) of spherical balloon, when air is pumped into it = 14 cm
Therefore, the ratio between the surface areas in these two cases is 1:4.
##### Video Solution for surface areas and volumes (Page: 225 , Q.No.: 4)
NCERT Solution for Class 9 maths - surface areas and volumes 225 , Question 4
#### Question 5:
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2.
Inner radius (r) of hemispherical bowl
Surface area of hemispherical bowl = 2πr2
Cost of tin-plating 100 cm2 area = Rs 16
Cost of tin-plating 173.25 cm2 area = Rs 27.72
Therefore, the cost of tin-plating the inner side of the hemispherical bowl is Rs 27.72.
##### Video Solution for surface areas and volumes (Page: 225 , Q.No.: 5)
NCERT Solution for Class 9 maths - surface areas and volumes 225 , Question 5
#### Question 6:
Find the radius of a sphere whose surface area is 154 cm2.
Let the radius of the sphere be r.
Surface area of sphere = 154
∴ 4πr2 = 154 cm2
Therefore, the radius of the sphere whose surface area is 154 cm2 is 3.5 cm.
##### Video Solution for surface areas and volumes (Page: 225 , Q.No.: 6)
NCERT Solution for Class 9 maths - surface areas and volumes 225 , Question 6
#### Question 7:
The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface area.
Let the diameter of earth be d. Therefore, the diameter of moon will be.
Radius of earth =
Radius of moon =
Surface area of moon =
Surface area of earth =
Required ratio
Therefore, the ratio between their surface areas will be 1:16.
##### Video Solution for surface areas and volumes (Page: 225 , Q.No.: 7)
NCERT Solution for Class 9 maths - surface areas and volumes 225 , Question 7
#### Question 8:
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Inner radius of hemispherical bowl = 5 cm
Thickness of the bowl = 0.25 cm
∴ Outer radius (r) of hemispherical bowl = (5 + 0.25) cm
= 5.25 cm
Outer CSA of hemispherical bowl = 2πr2
Therefore, the outer curved surface area of the bowl is 173.25 cm2.
##### Video Solution for surface areas and volumes (Page: 225 , Q.No.: 8)
NCERT Solution for Class 9 maths - surface areas and volumes 225 , Question 8
#### Question 9:
A right circular cylinder just encloses a sphere of radius r (see figure). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).
(i) Surface area of sphere = 4πr2
(ii) Height of cylinder = r + r = 2r
Radius of cylinder = r
CSA of cylinder = 2πrh
= 2πr (2r)
= 4πr2
(iii)
Therefore, the ratio between these two surface areas is 1:1.
##### Video Solution for surface areas and volumes (Page: 225 , Q.No.: 9)
NCERT Solution for Class 9 maths - surface areas and volumes 225 , Question 9
#### Question 1:
A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes?
Matchbox is a cuboid having its length (l), breadth (b), height (h) as 4 cm, 2.5 cm, and 1.5 cm.
Volume of 1 match box = l × b × h
= (4 × 2.5 × 1.5) cm3 = 15 cm3
Volume of 12 such matchboxes = (15 × 12) cm3
= 180 cm3
Therefore, the volume of 12 match boxes is 180 cm3.
##### Video Solution for surface areas and volumes (Page: 228 , Q.No.: 1)
NCERT Solution for Class 9 maths - surface areas and volumes 228 , Question 1
#### Question 2:
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1 m3 = 1000l)
The given cuboidal water tank has its length (l) as 6 m, breadth (b) as 5 m, and height (h) as 4.5 m.
Volume of tank = l × b × h
= (6 × 5 × 4.5) m3 = 135 m3
Amount of water that 1 m3 volume can hold = 1000 litres
Amount of water that 135 m3 volume can hold = (135 × 1000) litres
= 135000 litres
Therefore, such tank can hold up to 135000 litres of water.
##### Video Solution for surface areas and volumes (Page: 228 , Q.No.: 2)
NCERT Solution for Class 9 maths - surface areas and volumes 228 , Question 2
#### Question 3:
A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?
Let the height of the cuboidal vessel be h.
Length (l) of vessel = 10 m
Width (b) of vessel = 8 m
Volume of vessel = 380 m3
l × b × h = 380
[(10) (8) h] m2= 380 m3
m
Therefore, the height of the vessel should be 4.75 m.
##### Video Solution for surface areas and volumes (Page: 228 , Q.No.: 3)
NCERT Solution for Class 9 maths - surface areas and volumes 228 , Question 3
#### Question 4:
Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs 30 per m3.
The given cuboidal pit has its length (l) as 8 m, width (b) as 6 m, and depth (h)as 3 m.
Volume of pit = l × b × h
= (8 × 6 × 3) m3 = 144 m3
Cost of digging per m3 volume = Rs 30
Cost of digging 144 m3 volume = Rs (144 × 30) = Rs 4320
##### Video Solution for surface areas and volumes (Page: 228 , Q.No.: 4)
NCERT Solution for Class 9 maths - surface areas and volumes 228 , Question 4
#### Question 5:
The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Let the breadth of the tank be b m.
Length (l) and depth (h) of tank is 2.5 m and 10 m respectively.
Volume of tank = l × b × h
= (2.5 × b × 10) m3
= 25b m3
Capacity of tank = 25b m3 = 25000 b litres
∴ 25000 b = 50000
b = 2
Therefore, the breadth of the tank is 2 m.
##### Video Solution for surface areas and volumes (Page: 228 , Q.No.: 5)
NCERT Solution for Class 9 maths - surface areas and volumes 228 , Question 5
#### Question 6:
A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last?
The given tank is cuboidal in shape having its length (l) as 20 m, breadth (b) as 15 m, and height (h) as 6 m.
Capacity of tank = l × b× h
= (20 × 15 × 6) m3 = 1800 m3 = 1800000 litres
Water consumed by the people of the village in 1 day = (4000 × 150) litres
= 600000 litres
Let water in this tank last for n days.
Water consumed by all people of village in n days = Capacity of tank
n × 600000 = 1800000
n = 3
Therefore, the water of this tank will last for 3 days.
##### Video Solution for surface areas and volumes (Page: 228 , Q.No.: 6)
NCERT Solution for Class 9 maths - surface areas and volumes 228 , Question 6
#### Question 7:
A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.
The godown has its length (l1) as 40 m, breadth (b1) as 25 m, height (h1) as 15 m, while the wooden crate has its length (l2) as 1.5 m, breadth (b2) as 1.25 m, and height (h2) as 0.5 m.
Therefore, volume of godown = l1 × b1 × h1
= (40 × 25 × 15) m3
= 15000 m3
Volume of 1 wooden crate = l2 × b2 × h2
= (1.5 × 1.25 × 0.5) m3
= 0.9375 m3
Let n wooden crates can be stored in the godown.
Therefore, volume of n wooden crates = Volume of godown
0.9375 × n = 15000
Therefore, 16,000 wooden crates can be stored in the godown.
##### Video Solution for surface areas and volumes (Page: 228 , Q.No.: 7)
NCERT Solution for Class 9 maths - surface areas and volumes 228 , Question 7
#### Question 8:
A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
Side (a) of cube = 12 cm
Volume of cube = (a)3 = (12 cm)3 = 1728 cm3
Let the side of the smaller cube be a1.
Volume of 1 smaller cube
a1 = 6 cm
Therefore, the side of the smaller cubes will be 6 cm.
Ratio between surface areas of cubes
Therefore, the ratio between the surface areas of these cubes is 4:1.
##### Video Solution for surface areas and volumes (Page: 228 , Q.No.: 8)
NCERT Solution for Class 9 maths - surface areas and volumes 228 , Question 8
#### Question 9:
A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Rate of water flow = 2 km per hour
Depth (h) of river = 3 m
Width (b) of river = 40 m
Volume of water flowed in 1 min = 4000 m3
Therefore, in 1 minute, 4000 m3 water will fall in the sea.
##### Video Solution for surface areas and volumes (Page: 228 , Q.No.: 9)
NCERT Solution for Class 9 maths - surface areas and volumes 228 , Question 9
#### Question 1:
The circumference of the base of cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm3 = 1l)
Let the radius of the cylindrical vessel be r.
Height (h) of vessel = 25 cm
Circumference of vessel = 132 cm
r = 132 cm
Volume of cylindrical vessel = πr2h
= 34650 cm3
= 34.65 litres
Therefore, such vessel can hold 34.65 litres of water.
##### Video Solution for surface areas and volumes (Page: 230 , Q.No.: 1)
NCERT Solution for Class 9 maths - surface areas and volumes 230 , Question 1
#### Question 2:
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g.
Inner radius (r1) of cylindrical pipe =
Outer radius (r2) of cylindrical pipe =
Height (h) of pipe = Length of pipe = 35 cm
Volume of pipe =
Mass of 1 cm3 wood = 0.6 g
Mass of 5720 cm3 wood = (5720 × 0.6) g
= 3432 g
= 3.432 kg
##### Video Solution for surface areas and volumes (Page: 230 , Q.No.: 2)
NCERT Solution for Class 9 maths - surface areas and volumes 230 , Question 2
#### Question 3:
A soft drink is available in two packs − (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?
The tin can will be cuboidal in shape while the plastic cylinder will be cylindrical in shape.
Length (l) of tin can = 5 cm
Breadth (b) of tin can = 4 cm
Height (h) of tin can = 15 cm
Capacity of tin can = l × b × h
= (5 × 4 × 15) cm3
= 300 cm3
Radius (r) of circular end of plastic cylinder =
Height (H) of plastic cylinder = 10 cm
Capacity of plastic cylinder = πr2H
Therefore, plastic cylinder has the greater capacity.
Difference in capacity = (385 − 300) cm3 = 85 cm3
##### Video Solution for surface areas and volumes (Page: 230 , Q.No.: 3)
NCERT Solution for Class 9 maths - surface areas and volumes 230 , Question 3
#### Question 4:
If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then find (i) radius of its base (ii) its volume. [Use π = 3.14]
(i) Height (h) of cylinder = 5 cm
Let radius of cylinder be r.
CSA of cylinder = 94.2 cm2
rh = 94.2 cm2
(2 × 3.14 × r × 5) cm = 94.2 cm2
r = 3 cm
(ii) Volume of cylinder = πr2h
= (3.14 × (3)2 × 5) cm3
= 141.3 cm3
##### Video Solution for surface areas and volumes (Page: 230 , Q.No.: 4)
NCERT Solution for Class 9 maths - surface areas and volumes 230 , Question 4
#### Question 5:
It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per m2, find
(i) Inner curved surface area of the vessel
(ii) Radius of the base
(iii) Capacity of the vessel
(i) Rs 20 is the cost of painting 1 m2 area.
Rs 2200 is the cost of painting =
= 110 m2 area
Therefore, the inner surface area of the vessel is 110 m2.
(ii) Let the radius of the base of the vessel be r.
Height (h) of vessel = 10 m
Surface area = 2πrh = 110 m2
(iii) Volume of vessel = πr2h
= 96.25 m3
Therefore, the capacity of the vessel is 96.25 m3 or 96250 litres.
##### Video Solution for surface areas and volumes (Page: 231 , Q.No.: 5)
NCERT Solution for Class 9 maths - surface areas and volumes 231 , Question 5
#### Question 6:
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?
Let the radius of the circular end be r.
Height (h) of cylindrical vessel = 1 m
Volume of cylindrical vessel = 15.4 litres = 0.0154 m3
r = 0.07 m
Therefore, 0.4708 m2 of the metal sheet would be required to make the cylindrical vessel.
##### Video Solution for surface areas and volumes (Page: 231 , Q.No.: 6)
NCERT Solution for Class 9 maths - surface areas and volumes 231 , Question 6
#### Question 7:
A lead pencil consists of a cylinder of wood with solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
Radius (r1) of pencil == 0.35 cm
Radius (r2) of graphite = = 0.05 cm
Height (h) of pencil = 14 cm
Volume of wood in pencil =
##### Video Solution for surface areas and volumes (Page: 231 , Q.No.: 7)
NCERT Solution for Class 9 maths - surface areas and volumes 231 , Question 7
#### Question 8:
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Radius (r) of cylindrical bowl =
Height (h) of bowl, up to which bowl is filled with soup = 4 cm
Volume of soup in 1 bowl = πr2h
= (11 × 3.5 × 4) cm3
= 154 cm3
Volume of soup given to 250 patients = (250 × 154) cm3
= 38500 cm3
= 38.5 litres.
##### Video Solution for surface areas and volumes (Page: 231 , Q.No.: 8)
NCERT Solution for Class 9 maths - surface areas and volumes 231 , Question 8
#### Question 1:
Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm
(i) Radius (r) of cone = 6 cm
Height (h) of cone = 7 cm
Volume of cone
Therefore, the volume of the cone is 264 cm3.
(ii) Radius (r) of cone = 3.5 cm
Height (h) of cone = 12 cm
Volume of cone
Therefore, the volume of the cone is 154 cm3.
##### Video Solution for surface areas and volumes (Page: 233 , Q.No.: 1)
NCERT Solution for Class 9 maths - surface areas and volumes 233 , Question 1
#### Question 2:
Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm
(i) Radius (r) of cone = 7 cm
Slant height (l) of cone = 25 cm
Height (h) of cone
Volume of cone
Therefore, capacity of the conical vessel
=
= 1.232 litres
(ii) Height (h) of cone = 12 cm
Slant height (l) of cone = 13 cm
Radius (r) of cone
Volume of cone
Therefore, capacity of the conical vessel
=
= litres
##### Video Solution for surface areas and volumes (Page: 233 , Q.No.: 2)
NCERT Solution for Class 9 maths - surface areas and volumes 233 , Question 2
#### Question 3:
The height of a cone is 15 cm. If its volume is 1570 cm3, find the diameter of its base. [Use π = 3.14]
Height (h) of cone = 15 cm
Let the radius of the cone be r.
Volume of cone = 1570 cm3
r = 10 cm
Therefore, the diameter of the base of cone is $10×2=20$ cm.
##### Video Solution for surface areas and volumes (Page: 233 , Q.No.: 3)
NCERT Solution for Class 9 maths - surface areas and volumes 233 , Question 3
#### Question 4:
If the volume of a right circular cone of height 9 cm is 48π cm3, find the diameter of its base.
Height (h) of cone = 9 cm
Let the radius of the cone be r.
Volume of cone = 48π cm3
Diameter of base = 2r = 8 cm
##### Video Solution for surface areas and volumes (Page: 233 , Q.No.: 4)
NCERT Solution for Class 9 maths - surface areas and volumes 233 , Question 4
#### Question 5:
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Radius (r) of pit
Height (h) of pit = Depth of pit = 12 m
Volume of pit
= 38.5 m3
Thus, capacity of the pit = (38.5 × 1) kilolitres = 38.5 kilolitres
##### Video Solution for surface areas and volumes (Page: 233 , Q.No.: 5)
NCERT Solution for Class 9 maths - surface areas and volumes 233 , Question 5
#### Question 6:
The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone
(i) Radius of cone =
Let the height of the cone be h.
Volume of cone = 9856 cm3
h = 48 cm
Therefore, the height of the cone is 48 cm.
(ii) Slant height (l) of cone
Therefore, the slant height of the cone is 50 cm.
(iii) CSA of cone = πrl
= 2200 cm2
Therefore, the curved surface area of the cone is 2200 cm2.
##### Video Solution for surface areas and volumes (Page: 233 , Q.No.: 6)
NCERT Solution for Class 9 maths - surface areas and volumes 233 , Question 6
#### Question 7:
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
When right-angled ΔABC is revolved about its side 12 cm, a cone with height (h) as 12 cm, radius (r) as 5 cm, and slant height (l) 13 cm will be formed.
Volume of cone
= 100π cm3
Therefore, the volume of the cone so formed is 100π cm3.
##### Video Solution for surface areas and volumes (Page: 233 , Q.No.: 7)
NCERT Solution for Class 9 maths - surface areas and volumes 233 , Question 7
#### Question 8:
If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
When right-angled ΔABC is revolved about its side 5 cm, a cone will be formed having radius (r) as 12 cm, height (h) as 5 cm, and slant height (l) as 13 cm.
Volume of cone
Therefore, the volume of the cone so formed is 240π cm3.
Required ratio
##### Video Solution for surface areas and volumes (Page: 233 , Q.No.: 8)
NCERT Solution for Class 9 maths - surface areas and volumes 233 , Question 8
#### Question 9:
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Radius (r) of heap
Height (h) of heap = 3 m
Volume of heap
Therefore, the volume of the heap of wheat is 86.625 m3.
Area of canvas required = CSA of cone
Therefore, 99.825 m2 canvas will be required to protect the heap from rain.
##### Video Solution for surface areas and volumes (Page: 233 , Q.No.: 9)
NCERT Solution for Class 9 maths - surface areas and volumes 233 , Question 9
#### Question 1:
Find the volume of a sphere whose radius is
(i) 7 cm (ii) 0.63 m
(i) Radius of sphere = 7 cm
Volume of sphere =
Therefore, the volume of the sphere is 1437 cm3.
(ii) Radius of sphere = 0.63 m
Volume of sphere =
Therefore, the volume of the sphere is 1.05 m3 (approximately).
##### Video Solution for surface areas and volumes (Page: 236 , Q.No.: 1)
NCERT Solution for Class 9 maths - surface areas and volumes 236 , Question 1
#### Question 2:
Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm (ii) 0.21 m
(i) Radius (r) of ball =
Volume of ball =
Therefore, the volume of the sphere is cm3.
(ii)Radius (r) of ball = = 0.105 m
Volume of ball =
Therefore, the volume of the sphere is 0.004851 m3.
##### Video Solution for surface areas and volumes (Page: 236 , Q.No.: 2)
NCERT Solution for Class 9 maths - surface areas and volumes 236 , Question 2
#### Question 3:
The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3?
Radius (r) of metallic ball =
Volume of metallic ball =
Mass = Density × Volume
= (8.9 × 38.808) g
= 345.3912 g
Hence, the mass of the ball is 345.39 g (approximately).
##### Video Solution for surface areas and volumes (Page: 236 , Q.No.: 3)
NCERT Solution for Class 9 maths - surface areas and volumes 236 , Question 3
#### Question 4:
The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Let the diameter of earth be d. Therefore, the radius of earth will be .
Diameter of moon will be and the radius of moon will be .
Volume of moon =
Volume of earth =
Therefore, the volume of moon is of the volume of earth.
##### Video Solution for surface areas and volumes (Page: 236 , Q.No.: 4)
NCERT Solution for Class 9 maths - surface areas and volumes 236 , Question 4
#### Question 5:
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Radius (r) of hemispherical bowl = = 5.25 cm
Volume of hemispherical bowl =
= 303.1875 cm3
Capacity of the bowl =
Therefore, the volume of the hemispherical bowl is 0.303 litre.
##### Video Solution for surface areas and volumes (Page: 236 , Q.No.: 5)
NCERT Solution for Class 9 maths - surface areas and volumes 236 , Question 5
#### Question 6:
A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
Inner radius (r1) of hemispherical tank = 1 m
Thickness of hemispherical tank = 1 cm = 0.01 m
Outer radius (r2) of hemispherical tank = (1 + 0.01) m = 1.01 m
##### Video Solution for surface areas and volumes (Page: 236 , Q.No.: 6)
NCERT Solution for Class 9 maths - surface areas and volumes 236 , Question 6
#### Question 7:
Find the volume of a sphere whose surface area is 154 cm2.
Let radius of sphere be r.
Surface area of sphere = 154 cm2
⇒ 4πr2 = 154 cm2
Volume of sphere =
Therefore, the volume of the sphere is cm3.
##### Video Solution for surface areas and volumes (Page: 236 , Q.No.: 7)
NCERT Solution for Class 9 maths - surface areas and volumes 236 , Question 7
#### Question 8:
A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs 498.96. If the cost of white-washing is Rs 2.00 per square meter, find the
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.
(i) Cost of white-washing the dome from inside = Rs 498.96
Cost of white-washing 1 m2 area = Rs 2
Therefore, CSA of the inner side of dome =
= 249.48 m2
(ii) Let the inner radius of the hemispherical dome be r.
CSA of inner side of dome = 249.48 m2
r2 = 249.48 m2
r = 6.3 m
Volume of air inside the dome = Volume of hemispherical dome
= 523.908 m3
= 523.9 m3 (approximately)
Therefore, the volume of air inside the dome is 523.9 m3.
##### Video Solution for surface areas and volumes (Page: 236 , Q.No.: 8)
NCERT Solution for Class 9 maths - surface areas and volumes 236 , Question 8
#### Question 9:
Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the
(i) radius r' of the new sphere, (ii) ratio of S and S'.
(i)Radius of 1 solid iron sphere = r
Volume of 1 solid iron sphere
Volume of 27 solid iron spheres
27 solid iron spheres are melted to form 1 iron sphere. Therefore, the volume of this iron sphere will be equal to the volume of 27 solid iron spheres. Let the radius of this new sphere be r'.
Volume of new solid iron sphere
(ii) Surface area of 1 solid iron sphere of radius r = 4πr2
Surface area of iron sphere of radius r' = 4π (r')2
= 4 π (3r)2 = 36 πr2
##### Video Solution for surface areas and volumes (Page: 236 , Q.No.: 9)
NCERT Solution for Class 9 maths - surface areas and volumes 236 , Question 9
#### Question 10:
A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule?
Radius (r) of capsule
Volume of spherical capsule
=
= 22.458 mm3
= 22.46 mm3 (approximately)
Therefore, the volume of the spherical capsule is 22.46 mm3.
##### Video Solution for surface areas and volumes (Page: 236 , Q.No.: 10)
NCERT Solution for Class 9 maths - surface areas and volumes 236 , Question 10
#### Question 1:
A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm (see the given figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm2 and the rate of painting is 10 paise per cm2, find the total expenses required for polishing and painting the surface of the bookshelf.
External height (l) of book self = 85 cm
External breadth (b) of book self = 25 cm
External height (h) of book self = 110 cm
External surface area of shelf while leaving out the front face of the shelf
= lh + 2 (lb + bh)
= [85 × 110 + 2 (85 × 25 + 25 × 110)] cm2
= (9350 + 9750) cm2
= 19100 cm2
Area of front face = [85 × 110 − 75 × 100 + 2 (75 × 5)] cm2
= 1850 + 750 cm2
= 2600 cm2
Area to be polished = (19100 + 2600) cm2 = 21700 cm2
Cost of polishing 1 cm2 area = Rs 0.20
Cost of polishing 21700 cm2 area Rs (21700 × 0.20) = Rs 4340
It can be observed that length (l), breadth (b), and height (h) of each row of the book shelf is 75 cm, 20 cm, and 30 cm respectively.
Area to be painted in 1 row = 2 (l + h) b + lh
= [2 (75 + 30) × 20 + 75 × 30] cm2
= (4200 + 2250) cm2
= 6450 cm2
Area to be painted in 3 rows = (3 × 6450) cm2 = 19350 cm2
Cost of painting 1 cm2 area = Rs 0.10
Cost of painting 19350 cm2 area = Rs (19350 × 0.1)
= Rs 1935
Total expense required for polishing and painting = Rs (4340 + 1935)
= Rs 6275
Therefore, it will cost Rs 6275 for polishing and painting the surface of the bookshelf.
##### Video Solution for surface areas and volumes (Page: 236 , Q.No.: 1)
NCERT Solution for Class 9 maths - surface areas and volumes 236 , Question 1
#### Question 2:
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in the given figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2.
Radius (r) of wooden sphere =
Surface area of wooden sphere = 4πr2
Radius (r1) of the circular end of cylindrical support = 1.5 cm
Height (h) of cylindrical support = 7 cm
CSA of cylindrical support = 2πrh
Area of the circular end of cylindrical support = πr2
= 7.07 cm2
Area to be painted silver = [8 × (1386 − 7.07)] cm2
= (8 × 1378.93) cm2 = 11031.44 cm2
Cost for painting with silver colour = Rs (11031.44 × 0.25) = Rs 2757.86
Area to be painted black = (8 × 66) cm2 = 528 cm2
Cost for painting with black colour = Rs (528 × 0.05) = Rs 26.40
Total cost in painting = Rs (2757.86 + 26.40)
= Rs 2784.26
Therefore, it will cost Rs 2784.26 in painting in such a way.
##### Video Solution for surface areas and volumes (Page: 237 , Q.No.: 2)
NCERT Solution for Class 9 maths - surface areas and volumes 237 , Question 2
#### Question 3:
The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?
Let the diameter of the sphere be d.
Radius (r1) of sphere
CSA (S1) of sphere =
CSA (S2) of sphere when radius is decreased =
Decrease in surface area of sphere = S1S2
##### Video Solution for surface areas and volumes (Page: 237 , Q.No.: 3)
NCERT Solution for Class 9 maths - surface areas and volumes 237 , Question 3
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\end{align*}. We have learnt how to determine the average gradient of a curve and how to determine the gradient of a curve at a given point. Determine the velocity of the ball after $$\text{1,5}$$ $$\text{s}$$. When average rate of change is required, it will be specifically referred to as average rate of change. For example we can use algebraic formulae or graphs. TABLE OF CONTENTS TEACHER NOTES . \end{align*}, We also know that acceleration is the rate of change of velocity. \text{Instantaneous velocity } &= \text{Instantaneous rate of change } \\ A wooden block is made as shown in the diagram. Navigation. &= 4xh + x^2 + 2x^2 \\ a &= 3t Principles of Mathematics, Grades 11–12. Chapter 9 Differential calculus. Chapter 5. Burnett Website; BC's Curriculum; Contact Me. Questions and Answers on Functions. The vertical velocity with which the ball hits the ground. \begin{align*} \begin{align*} 5. The velocity after $$\text{4}$$ $$\text{s}$$ will be: The ball hits the ground at a speed of $$\text{20}\text{ m.s^{-1}}$$. Aims and outcomes of tutorial: Improve marks and help you achieve 70% or more! Determine the velocity of the ball after $$\text{3}$$ seconds and interpret the answer. \text{where } D &= \text{distance above the ground (in metres)} \\ %�쏢 ADVANCED PLACEMENT (AP) CALCULUS BC Grades 11, 12 Unit of Credit: 1 Year Pre-requisite: Pre-Calculus Course Overview: The topic outline for Calculus BC includes all Calculus AB topics. A soccer ball is kicked vertically into the air and its motion is represented by the equation: \text{Reservoir empty: } V(d)&=0 \\ Relations and Functions Part -1 . %PDF-1.4 Just because gravity is constant does not mean we should necessarily think of acceleration as a constant. &= \frac{3000}{x}+ 3x^2 \text{Hits ground: } D(t)&=0 \\ 10. Calculate the average velocity of the ball during the third second. The speed at the minimum would then give the most economical speed. Substituting $$t=2$$ gives $$a=\text{6}\text{ m.s^{-2}}$$. Ontario. D'(\text{6,05})=18-5(\text{6,05})&= -\text{18,3}\text{ m.s$^{-1}$} Let the two numbers be $$a$$ and $$b$$ and the product be $$P$$. Mathemaics Download all Formulas and Notes For Vlass 12 in pdf CBSE Board . The ball has stopped going up and is about to begin its descent. A(x) &= \frac{3000}{x}+ 3x^2 \\ Calculate the width and length of the garden that corresponds to the largest possible area that Michael can fence off. v &=\frac{3}{2}t^{2} - 2 \\ Germany. Handouts. &=\frac{8}{x} - (-x^{2}+2x+3) \\ Sitemap. Interpretation: this is the stationary point, where the derivative is zero. The use of different . t&= \text{ time elapsed (in seconds)} &= \text{0}\text{ m.s$^{-1}$} Grade 12 Biology provides students with the opportunity for in-depth study of the concepts and processes associated with biological systems. A rectangular juice container, made from cardboard, has a square base and holds $$\text{750}\text{ cm}^{3}$$ of juice. Differential Calculus - Grade 12 Rory Adams reeF High School Science Texts Project Sarah Blyth This work is produced by The Connexions Project and licensed under the Creative Commons Attribution License y Chapter: Di erential Calculus - Grade 12 1 Why do I have to learn this stu ? \text{Substitute } h &= \frac{750}{x^2}: \\ Matrix . Calculus Concepts Questions. We know that the area of the garden is given by the formula: The fencing is only required for $$\text{3}$$ sides and the three sides must add up to $$\text{160}\text{ m}$$. Start by finding an expression for volume in terms of $$x$$: Now take the derivative and set it equal to $$\text{0}$$: Since the length can only be positive, $$x=10$$, Determine the shortest vertical distance between the curves of $$f$$ and $$g$$ if it is given that: Foundations of Mathematics, Grades 11–12. Is this correct? Therefore, $$x=\frac{20}{3}$$ and $$y=20-\frac{20}{3} = \frac{40}{3}$$. Grade 12 introduction to calculus (45S) [electronic resource] : a course for independent study—Field validation version ISBN: 978-0-7711-5972-5 1. Distance education—Manitoba. \end{align*}. We will therefore be focusing on applications that can be pdf download done only with knowledge taught in this course. \end{align*}. View Pre-Calculus_Grade_11-12_CCSS.pdf from MATH 122 at University of Vermont. 3978 | 12 | 1. \therefore d = 16 \text{ or } & d = -\frac{4}{3} \\ \text{Average velocity } &= \text{Average rate of change } \\ 2 + 3 (10 marks) a) Determine the slope of the secant lines PR, PS, and PT to the curve, given the coordinates P(1, 1), R(4, -29), S(3, -15), T(1.1, 0.58). Calculus—Study and teaching (Secondary)—Manitoba. In other words, determine the speed of the car which uses the least amount of fuel. Calculus—Study and teaching (Secondary). The diagram shows the plan for a verandah which is to be built on the corner of a cottage. Calculus Concepts Questions application of calculus grade 12 pdf application of calculus grade 12 pdf Questions application of calculus grade 12 pdf and Answers on Functions. The length of the block is $$y$$. Click below to download the ebook free of any cost and enjoy. Mathematically we can represent change in different ways. Therefore, acceleration is the derivative of velocity. \therefore 64 + 44d -3d^{2}&=0 \\ Exploring the similarity of parabolas and their use in real world applications. During an experiment the temperature $$T$$ (in degrees Celsius) varies with time $$t$$ (in hours) according to the formula: $$T\left(t\right)=30+4t-\frac{1}{2}{t}^{2}, \enspace t \in \left[1;10\right]$$. 12. \end{align*}. 14. Test yourself and learn more on Siyavula Practice. Lessons. To draw a rough sketch of the graph we need to calculate where the graph intersects with the axes and the maximum and minimum function values of the turning points: Note: the above diagram is not drawn to scale. We are interested in maximising the area of the garden, so we differentiate to get the following: To find the stationary point, we set $${A}'\left(l\right)=0$$ and solve for the value(s) of $$l$$ that maximises the area: Therefore, the length of the garden is $$\text{40}\text{ m}$$. Related. This means that $$\frac{dv}{dt} = a$$: Is the volume of the water increasing or decreasing at the end of $$\text{8}$$ days. 36786 | 185 | 8. Password * The important areas which are necessary for advanced calculus are vector spaces, matrices, linear transformation. \end{align*}, \begin{align*} Integrals . 13. Handouts. v &=\frac{3}{2}t^{2} - 2 We set the derivative equal to $$\text{0}$$: Unit 6 - Applications of Derivatives. A survey involves many different questions with a range of possible answers, calculus allows a more accurate prediction. V & = x^2h \\ Apart from whole-class teaching, teachers can utilise pair and group work to encourage peer interaction and to facilitate discussion. 0 &= 4 - t \\ We use this information to present the correct curriculum and Determine an expression for the rate of change of temperature with time. &=\text{9}\text{ m.s$^{-1}$} Calculate the maximum height of the ball. Grade 12 Page 1 DIFFERENTIAL CALCULUS 30 JUNE 2014 Checklist Make sure you know how to: Calculate the average gradient of a curve using the formula Find the derivative by first principles using the formula Use the rules of differentiation to differentiate functions without going through the process of first principles. Calculate the dimensions of a rectangle with a perimeter of 312 m for which the area, V, is at a maximum. Nelson Mathematics, Grades 7–8. If $$f''(a) > 0$$, then the point is a local minimum. \end{align*}. Statisticianswill use calculus to evaluate survey data to help develop business plans. Xtra Gr 12 Maths: In this lesson on Calculus Applications we focus on tangents to a curve, remainder and factor theorem, sketching a cubic function as well as graph interpretation. \begin{align*} Make $$b$$ the subject of equation ($$\text{1}$$) and substitute into equation ($$\text{2}$$): We find the value of $$a$$ which makes $$P$$ a maximum: Substitute into the equation ($$\text{1}$$) to solve for $$b$$: We check that the point $$\left(\frac{10}{3};\frac{20}{3}\right)$$ is a local maximum by showing that $${P}''\left(\frac{10}{3}\right) < 0$$: The product is maximised when the two numbers are $$\frac{10}{3}$$ and $$\frac{20}{3}$$. V'(8)&=44-6(8)\\ What is the most economical speed of the car? If $$AB=DE=x$$ and $$BC=CD=y$$, and the length of the railing must be $$\text{30}\text{ m}$$, find the values of $$x$$ and $$y$$ for which the verandah will have a maximum area. $A (\text{in square centimetres}) = \frac{\text{3 000}}{x} + 3x^{2}$. Sign in with your email address. \end{align*}. \text{where } V&= \text{ volume in kilolitres}\\ Calculus Questions, Answers and Solutions Calculus questions with detailed solutions are presented. Home; Novels. 339 12.1 Introduction 339 12.2 Concept of Logarithmic 339 12.3 The Laws of Exponent 340 12… Applications of Derivatives ... Calculus I or needing a refresher in some of the early topics in calculus. 5 0 obj Velocity is one of the most common forms of rate of change: Velocity refers to the change in distance ($$s$$) for a corresponding change in time ($$t$$). Determine the acceleration of the ball after $$\text{1}$$ second and explain the meaning of the answer. \end{align*}. The time at which the vertical velocity is zero. This book is a revised and expanded version of the lecture notes for Basic Calculus and other similar courses o ered by the Department of Mathematics, University of Hong Kong, from the first semester of the academic year 1998-1999 through the second semester of 2006-2007. Siyavula's open Mathematics Grade 12 textbook, chapter 6 on Differential calculus covering Applications of differential calculus 14. When we mention rate of change, the instantaneous rate of change (the derivative) is implied. Lessons. To find the optimised solution we need to determine the derivative and we only know how to differentiate with respect to one variable (more complex rules for differentiation are studied at university level). \text{After 8 days, rate of change will be:}\\ \begin{align*} How long will it take for the ball to hit the ground? Inverse Trigonometry Functions . A & = \text{ area of sides } + \text{ area of base } + \text{ area of top } \\ \therefore t&=-\text{0,05} \text{ or } t=\text{6,05} The interval in which the temperature is increasing is $$[1;4)$$. SESSION TOPIC PAGE . The interval in which the temperature is dropping is $$(4;10]$$. Unit 7 - Derivatives of Trigonometric Functions. Lessons. \end{align*}. \begin{align*} We have seen that differential calculus can be used to determine the stationary points of functions, in order to sketch their graphs. We have seen that the coordinates of the turning point can be calculated by differentiating the function and finding the $$x$$-coordinate (speed in the case of the example) for which the derivative is $$\text{0}$$. Calculus—Programmed instruction. The total surface area of the block is $$\text{3 600}\text{ cm^{2}}$$. Math Focus, Grades 7–9. 9. \end{align*}. 12 Exponential Form(s) of a Positive Real Number and its Logarithm(s): Pre-Requisite for Understanding Exponential and Logarithmic Functions (What must you know to learn Calculus?) The coefficient is negative and therefore the function must have a maximum value. The cardboard needed to fold the top of the container is twice the cardboard needed for the base, which only needs a single layer of cardboard. Between 09:01 and 09:02 it … One of the numbers is multiplied by the square of the other. Module 2: Derivatives (26 marks) 1. High marks in maths are the key to your success and future plans. Unit 1 - Introduction to Vectors > Homework Solutions. We start by finding the surface area of the prism: Find the value of $$x$$ for which the block will have a maximum volume. A set of questions on the concepts of a function, in calculus, are presented along with their answers and solutions. \end{align*}, \begin{align*} We can check this by drawing the graph or by substituting in the values for $$t$$ into the original equation. Let $$f'(x) = 0$$ and solve for $$x$$ to find the optimum point. University Level Books 12th edition, math books, University books Post navigation. We get the following two equations: Rearranging the first equation and substituting into the second gives: Differentiating and setting to $$\text{0}$$ gives: Therefore, $$x=20$$ or $$x=\frac{20}{3}$$. Those in shaded rectangles, e. TEACHER NOTES . Mathematics for Knowledge and Employability, Grades 8–11. Title: Grade 12_Practical application of calculus Author: teacher Created Date: 9/3/2013 8:52:12 AM Keywords () If we draw the graph of this function we find that the graph has a minimum. Notice that the sign of the velocity is negative which means that the ball is moving downward (a positive velocity is used for upwards motion). PreCalculus 12 > PreCalc 12 Notes. We think you are located in Creative Commons Attribution License. Calculating stationary points also lends itself to the solving of problems that require some variable to be maximised or minimised. Chapter 1. We use the expression for perimeter to eliminate the $$y$$ variable so that we have an expression for area in terms of $$x$$ only: To find the maximum, we need to take the derivative and set it equal to $$\text{0}$$: Therefore, $$x=\text{5}\text{ m}$$ and substituting this value back into the formula for perimeter gives $$y=\text{10}\text{ m}$$. \end{align*}, \begin{align*} If the length of the sides of the base is $$x$$ cm, show that the total area of the cardboard needed for one container is given by: \text{and } g(x)&= \frac{8}{x}, \quad x > 0 \begin{align*} The additional topics can be taught anywhere in the course that the instructor wishes. 750 & = x^2h \\ If we set $${f}'\left(v\right)=0$$ we can calculate the speed that corresponds to the turning point: This means that the most economical speed is $$\text{80}\text{ km/h}$$. Unit 8 - Derivatives of Exponential Functions. The volume of the water is controlled by the pump and is given by the formula: Primary Menu. Advanced Calculus includes some topics such as infinite series, power series, and so on which are all just the application of the principles of some basic calculus topics such as differentiation, derivatives, rate of change and o on. &\approx \text{12,0}\text{ cm} some of the more challenging questions for example question number 12 in Section A: Student Activity 1. I’ve tried to make these notes as self contained as possible and so all the information needed to read through them is either from an Algebra or Trig class or contained in other sections of the notes. The container has a specially designed top that folds to close the container. Chapter 3. 2. &= 18-6(3) \\ 11. Data Handling Transformations 22–32 16 Functions 33–44 17 Calculus 45 – 53 18 54 - 67 19 Linear Programming Trigonometry 3 - 21 2D Trigonometry 3D Trigonometry 68 - 74 75 - 86. Acceleration is the change in velocity for a corresponding change in time. \end{align*}, \begin{align*} f(x)&= -x^{2}+2x+3 \\ Notice that this formula now contains only one unknown variable. Effective speeds over small intervals 1. &=\frac{8}{x} +x^{2} - 2x - 3 [email protected] 604-668-6478 . A'(x) &= - \frac{3000}{x^2}+ 6x \\ This implies that acceleration is the second derivative of the distance. Therefore the two numbers are $$\frac{20}{3}$$ and $$\frac{40}{3}$$ (approximating to the nearest integer gives $$\text{7}$$ and $$\text{13}$$). The questions are about important concepts in calculus. \text{Initial velocity } &= D'(0) \\ \begin{align*} Mathematics / Grade 12 / Differential Calculus. Interpretation: the velocity is decreasing by $$\text{6}$$ metres per second per second. \end{align*}. \begin{align*} Michael wants to start a vegetable garden, which he decides to fence off in the shape of a rectangle from the rest of the garden. We look at the coefficient of the $$t^{2}$$ term to decide whether this is a minimum or maximum point. to personalise content to better meet the needs of our users. It contains NSC exam past papers from November 2013 - November 2016. MALATI materials: Introductory Calculus, Grade 12 5 3. One of the numbers is multiplied by the square of the other. Determine the rate of change of the volume of the reservoir with respect to time after $$\text{8}$$ days. Revision Video . Homework. �np�b!#Hw�4 +�Bp��3�~~xNX\�7�#R|פ�U|o�N��6� H��w���1� _*�B #����d���2I��^A�T6�n�l2�hu��Q 6(��"?�7�0�՝�L���U�l��7��!��@�m��Bph����� If the displacement $$s$$ (in metres) of a particle at time $$t$$ (in seconds) is governed by the equation $$s=\frac{1}{2}{t}^{3}-2t$$, find its acceleration after $$\text{2}$$ seconds. T'(t) &= 4 - t Continuity and Differentiability. \therefore 0 &= - \frac{3000}{x^2}+ 6x \\ Grade 12 | Learn Xtra Lessons. & \\ Explain your answer. GRADE 12 . (16-d)(4+3d)&=0\\ Grade 12 Mathematics Mobile Application contains activities, practice practice problems and past NSC exam papers; together with solutions. Grade 12 Introduction to Calculus. Chapter 7. V(d)&=64+44d-3d^{2} \\ Homework. If each number is greater than $$\text{0}$$, find the numbers that make this product a maximum. Given: g (x) = -2. x. Determine the following: The average vertical velocity of the ball during the first two seconds. The ball hits the ground after $$\text{4}$$ $$\text{s}$$. (Volume = area of base $$\times$$ height). from 09:00 till 09:01 it travels a distance of 7675 metres. It can be used as a textbook or a reference book for an introductory course on one variable calculus. Common Core St at e St andards: Mat hemat ics - Grade 11 Mat hemat ics Grade: 11 CCSS.Math.Content.HSA A &= 4x\left( \frac{750}{x^2} \right) + 3x^2 \\ Chapter 4. &\approx \text{7,9}\text{ cm} \\ Mathematics for Apprenticeship and Workplace, Grades 10–12. &= 4xh + 3x^2 \\ Graphs give a visual representation of the rate at which the function values change as the independent (input) variable changes. \therefore x &= \sqrt[3]{500} \\ 4. Thomas Calculus 12th Edition Ebook free download pdf, 12th edition is the most recomended book in the Pakistani universities now days. > Grade 12 – Differential Calculus. Fanny Burney. t&=\frac{-18\pm\sqrt{336}}{-6} \\ CAMI Mathematics: :: : Grade 12 12.5 Calculus12.5 Calculus 12.5 Practical application 12.5 Practical application A. stream \end{align*} The fuel used by a car is defined by $$f\left(v\right)=\frac{3}{80}{v}^{2}-6v+245$$, where $$v$$ is the travelling speed in $$\text{km/h}$$. 1. Velocity after $$\text{1,5}$$ $$\text{s}$$: Therefore, the velocity is zero after $$\text{2}\text{ s}$$, The ball hits the ground when $$H\left(t\right)=0$$. Related Resources. Calculus 12. \text{Let the distance } P(x) &= g(x) - f(x)\\ R�nJ�IJ��\��b�'�?¿]|}��+������.�)&+��.��K�����)��M��E�����g�Ov{�Xe��K�8-Ǧ����0�O�֧�#�T���\�*�?�i����Ϭޱ����~~vg���s�\�o=���ZX3��F�c0�ïv~�I/��bm���^�f��q~��^�����"����l'���娨�h��.�t��[�����t����Ն�i7�G�c_����_��[���_�ɘ腅eH +Rj~e���O)MW�y �������~���p)Q���pi[���D*^����^[�X7��E����v���3�>�pV.����2)�8f�MA���M��.Zt�VlN\9��0�B�P�"�=:g�}�P���0r~���d�)�ǫ�Y����)� ��h���̿L�>:��h+A�_QN:E�F�( �A^$��B��;?�6i�=�p'�w��{�L���q�^���~� �V|���@!��9PB'D@3���^|��Z��pSڍ�nݛoŁ�Tn�G:3�7�s�~��h�'Us����*鐓[��֘��O&�`���������nTE��%D� O��+]�hC 5��� ��b�r�M�r��,R�_@���8^�{J0_�����wa���xk�G�1:�����O(y�|"�פ�^�w�L�4b�$��%��6�qe4��0����O;��on�D�N,z�i)怒������b5��9*�����^ga�#A \begin{align*} If $$x=20$$ then $$y=0$$ and the product is a minimum, not a maximum. Embedded videos, simulations and presentations from external sources are not necessarily covered D'(0) =18 - 6(0) &=\text{18}\text{ m.s$^{-1}$} To check whether the optimum point at $$x = a$$ is a local minimum or a local maximum, we find $$f''(x)$$: If $$f''(a) < 0$$, then the point is a local maximum. \begin{align*} Application on area, volume and perimeter A. 3. Determine the velocity of the ball when it hits the ground. Michael has only $$\text{160}\text{ m}$$ of fencing, so he decides to use a wall as one border of the vegetable garden. Applied Mathematics 9. T(t) &=30+4t-\frac{1}{2}t^{2} \\ &= -\text{4}\text{ kℓ per day} \begin{align*} A rectangle’s width and height, when added, are 114mm. 1. \text{Velocity after } \text{6,05}\text{ s}&= D'(\text{6,05}) \\ We can check that this gives a maximum area by showing that $${A}''\left(l\right) < 0$$: A width of $$\text{80}\text{ m}$$ and a length of $$\text{40}\text{ m}$$ will give the maximum area for the garden. by this license. \therefore \text{ It will be empty after } \text{16}\text{ days} The ends are right-angled triangles having sides $$3x$$, $$4x$$ and $$5x$$. It is very useful to determine how fast (the rate at which) things are changing. \text{Velocity } = D'(t) &= 18 - 6t \\ t&=\frac{-18 \pm\sqrt{(18^{2}-4(1)(-3)}}{2(-3)} \\ Very useful to determine how fast ( the rate of change is described by the square the... Of its journey, i.e determine an expression for the area in terms of only one variable! Used to determine how fast ( the derivative fence off ) to find the numbers \... Math 122 at University of Vermont independent ( input ) variable changes determine how fast ( the rate at the! Use this information to present the correct Curriculum and to personalise content to better meet the needs our... { x } \ ) topics in calculus, Grade 12 12.5 Calculus12.5 calculus 12.5 Practical application a fence.... Useful to determine how fast ( the rate of change and the product be \ ( P\ ) \. Of change used as a textbook or a reference book for an Introductory course on variable... The derivative know that velocity is zero when it hits the ground largest. Topics that are BC topics are found in paragraphs marked with a perimeter of the graph this! Michael can fence off maximised must be expressed in terms of only one variable diagram shows the for! At University of Vermont pdf download done only with knowledge taught in course! 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Is zero the water increasing or decreasing at the minimum would then give the most speed! Local minimum was developed from algebra and geometry verandah which is to be or! Function, in order to sketch their graphs seconds and interpret the answer the... Field of calculus education makes it difficult to produce an exhaustive state-of-the-art summary will the amount fuel... Nsc exam papers ; together with solutions which is to be built on the corner of a,... Website ; BC 's Curriculum ; Contact Me then \ ( a\ ) and \ ( a\ ) \! An expression for the area and modified perimeter of the verandah and conduct investigations in the diagram for. And population dynamics sources are not necessarily covered by this License diversity of block! Metabolic processes, molecular genetics, homeostasis, evolution, and population dynamics be determined by calculating the derivative zero... Companiesuse calculus to set the minimum payments due on Credit card companiesuse calculus set! ^ { -2 } \$ } \ ) average velocity of the at... ) is implied, evolution, and population dynamics download done only with knowledge taught this. Taught in this chapter we will therefore be determined by calculating the derivative is zero visual of! Conduct investigations in the diagram shows the plan for a verandah which to! Will study theory and conduct investigations in the values for \ ( )... { 3 } \ ) \ ), then the point is a minimum not! Improve marks and help you achieve 70 % or more the optimum point 0\ ), the... A function, in calculus, Grade 12 Biology provides students with the opportunity for in-depth study the... = area of the ball to hit the ground is decreasing by \ ( f '' a. Topics that are BC topics are found in paragraphs marked with a perimeter of m! Presented along with their answers and solutions calculus questions, answers and calculus. Shows the plan for a corresponding change in velocity for a corresponding change in velocity for verandah! Minimum payments due on Credit card companiesuse calculus to evaluate survey data to help develop business plans systems..., when added, are presented along with their answers and solutions coefficient is negative, so the function decreasing! 20 } \ ) to evaluate survey data to help develop business plans m which... And group work to encourage peer interaction and to facilitate discussion graph has a specially designed top folds! 3X\ ), then the point is a local minimum ends are triangles. Burnett Website ; BC 's Curriculum ; Contact Me of information given are related to the possible... Change is negative and therefore the function is decreasing with biological systems first two seconds is! Major applications of Derivatives... calculus I or needing a refresher in some of the that! Need to determine the velocity is the rate at which ) things are changing P\ ) we find the... Function must have a maximum, molecular genetics, homeostasis, evolution, and population dynamics I or a! When it hits the ground use this information to present the correct Curriculum and to personalise to! Of fuel evaluate survey data to help develop business plans a distance of 7675 metres to better the! ’ s width and length of the cardboard used is minimised card statements at the it. When will the amount of water be at a maximum average vertical of... For the area in terms of only one variable: the average vertical velocity of the other developed from and... ) is implied application a from 09:00 till 09:01 it travels a distance 7675... Be focusing on applications that can be used to determine an expression for the ball after (. Mathematics and was developed from algebra and geometry, evolution, and population.. ( x\ ) to find the numbers that make this product a maximum and modified perimeter of m! { 1 } \ ) rates of change must have a maximum it will be referred... Of Vermont negative, so the function values change as the independent ( input ) variable changes (! Water be at a maximum value quantity that is to be constructed around the four of... With functions and their use in real world applications aims and application of calculus grade 12 pdf of:! The corner of a Creative Commons Attribution License the diversity of the which. Functions and their use in real world applications independent ( input ) changes... Past papers from November 2013 - November 2016 that \ ( ABCDE\ ) is to be constructed around four...: this is the second derivative of the research in the values for \ ( x\ to... Which is to be constructed around the four edges of the ball hits the ground after \ y\... Tutorial: Improve marks and help you achieve 70 % or more to your and... Not mean we should necessarily think of acceleration as a textbook or a book! Decreasing at the exact time the statement is processed graph of this function we find that the area of \. Is constant does not mean we should necessarily think of acceleration as a textbook or a reference book for Introductory! ’ s width and length of the concepts of a function, in,! Derivative of the more challenging questions for example we can check this by drawing the graph has a minimum pair... { 6 } \ ) for in-depth study of the car which the. The two numbers be \ ( a=\text { 6 } \text application of calculus grade 12 pdf 3 } \ \. Numbers that make this product a maximum: the velocity of the major applications of...... This course an expression for the rate of change is required, it be. That are BC topics are found in paragraphs marked with a perimeter of the car which uses the least of!
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# How do you differentiate g(x) = (5x^6 - 4)(x^2-4) using the product rule?
Feb 25, 2017
$g ' \left(x\right) = 40 {x}^{7} - 120 {x}^{5} - 8 x$
#### Explanation:
$\text{Given "g(x)=f(x).h(x)" then}$
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{g ' \left(x\right) = f \left(x\right) h ' \left(x\right) + h \left(x\right) f ' \left(x\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}} \leftarrow \text{ product rule}$
$\text{here } f \left(x\right) = 5 {x}^{6} - 4 \Rightarrow f ' \left(x\right) = 30 {x}^{5}$
$\text{and } h \left(x\right) = {x}^{2} - 4 \Rightarrow h ' \left(x\right) = 2 x$
$\Rightarrow g ' \left(x\right) = \left(5 {x}^{6} - 4\right) . \left(2 x\right) + \left({x}^{2} - 4\right) . \left(30 {x}^{5}\right)$
$\textcolor{w h i t e}{\Rightarrow g ' \left(x\right)} = 10 {x}^{7} - 8 x + 30 {x}^{7} - 120 {x}^{5}$
$\textcolor{w h i t e}{\Rightarrow g ' \left(x\right)} = 40 {x}^{7} - 120 {x}^{5} - 8 x$
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# The diagonals of a rectangle ABCD intersect at O, if $\angle BOC = 70^\circ$ . Find $\angle ODA$
Last updated date: 24th Feb 2024
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Hint:
Here, we have to find the measure of a given angle. We will use the properties of angles in a rectangle and triangle. We will first find the measure of the opposite angle of the given angle using the property vertically opposite angle. Then we will use the property of isosceles triangle and sum property of a triangle to find the measure of the required angle.
Complete step by step solution:
Let ABCD be a rectangle. The diagonals of a rectangle ABCD intersect at O.
We are given that $\angle BOC = 70^\circ$.
We know that in a rectangle, vertically opposite angles are equal.
By using this property, we get
$\angle BOC = 70^\circ = \angle AOD$ ………………………………………………………………………….$\left( 1 \right)$
We know that the diagonals of a rectangle are equal and bisect each other.
So, in $\Delta AOD$ , we get
$AO = OD$
We know that angles opposite to equal sides of an isosceles triangle are equal.
By using this property, we get
$\angle OAD = \angle ODA$ ……………………………………………………………………………………………$\left( 2 \right)$
We know that the sum of the angles of a triangle is $180^\circ$.
By using this property in $\Delta AOD$, we get
$\angle OAD + \angle AOD + \angle ODA = 180^\circ$
By substituting equation $\left( 1 \right)$ in the above equation, we get
$\Rightarrow \angle ODA + \angle AOD + \angle ODA = 180^\circ$
By adding the equal angles, we get
$\Rightarrow 2\angle ODA + \angle AOD = 180^\circ$
By substituting equation $\left( 1 \right)$ in the above equation, we get
$\Rightarrow 2\angle ODA + 70^\circ = 180^\circ$
Subtracting $70^\circ$ from both side, we get
$\Rightarrow 2\angle ODA = 180^\circ - 70^\circ$
$\Rightarrow 2\angle ODA = 110^\circ$
Dividing by 2 on both the sides, we get
$\Rightarrow \angle ODA = \dfrac{{110^\circ }}{2}$
$\Rightarrow \angle ODA = 55^\circ$
Therefore, $\angle ODA$ is $55^\circ$
Note:
We know that the sum of all interior angles of a rectangle is 360 degrees. The diagonals of a rectangle bisect each other so that the lengths of the diagonals are equal in length. Since the diagonal is a straight angle the diagonals of a rectangle bisect each other at different angles where one angle is an acute angle and the other angle is an obtuse angle. We should also remember that if the diagonals bisect each other at right angles, then it is a square.
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# Investigation: Constructions with a compass
Lesson
Every profession has particular tools that they use all the time, a chef has their knives, the sewer a machine and the mathematician, well they have their construction tools.
No, not hammers and hard hats - Mathematicians use a compass, pencil and a straight edge.
From just these three things we can create (nearly) all of what we require in geometry.
Notice how I used the word straight edge and not ruler. Well that's because all we really need for the constructions is a straight edge, we often don't even need numbers!
A great Greek mathematician named Euclid, who is credited to have written the first mathematics textbook over 2000 years ago, went to great lengths to detail many of the mathematical constructions we will look at today. Geometrical constructions were so important to mathematics at the time because most problems were solved graphically, not arithmetically.
### Copy a segment
To construct congruent line segments with a compass and straightedge, follow the instructions below. Play and pause the video at each step to help you.
#### Steps for construction
1. Start with the line segment $\overline{AB}$AB on the page that you want to copy.
2. Mark a point on the page where you want the start of the copied line segment. Call this point $C$C.
3. Set the compass width to the length of $\overline{AB}$AB.
4. Without changing the compass width, move the compass to $C$C
5. Draw an arc. The endpoint of the new line segment can be anywhere on the arc.
6. Choose a point ($D$D) on the arc and draw $\overline{CD}$CD with the straightedge.
### Copy an angle
As we have seen, congruent means same, so congruent angles are 2 (or more) angles that are exactly the same size. They can be facing in any direction, but if they are the same size then the angles are congruent. You can think of congruent angles as copies of each other.
To construct congruent angles you need a compass and a straight edge.
#### Steps for construction
1. Start with the angle $BAC$BAC ($\angle BAC$BAC) on the page that you want to copy.
2. Draw a ray, $\overrightarrow{PQ}$PQ, on the page.
3. Position the compass on point $A$A, adjust to any width.
4. Draw an arc that crosses both legs of $\angle BAC$BAC, name the points where the arc intersects the rays $F$F and $G$G
5. Position the compass on point on point $P$P, and draw another arc. Where it crosses $\overrightarrow{PQ}$PQ call it $M$M.
6. Measure the distance $FG$FG with the compass.
7. Place the compass at point $M$M, and cross the arc. Call this intersection $N$N.
8. Draw in a ray from $P$P, through $N$N. Call it $\overrightarrow{PR}$PR
9. Now you have created $\angle RPQ$RPQ.
10. $\angle RPQ\equiv\angle BAC$RPQBAC
### Outcomes
#### 7.G.A.2
Draw geometric shapes with given conditions using a variety of methods. Focus on constructing triangles from three measures of angles or sides, noticing when the conditions determine a unique triangle, more than one triangle, or no triangle.
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# Question: What Is Arithmetic Calculation?
## What are the examples of arithmetic sequence?
Mathwords: Arithmetic Sequence.
A sequence such as 1, 5, 9, 13, 17 or 12, 7, 2, –3, –8, –13, –18 which has a constant difference between terms.
The first term is a1, the common difference is d, and the number of terms is n..
## How do you write an arithmetic equation?
An arithmetic sequence is a sequence where the difference between each successive pair of terms is the same. The explicit rule to write the formula for any arithmetic sequence is this: an = a1 + d ( n – 1 )
## What is basic operation in algorithm?
Basic Operations A basic operation is an operation whose execution time is bounded above by a constant. We are only concerned with execution time of an algorithm defined within a multiplicative constant, thus it is the number of basic operations executed that matters in the analysis, and not the exact time required.
## What are the types of arithmetic operators?
These operators are + (addition), – (subtraction), * (multiplication), / (division), and % (modulo).
## How many types of arithmetic are there?
History of Arithmetic Arithmetic is another name given to number theory. The four elementary operations in arithmetic are addition, subtraction, multiplication and division. All these operations are discussed in brief here.
## Why is it called arithmetic sequence?
Sequences with such patterns are called arithmetic sequences. In an arithmetic sequence, the difference between consecutive terms is always the same. For example, the sequence 3, 5, 7, 9 … is arithmetic because the difference between consecutive terms is always two.
## What is arithmetic principle?
The sum of two natural numbers must be greater than either of the numbers. Adding a number and then subtracting that same number leaves the original value unchanged. Regularities such as these are principles that apply in the domain of arithmetic.
## What’s the difference between arithmetic and math?
When you’re referring to addition, subtraction, multiplication and division, the proper word is “arithmetic,” maintains our math fan. “Math,” meanwhile, is reserved for problems involving signs, symbols and proofs — algebra, calculus, geometry and trigonometry. (In which case, we’ve got to side with Barbie.
## What are the 4 types of sequences?
What are Some of the Common Types of Sequences?Arithmetic Sequences.Geometric Sequences.Harmonic Sequences.Fibonacci Numbers.
## Who is the father of mathematics?
mathematician ArchimedesGreek mathematician Archimedes is widely considered by many to be the “father of mathematics.” He is regarded as one of the leading scientists in classical antiquity and is credited with designing numerous innovative machines, including the screw pump and siege engines.
## Who invented minus sign?
Johannes WidmannThis article contains Unicode mathematical symbols.SymbolNameFirst author to use+plus signNicole Oresme−minus signJohannes Widmann√radical symbol (for square root)Christoff Rudolff(…)parentheses (for precedence grouping)Michael Stifel58 more rows
## What are the four arithmetic operations?
1 Numerical Calculations. The basic arithmetic operations (addition, subtraction, multiplication, division, and exponentiation) are performed in the natural way with Mathematica.
## Who is the father of arithmetic?
BrahmaguptaThe 7th Century Indian mathematician and astronomer Brahmagupta is the father of arithmetic.
## Which topics comes under arithmetic?
Topics in Arithmetic include whole numbers, place values, addition, subtraction, multiplication, division, factoring, fractions, decimals, exponents, scientific notations, percents, integers, proportions and word problems.
## What is arithmetic operation in computer?
An arithmetic operator is a mathematical function that takes two operands and performs a calculation on them. They are used in common arithmetic and most computer languages contain a set of such operators that can be used within equations to perform a number of types of sequential calculation.
## What is simple arithmetic?
Arithmetic is a name for working with numbers. It is a part of mathematics. The four basic arithmetic operations are addition, subtraction, multiplication, and division. … Arithmetic is needed in all areas of mathematics, science, and engineering. A calculator can be used to do arithmetic.
## What is the syllabus of arithmetic?
Arithmetic is a branch of mathematics which deals with problems relating to Number system, Decimals, Fractions, Simplification, HCF and LCM, Ratio and Proportion, Percentage, Partnership, Average, Profit and Losses, Simple Interest and Compound Interest, Mensuration, Time and Work, Time and Distance, Data …
## What are the basic rules of arithmetic?
The order of operations is as follows: 1) simplify terms inside parentheses or brackets, 2) simplify exponents and roots, 3) perform multiplication and division, 4) perform addition and subtraction. Multiplication and division are given equal priority, as are addition and subtraction.
## What is arithmetic and example?
The definition of arithmetic refers to working with numbers by doing addition, subtraction, multiplication, and division. An example of arithmetic is adding two and two together to make four.
## What is included in arithmetic?
The basic arithmetic operations are addition, subtraction, multiplication and division, although this subject also includes more advanced operations, such as manipulations of percentages, square roots, exponentiation, logarithmic functions, and even trigonometric functions, in the same vein as logarithms ( …
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Categories
# Probability in Game | AMC-10A, 2005 | Problem 18
Try this beautiful problem based on Probability in game from AMC-10A, 2005. You may use sequential hints to solve the problem.
Try this beautiful problem from AMC 10A, 2005 based on Probability in Game.
## Probability in Game – AMC-10A, 2005- Problem 18
Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?
• $\frac{1}{4}$
• $\frac{1}{6}$
• $\frac{1}{5}$
• $\frac{2}{3}$
• $\frac{1}{3}$
### Key Concepts
Probability
combinatorics
Answer: $\frac{1}{5}$
AMC-10A (2005) Problem 18
Pre College Mathematics
## Try with Hints
Given that The first team to win three games wins the series, team B wins the second game and team A wins the series. So the Total number of games played=$5$. Now we have to find out the possible order of wins…..
Can you now finish the problem ……….
Possible cases :
If team B won the first two games, team A would need to win the next three games. Therefore the possible order of wins is BBAAA.
If team A won the first game, and team B won the second game, the possible order of wins is $A B B A A, A B A B A,$ and $A B A A X,$ where $X$ denotes that the 5th game wasn’t played.
since ABAAX is dependent on the outcome of 4 games instead of 5, it is twice as likely to occur and can be treated as two possibilities.
According to the question, there is One possibility where team $\mathrm{B}$ wins the first game and 5 total possibilities, Therefore the required probability is $\frac{ 1}{5}$
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# CBSE Class 10 Maths Case Study Questions for Chapter 10 - Circles (Published By CBSE)
Check CBSE Class 10 Maths Case Study Questions for Chapter 10 - Circles. These questions have been published by the board for class 10 Mathematics.
Created On: Apr 20, 2021 18:50 IST
CBSE Class 10 Maths Case Study Questions for Chapter 10
CBSE: Case study questions for CBSE Class 10 Maths Chapter 10 - Circles are provided here which students can practice to get familiarised with the new format of questions. These questions have been published by CBSE itself. Answers to all the questions have been provided for the convenience of students. Case study questions are helpful for the preparation of the Class 10 Maths Exam 2021-2022.
Case Study Questions for Class 10 Maths Chapter 10 - Circles
CASE STUDY 1:
A Ferris wheel (or a big wheel in the United Kingdom) is an amusement ride consisting of a rotating upright wheel with multiple passenger-carrying components (commonly referred to as passenger cars, cabins, tubs, capsules, gondolas, or pods) attached to the rim in such a way that as the wheel turns, they are kept upright, usually by gravity.
After taking a ride in Ferris wheel, Aarti came out from the crowd and was observing her friends who were enjoying the ride . She was curious about the different angles and measures that the wheel will form. She forms the figure as given below.
1. In the given figure find ∠ROQ
a) 60
b) 100
c) 150
d) 90
2. Find ∠RQP
a) 75
b) 60
c) 30
d) 90
3. Find ∠RSQ
a) 60
b) 75
c) 100
d) 30
4. Find ∠ORP
a) 90
b) 70
c) 100
d) 60
CASE STUDY 2:
Varun has been selected by his School to design logo for Sports Day T-shirts for students and staff . The logo design is as given in the figure and he is working on the fonts and different colours according to the theme. In given figure, a circle with centre O is inscribed in a ΔABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. The lengths of sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively.
1. Find the length of AD
a) 7
b) 8
c) 5
d) 9
2. Find the Length of BE
a) 8
b) 5
c) 2
d) 9
3. Find the length of CF
a) 9
b) 5
c) 2
d) 3
4. If radius of the circle is 4cm, Find the area of ∆OAB
a) 20
b) 36
c) 24
d) 48
5. Find area of ∆ABC
a) 50
b) 60
c) 100
d) 90
Also Check:
Tips to Solve Case Study Based Questions Accurately
CBSE Class 10 Maths Best Study Material for Preparation of Board Exam 2021
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Session 5, Part C:
The Midline Theorem
In This Part: The Midline Cut | Some Geometry Facts | Proving the Midline Theorem
The midline theorem claims that cutting along the midline of a triangle creates a segment that is parallel to the base and half as long. Does that seem reasonable? To prove the midline cut works, you need to use some geometry facts that you may already have encountered. If not, take some time to consider why these statements are true. Note 5 Fact 1: Vertical angles (the angles opposite each other when two lines intersect) are congruent (they have the same measure). Why: We can show why, for example, m1 = m3: m1 + m2 = 180° and m2 + m3 = 180°, since in both cases the two angles together create a "straight angle." So m1 + m2 = m2 + m3 = 180°. Subtracting m2 from each part of the equation, we see that m1 = m3 = 180° - m2.
Fact 2: If two triangles have two sides that are the same length, and the angle between those two sides has the same measure, then the two triangles are congruent. The two triangles must have the same size and shape, so all three sides have the same length, and all three angles have the same measure. This is known as SAS (side-angle-side) congruence. The single tick indicates the two sides that are the same length. The double tick indicates thetwo sides that are the same length. The angle markings indicate that those two angles have the same measure. Why: The easiest way to be convinced of the fact that the two triangles are congruent is to draw some triangles. Draw a segment 2 inches long and a segment 3 inches long, with a 60° angle between them. Is there more than one way to complete the triangle? Come up with other cases to try.
Fact 3: There are many equivalent definitions of "parallelogram":
• A quadrilateral with both pairs of opposite sides parallel • A quadrilateral with both pairs of opposite angles congruent • A quadrilateral with both pairs of opposite sides congruent • A quadrilateral with one pair of opposite sides both congruent and parallel • The last definition is the one that will come in handy here. Go through the steps of understanding a definition in Session 3 if you're not sure why it works.
You may want to print this page for use in the problems that follow.
Session 5: Index | Notes | Solutions | Video
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# Demystifying the Distributive Property
Okay, raise your hand if you’re an elementary teacher and just the mention of the distributive property makes you break out in a cold sweat. Go ahead…no one can see you. Hmmmm, that’s what I thought. ๐
The Common Core standards are all about the properties, and that’s very new to most elementary teachers. It’s okay to admit that you had to go back and brush up on “big” math.
CCSSM 3.OA.5 reads, in part: Knowing that 8 x 5 = 40 and 8 x 2 = 16, one can find 8 x 7 as 8 x (5 + 2) = (8 x 5) + (8 x 2) = 40 + 16 = 56. Whew! That’s a mouthful! To put it in elementary terms, you split 7 (one of the factors) into 5 + 2 (friendlier numbers), multiply 8 x 5 and 8 x 2, and add the products together. Not so bad, really. Now I don’t know about you, but 8 x 7 was a hard fact for me, and maybe this strategy might have helped!
As with any other math concept, it’s important to take this skill through the concrete (manipulatives) and representational (drawing) stages beforeย the abstract (purely symbolic) stage.
So, first, there is a mat that kiddos can use to build their array and then split it into two parts. This is the concrete learning. They write equations for both the original array and the split array (requires parenthesis). This connects the concrete to the abstract. Please note that there is no right way to split the array. For example, I split the 8 x 3 array into (4 x 3) and (4 x 3). Another student might split it into (6 x 3) and (2 x 3). That’s the cool thing! Students will see that the array can be split more than one way.
Next, I have a little game that students can play to practice representing arrays and connecting them to equations.
## Similar Posts
1. Cheryl says:
Oh my goodness….thank you so much for posting this tonight. You must have been reading my mind because I was struggling with a way to make this kid friendly…your activity fits the bill. Looking forward to using this with my class soon!
1. Donna Boucher says:
Ha ha…I LOVE it when my mind-reading powers kick in! So glad this is timely for you. ๐
2. jen ormsbee says:
Hi Donna!
Thank you so much for these activities ๐
With the common core standards being as rigorous as they are, I find that my students lack the basic number sense to tackle newer, more difficult problems. Your activities really help students dig deeper into the concepts and provide students with a deeper mathematical understanding! I really appreciate all the freebies and so do my students ๐
–jen
teaching, life, and everything in between
1. Donna Boucher says:
You are so right about number sense! So many kiddos are really lacking it. I’m glad my activities help. ๐
3. Terry says:
Thank you so much Donna! It’s exactly what I needed!
1. Donna Boucher says:
So glad to hear it, Terry. ๐
4. Michele Tuvell says:
This is what I have been reading about as a better way to teach multiplication facts. I was also trying to think of a way to make this kid friendly and accessible to my third graders, but you have done a better job than my attempts. Thank you for sharing.
1. Donna Boucher says:
You are so welcome, Michele! Glad you find it useful. ๐
5. Melissa says:
Thanks so much for sharing this! I am teaching this very concept next week so your timing was perfect!
1. Donna Boucher says:
Perfect! I hope it works well for you! ๐
6. Jennifer Polk says:
Thank you, I teach 4th grade and have been teaching this concept for a couple weeks. I still have some that aren’t quite getting this. I am going to put this in my game/manip center for rotations.
1. Donna Boucher says:
Sounds like a great idea, Jennifer!
7. Great post (and yes, I know, I am late to the party)! Just think of all the other things students can do by understanding the distributive property! This is actually one of my favorite things to teach in upper level math. And I love the moment when they are able to use this for doing mental math and discuss it during our number talks! ๐
Janaye
1. Donna Boucher says:
No such thing as late in the blogoshpere! I totally agree with how exciting the “ah ha” moment can be!
8. Teacher and Life Long Learner says:
This is adult and student friendly! Your resources and teaching strategies amaze me! Is it to early to nominate you for Blogger of the Year?
1. Donna Boucher says:
How did I miss this comment? Ha ha. So sweet!
9. Jenn Miron says:
What a great activity! Thanks! I’m a new follower!
Jenn
1. Donna Boucher says:
Welcome, Jenn! Glad to hear you’re finding some information that’s useful to you. ๐
10. Farrah Kilgo says:
Thanks for sharing, Donna! I plan to do this tomorrow to help my kiddos really “get” it.
Farrah
1. Donna Boucher says:
Awesome, Farrah! I hope it went well! ๐
11. Johor Property says:
You got a really useful blog I have been here reading for about an hour.
1. Donna Boucher says:
12. Johor Property says:
Thanks for sharing! I look forward in viewing more of these articles.
13. tricia b says:
Thank you! I will share this with my grade level and your great blog! So many teachers worry about students knowing the name of the property instead of knowing how the properties can help them.
14. Chaya Phillips says:
What a great resource you are! Thank you for everything you have shared with me. My students love the games.
1. Donna Boucher says:
Always nice to hear that kids are loving math, Chaya! ๐
15. Roberta Haren says:
You came to my rescue! I am working after school with a group of our t risk 5th graders who lack a fundamental number sense. I was beginning to sense that the properties were the way to reach them and you have given me a valuable tool that will be used this week. THANKS!
1. Donna Boucher says:
Oh, perfect, Roberta! I hope it helps them out! ๐
16. LizB says:
Perfect timing in more ways than one. First, we’re just now getting to the the larger, more difficult factors where my third graders can benefit from decomposing one of them into friendlier numbers. And secondly, I’m watching my Seahawks on Monday Night Football tonight!
1. Donna Boucher says:
Ha ha, too funny, Liz!
17. Let Freedom Ring says:
Oh my goodness, I teach 6th grade and we still are on regular state standards until next year but the district is full throttle common core. Since we aren’t going to be tested on common core this year we have to fill in the gaps. This is one of them. I have had a hard time presenting this to students in the past but I think your activities will help a lot!
18. Caren Laundree says:
This is such a great activity! Thank you so much for sharing!
19. Anonymous says:
Fabulous! Thank You!
20. Anonymous says:
Thank you for providing an activity that students can do independently that requires mathematical thinking!
21. Anonymous says:
Your game is excellent! It really helped my students get a better feel for the Distributive Property! Thanks for sharing!
22. Roseline Louis says:
Hi I’m not a teacher technically. I pulled my 4th grader out mid year because common core was kicking his bottom and the Teacher just couldn’t get though to him. Anyway I do not remember him being taught this at all. Can you tell me what grade common core introduces the distributive property concept?
23. I teach 6th grade math and I am so using this!! Just what I was looking for…something fun and visual for the kids! Thank you SOOO much for sharing!
1. Donna Boucher says:
You’re welcome, Kendra! ๐
24. Peg Wimmer says:
Thank you for a great resource! I’m using it as extra practice for my 9th graders taking a Math 180 course!
25. Anonymous says:
Glad to see someone (above) using this for higher grades. I don’t know what’s happened before hand to my 8th graders but I am certainly going to use this for some of my strugglers. Thank you!
26. Anonymous says:
I’m new to teaching Math/Science at my school. In fact, I have a degree in History! Looking at our year at a glance and where were supposed to be. I was thinking how in the world do I even begin to explain the distributive property. I think of algebra as soon as see this in action. This will definitely help! thank you!
27. Anonymous says:
This is excellent!! Thank you for sharing
28. Maria says:
This is great! Thanks! Keep posting 3rd grade stuff. I need all the help I can get! My students have a hard time with math!
29. kat says:
I’m just about to move on to area models for multiplication… great timing to find another way to do it! Thanks so much for sharing. You have such awesome ideas and activities that you present here. =)
30. Melissa Costello says:
Thank you for sharing! Great activity.
1. Donna Boucher says:
My pleasure, Melissa! ๐
31. Leila says:
Thanks for sharing. I am doing this lesson for my evaluation and I needed a graphic organizer that would assist my ELL’s grasp and organize the process. I can’t wait to try it!
32. Brenda says:
This is a great activity and so simple! Thank you.
33. Leslie says:
We have been working on this concept in intervention. Can’t wait to try these activities out! Thanks a bunch!
34. Nancy says:
Thanks so much for sharing! Great ideas!
35. Sarah says:
Excited to use this activity tomorrow…I think it will really enhance my distributive property lessons from the previous days! Thank you!
36. Catherine Rollins says:
Thank you for sharing this resource! I really like the simplicity of it!
37. Cycle 1 Week 23 FREE Resources | Homeschooling Women of God says:
[…] Distributive Property Card Game […]
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# Class 12 NCERT Solutions- Mathematics Part I – Chapter 6 Application of Derivatives -Exercise 6.3 | Set 2
• Last Updated : 06 Apr, 2021
### Question 14. Find the equations of the tangent and normal to the given curves at the indicated points:
(i) y = x4– 6x3 + 13x2 – 10x+ 5 at (0, 5)
(ii) y = x4 – 6x3 + 13x2 – 10x + 5 at (1, 3)
(iii) y = x3 at(1, 1)
(iv) y = x2 at(0, 0)
(v) x = cos t, y = sin t at t = π/4
Solution:
(i) Given curve
y = x4 – 6x3 + 13x2 – 10x + 5
Given point, (0, 5)
dy/dx = 4x3 – 18x2 + 26x – 10
dy/dx = 4(0)3 – 18(0)2 + 26(0) – 10
dy/dx = -10,
-dx/dy = 1/10
Now, with the help of points slope form
y – y1 = m(x – x1)
y – 5 = -10(x – 0)
y + 10x = 5 is the required equation of the tangent
For equation of normal,
y – y1 =
y – 5 =
10y – x – 50 is the equation of normal.
(ii) Given curve: y = x4 – 6x3 + 13x2 – 10x + 5
Given point is (1, 3)
dy/dx = 4x3 – 18x2 + 26x – 10
dy/dx = 4(1)3 – 18(1)2 + 26(1) – 10
dy/dx = 4 – 18 + 26 – 10 = 2
dy/dx = 2
-dx/dy = -1/2
Using point slope form, equation of tangent is
y – y1 = dy/dx(x – x1)
y – 3 = 2(x – 1)
y – 2x = 1 is the equation of tangent.
Using point slope form, equation of normal is
y – y1 = -dx/dy(x – 1)
y – 3 = -1/2(x – 1)
2y – 6 = -x + 1
2y + x = 7 is the equation of normal.
(iii) Given curve : y = x3
Given point is (1, 1)
dy/dx = 3x2
dy/dx = 3(1)2 = 3
dy/dx = 3 & -dx/dy = -1/3
Using point slope form, equation of tangent is y – y1 = dy/dx(x – x1)
y – y1 = dy/dx(x – x1)
y – 1 = 3(x – 1)
y – 3x + 2 = 0 is the equation of tangent
Using point slope form, equation of normal is
y – y1 =
y – 1 =
3y – 3 = -x + 1
3y + x = 4 is the equation of normal.
(iv) Given curve: y = x2
Given point (0, 0)
dy/dx = 2x
dy/dx = 0
dy/dx = 0 & -dx/dy = not defined is
y – y1 = dy/dx(x – x1)
y – 0 = 0(x – 0)
y = 0 is the equation of tangent.
Using point slope form, equation of normal is
y – y1 -(1)
-dx\dy is undefined, so we can write eq(1) as
Now putting dy/dx = 0 we get
0(y – 0) = x-0
x = 0 is the equation of normal.
(v) Equation of curve: x = cos t and y = sin t
Point t = π/4
-(1)
On putting these values in eq(1), we get
dy/dx = -1 & -dx/dy = 1
Now for t = π/4,
y1 = sin t = sin(π/4) = 1/√2
x1 = cos t = cos(π/4) = 1/√2
The point is (1/√2, 1/√2)
y – y1
y – (1/√2) = -1(x – 1/√2)
y – 1/√2 = -x + 1/√2
x + y = √2 is the equation of normal is
y – 1/√2 = 1(x – 1/√2)
x = y is the equation of normal.
### Question 15. Find the equation of the tangent line to the curve y = x2 – 2x + 7 which is
(i) Parallel to line 2x – y + 9 = 0
(ii) Perpendicular to the line 5y – 15x = 13
Solution:
Given curve: y = x2 – 2x + 7
On differentiating w.r.t. x, we get
dy/dx = 2x – 2 -(1)
(i) Tangent is parallel to 2x – y + 9 = 0 that means,
Slope of tangent = slope of 2x – y + 9 = 0
y = 2x + 9
Slope = 2 -(Comparing with y = mx + e)
dy/dx = slope = 2
2x – 2 = 2
x1 = 2
Corresponding to x1 = 2,
y1 = x12 – 2x1 + 7
y1 = (2)2 – 2(2) + 7
y1 = 7
The point of contact is (2, 7).
Using point slope form, equation of tangent is
y – y1
y – 7 = 2(x – 2)
y – 2x = 3 is the equation of tangent.
(ii) Tangent is perpendicular to the line 5y – 15x = 13
That means (slope of tangent) x (slope of line) = -1
For, slope of line 5y – 15x = 13
5y = 15x + 13
y = 3x + 13/15
Slope = 3
(Slope of tangent) x 3 = -1
Slope of tangent =-1/3
Now, y = x2 – 2x + 7
dy/dx = 2x – 2
On comparing dy/dx with slope, we get
2x – 2 = -1/3
6x – 6 = -1
6x = 5
x1 = 5/6
For x1 = 5/6,
y1 = x12 – 2x1 + 7
y1 = (5/6)12 – 2(5/6)1 + 7
y1 = 217/36
Now using point slope form, equation of tangent is
y – y1 = m(x – x1)
36y – 217 = -12x + 10
36y + 12x = 227 is the required equation of tangent.
### Question 16. Show that the tangent to the curve y = 7x3 + 11 at the points where x = 2 and x = -2 are parallel.
Solution:
Given curve: y = 7x3 + 11
On differentiating w.r.t. x, we get
dy/dx = 21x2
dy/dx = 21(2)2 = 84
The slopes at x – 2 & -2 are the same,
Hence the tangent will be parallel to each other.
### Question 17. Find the points on the curve y = x3 at which the slope of the tangent is equal to the y-coordinates of the point.
Solution:
Given curve: y = x3
On differentiating w.r.t. x, we get
dy/dx = 3x2 -(1)
Now let us assume that the point is (x1, y1)
dy/dx = 3x12
Also, slope of tangent at (x1, y1) is equal to y1.
So, 3x12 = y1 -(2)
Also, (x1, y1) lies on y = x3 x3,so
y1 = x13 -(3)
From eq(2) & (3)
3x12 = x13
3x12 = x13 = 0
x12(3 – x1) = 0
For x1 = 0, y1 = x13 = (0)3 = 0
One such point is (0, 0)
For x1 = 3, y1. = (3)3 = 27
Second point is (3, 27)
### Question 18. For the curve y = 4x3 – 2x5, find all the points at which the tangent passes through the origin.
Solution:
Given curve: y = 4x3 – 3x5
Clearly at x = 0, y = 0, i.e the curve passes through origin.
Now the tangent also passes through origin.
Equation of a line passing through origin is y = mx.
Now tangent is touching the curve, so
y = mx will satisfy in curve.
mx = 4x3 – 2x5 -(1)
Now dy/dx = 12x2 – 10x4
Also m is the slope of tangent, so
m = 12x2 – 10x4 -(2)
From eq(1) & (2),
(12x2 – 10x4)x = 4x3 – 3x5
x3(12 – 10x2) = x3(4 – 2x2) -(3)
For the first point, x = 0
For x1 = 0, y1 = 4x13 – 2x15 = 0
So, (0, 0) is one such point
Now for other roots of 3
12 – 10x2 = 4 – 2x2
8 = 8x2
x2 = 1
x = ±1
For x2 = 1, y2 = 4x23 – 2x25 = 4(1)3 – 2(1)5 = 2
For x3 = -1, y3 = 4x33 – 2x35 = 4(-1)3 – 2(-1)5 = -2
The other points are(1, 2) & (-1, -2)
### Question 19. Find the points on the curve x2 + y2 – 2x – 3 = 0 at which the tangents are parallel to the x-axis.
Solution:
Given curve: x2 + y2 – 2x – 3 = 0
On differentiating w.r.t. x, we get
3x + 2y(dy/dx) – 2 – 0 = 0
Given that the tangent are parallel to x-axis,
So, dy/dx = slope = 0
1 – x/y = 0
For x1 = 1,
x12 + y12 – 2x1 – 3 = 0
(1)2 + y12 – 2(1) – 3 = 0
y12 = 4
y12 = ≠2
The points are (1, 2) & (1, -2)
### Question 20. Find the equation of the normal at the point (am2, am3) for the curve ay2 = x3.
Solution:
Given curve: ay2 = x3
On differentiating w.r.t. x, we get
2ay.dy/dx = 3x2
&
So, by point slope form, equation of normal is,
3my – 3am4 = -2x + 2am2
Hence, 3my + 2x = 2am2 + 3am4 is the required equation of normal.
### Question 21. Find the equation of the normals to the curve y = x3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0.
Solution:
Given curve: y = x3 + 2x + 6
On differentiating w.r.t. x, we get
-(Slope of normal)
Now, the normal are parallel to x + 14y + 4 = 0
13 = 3x2 + 2
3x2 = 12 ⇒ x2 = 4
x = ±2
x1 = 2 & x2 = -2
For x1 = 2; y1 = (2)3 + 2(2) + 6 = 18
For x2 = -2; y2 = (-2)3 + 2(-2) + 6 = -6
Normal through (2,18) is
y – 18 =
14y – 252 = -x + 2
14y + x = 254 is one such equation.
Normal through (-2, -6) is
y + 6 =
14y + 84 = -x – 2
14y + x + 86 = 0 is the other equation of normal.
### Question 22. Find the equations of tangent and normal to the parabola y2 = 4ax at the point (at2, 2at).
Solution:
Given parabola: y2 = 4ax
On differentiating w.r.t. x, we get
Now, by point slope form, equation of tangent is,
y – y1
ty = x + at2 is the equation of tangent to the parabola y2 = 4ax at (at2, 2at)
Now
Now, by point slope form, equation of normal is,
y – y1
y – 2at = -t(x – at2)
y + xt = 2at + at3 is the equation of normal to the parabola y2 = 4ax at (at2, 2at)
### Question 23. Prove that the curves x = y2 and xy = k cut at right angles if 8k2 = 1.
Solution:
Given curves: x = y2 & xy = k
Two curves intersect at right angles f the tangents through their point
intersection is perpendicular to each other.
Now if tangents are perpendicular their product of their slopes will be equal to -1.
Curve 1: x = y2
1 = 2y.dy/dx
dy/dx = 1/2 y = m1 -(1)
Curve 2: xy = k
y = k/x
Let’s find their point of intersection
x = y2 & xy = k
k/y = y2
y = k1/3
x = y2
x = k2/3
The point is (k2/3, k1/3)
m1 = 1/2k1/3
For curves to be intersecting each other at right angles,
m1m2 = -1
8k2 = 1
Hence proved
### Question 24. Find the equations of the tangent and normal to the hyperbola at the point (xo, yo).
Solution:
Given curve:
On differentiating both sides with respect to x,
Now,
Equation of tangent by point slope form is,
-((x0,y0) lie on )
is the equation of tangent.
Now,
Equation of normal by point slope form is,
x0b2y – x0b2y0 = -y0a2x + y0a2x0
x0b2y + y0a2x = x0y0(a2 + b2) is the equation of normal
### Question 25. Find the equation of the tangent to the curve y = which is parallel to the line 4x – 2y + 5 = 0.
Solution:
Given curve:
On differentiating w.r.t. x, we get
Now, it is given that the tangent is parallel to the line 4x – 2y + 5 = 0,
so their slopes must be equal.
Slope of 2y = 4x + 5 is 2.
So,
9/16 = 3x – 2
x1 = 41/48
Now,
The point is (41/48, 3/4)
Now by point slope form, the equation of tangent will be
y – y1 = m(x – x1)
24y – 48x + 23 = 0 is the required equation of tangent.
### (A) 3 (B) 1/3 (C)-3 (D) -1/3
Solution:
Given curve: y = 2x2 + 3 sin x
On differentiating w.r.t. x, we get
dy/dx = 4x + 3cos x
-(Slope of normal)
Hence, the correct option is D.
### (A) (1, 2) (B) (2, 1) (C) (1, -2) (D) (-1, 2)
Solution:
Given curve: y2 = 4x
On differentiating w.r.t. x, we get
y = x + 1 is tangent, slope is 1, so, 2/y = 2
y1 = 2,
y12 = 4x1
x1
x1 = 1
So, (1, 2) is the point.
Hence, the correct option is A.
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# How many numbers are there between 1032 and 1209?
Wiki User
2012-08-17 07:53:47
There are infinitely many numbers between 1032 and 1209. Let me begin to explain my answer by presenting a modified version of the original question:
"How many whole numbers are there between 1032 and 1209?"
The answer? There are 76 whole numbers between 1032 and 1209.
So, I would say there are two possibly-confusing things going on here. One of them is the distinction between "numbers" and "whole numbers"; and the other is the distinction between "the difference between two numbers" and "the number of numbers between two numbers."
"Whole numbers" are the number 0, the number 1, and any number that is the sum of the number one added to itself, however many times. So, 0, 1, 2, 3, 4, 5, and so on, are all whole numbers.
There are infinitely many whole numbers (because there's no end to the numbers you can get by adding 1 to the preceding number), but the number of whole numbers between any two, non-negative numbers (not including Infinity) is always finite. Between 2 and 5, for example, there are two whole numbers: 3 and 4. Between 2 and 1,000,005 there are 1,000,002 whole numbers, which you can name by starting with the number 3, and counting up -- by ones -- to the number 1,000,004. That's the number you would stop with because we are naming (and counting) the whole numbers between 2 and 1,000,005.
Between 1032 and 1209 there are 76 whole numbers. To get the number of whole numbers between two non-infinite, non-negative numbers, you take the difference of the two numbers (the smaller number subtracted from the larger number) and subtract 1. The difference between 2 and 5 is 3; and the number of whole numbers between 2 and 5 is 2 -- namely, 3 and 4. The difference between 1032 and 1209 is 77, and the number of whole numbers between them is 76.
But when you ask about "numbers" instead of "whole numbers" you are including the infinitely many intervening numbers between any two whole numbers. There are even infinitely many intervening numbers between 0 and 1: 0.1, 0.01, 0.001, 0.00230791502, 0.00000000000000000100000000010341010984050505326, and so on. Without the restriction to whole numbers, all the infinite divisions of the number 1 would have to be named and counted, if you wanted to say "how many numbers there are between" any two numbers starting with zero and up to any positive number other than infinity -- say, for example, between 1032 and 1209. So infinitely many numbers is how many numbers there are between 1032 and 1209.
Wiki User
2012-08-17 07:53:47
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Ex 1.2
Chapter 1 Class 12 Relation and Functions
Serial order wise
### Transcript
Ex 1.2 , 7 In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer. f: R → R defined by f(x) = 3 − 4x f(x) = 3 – 4x Checking one-one f (x1) = 3 – 4x1 f (x2) = 3 – 4x2 Putting f(x1) = f(x2) 3 – 4x1 = 3 – 4x2 Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 –4x1 = –4x2 x1 = x2. Hence, if f(x1) = f(x2) , Then x1 = x2 ∴ function f is one-one Onto f(x) = 3 − 4x Let f(x) = y , such that y ∈ R 3 – 4x = y –4x = y – 3 x = (𝑦 − 3)/(−4) Now, Checking for y = f(x) Putting value of x in f(x) f(x) = f((𝑦 − 3)/(−4)) = 3 − 4((𝑦 − 3)/(−4)) = 3 + (y − 3) = y Thus, for every y ∈ R, there exists x ∈ R such that f(x) = y Hence, f is onto
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Illinois State University Mathematics Department
MAT 312: Probability and Statistics for Middle School Teachers Dr. Roger Day ([email protected])
### One-Variable Data Analysis: Initial Explorations
Here are the semester exam score for 21 students in a college statistics course:
Final Exam Scores: College Statistics Course 51 46 31 35 37 51 56 51 43 48 52 33 42 37 27 57 65 36 37 55 42
What do these data tell us about the performance of the group? By looking at the raw data, in the form presented here, it is not easy to make statements that describe group performance. Without a more careful look, we can't be sure of the highest or lowest score in the group. In its raw form, it's difficult to determine whether there are clusters of scores near certain values or test scores that are evenly distributed from low to high.
Here's where data analysis provides a helping hand in our attempts to make sense of these data. Using techniques of data analysis, we can manipulate these data to better address questions about the group's performance. The manipulations include carrying out calculations, showing the data in different formats, making comparisons within and beyond these data, and similar techniques. Our goal here is to describe, illustrate, practice, apply, and expand these techniques.
One of the first techniques you might apply is to order the test scores from greatest to least. The table to the right shows the ordered list of scores. What is revealed by ordering these scores?
At least three characteristics of the data can be revealed by examining an ordered list:
1. The largest and the smallest data values are found at the top and bottom of an ordered list. Here, the score of 65 points is the largest value in the data set and the score of 27 points is the smallest value in the data set.
2. Repeated data values are easier to identify in an ordered list. Here, a scan of the list shows that a score of 51 points appears three times, a score of 42 appears twice, and a score of 37 appears three times.
3. Carrying out a bit of comparision as we scan an ordered list, we can identify gaps between values. Here, a significant gap occurs between scores of 57 points and 65 points and a moderate gap exists between 37 points and 42 points.
Here is another unordered data set. The table shows the top 25 pitchers in NCAA Division I softball, as determined by earned run average (ERA), for the 2003 season. Concentrate on the column labeled Appearances. This tells the number of games in which each player appeared.
Order the Appearances data from least to greatest.
Use the ordered Appearances data to describe the three characteristics just discussed.
Here are the Appearances data, ordered from least to greatest.
21 27 29 29 30 32 32 32 33 33 36 38 38 39 40 40 41 43 44 46 48 48 48 51 52
1. We see that 21 is the least number of appearances and 52 is the greatest number of appearances among those in the top 25 ERA rankings.
2. Three of the top 25 pitchers appeared in 32 games while three others appeared in 48 games. Two players each appeared in 29 games, 33 games, 38 games, and 40 games. These are the data values that repeat in this data set.
3. The only significant gap that occurs is between the player with 21 appearances and the player with 27 appearances.
These examples illustrate several important characteristics of one-variable data sets. Some of these characteristics have been named or defined so we can communicate more effectively about them. Some related characteristics are described here as well.
• Clusters are isolated groups of points.
• Gaps are large spaces between data points.
• Outliers are data values substantially larger or smaller than any other data points.
• The maximum value of a data set is the largest value in the set. It is also called the upper extreme.
• The minimum value of a data set is the smallest value in the set. It is also called the lower extreme.
• The range of a data set is the difference between the maximum and minimum values of a data set.
• The mode of a data set is the value that occurs most often in the data set. A data set can have more than one mode, and if no data value occurs more often than any other, there is no mode.
Return to the NCAA Division I Softball data on pitchers. For the column labeled Runs:
1. Order the data set from least to greatest.
2. Identify any clusters, gaps, and outliers.
3. Determine the minimum and maximum values of this set and calculate the range of these data.
65 57 56 55 52 51 51 51 48 46 43 42 42 37 37 37 36 35 33 31 27
In the next few sections, we look more formally at visual and numerical tools for one-variable data analysis.
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CHAPTER 07.01: PRIMER ON INTEGRAL CALCULUS: Accumulation Function: Example In this segment, we're going to talk about accumulation function. To talk about accumulation function is important for two reasons. So one is that it reviews what you have learned in your integral calculus. It also forms . . . gives you a good idea of the first fundamental theorem of calculus but more importantly it shows you that how the numerical methods have been introduced to you in the integral calculus class itself, and in fact, that becomes your first introduction to integration, as opposed to just what we talk about in analytical calculus. So let's take an example here. What we're going to do is we're going to show that if you are going to integrate t squared from 0 to x, you're going to get x cubed, divided by 3, and you have see these equations given all the time, that you are given tables of integrals, and in one of those tables of integrals, this might be one of the integrals which is given to you. But how do we really get . . . how do we go from here to here? That's what we're going to show by the accumulation function example in this case. So this is how it's going to work. If we have this function, f of t equal to t squared, as a function of t, and what I want to do is I want to go all the way up to x, I'm going to start from 0 and go to x, I want to find out what the area under the curve is, that's what I want to be able to do. And the way I'm going to do it is I'm going to break this . . . I'm going to take this 0, comma, x, I'm going to partition into n equal segments, that's what I'm going to do first. So I'm going to partition it into n equal segments, which basically tells me that the delta t which I have is x divided by n. So if I'm going to break it up into n equal segments, I'm going to get delta t equal to x divided by n, because that's the length of the segment . . . total interval, and I'm dividing it by n. So what that means is that if this is 0, this will be . . . this will be t0, and this will be t1, and all the way up to, this will be tn, that's what I'm going to get. And what I'm going to do is I'm going to use the right-hand Riemann sum. I'm going to use the right-hand Riemann sum to calculate my areas under the curve, so the last one will look like this. And again, t0, for example, is 0, t1 is nothing but x divided by n, then t2 is 2 x, divided by n, all the way up to tn, which is nothing but x itself, so because they are equally space, so the space between t0 and t1 is x divided by n, between t1 and t2 is x divided by n, and what I am doing is that in order to find out what the sum . . . what the area under the curve is, I'm going to use the right-hand Riemann sum. So I'm basically going to calculate this area, I'm going to calculate this area, and I'll have t3 here, for example, I'm going to calculate that area, and then the last area which I will have, and that I'm going to . . . that I can use as an approximate value of the integral going from 0 to x, for this particular function of t squared, but if I make infinite such rectangles, so if I choose n going to, approaching infinity, then the value of the area under the curve will be exactly the same as the actual area under the curve. So that's the whole point what I wanted to show, but what we have to appreciate here is that if n is a finite number, I am basically doing a numerical method here of finding the area under the curve. So it's by choosing n going to infinity that I'll be able to get the exact value of the area under the curve, and that's what forms the analytical form of that particular part of the calculus. So let's go ahead and write down what the total area is. So if I'm going to look at the area, it'll be equal to the value of the function at t1 times delta t, so that will be the first area which I am calculating, because all I am doing is calculating the value of the function at t1, because I am using the right-hand Riemann sum, and multiplying by the width, so that becomes the area of the rectangle. The area of the next rectangle will be f of t2 times delta t, and the area of the last rectangle will be f of tn times delta t. So again, keep in mind that when we are using this right-hand Riemann sum, we are choosing the right point as the sample point, and also that we are partitioning it in equal segments, so those are two things which you've got to think about, because in Riemann sum, you can have sample point anywhere in the partition, and all the partitions do not need to be of equal width. So, but in the right-hand Riemann sum, the way we are doing it is that we are choosing the right point of the segment, of the partition, and then also we are choosing equal widths there with delta t, and also we know that delta t is x divided by n. So this I can write as a summation, i is equal to 1 to n, f of ti times delta t. But what is ti? I know that ti is like t0 is 0, t1 is x divided by n, t2 is 2 x, divided by n, it is nothing but i x, divided by n, times delta t, and what is delta t? delta t is nothing but x divided by n. So because I know that t0 is 0 times x, divided by n, t1 is 1 times x, divided by n, and so on and so forth, so that's what the argument of the function f is, and then x divided by n is simply the value of delta t. So here, we're going to get summation, i is equal to 1 to n, and what is the function? The function is simply the square of the argument, which will be i times x, divided by n squared, times x divided by n. Now, keep in mind that I could have straightaway gone to this particular step right here. All I'm trying to show you is that you can do the same process for any other function f, like for example, if you had integral of t cubed, or sine x, or whatever it is, you could do that. The only is that you have to be able then be able to sum the series exactly, but this kind of procedure will work for integrating any kind of polynomial function which somebody throws at you in order to be able to get what the exact . . . exact expression for the integral is. So from here, what I am getting is summation of, i is equal to 1 to n, I'm getting i squared x squared, divided by n squared, x divided by n, and since x and n are constants in this case, i squared is the only thing which will stay inside the summation. So in this case, I will get equal to x cubed, divided by n cubed, summation, i is equal to 1 to n, i squared. And you can find out this summation from any handbook, and you will find out that this summation here will turn out to be n times n plus 1 times 2 n plus 1, divided by n cubed . . . divided by 6 . . . and divided by 6, that's what you get for the summation of i squared. What I can also do is I can now see that, hey, this is what I'm going to get for the value of the area under the curve if n is a finite number. So if somebody told me that I'm going to break 0 to x into, let's suppose, five segments, all I have to choose is n equal to 5, and I will get some approximation of the area under the curve. But as we said that if I'm going to choose n to be infinity, that means that I'm going to have these infinite rectangles, then in that case, the exact value of the integral is simply the limit of an approaching infinity of this summation, i is equal to 1 to n, the value of the function at particular sample points, in this case it is the right sample point at the end of the partition, f of ti times delta t. So I already calculated this for our integral right here, which is this, so this limit of n approaching infinity, x cubed, divided by n cubed, times n times n plus 1, 2 n plus 1, divided by 6. Since the limit is being taken on n, I can take x cubed, divided by 6 outside, so x cubed, divided by 6 I can take outside, and I'm going to say limit of n approaching infinity, now, what is inside? I'm going to expand this factored form, I get 2 n cubed, plus 3 n squared, plus n, that's what I get from the numerator here, the denominator is n cubed. And, so now I'm going to take the limit of this expression as n approaches infinity. So I'm going to get equal to x cubed divided by 6, limit of n approaching infinity, I'm going to divide the numerator by n cubed and the denominator by n cubed, so I'm going to get 2, plus 3 by n, plus 1 by n squared, that's what I'm going to get. So as n approaches infinity, this part is going to become 0, and this part is going to become 0, so I'll be left with what? x cubed, divided by 6, times 2, and that gives me x cubed, divided by 3. So that's how you are showing that the integral of 0 to x, t squared dt is equal to x cubed, divided by 3. And you can do this for any other polynomial function in a similar fashion, or any other function, but you may have problems with getting the exact value of the summations, you may have to use Taylor series for trigonometric and some transcendental functions to be able to follow the same procedure. It might be algebraically a little bit more complicated, but it can be done. And that is the end of this segment.
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Add Fractions Calculator Add Fractions Calculator Add Fractions Calculator is a special tool which helps us to learn about the procedure followed to learn addition of fraction numbers. Now we first define fraction numbers. The numbers which can be expressed in the form of a / b are called fraction numbers, where we have a and b as the whole numbers and b <> 0. All the whole numbers are also called fraction numbers as any whole number n can be written as n / a, which is in the form of numerator / denominator. Let us assume that any fraction number a/b and another number 0 are added, then we get the sum of the two fraction numbers as a/ b + 0 = a / b Now let us take two like fractions such that the denominators of the two fraction numbers are equal.
Know More About :- Derivative of Inverse
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Then we say that the sum will have the sum of the two numerators and the denominator will remain same. So if a / b and c / b are any two fraction numbers, then we say that the sum will be: a/b + c / b = ( a + c ) / b In another situation, we take the two unlike fraction numbers. In this case, we will first convert the two unlike fraction numbers into like fraction numbers. For this we will find the LCM of the denominators of the two given fraction numbers and then the two fraction numbers will be converted into their equivalent form such that the denominator of both the fraction numbers becomes equal to the LCM. After this the problem converts into the addition of simple fraction numbers with the same denominator. Thus we say that if we have the following fractions to be added: 3/5 + 5/4 So LCM of denominators 4 and 5 will be 20 and so we will write them in the equivalent form by making the denominators of both the fraction numbers as 20. We get = ( 12 / 20 ) + ( 25 / 20 ) Now we will add the numerators and we will get the fraction as follows : ( 12 + 25)/ 20 = 37 / 20 This resultant fraction is improper fraction number and so we will convert the result into improper fraction number. For this we will divide the numerator with the denominator and get :
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1 whole 17 / 20 Ans If we are given any two mixed fraction numbers to be added, then we can solve the given expression in two ways. First is we simply add the two whole parts of the mixed fractions and then we will add the fraction parts of the two numbers. On the other hand, we say that the two mixed fraction numbers can be converted into the improper fraction numbers and then the sum of the two fraction numbers can be added up to get the result. Every time we need to check the resultant fraction numbers and then we will convert it to the mixed fraction number in case the resultant fraction number is in the form of improper fraction number and get the result.
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Let Quotient Rule Formula In calculus, Quotient rule is helps govern the derivative of a quotient with existing derivatives. Then, if $$v\left( x \right) \ne 0$$, the derivative of the quotient of these functions is calculated by the formula Quotient Rule: The quotient rule is a formula for taking the derivative of a quotient of two functions. }$$The quotient rule states that the derivative of$${\displaystyle f(x)} is The g(x) function, the LO, is x^4. just create an account. . . There's a differentiationlaw that allows us to calculatethe derivatives of quotients of functions.Oddly enough, it's called the Quotient Rule. {\displaystyle f(x)} f To find the derivative of this function, we only need to remember that a quotient is in reality a product. This can also be written as . x ( Apply the quotient rule first. f To evaluate the derivative in the second term, apply the power rule along with the chain rule: Finally, rewrite as fractions and combine terms to get, Implicit differentiation can be used to compute the nth derivative of a quotient (partially in terms of its first n − 1 derivatives). \$1 per month helps!! = By the Quotient Rule, if f (x) and g(x) are differentiable functions, then d dx f (x) g(x) = g(x)f (x)− f (x)g (x) [(x)]2. h ( ) {\displaystyle f(x)={\frac {g(x)}{h(x)}}=g(x)h(x)^{-1}.} h Now, let's take the derivative of each function. f The quotient rule is a formula for taking the derivative of a quotient of two functions. Plus, get practice tests, quizzes, and personalized coaching to help you ( / ) ) The quotient rule is a formal rule for differentiating of a quotient of functions.. Let $$u\left( x \right)$$ and $$v\left( x \right)$$ be again differentiable functions. So for example if I have some function F of X and it can be expressed as the quotient of two expressions. credit-by-exam regardless of age or education level. where both Anyone can earn So, df (x) means the derivative of function f and dg (x) means the derivative of function g. The formula states that to find the derivative of f (x) divided by g (x), you must: 2. + So let's say U of X over V of X. Log in here for access. Remember the rule in the following way. h ( + The quotient rule is used to determine the derivative of one function divided by another. | {{course.flashcardSetCount}} She has over 10 years of teaching experience at high school and university level. first two years of college and save thousands off your degree. Find the value of h'(1). Using the quotient rule, dy/dx = (x + 4)(3x²) - x³(1) = 2x³ + 12x² (x + 4)² (x + 4)² Differiente the function y = \frac{cosx}{1 - sinx}. 2 ( ′ b f (x) = (6x3 −x)(10−20x) f (x) = (6 x 3 − x) (10 − 20 x) Show Solution Let’s now work an example or two with the quotient rule. ) g This rule states that: The derivative of the quotient of two functions is equal to the denominator multiplied by the derivative of the numerator minus the numerator multiplied by the derivative of the denominator, all divided by … If you have function f(x) in the numerator and the function g(x) in the denominator, then the derivative is found using this formula: In this formula, the d denotes a derivative. Use the quotient rule to find the derivative of f. Then (Recall that and .) ) are differentiable and For example – $\ \frac{d}{dx}(\frac{u}{v}) = \frac{v \frac{du}{dx} – u \frac{dv}{dx}}{v^2}$ The quotient rule can be used to differentiate tan(x), because of a basic quotient identity, taken from trigonometry: tan(x) = sin(x) / cos(x). {\displaystyle f(x)={\frac {g(x)}{h(x)}},} In short, quotient rule is a way of differentiating the division of functions or the quotients. The quotient rule is a formula for differentiation problems where one function is divided by another. ) Simplify number 1 as much as possible. This discussion will focus on the Quotient Rule of Differentiation. ) . Providing each function has a derivative, simply substitute the values into the quotient rule formula for the answer. ( For example, differentiating The product rule then gives ) ) ) ″ Imagine a frog yodeling, 'LO dHI less HI dLO over LO LO.' = x In this mnemonic device, LO refers to the denominator function and HI refers to the numerator function. h / - How-To & Steps, Finding the Derivative of the Square Root of x, When to Use the Quotient Rule for Differentiation, Implicit Differentiation: Examples & Formula, Glencoe Math Course: Online Textbook Help, CUNY Assessment Test in Math: Practice & Study Guide, GED Math: Quantitative, Arithmetic & Algebraic Problem Solving, SAT Subject Test Mathematics Level 1: Practice and Study Guide, SAT Subject Test Mathematics Level 2: Practice and Study Guide, Introduction to Statistics: Help and Review, Introduction to Statistics: Tutoring Solution, NY Regents Exam - Integrated Algebra: Help and Review, High School Geometry: Homework Help Resource. LO LO means take the denominator times itself: g(x) squared. ″ x {\displaystyle f(x)=g(x)/h(x).} Applying the definition of the derivative and properties of limits gives the following proof. Let's define the functions for the quotient rule formula and the mnemonic device. All rights reserved. The quotient rule is actually the product rule in disguise and is used when differentiating a fraction. ≠ There is a formula we can use to differentiate a quotient - it is called thequotientrule. Some problems call for the combined use of differentiation rules: If that last example was confusing, visit the page on the chain rule. SOLUTION 9 : Consider the function . ( {\displaystyle g} ) In this scenario let’s consider a function which is equal to one function divided by another function i.e.h To solve such functions we use the quotient rule which is defined by the formula: The derivative of the quotient of two functions is equal to the derivative of the function in the numerator multiplied by the function in the denominator minus the function in the numerator multiplied by the derivative of the function in the denominator and then divide this whole expression by the square of the function in the denominat… If h (2) = 3 and h' (2) = -4, find d / dx (h (x) / x)|_{x = 2}. f Students will also use the quotient rule to show why the derivative of tangent is secant squared. Not sure what college you want to attend yet? f flashcard set{{course.flashcardSetCoun > 1 ? ) The quotient rule states that for two functions, u and v, (See if you can use the product rule and the chain rule on y = uv-1 to derive this formula.) ( The quotient rule is a formal rule for differentiating problems where one function is divided by another. x = In the following practice problems, students will use the quotient rule to find the derivatives of various functions. x x Calculating the limit of product/quotient or sum/differences in math is as simple as bringing the operations outside of the limit function. {\displaystyle g(x)=f(x)h(x).} ( Finally, (Recall that and .) f The f (x) function (the HI) is x ^3 - x + 7. The limit of … Already registered? And lastly, after applying the formula, you may still need to simplify the resulting expression. g The quotient rule, is a rule used to find the derivative of a function that can be written as the quotient of two functions. = x x Log in or sign up to add this lesson to a Custom Course. © copyright 2003-2020 Study.com. and substituting back for f Quotient Rule Formula. = x − = Let the given … {\displaystyle f(x)=g(x)/h(x),} 4 • (x 3 +5) 2 = 4x 6 + 40 x 3 + 100 derivative = 24x 5 + 120 x 2. Find the derivative of f(x) = \frac{e^x}{x^2 + x}. f {{courseNav.course.topics.length}} chapters | The engineer's function brick(t)=3t6+52t2+7 involves a quotient of the functions f(t)=3t6+5 andg(t)=2t2+7. h Always start with the bottom'' function and end with the bottom'' function squared. is. x + Select a subject to preview related courses: Now, we can arrange those pieces into either the formula or the mnemonic device to find the derivative, which as you can see is: Then, you can multiply out the terms in the numerator and combine the like terms to get your final derivative, which, as you can see, is: Let's do another example. ( b) Find the derivative by dividing the expressions first. For example, y = cosx x2 We write this as y = u v where we identify u as cosx and v as x2. The quotient rule is as follows: Plug f (x) and g (x) into the quotient rule formula: See also derivatives, product rule, chain rule. If F(x) = cot(x) , prove F'(x) = -csc^2(x) . It follows from the limit definition of derivative and is given by . h ) Study.com has thousands of articles about every succeed. In this unit we will state and use the quotient rule. ) Then the quotient rule tells us that F prime of X is going to be equal to and this is going to look a little bit complicated but once we apply it, you'll hopefully get a little bit more comfortable with it. A formula for the quotient rule to find the right school LO LO means take denominator. And has a master 's degree in Curriculum and Instruction and exams calculate the derivative dividing. ' ( x ) =g ( x ) / h ( x \neq... Of x over v of x v ( du/dx ) - u ( dv/dx ) dx.! That we had to be followed for finding out the derivative quotient rule formula this function the. Following proof high school and university level in the previous section, have... Squared, first rewrite tangent in terms of sine and cosine LO to. Problems, students will also use the quotient rule states that the derivative of quotient! Quotient - it is more prac… SOLUTION 9: consider the function y \frac. Is 3x^2 - 1. dg ( x ) = g ( x ). years college! Or quotients why the derivative by dividing the expressions first the answer experience high... Degree in Curriculum and Instruction function divided by another … functions often come as quotients, which! Resulting expression of derivatives and cosine can test out of the two functions, frog 's yodel into... \Displaystyle f ( x ) = -csc^2 ( x ) = -csc^2 ( x ) h ( ). Unbiased info you need to find the derivatives of rational functions a master degree! May still need to simplify the resulting expression differentiate rational functions = g ( x =f! And. we will state and use the quotient rule is a formal rule for differentiating where. Customer support LO ) is x^2 - 3 of h ' ( 1 ). the! Imagine a frog yodeling, 'LO dHI less HI dLO over LO LO means the... Rule Date_____ Period____ differentiate each function -csc^2 ( x ) function, the quotient rule can used... Is called thequotientrule a frog yodeling, 'LO dHI less HI dLO over LO LO. use the quotient is. Function divided by another Ph.D. in Mathematics from UW-Milwaukee in 2019 in calculus, quotient rule page or! Of each function has a master 's degree in Curriculum and Instruction of enough! ) = \frac { cosx } { x^2 + x } has a derivative simply... And the bottom term g ( x ). ) find the derivative of f. (. Math is as simple as bringing the operations outside of the given function + ). The following quotient: we start by defining the functions for the quotient rule is a way differentiating... } { 1 - sinx } x ) times dg ( x ) function, LO! A method of finding the derivatives of rational functions consider two expressions with is in reality a product is! Problems where one function is divided by another what college you want to yet. Enrolling in a Course lets you earn progress by passing quizzes and exams, visit our Credit. Dlo over LO LO. say u of x a couple of examples where we have similar to the rule. Sure what college you want to attend yet the top term f ( x ), f... Terms of sine and cosine sum/differences in math is as simple as bringing the operations outside the! That is the Difference Between Blended Learning & Distance Learning and then the! To determine the derivative of a quotient is in form q is given by the answer function the... Regardless of age or education level functions for the quotient rule formula in calculus, quotient rule is similar the! Grad shows an easy way to use the quotient rule can be used to calculate the derivative the. Copyrights are the property of their respective owners the division: help & Review page to learn,! S take a look at this in action -csc^2 ( x ) { \displaystyle f x... Uw-Milwaukee in 2019 may still need to find the value of h ' ( 1.. Can earn credit-by-exam regardless of age or education level with is in reality a product taught and! On the quotient rule states that the derivative of the numerator function consider two with... Differentiating the division: help & Review page to learn more you earn progress passing! By passing quizzes and exams unit we will state and use the quotient rule show!, simply substitute the values into the quotient rule: the quotient rule calculate the derivative of the of. Given as quotient rule to differentiate rational functions multiply this out and then take the derivative of f x! Easy way to use the quotient rule easy way to use the quotient rule is formula. The chain rule, and remembering that the derivative of a quotient is in reality a product two expressions is... Hi dLO over LO LO means take the denominator: f ( ). V ( du/dx ) - u ( dv/dx ) dx v² ) or... The bottom '' function and end with the bottom '' and... Be careful when differentiating products or quotients the definition of the terms derivative dividing... Section, we only need to find the quotient rule formula of rational functions and a to! In reality a product finding the derivatives of various functions high-school math for over years! = g ( x ) } and university level function y = and. Off your degree examples where we have itself: g ( x ) prove! ( the HI ) is x ^2 - 3 the unbiased info you need to the... Lesson you must be a Study.com Member always start with the bottom '' function squared for 30 days just! Or dLO, is cos x. dg ( x ), or dHI, is x^4 given two functions! Differentiate rational functions Learning & Distance Learning and HI refers to the rule. Unlock this lesson to a Custom Course value of h ' ( x ) } is s time... You earn progress by passing quizzes and exams the bottom term g ( x ). by dividing expressions! =G ( x ) function, the quotient rule is a method of finding the derivative of then. State and use the quotient rule is a formula for the quotient rule similar! First rewrite tangent in terms of sine is cosine, we noted that we had be... And. in the previous section, we noted that we had to be for... That and. it makes it somewhat easier to keep track of all of the.! Her Ph.D. in Mathematics from UW-Milwaukee in 2019 in calculus, quotient rule is for... For finding the derivatives of quotients of functions.Oddly enough, it 's called the rule. The f ( x ). ) =f ( x ). du/dx ) - u ( dv/dx dx. A function that is the Difference Between Blended Learning & Distance Learning is similar to the product rule ) (! ) /h ( x ). derivative by dividing the expressions first you may still to. Providing each function where one function divided by another university level support me on Patreon education.. Denominator quotient rule formula itself: g ( x + 4 ). dy/dx x 7... Division of functions or the quotients Learning & Distance Learning thousands off your degree or the quotients x... + 7 - it is more prac… SOLUTION 9: consider the function y = \frac { x f x. Dhi less HI dLO over LO LO means take the derivative and given... Click here to return to the list of problems to help you remember formula. Get practice tests, quizzes, and personalized coaching to help you succeed of differentiating division! The LO, is 4x^3 4 ). ( dv/dx ) dx v² unbiased info need... It follows from the limit of product/quotient or sum/differences in math is as simple as bringing the outside! To the denominator function and end with the bottom '' function and end with the bottom function... After applying the definition of derivative and is given as quotient rule to differentiate rational functions 1. (! ) use the quotient rule formula that can be used to calculate derivative... Two years of teaching experience at high school and university level 's degree in Curriculum Instruction! Still need to find the derivative Ph.D. in Mathematics from UW-Milwaukee in 2019 of f ( x \neq... Top term f ( x ) } { 1 - sinx } & 39. You may still need quotient rule formula remember that a quotient - it is called thequotientrule take a look at a of... It ’ s now time to … Thanks to all of you who support me on.. High-School math for over 10 years of college and save thousands off your degree high school and university.... Help you remember the formula for the quotient rule this function, we have to apply the rule. Chain rule, and remembering that the derivative of the denominator function and end with the bottom function! First two years of teaching experience at high school and university level: we by. Unit we will state and use the quotient rule is similar to the numerator: g ( x function... Of functions or the quotients this mnemonic device functions or the quotients finding the of! X³, find dy/dx x + 4 differentiating problems where one function is divided by another may! Or the quotients in or sign up to add this lesson, you may still need to the. \Displaystyle h ( x ). is called thequotientrule unit we will state and the... { 1 - sinx } get practice tests, quizzes, and that!
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### Pumpkin Pie Problem
Peter wanted to make two pies for a party. His mother had a recipe for him to use. However, she always made 80 pies at a time. Did Peter have enough ingredients to make two pumpkin pies?
### The Tree and the Greenhouse
Mum, Dad, Tim and Jill moved into a new house. If they cut down the big tree, will it miss the greenhouse?
### Orange Drink
A 750 ml bottle of concentrated orange squash is enough to make fifteen 250 ml glasses of diluted orange drink. How much water is needed to make 10 litres of this drink?
# Cherry Buns
##### Stage: 2 Challenge Level:
There were several different ways to approach this problem. We received many solutions from children at Ardingly College who all tackled it in a similar way. Here is Jess' reasoning:
I wrote out the ingredients of the cherry buns and showed it like this:
egg = $2$
flour = $2$ eggs
sugar = $2$ eggs
butter = $2$ eggs
cherries = $1$ egg
total = $9$ eggs
First I did $12 \times 45$g = $540$g (the total weight of the mixture)
$540/9$ = $60$g so one egg weighs $60$g.
Alistair from Histon Junior School wrote Jess' solution in a slightly shorter way:
If e = $1$ egg, there are $9$e in the recipe.
I multiplied $45$ by $12$ to get the total weight of mixture. $45x12 =540$
So an egg would be $540/9$ which is $60$, (then turn it into grams) making e = $60$g
Pupils from Oakwood Junior School did it a slightly different way. This is what Sophie wrote:
First I found out how much mixture there was by multiplying $45$g by $12$ paper cases. This gave me an answer of $540$g.
Then next I worked out how much of each ingredient there was in each case.
Eggs $10$g
Flour $10$g
Sugar $10$g
Butter $10$g
Cherries $5$g
Then I worked out how much mixture there was altogether for the 12 cakes:
Eggs $120$g
Flour $120$g
Sugar $120$g
Butter $120$g
Cherries $60$g
After this I halved the amount for the eggs and this gave me $60$g for one egg.
Davis from Berkeley Preparatory School used a trial and improvement approach:
First, my teacher and I found out how many grams the batter weighed by multiplying $45$grams times $12$ paper cake cases.
That means the total batter weighed $540$ grams.
Then, we wrote a formula:
Eggs + flour + sugar + butter + cherries = $540$ grams.
Since the eggs, flour, sugar, and butter all weighed exactly the same, at first we guessed that each ingredient weighed $100$ grams.
That would mean $100$g + $100$g + $100$g + $100$g + cherries (which weigh as much as half of the eggs...which would be $50$g)
However, when we added that together, it only equalled $450$g.
That told me that each ingredient had to weigh more than $100$ grams. So I decided to try $120$ grams.
$120$g + $120$g + $120$g + $120$g + cherries ($60$g) = $540$grams
Now that I know that TWO eggs equals the same as $120$g, ONE egg would equal $60$ grams.
Thank you Davis. Beth, Jennie and Henry found another way to answer the problem:
We set about solving it like this:
She put $45$g in each of $12$ cake cases. That is $12 \times 45$g = $540$g.
So the total mixture weighs $540$g.
Then we listed the ingredients:
$2$ eggs
flour
sugar
butter
glace cherries
The first $4$ weigh the same but the last one weighs only half.
So we need $540/4.5$. This is the weight of each of the first four ingredients.
$540/4.5$ = $120$ (we found this out by trial and improvement)
So $2$ eggs weigh $120$g and the weight of one egg is $60$g.
We checked our solution by writing out the ingredient list again with the weights and checking that the total was $540$g.
Thank you to everyone.
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sodni3
2021-03-18
Determine , if possible , the value of x which will male the following true:
$\left[\begin{array}{cc}4& 3\\ 1& x\end{array}\right]+2\left[\begin{array}{c}2\\ 1\end{array}\right]\left[\begin{array}{cc}3& x\end{array}\right]=\left[\begin{array}{cc}16& 11\\ 7& 6\end{array}\right]$
Step 1
Product of two matrices:
$\left[\begin{array}{c}a\\ b\end{array}\right]\left[\begin{array}{cc}c& d\end{array}\right]=\left[\begin{array}{cc}ab& ad\\ bc& bd\end{array}\right]$
Product of a scalar and a matrix:
$k\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}ka& kb\\ kc& kd\end{array}\right]$
If two matrices are equal then the corresponding elements of the matrices are equal.
Step 2
Given matrix equation is:
$\left[\begin{array}{cc}4& 3\\ 1& x\end{array}\right]+2\left[\begin{array}{c}2\\ 1\end{array}\right]\left[\begin{array}{cc}3& x\end{array}\right]=\left[\begin{array}{cc}16& 11\\ 7& 6\end{array}\right]$
We need to find the value of x.
First, we need to simplify the left-hand side of the equation, and then we have to compare the elements of the left-hand side matrix and the right-hand side matrix.
$\left[\begin{array}{cc}4& 3\\ 1& x\end{array}\right]+2\left[\begin{array}{cc}2×3& 2×x\\ 1×3& 1×x\end{array}\right]=\left[\begin{array}{cc}16& 11\\ 7& 6\end{array}\right]$
$\left[\begin{array}{cc}4& 3\\ 1& x\end{array}\right]+2\left[\begin{array}{cc}6& 2x\\ 3& x\end{array}\right]=\left[\begin{array}{cc}16& 11\\ 7& 6\end{array}\right]$
$\left[\begin{array}{cc}4& 3\\ 1& x\end{array}\right]+\left[\begin{array}{cc}2\left(6\right)& 2\left(2x\right)\\ 2\left(3\right)& 2\left(x\right)\end{array}\right]=\left[\begin{array}{cc}16& 11\\ 7& 6\end{array}\right]$
Step 3
Now,
$\left[\begin{array}{cc}4& 3\\ 1& x\end{array}\right]+\left[\begin{array}{cc}12& 4x\\ 6& 2x\end{array}\right]=\left[\begin{array}{cc}16& 11\\ 7& 6\end{array}\right]$
$\left[\begin{array}{cc}4+12& 3+4x\\ 1+6& x+2x\end{array}\right]=\left[\begin{array}{cc}16& 11\\ 7& 6\end{array}\right]$
$\left[\begin{array}{cc}16& 3+4x\\ 7& 3x\end{array}\right]=\left[\begin{array}{cc}16& 11\\ 7& 6\end{array}\right]$
On comparing both sides, we get
Step 4
Answer: The value of x is 2.
Jeffrey Jordon
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# [T03] The game show puzzle
As a final test of your understanding of probability, try the following rather famous puzzle.
Imagine that you are a contestant on a television game show. You are shown three large doors. Behind one of the doors is a new car, and behind each of the other two is a goat. To win the car, you simply have to choose which door it is behind. When you choose a door, the host of the show opens one of the doors you have not chosen, and shows you that there is a goat behind it. You are then given a choice; you may stick with your original choice, or you may switch to the remaining closed door.
What should you do to maximize your chances of winning the car? Think about it for a while, and when you have decided, read the two arguments below and decide which is right.
Argument 1. No need to switch: Suppose you choose door number 1. The probability that the car is behind door 1 is initially 1/3 (since there are three doors, and the car has an equal chance of being behind each). Then suppose the host opens door number 3 and shows you that there is a goat behind it. We then need to calculate a conditional probability--the probability that the car is behind door 1, given that there is a goat behind door 3. Since there are only two doors left, and there is an equal chance that the car is behind each of them, this probability is 1/2. But similarly, the probability that the car is behind door 2, given that there is a goat behind door three, is also 1/2. So whether you stick with door 1 or switch to door 2, your chance of winning is 1/2. So it really makes no difference whether you switch or not.
Argument 2. You should switch: Suppose you choose door number 1. There are three possibilities; either the car is behind door 1, or door 2, or door 3. Each of these possibilities has the same probability (1/3). In each of the three cases, consider which door the host will open. If the car is behind door 1, the host could open either door 2 or door 3. In this case, if you stick with your original choice you win the car, but if you switch to the remaining door you lose. If the car is behind door 2, the host will open door 3. In this case, if you stick with your original choice you lose, but if you switch, you win. Finally, if the car is behind door 3, the host will open door 2. Again, if you stick with your original choice you lose, but if you switch, you win. Remember that each of the three possibilities has a probability of 1/3, and note that they are mutually exclusive (the car is only behind one door). If you switch, you will win in two cases out of three (probability 2/3), but if you stick you will only win in one case out of three (probability 1/3). So you should switch doors, since it doubles your chance of winning.
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# How to Calculate Equilibrant & Resultant
Save
Students encounter the equation f=ma soon after they begin to study physics. If an object experiences a net force, it will experience a corresponding acceleration proportional to the magnitude of that net force. When multiple forces simultaneously act upon the object, you'll need to add the different forces together to get the resultant force. A force identical in magnitude but in the opposite direction is the equilibrant force.
• Identify all the different forces acting upon the object. These are vector quantities, which means they will have both magnitude and direction. Some forces, such as air resistance, may be less obvious than others.
• Add all the different vectors together, using the rules for vector addition. If all the different forces are in the same direction, or exactly in the opposite direction, then these calculations are elementary. If someone is exerting 10 lbs. of force pushing a lead weight up while gravity is exerting 5 lbs. of pressure pushing it down, the net force is clearly 5 lbs. of force upward. If you are exerting 10 lbs. of force pushing a car north, and someone else is pushing in the same direction with 5 lbs. of force, the net force is 15 lbs. of force in the north direction.
• Construct a right triangle to represent two different forces acting at right angles. If you are exerting 1 lb. of force pushing an object north and someone else is exerting 1 lb. of pressure pushing the same object east, then two of the sides of the triangle are each 1 unit long. The hypotenuse of this triangle represents the vector addition of the two forces. The Pythagorean theorem tells us that the square of the hypotenuse is equal to the sum of the squares of the two sides, so since each side is 1 unit long and its square is also 1, the hypotenuse equals the square root of 2 or 1.41421356. Elementary trigonometry will also reveal the angle between the hypotenuse and the two sides, which indicates the direction of the resultant force. For this example, that's 45 degrees. The object would experience a resultant force in the northeast direction.
• Remember that you can represent any force as the sum of two different vectors. If an object is experiencing two different forces, but they are not at right angles to one another, represent one of the forces as a sum of two vectors, one in the same direction as the other force, and the other at a right angle to it. If one force is pushing an object due north and another is pushing it directly northeast, represent the northeast force as a north vector and an east vector. Simply add the two north vectors and use the Pythagorean theorem to add the east vector.
• Take the sum of all the vectors. That's the resultant force. A force equal in quantity but oriented in the exact opposite direction is the equilibrant force.
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# How do you find lim (sqrt(x+1)+1)/(sqrt(x+1)-1) as x->0^+ using l'Hospital's Rule or otherwise?
##### 2 Answers
Feb 16, 2017
This is not an indeterminate form. See below.
#### Explanation:
As $x \rightarrow 0$, the numerator goes to $2$. Therefore, the form is not indeterminate and it would be an error to use l'Hospital's Rule.
As $x \rightarrow 0$, the denominator goes to $0$, so the limit from the right is either $\infty$ or $- \infty$.
To determine which, consider an $x$, just a bit greater than $0$.
Then $x + 1$ is a bit more than $1$, and $\sqrt{x + 1}$ is a bit greater than $1$. So the denominator is a positive number near $0$.
We can write the form of the limit as $\frac{2}{0} ^ +$ which tells us that, as $x \rightarrow {0}^{+}$, the quotient increases without bound.
We write
${\lim}_{x \rightarrow {0}^{+}} \frac{\sqrt{x + 1} + 1}{\sqrt{x + 1} - 1} = \infty$
Feb 16, 2017
There is no limit.
#### Explanation:
Generally:
${\lim}_{x \to 0} \frac{\sqrt{x + 1} + 1}{\sqrt{x + 1} - 1}$
$= {\lim}_{x \to 0} \frac{\left(1 + \frac{1}{2} x + \ldots\right) + 1}{\left(1 + \frac{1}{2} x + \ldots\right) - 1}$
$= {\lim}_{x \to 0} \frac{\left(2 + \frac{1}{2} x + \ldots\right)}{\left(\frac{1}{2} x + \ldots\right)}$
For small positive $x$, this is always going to be positive. For small negative $x$, this is always going to be negative.
So there is no limit :(
We have a 2-sided limit.
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# Add a Two-Digit Number and a Single-Digit Number Mentally
This is a complete lesson for 2nd grade math where children learn to add a 2-digit number and a single-digit number mentally, without crossing to the next ten, such as 35 + 4 or 61 + 5. They add "within" the same ten. The lesson contains both instruction and plenty of exercises, and even a 'bonus' puzzle problem in the end.
+ 25 + 3 = 28
The 2 tens do not change.
+ 12 + 7 = 19
The ten does not change.
+ 34 + 4 = 38
The 3 tens do not change.
1. Write an addition sentence for each picture.
a. +
_______ + _______ = ______
b. +
_______ + _______ = ______
c. +
_______ + _______ = ______
d. +
_______ + _______ = ______
2. Add. Compare the problems. The top problem helps you solve the bottom one!
a. 5 + 2 = _____ 35 + 2 = _____ b. 4 + 5 = _____ 64 + 5 = _____ c. 3 + 6 = _____ 93 + 6 = _____
3. Add. For each problem, think about a “helping” problem with numbers less than 10.
a. 52 + 7 = _____ 2 + 7 = _____ b. 33 + 1 = _____ ____ + ____ = _____ c. 11 + 5 = _____ _____ + ____ = _____
4. The numbers are written in boxes! Add ones in their own column.
Copy the tens number down below.
a. 35 + 3
tens ones 3 5 + ↓ 3 3 8
b. 12 + 6
tens ones 1 2 + ↓ 6
c. 57 + 1
tens ones 5 7 + ↓ 1
d. 64 + 3
tens ones 6 4 + ↓ 3
5. Now you write the numbers in the boxes. Add ones in their own column.
a. 26 + 3
tens ones + ↓
b. 72 + 4
tens ones + ↓
c. 65 + 4
tens ones + ↓
d. 81 + 4
tens ones + ↓
a.6 + 2 = ______ 16 + 2 = ______ 36 + 2 = ______ b.4 + 3 = ______ 24 + 3 = ______ 34 + 3 = ______ c. 5 + 4 = ______ 45 + 4 = ______ 65 + 4 = ______ d. 11 + 7 = ______ 61 + 7 = ______ 41 + 7 = ______
a. 20 + 5 + 2 = ______ 44 + 2 + 2 = ______ b. 93 + 1 + 5 = ______ 83 + 4 + 3 = ______ c. 100 + 5 + 4 = ______ 52 + 4 + 2 = ______
Remember how to break
a number into its
TENS and ONES?
23 = 20 + 3 tens ones
47 = 40 + 7 tens ones
8. Break the numbers into tens and ones or do it the other way around.
a. 18 = 10 + 8 25 = _____ + _____ 55 = _____ + _____ b. 32 = _____ + _____ 95 = _____ + _____ ______ = 40 + 9 c. _______ = 60 + 6 _______ = 9 + 80 _______ = 8 + 70
9. Compare. Write < > , or = .
a. 24 + 3 24 + 5 b. 83 + 5 85 + 3
c. 17 + 2 19 + 2 d. 36 + 4 46 + 4
e. 58 8 + 51 f. 66 5 + 61
is to compare without knowing the mystery number!
For example, which is more, + 2 or + 7?
Write < or > in the boxes. Note: t
here is one comparison you cannot do without knowing the mystery number. Can you find it?
+ 5 + 4 – 5 – 4 – 5 + 2 + 7 – 5 – 6 + + 20
This lesson is taken from my book Math Mammoth Add & Subtract 2B. The lesson addresses the following Common Core standards for 2nd grade: 2.OA.1 and 2.NBT.5.
#### Math Mammoth Add & Subtract 2-B
A self-teaching worktext for 2nd grade that covers mental addition and subtraction with two-digit numbers, and regrouping in addition and subtraction (carrying & borrowing).
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# 2001 AMC 10 Problems/Problem 21
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter $10$ and altitude $12$, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.
$\textbf{(A)}\ \frac{8}3\qquad\textbf{(B)}\ \frac{30}{11}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ \frac{25}{8}\qquad\textbf{(E)}\ \frac{7}{2}$
## Solution 2
$[asy] draw((5,0)--(-5,0)--(0,12)--cycle); unitsize(.75cm); draw((-30/11,0)--(-30/11,60/11)); draw((-30/11,60/11)--(30/11,60/11)); draw((30/11,60/11)--(30/11,0)); draw((0,0)--(0,12)); label("2r",(0,30/11),E); label("12-2r",(0,80/11),E); label("2r",(0,60/11),S); label("10",(0,0),S); label("A",(0,12),N); label("B",(-5,0),SW); label("C",(5,0),SE); label("D",(-30/11,60/11),W); label("E",(30/11,60/11),E); [/asy]$
Let the diameter of the cylinder be $2r$. Examining the cross section of the cone and cylinder, we find two similar triangles. Hence, $\frac{12-2r}{12}=\frac{2r}{10}$ which we solve to find $r=\frac{30}{11}$. Our answer is $\boxed{\textbf{(B)}\ \frac{30}{11}}$.
## Solution 3 (Very similar to solution 2 but explained more)
We are asked to find the radius of the cylinder, or $r$ so we can look for similarity. We know that $\angle BEF = \angle BDA$ and $\angle FBE = \angle ABD$, thus we have similarity between $\triangle BFE$ and $\triangle BAD$ by $AA$ similarity.
Therefore, we can create an equation to find the length of the desired side. We know that:
$\frac{BE}{BD}=\frac{FE}{AD}.$
Plugging in yields:
$\frac{12-2r}{12}=\frac{r}{5}.$
Cross multiplying and simplifying gives:
$5(12-2r)=12r$
$\Downarrow$
$r=\frac{30}{11}.$
Since the problem asks us to find the radius of the cylinder, we are done and the radius of the cylinder is $\boxed{\textbf{(B)}\ \frac{30}{11}}$.
~etvat
## Solution 4 (graphical)
Assume that a point on a given diameter of the cone is the point $(0,0)$ on a two-dimensional representation of the cone as shown in Solution 2. The top point of the cone is thus $(5,12)$ and the line that goes through both points is $y=\frac{12}{5}x$.
Now we create a second equation. We must choose some point $(x,y)$ on the line $y=\frac{12}{5}x$ such that $y=10-2x$, which implies that the cylinder’s diameter, $10-2x$, must be equal to its height, $y$. Solving yields $x=\frac{25}{11}$, and the radius is thus $\frac{10-2x}{2}=\frac{\frac{60}{11}}{2}=\boxed{\textbf{(B)}\ \frac{30}{11}}$.
## Solution 5 (Without similar triangles)
Like in Solution 2, we draw a diagram.
$[asy] draw((5,0)--(-5,0)--(0,12)--cycle); unitsize(.75cm); draw((-30/11,0)--(-30/11,60/11)); draw((-30/11,60/11)--(30/11,60/11)); draw((30/11,60/11)--(30/11,0)); draw((0,0)--(0,12)); label("2x",(0,30/11),E); label("2x",(0,60/11),S); label("H",(0,0),S); label("A",(0,12),N); label("B",(-5,0),SW); label("C",(5,0),SE); label("D",(-30/11,60/11),W); label("E",(30/11,60/11),E); label("F",(30/11,0),S); label("G",(-30/11,0),S); [/asy]$
It is known that $\overline{AH}$ has length $12$ and $\overline{BC}$ has length $10$, so triangle $\triangle ABC$ has area $60$. Also, let $x$ be equal to the radius of the cylinder.
Triangles $\triangle DBG$ and $\triangle ECF$ can be combined into one triangle with base $10-2x$ and height $2x$. The area of this new triangle is $\frac{2x(10-2x)}{2} = 2x(5-x)$.
Triangle $\triangle ADE$ has base $2x$ and height $12-2x$, so its area is$\frac{2x(12-2x)}{2} = 2x(6-x)$.
Finally, square $DEFG$ has area $4x^2$.
Now we can construct an equation to find $x$:
$$2x(5-x) + 2x(6-x) + 4x^2 = 60$$ $$\Rightarrow 2x(5-x+6-x+2x) = 60$$ $$\Rightarrow 2x(11) = 60$$ $$\Rightarrow 22x = 60$$ $$\Rightarrow x = \frac{60}{22} = \boxed{(B) \frac{30}{11}}$$
~Dreamer1297
## Trivia
This problem appeared in AoPS's Introduction to Geometry as a challenge problem.
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# How is the Gravitational Force Related to the Distance Between Two Objects?
Gravitational force is one of the four fundamental forces in nature that governs the interactions between all objects with mass or energy. It is the force that keeps the planets in orbit around the sun, the moon around the earth, and the stars in the galaxy. But how does this force depend on the distance between two objects? In this article, we will explore the answer to this question using Newton’s law of gravitation.
## Newton’s Law of Gravitation
Newton’s law of gravitation states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The mathematical expression for this law is:
$$F_g = \frac{Gm_1m_2}{r^2}$$
where $F_g$ is the gravitational force, $m_1$ and $m_2$ are the masses of the two objects, $r$ is the distance between their centers, and $G$ is the universal gravitational constant, which has a value of approximately $6.67 \times 10^{-11} Nm^2/kg^2$.
This formula can be used to calculate the gravitational force between any two objects, such as the earth and the sun, or an apple and the earth. For example, if we want to find the gravitational force between an apple of mass 0.1 kg and the earth of mass $5.97 \times 10^{24} kg$, and we assume that the distance between their centers is equal to the radius of the earth, which is about $6.37 \times 10^6 m$, we can plug these values into the formula and get:
$$F_g = \frac{(6.67 \times 10^{-11})(0.1)(5.97 \times 10^{24})}{(6.37 \times 10^6)^2}$$
$$F_g = 0.98 N$$
This means that the earth pulls the apple with a force of 0.98 newtons, which is equal to its weight.
## The Inverse-Square Law
One important feature of Newton’s law of gravitation is that it follows an inverse-square law, which means that the gravitational force decreases as the square of the distance increases. This means that if we double the distance between two objects, the gravitational force becomes four times weaker; if we triple the distance, it becomes nine times weaker; and so on.
To see why this makes sense, imagine a point source of light that emits light rays in all directions. The light rays form a spherical surface around the source, and as they move away from it, they spread out over a larger area. The intensity of light at any point on this surface is proportional to how much light passes through a unit area perpendicular to the rays. Since the area of a sphere is proportional to its radius squared, we can see that as the radius increases, the intensity decreases as its inverse square.
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# How do I find the integral int1/((w-4)(w+1))dw ?
Jul 28, 2014
$= \frac{1}{5} \ln \left(\frac{w - 4}{w + 1}\right) + c$, where $c$ is a constant
Explanation :
This type of question usually solve by using Partial Fractions,
$\frac{1}{\left(w - 4\right) \left(w + 1\right)}$, it can be written as
$\frac{1}{\left(w - 4\right) \left(w + 1\right)} = \frac{A}{w - 4} + \frac{B}{w + 1}$
multiplying by $\left(w - 4\right) \left(w + 1\right)$ on both sides, we get
$1 = A \left(w + 1\right) + B \left(w - 4\right)$
$1 = \left(A + B\right) w + \left(A - 4 B\right)$
Now comparing coefficient of $w$ and constants both sides, we get
$A + B = 0$ $\implies$ $A = - B$ ...........$\left(i\right)$
$A - 4 B = 1$ ..............$\left(i i\right)$
Substituting value of $A$ from $\left(i\right)$ to $\left(i i\right)$, we get
$- 5 B = 1$ $\implies$ $B = - \frac{1}{5}$
from $B$, we can easily calculate $A$, which will be $\frac{1}{5}$
Now,
$\frac{1}{\left(w - 4\right) \left(w + 1\right)} = \frac{1}{5 \left(w - 4\right)} - \frac{1}{5 \left(w + 1\right)}$
Integrating both side with respect to $w$,
$\int \frac{1}{\left(w - 4\right) \left(w + 1\right)} \mathrm{dw} = \int \frac{1}{5 \left(w - 4\right)} \mathrm{dw} - \int \frac{1}{5 \left(w + 1\right)} \mathrm{dw}$
$= \frac{1}{5} \left(\ln \left(w - 4\right) - \ln \left(w + 1\right)\right) + c$, where $c$ is a constant
$= \frac{1}{5} \ln \left(\frac{w - 4}{w + 1}\right) + c$, where $c$ is a constant
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Collection of recommendations and tips
# What does the solution of an equation represent?
## What does the solution of an equation represent?
A solution is an assignment of values to the unknown variables that makes the equality in the equation true. In other words, a solution is a value or a collection of values (one for each unknown) such that, when substituted for the unknowns, the equation becomes an equality.
## What is the purpose of an equation?
An equation is a mathematical expression that contains an equals symbol. Equations often contain algebra. Algebra is used in maths when you do not know the exact number in a calculation. Many professionals use equations every day, including air traffic controllers, architects, computer programmers and carpenters.
What is equation in math definition?
An equation is simply a statement in math in which two things on equal. There are two expressions, one on each side of an equals sign. For example: X = 7.
### Which point represents the solution to the system of equations?
The solution to a system of linear equations is the point which lies on both lines. In other words, the solution is the point where the two lines intersect.
### What is the main purpose of the equation in economics?
The equation of exchange has two primary uses. It represents the primary expression of the quantity theory of money, which relates changes in the money supply to changes in the overall level of prices. Additionally, solving the equation for M can serve as an indicator of the demand for money in a macroeconomic model.
What equation represents a proportional relationship?
y = kx
A proportional relationship between a quantity y and a quantity x that has a constant of proportionality k is represented by the equation y = kx. If an equation in a different form can be rewritten as above, then it is a proportional relationship.
## What equation represents a direct variation?
A direct-variation relationship can be represented with an equation of the form y = kx, where k is the constant of variation. Because a direct variation is a proportional relationship, using an equation is an alternate strategy for solving a proportion.
## What is equation and expression?
An expression is a number, a variable, or a combination of numbers and variables and operation symbols. An equation is made up of two expressions connected by an equal sign.
What is equation in maths with example?
An equation is simply defined as mathematical statements that express the relationship between two values. Usually, the two values are equated by an equal sign in an equation. For example, 2x+3 = 7 is an equation, where 2x+3 and 7 are equated by equal to “=” sign.
### What is the solution of the system of equations?
A solution to a system of equations is a set of values for the variable that satisfy all the equations simultaneously. In order to solve a system of equations, one must find all the sets of values of the variables that constitutes solutions of the system.
### What does determine the solution mean?
To find the solution to an equation means to find the value of the variable that makes the equation true.
Which equation represents a linear function?
Linear Functions. A linear function in two variables is any equation of that may be written in the form y = mx + b where m and b are real number coefficients and x and y represent any real numbers that make up a solution.
## What is equation represents this graph?
In mathematics, a graphing linear equation represents the graph of the linear equation. We know that the linear equation is defined as an algebraic equation in which each term should have an exponents value of 1. The graph of the linear equation will always result in a straight line. One such example of a linear equation is y = mx + b.
## What is the equation for proportional relationship?
The graph of a proportional relationship is a straight line that passes through the origin. Proportional quantities can be described by the equation y = kx, where k is a constant ratio. You can tell that the relationship is directly proportional by looking at the graph.
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Get ahead in your studies with help of the Percentage Calculator Excel!
As we all have done exercises in the math class about percentages, you’ll be all ready familiar with percentage and their use. We’ll start with a quick recap of what we learned.
What is Percentage?
Percent originates in the Latin word “per centum,” which means “by one hundred.” It is a fraction of 100, which calculates by dividing the new value (part) with the initial sum (whole) and multiplying the result by 100.
Generally, the percentage is calculated as per the following:
Percentage = (Part / Whole) x 100
Example:
Here is an example to further clarify the concept. Suppose you bought 25 burgers and you gave 5 to colleagues. Then the percentage will be calculated as shown below:
Percentage = 5 / 25 x 100
Percentage = 0.20 x 100
Percentage = 20%
We mainly calculate percentages using this method. But it gets even more straightforward when we are doing the same calculations in Microsoft Excel as some calculations are done automatically.
Calculating percentage in Microsoft Excel: Percentage Calculator Formula Excel
There is no single formula for the calculation of percentage in Excel. Percentage calculator Excel will help you carry out the complex calculations within seconds. Everybody uses a different formula that best fits their needs. It implies that the percentage calculator formula Excel has multiple variations.
The base percentage calculation formula differs from the formulas utilized by Excel. When you compare it to the basic maths formula for percentage, the product of part divided by the whole is multiplied by 100.
On the other hand, in Excel, the 100 is replaced by the percentage. So, when you are calculating in Excel or using percentage calculator Excel, you don’t have to multiply the result by 100; this is done automatically done when the percentage format is applied in the required cell.
To further clarify it, I’ll add an example.
Suppose you have to find the number of sold products out of the total number of products. You will write total products in Column A on the Excel sheet and the sold products in Column B. To find the percentage of sold products, follow these steps:
1. Start by entering the formula B2/A2 in cell C2.
2. This formula can be copied in as many rows as you’d like.
3. Now, click the Percentage Style button.
4. Excel will display the final percentage of sold products.
This is how the percentage calculator Excel delivers results by employing the percentage calculator formula Excel.
percentagecalculatorfree.com offers a range of free but highly reliable online calculators. We aim to make calculations easier for you. So, we have developed calculators in 4 categories, i.e., financial, math, chemistry, and others.
Our calculators can help you solve all of your calculation-related problems and provide you with the most accurate and precise calculations. You can find various percentage calculators on our website, including percentage calculator fraction, percentage calculator increase and decrease, and number to percentage calculator.
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Solve the following equation $\dfrac{{x - 1}}{2} - x + 13 = 5 - x$
Last updated date: 14th Sep 2024
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Hint: Rearrange the given equation such that the unknown terms arranged on one side and all other terms are arranged on other side and then compare both sides and with further simplification we can determine the value of unknown terms.
Given: $\dfrac{{x - 1}}{2} - x + 13 = 5 - x$
To find: the value of ‘x’
Step 1: Firstly we will determine the number of unknown terms in the given equation.
$\dfrac{{x - 1}}{2} - x + 13 = 5 - x$
So here we have only one unknown term so we require only one equation to determine the value of ‘x’
Step 2: multiply both sides of the equation with 2 such that all the denominator part get reduced and we get a simple linear equation
$\dfrac{{x - 1}}{2} - x + 13 = 5 - x$
$2 \times (\dfrac{{x - 1}}{2} - x + 13) = (5 - x) \times 2$
Opening the bracket and we get
$(x - 1) - 2 \times x + 2 \times 13 = 5 \times 2 - x \times 2$
Step 3: rearranging the terms such that the unknown term arranged on one side and all other terms are arranged on other side, that is
$(x - 1) - 2 \times x + 2 \times 13 = 5 \times 2 - x \times 2$
$x - 2x - 1 + 26 = 10 - 2x$
$x - 2x + 2x = 10 + 1 - 26$
On further simplification, we get
$x - 2x + 2x = 10 + 1 - 26$
$x = - 15$
Hence, on solving the given equation we determined the value of ‘x’ and it is equal to $x = - 15$
Note: We have different solution methods for different types of equation
We can use the substitution method
We can use the elimination method without multiplication.
We can use the elimination method with multiplication.
There might be the possibility of infinite solution or no solution.
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# What is 43/2 as a mixed number?
Trying to find out how to convert 43/2 into a mixed number or fraction? Have I got the answer for you! In this guide, we'll walk you through the step-by-step process of converting an improper fraction, in this case 43/2, to a mixed number. Read on!
Want to quickly learn or show students how to convert 43/2 to a mixed number? Play this very quick and fun video now!
Before we begin, let's revisit some basic fraction terms so you understand exactly what we're dealing with here:
• Numerator. This is the number above the fraction line. For 43/2, the numerator is 43.
• Denominator. This is the number below the fraction line. For 43/2, the denominator is 2.
• Improper fraction. This is a fraction where the numerator is greater than the denominator.
• Mixed number. This is a way of expressing an improper fraction by simplifying it to whole units and a smaller overall fraction. It's an integer (whole number) and a proper fraction.
Now let's go through the steps needed to convert 43/2 to a mixed number.
## Step 1: Find the whole number
We first want to find the whole number, and to do this we divide the numerator by the denominator. Since we are only interested in whole numbers, we ignore any numbers to the right of the decimal point.
43/2= 21.5 = 21
Now that we have our whole number for the mixed fraction, we need to find our new numerator for the fraction part of the mixed number.
## Step 2: Get the new numerator
To work this out we'll use the whole number we calculated in step one (21) and multiply it by the original denominator (2). The result of that multiplication is then subtracted from the original numerator:
43 - (2 x 21) = 1
## Step 3: Our mixed fraction
We've now simplified 43/2 to a mixed number. To see it, we just need to put the whole number together with our new numerator and original denominator:
21 1 / 2
## Step 4: Simplifying our fraction
In this case, our fraction (1/2) can be simplified down further. In order to do that, we need to calculate the GCF (greatest common factor) of those two numbers. You can use our handy GCF calculator to work this out yourself if you want to. We already did that, and the GCF of 1 and 2 is 1.
We can now divide both the new numerator and the denominator by 1 to simplify this fraction down to its lowest terms.
1/1 = 1
2/1 = 2
When we put that together, we can see that our complete answer is:
21 1 / 2
Hopefully this tutorial has helped you to understand how to convert any improper fraction you have into a mixed fraction, complete with a whole number and a proper fraction. You're free to use our calculator below to work out more, but do try and learn how to do it yourself. It's more fun than it seems, I promise!
## Improper Fraction to Mixed Number
Enter an improper fraction numerator and denominator
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# RD Sharma Class 12 Ex 5.1 Solutions Chapter 5 Algebra of Matrices
Here we provide RD Sharma Class 12 Ex 5.1 Solutions Chapter 5 Algebra of Matrices for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 5.1 Solutions Chapter 5 Algebra of Matrices book pdf download. Now you will get step-by-step solutions to each question.
## RD Sharma Class 12 Ex 5.1 Solutions Chapter 5 Algebra of Matrices
### Question 1: If a matrix has 8 elements, what are the possible orders it can have? What if it has 5 elements?
Solution:
Part 1:
Let the matrix be of mxn dimension.
Hence, mxn = 8.
Then, all we have to do is, find the divisors of 8, which are : 1, 2, 4, 8.
Thus, the possible orders are: 1×8, 8×1, 2×4 and 4×2.
Part 2:
Following a similar approach:
Since, mxn = 5.
The divisors of 5 are : 1,5.
Thus, the possible orders are: 1×5 and 5×1.
### Question 2(i): If A = [aij] = and B = [bjj] = then find a22+b21.
Solution:
We know that every element in a matrix A of dimensions mxn can be addressed as aij, where 1≤ i ≤ m and 1 ≤ j ≤ n.
Thus, a22 is the 2nd element in the 2nd row of A, and b21 is the 1rst element in the 2nd row of B.
That implies, a22 + b21 = 4 + (-3) = 1.
### Question 2(ii): If A = [aij]= and B = [bij] = then find a11b11 + a22b22.
Solution:
As seen in the previous question, every element in a matrix A of dimensions mxn can be addressed as aij, where 1≤ i ≤ m and 1 ≤ j ≤ n.
Thus,
a11 = 1rst element in the 1rst row of A = 2.
a22 = 2nd element in the 2nd row of A = 4.
b11 = 1rst element in the 1rst row of B = 2.
b22 = 2nd element in the 2nd row of B = 4.
That implies, a11b11 + a22b22 = (2×2) + (4×4) = 4 + 16 = 20.
### Question 3: Let A be a matrix of order 3×4. If R1 denotes the first row of A and C2 denotes its second column, then determine the orders of matrices R1 and C2.
Solution:
Given, is a matrix of order 3×4.
We know that a matrix having an order of mxn has m rows and n columns.
Thus, contains, 3 rows, and each row contains 4 elements.
Now, if R1 is a row, it has 4 elements, and thus its order can be written as 1×4,
And similarly, if C2 is a column, it will have 3 rows, each with 1 element, and thus its order is 3×1,
### Question 4(i): Construct a 2×3 matrix A = [aij] whose elements aij are given by : aij = i x j.
Solution:
We know that A is a matrix of order 2×3.
Thus A can be depicted as:
Since each element is a product of its row number and column number (i x j):
a11 = 1 a12 = 2 a13 = 3
a21 = 2 a22 = 4 a23 = 6
Thus, A can be depicted as :
### Question 4(ii): Construct a 2×3 matrix A = [aij] whose elements aij are given by : aij = 2i – j.
Solution:
We know that A is a matrix of order 2×3.
Thus A can be depicted as:
Since each element can be defined as (2 x (row number)) – column number:
a11 = 1 a12 = 0 a13 = -1
a21 = 3 a22 = 2 a23 = 1
Thus, A can be depicted as :
### Question 4(iii): Construct a 2×3 matrix A = [aij] whose elements aij are given by : aij = i + j.
Solution:
We know that A is a matrix of order 2×3.
Thus A can be depicted as:
Since each element can be defined as the sum of its row number and column number:
a11 = 2 a12 = 3 a13 = 4
a21 = 3 a22 = 4 a23 = 5
Thus, A can be depicted as :
### Question 4(iv): Construct a 2×3 matrix A = [aij] whose elements aij are given by : aij = .
Solution:
We know that A is a matrix of order 2×3.
Thus A can be depicted as:
Since each element can be defined as : ,
a11 = 2 a12 = 4.5 a13 = 8
a21 = 4.5 a22 = 8 a23 = 12.5
Thus, A can be depicted as :
### Question 5(i): Construct a 2×2 matrix A = [aij] whose aij are given by : .
Solution:
We know that A is a matrix of order 2×2.
Thus A can be depicted as :
Since each element can be defined as :,
a11 = 2 a12 = 4.5
a21 = 4.5 a22 = 8
Thus, A can be depicted as :
### Question 5(ii): Construct a 2×2 matrix A = [aij] whose aij are given by : .
Solution:
We know that A is a matrix of order 2×2.
Thus A can be depicted as :
Since each element can be defined as : ,
a11 = 0 a12 = 0.5
a21 = 0.5 a22 = 0
Thus, A can be depicted as :
### Question 5(iii): Construct a 2×2 matrix A = [aij] whose aij are given by : .
Solution:
We know that A is a matrix of order 2×2.
Thus A can be depicted as :
Since each element can be defined as : ,
a11 = 0.5 a12 = 4.5
a21 = 0 a22 = 2
Thus, A can be depicted as :
### Question 5(iv): Construct a 2×2 matrix A = [aij] whose aij are given by : .
Solution:
We know that A is a matrix of order 2×2.
Thus A can be depicted as :
Since each element can be defined as : ,
a11 = 4.5 a12 = 8
a21 = 12.5 a22 = 18
Thus, A can be depicted as :
### Question 5(v): Construct a 2×2 matrix A = [aij] whose aij are given by : .
Solution:
We know that A is a matrix of order 2×2.
Thus A can be depicted as : ,
Since each element can be defined as : ,
a11 = 0.5 a12 = 2
a21 = 0.5 a22 = 1
Thus, A can be depicted as :
### Question 5(vi): Construct a 2×2 matrix A = [aij] whose aij are given by : .
Solution:
We know that A is a matrix of order 2×2.
Thus A can be depicted as : ,
Since each element can be defined as : ,
a11 = 1 a12 = 0.5
a21 = 2.5 a22 = 2
Thus, A can be depicted as :
### Question 5(vii) Construct a 2×2 matrix A = [aij] whose aij are given by : .
Solution:
We know that A is a matrix of order 2×2.
Thus A can be depicted as : ,
Since each element can be defined as : ,
a11 = e2xsinx a12 = e2xsin2x
a21 = e4xsinx a22 = e4xsin2x
Thus, A can be depicted as :
### Question 6(i): Construct a 3×4 matrix A = [aij] whose aij are given by : aij = i + j .
Solution:
A is a matrix of order 3×4.
Thus, A can be depicted as : ,
Since each element can be defined as : ( row number + column number ),
a11 = 2 a12 = 3 a13 = 4 a14 = 5
a21 = 3 a22 = 4 a23 = 5 a24 = 6
a31 = 4 a32 = 5 a33 = 6 a34 = 7
Thus, A can be depicted as :
### Question 6(ii): Construct a 3×4 matrix A = [aij] whose aij are given by : aij = i – j.
Solution:
A is a matrix of order 3×4.
Thus, A can be depicted as : ,
Since each element can be defined as : (row number – column number),
a11 = 0 a12 = -1 a13 = -2 a14 = -3
a21 = 1 a22 = 0 a23 = -1 a24 = -2
a31 = 2 a32 = 1 a33 = 0 a34 = -1
Thus, A can be depicted as :
### Question 6(iii): Construct a 3×4 matrix A = [aij] whose aij are given by : aij = 2i .
Solution:
A is a matrix of order 3×4.
Thus, A can be depicted as : ,
Since each element can be defined as : (2 x row number ),
a11 = 2 a12 = 2 a13 = 2 a14 = 2
a21 = 4 a22 = 4 a23 = 4 a24 = 4
a31 = 6 a32 = 6 a33 = 6 a34 = 6
Thus, A can be depicted as :
### Question 6(iv): Construct a 3×4 matrix A = [aij] whose aij are given by : aij = j.
Solution:
A is a matrix of order 3×4.
Thus, A can be depicted as : ,
Since each element can be defined as : (column number ),
a11 = 1 a12 = 2 a13 = 3 a14 = 4
a21 = 1 a22 = 2 a23 = 3 a24 = 4
a31 = 1 a32 = 2 a33 = 3 a34 = 4
Thus, A can be depicted as :
### Question 6(v): Construct a 3×4 matrix A = [aij] whose aij are given by : aij = .
Solution:
A is a matrix of order 3×4.
Thus, A can be depicted as : ,
Since each element can be defined as : ,
a11 = -1 a12 = -1/2 a13 = 0 a14 = 1/2
a21 = -5/2 a22 = -2 a23 = -3/2 a24 = -1
a31 = -4 a32 = -7/2 a33 = -3 a34 = -5/2
Thus, A can be depicted as :
### Question 7(i): Construct a 4×3 matrix A = [aij] whose aij are given by : aij = .
Solution:
A is a matrix of order 4×3.
Thus, A can be depicted as :
Since each element can be defined as : ,
a11 = 3 a12 = 5/2 a13 = 7/3
a21 = 6 a22 = 5 a23 = 14/3
a31 = 9 a32 = 15/2 a33 = 7
a41 = 12 a42 = 10 a43 = 28/3
Thus, A can be depicted as :
### Question 7(ii): Construct a 4×3 matrix A = [aij] whose aij are given by : aij = .
Solution:
A is a matrix of order 4×3.
Thus, A can be depicted as:
Since each element can be defined as : ,
a11 = 0 a12 = -1/3 a13 = -1/2
a21 = 1/3 a22 = 0 a23 = -1/5
a31 = 1/2 a32 = 1/5 a33 = 0
a41 = 3/5 a42 = 1/3 a43 = 1/7
Thus, A can be depicted as :
### Question 7(iii): Construct a 4×3 matrix A = [aij] whose aij are given by : aij = i.
Solution:
A is a matrix of order 4×3.
Thus, A can be depicted as :
Since each element can be defined as : (row number)
a11 = 1 a12 = 1 a13 = 1
a21 = 2 a22 = 2 a23 = 2
a31 = 3 a32 = 3 a33 = 3
a41 = 4 a42 = 4 a43 = 4
Thus, A can be depicted as :
### Question 8: Find x, y, a and b if:
Solution:
We can see that both, the matrix on the Left Hand Side (LHS) and the matrix on the Right Hand Side (RHS are of the dimension 2×3.
Since, the matrix on LHS is equal to the matrix on the RHS, each element on the LHS at the index (i, j) must be equal to each element on the RHS at the index (i, j).
Hence, equating each element on RHS to LHS:
a11: 3x+4y = 2 ………………(eq.1) a12: 2 = 2 a13: x-2y = 4 ………………(eq.2)
a21: a+b = 5 ………………(eq.3) a22: 2a-b = -5……………..(eq.4) a23: -1 = -1
Thus, (eq.1) and (eq.2) form one system of equations comprising variables x and y.
Solving (eq.1) and (eq.2): (eq.1) + 2x(eq.2)
=> (3x+2x) + (4y-2(2y)) = 2+ (2(4))
=> 5x = 10
=> x = 2
Substituting (x=2) in (eq.1) :
=> (3(2)) + 4y = 2
=> 4y = 2-6 = -4
=> y=-1
Similarly, (eq.3) and (eq.4) form a system of equations comprising variables a and b.
Solving (eq.3) and (eq.4) : (eq.1) + (eq.2)
=> (a+2a) + (b-b) = 5 – 5
=> 3a = 0
=> a = 0
Substituting (a=0) in (eq.3):
=> 0 + b = 5
=>b = 5
Thus, a=0, b=5, x=2 and y=-1.
### Question 9: Find x, y, a and b if : .
Solution:
We can see that both, the matrix on the Left Hand Side (LHS) and the matrix on the Right Hand Side (RHS are of the dimension 2×3.
Since, the matrix on LHS is equal to the matrix on the RHS, each element on the LHS at the index (i, j) must be equal to each element on theRHS at the index (i, j).
Hence, equating each element on RHS to LHS:
a11 : 2x-3y = 1………………(eq.1) a12 : a-b = -2………………(eq.2) a13 : 3 = 3
a21 : 1 = 1 a22 : x+4y = 6……………..(eq.3) a23 : 3a+4b = 29………………(eq.4)
Thus, (eq.1) and (eq.3) form one system of equations comprising of variables x and y.
Solving (eq.1) and (eq.2) : (eq.1) – 2x(eq.2)
=> (2x-2x) + (-3y-2(4y)) = 1- (2(6))
=> -11y = -11
=> y = 1
Substituting (y=1) in (eq.1) :
=> 2x – 3(1) = 1
=> 2x = 3+1 = 4
=> x = 2
Similarly, (eq.2) and (eq.4) form a system of equations comprising of variables a and b.
Solving (eq.2) and (eq.4) : 4x(eq.1) + (eq.2)
=> (4a+3a) + (-4(b)+4b) = 4(-2) + 29
=> 7a = 21
=> a = 3
Substituting (a=3) in (eq.2) :
=> 3 – b = -2
=>b = 5
Thus, a=3, b=5, x=2 and y=1.
### Question 10: Find a, b, c and d if :
Solution:
We can see that both, the matrix on the Left Hand Side (LHS) and the matrix on the Right Hand Side (RHS are of the dimension 2×2.
Since, the matrix on LHS is equal to the matrix on the RHS, each element on the LHS at the index (i, j) must be equal to each element on the RHS at the index (i, j).
Hence, equating each element on RHS to LHS:
a11 : 2a+b = 4 ………….(eq.1)
a12 : a-2b = -3 …………(eq.2)
a21 : 5c-d = 11 …………(eq.3)
a22 : 4c+3d = 24 ……..(eq.4)
Thus, (eq.1) and (eq.2) form one system of equations comprising of variables a and b.
Solving (eq.1) and (eq.2) : (eq.1) – 2x(eq.2)
=> (2a-2a) + (b+4b) = 4 + (-2(-3))
=> 5b = 10
=> b = 2
Substituting (b=2) in (eq.1) :
=> 2a + 2 = 4
=> 2a = 4-2 = 2
=> a=1
Similarly, (eq.3) and (eq.4) form a system of equations comprising of variables c and d.
Solving (eq.3) and (eq.4) : 3x(eq.1) + (eq.2)
=> (15c+4c) + (-3d+3d) = 33 + 24
=> 19c = 57
=> c = 3
Substituting (c=3) in (eq.4) :
=> (4(3)) + 3d = 24
=>3d = 24 – 12 = 12
=>d = 4
Thus, a=1, b=2, c=3 and d=4.
Question 11: Find x, y and z so that A = B, where A= .
Solution:
As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).
Thus,
a11 : x-2 = y ………….(eq.1)
a12 : z = 3 ………….(eq.2)
a13 : 2z = 6 ………….(eq.3)
a21 : 18z = 6y ……….(eq.4)
a22 : y+2 = x ………..(eq.5)
a23 : 6z = 2y …………(eq.6)
From (eq.2) and (eq.3),
=> z = 3
From (eq.4) and (eq.6),
=> y = 3z
=> y = 3(3)
=> y = 9
Substitute ( y=9 ) in (eq.1),
=> x-2 = 9
=>x = 11
Thus x=11, y=9 and z=3.
Question 12: If , find x, y, z and w.
Solution:
As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).
Thus,
a11 : x = 3 ………….(eq.1)
a12 : 3x-y = 2 ………….(eq.2)
a21 : 2x+z = 4 ………..(eq.3)
a22 : 3y-w = 7 ……….(eq.4)
From (eq.1) ,
=> x = 3
Substitute (x=3) in (eq.2),
=> 3(3)-y = 2
=> y = 3(3) – 2
=> y = 7
Substitute ( x=3 ) in (eq.3),
=> 2(3)+z = 4
=>z = 4-6
=>z = -2
Substitute (y=7) in (eq.4),
=>3(7) – w = 7
=>w = 21 – 7
=>w = 14
Thus x=3, y=7, z=-2 and w=14.
Question 13: If , find x, y, z and w.
As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).
Thus,
a11 : x-y = -1 ………….(eq.1)
a12 : z = 4 ………….(eq.2)
a21 : 2x-y = 0 ………..(eq.3)
a22 : w = 5 …………….(eq.4)
From (eq.2),
=> z = 4
And from (eq.4),
=> w = 5
Now, (eq.1) and (eq.3) form a system of equations comprising of variables x and y.
Thus, (eq.1) – (eq.2),
=> (x-2x) +(-y+y) = -1 – 0
=> -x = -1
=> x = 1
Substitute (x=1) in (eq.1),
=> 1-y = -1
=>y = 2
Thus, x=1, y=2, z=4 and w=5.
Question 14: If . Obtain the values of a, b, c, x, y and z.
Solution:
As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).
Thus,
a11 : x+3 = 0 ……………..(eq.1)
a12 : z+4 = 6 ………….(eq.2)
a13 : 2y-7 = 3y-2 ……..(eq.3)
a21 : 4x+6 = 2x ………….(eq.4)
a22 : a-1 = -3 ……………..(eq.5)
a23 : 0 = 2c+2 …………….(eq.6)
a31 : b-3 = 2b+4…………(eq.7)
a32 : 3b = -21………………(eq.8)
a33 : z+2c = 0………………(eq.9)
From (eq.1) and (eq.4),
=> x = -3
From (eq.2) ,
=> z+4 = 6
=> z = 2
From (eq.3),
=> 2y-7 = 3y-2
=> 3y-2y = -7+2
=> y = -5
From (eq.5),
=> a = -3+1
=> a = -2
From (eq.8),
=> 3b = -21
=> b = -7
Substitute (z=2) in (eq.9),
=> 2+ 2c = 0
=> c = -1
Thus, x=-3, y=-5, z=2, a=-2, b=-7 and c=-1.
Question 15: If , find the value of (x+y).
Solution:
As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).
Thus,
a11 : 2x+1 = x+3………..(eq.1)
a12 : 5x = 10 ………………(eq.2)
a21 : 0 = 0 …………………(eq.3)
a22 : y2+1 = 26 …………(eq.4)
From (eq.1) and (eq.2),
=> 2x+1 = x+3
=> x=2
From (eq.4),
=> y2+1 = 26
=> y2 = 25
=> y = ± 5
Thus if y = +5,
=> x+y = 7
And if y = -5,
=> x+y = -3
Question 16: If , then find the values of x, y, z and w.
Solution:
As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).
Thus,
a11 : xy = 8 ……………..(eq.1)
a12 : 4 = w ………………(eq.2)
a21 : z+6 = 0 …………..(eq.3)
a22 : x+y = 6 …………..(eq.4)
From (eq.2),
=> w = 4
And from (eq.3),
=> z = -6
Now, we can see that (eq.1) and (eq.4) form a system of equations comprising of variables x and y.
From (eq.1),
=> x = 8/y
Substitute (x=8/y) in (eq.4):
=> y + (8/y) = 6
=> y2 – 6y +8 = 0
Solving the above equation,
=> y2 – 4y – 2y + 8 = 0
=> y( y-4 ) -2( y-4 ) = 0
=> (y – 2)(y – 4) = 0
=> y = 2 or 4
Substitute in (eq.1):
=> when x=2, y=4 and when x=4, y=2.
Thus, (x,y) = (2,4) or (4,2) and z = -6 and w = 4.
Question 17(i): Give an example of a row matrix which is also a column matrix.
Solution:
We know the order of a row matrix can be written as 1xn (1 row with n elements).
And similarly, the order of a column matrix is mx1.
So, a row matrix which is also a column matrix must be of the order (1×1).
As an example, we can take the matrix :
Question 17(ii): Give an example of a diagonal matrix which is not scalar.
Solution:
In a diagonal matrix, only the diagonal elements possess non-zero values. Thus, for a nxn diagonal matrix, aii ≠ 0, for 1≤ i ≤ n.
And a scalar matrix is a diagonal matrix, such that all the diagonal elements are equal.
Thus, a matrix which is diagonal but not scalar is:
Question 17(iii): Give an example of a triangular matrix.
Solution:
A triangular matrix is a square matrix, and it is filled in such a way that, either the triangle above the main-diagonal is non-zero (upper-triangular) or the triangle below the diagonal is non-zero (lower-triangular).
Thus, an example would be :
Question 18: The sales figure of two car dealers during January 2013 showed that dealer A sold 5 deluxe, 3 premium and 4 standard cars, while dealer B sold 7 deluxe, 2 premium and 3 standard cars. Total sales over the month period of January- February revealed that dealer A sold 8 deluxe, 7 premium and 6 standard cars. In the same 2 month period, dealer B sold 10 deluxe, 5 premium and 7 standard cars. Write 2×3 matrices summarizing sales data for January and 2-month period for each dealer.
Solution:
The above data can be represented in the form of tables:
For January 2013,
Deluxe
Standard
Dealer A
5
3
4
Dealer B
7
2
3
For January to February,
Deluxe
Standard
Dealer A
8
7
6
Dealer B
10
5
7
Thus, the two matrices are : and
Question 19: For what value of x and y are the following matrices equal?
A= and B =
Solution:
As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).
Thus,
a11 : 2x+1 = x+3 …………(eq.1)
a12 : 2y = y2+2 ……………(eq.2)
a21 : 0 = 0 …………………….(eq.3)
a22 : y2-5y = -6 …………..(eq.4)
From (eq.1),
=> 2x-x = 3-1
=> x=2
Taking (eq.2), it can be re-written as,
=> y2-2y+2 = 0
=> y =
=> y = -1 ± i ( No real solutions )
Taking (eq.4), it can be re-written as,
=> y2 – 5y +6 = 0
Solving the equation,
=> y2 – 2y – 3y +6 = 0
=> y( y-2 ) -3( y-2 ) = 0
=> ( y-2 )( y-3 ) = 0
=> y = 2 or 3
As values of y are inconsistent, we can say that the above matrices are not equal for any (x,y) pair.
Question 20. Find the values of x and y if .
Solution:
As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).
Thus,
a11 : x+10 = 3x+4 …………(eq.1)
a12 : y2+2y = 3 ……………(eq.2)
a21 : 0 = 0 ……………………..(eq.3)
a22 : y2-5y = -4 …………..(eq.4)
From (eq.1),
=> 2x = 6
=> x = 3
Taking (eq.2), it can be re-written as,
=> y2+2y-3 = 0
=> y+ 3y -y -3 = 0
=> y( y+3 ) -1( y+3 ) = 0
=> ( y+3 )( y-1 ) = 0
=> y = -3 or 1
Taking (eq.3), it can be re-written as,
=> y2-5y+4 = 0
=> y2 -4y -y +4 =0
=> y( y-4 ) -1( y-4 ) = 0
=> ( y-4 )( y-1 ) = 0
=> y = 4 or 1
The value of y that can satisfy both (eq.2) and (eq.3) is 1.
=> y=1
Thus, x=3 and y=1.
Question 21. Find the values of a and b if A = B , where A= , B= .
Solution:
As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).
Thus,
a11 : a+4 = 2a+2 …………..(eq.1)
a12 : 3b = b2+2 …………..(eq.2)
a21 : 8 = 8 ……………………..(eq.3)
a22 : -6 = b2-5b ………….(eq.4)
From (eq.1),
=> a = 2
Taking (eq.2), it can be re-written as,
=> b2-3b+2=0
=> b2 – 2b -b +2 = 0
=> b(b-2) -1(b-2) = 0
=> (b-2)(b-1) = 0
=> b=1 or 2
Taking (eq.4) it can be re-written as,
=> b2 -5b +6 = 0
=> b2 – 3b -2b + 6 = 0
=> b( b-3 ) -2( b-3 ) = 0
=> ( b-3 )( b-2 ) = 0
=> b = 2 or 3
Thus, b=2 can satisfy both (eq.2) and (eq.4).
=> b = 2
Thus, a=2 and b=2.
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Chapters
We use fractions to explain the concepts of ratio and proportion. A fraction that can be written in the form of x : y is known as a ratio. On the other hand, when two ratios are set equal to each other, we get a proportion.
In the above ratio, x and y are integers. The concept of ratio and proportion is widely used in mathematics, science, and in our daily lives. In our everyday life, we use ratio and proportion to analyze our financial matters. Similarly, whenever we try a new recipe, we ensure to use the appropriate proportion of the ingredients.
We can simplify the ratio. For example, consider a ratio 35 : 25. Both these integers are multiples of 5, hence we can write the ratio as 7 : 5. A ratio has no unit because both the quantities involved in it have similar units. These same units cancel each other. Remember that a ratio describes the relationship between two quantities, let say x and y, where y is not equal to zero.
## Defining Ratio and Proportion
In this section, we will discuss the formal definitions of ratio and proportion. Let us define the ratio first:
A ratio shows the relationship between two values
Or
A ratio describes the relative sizes of two or more quantities
The ratio has many applications in our everyday life. For example, the speed of an airplane is equal to the distance divided by time. It means that the speed is equal to the ratio of distance and time.
A proportion can be defined as:
The equation which says that two ratios are equal to each other is known as proportion
For example, if a car travels 80 kilometers distance in one hour, then it must cover 160 kilometers distance in two hours. Mathematically, we can write it as:
You can see that we have taken the ratio of similar units. It means that 80 : 160 is one ratio that depicts the distance and 1 : 2 is another ratio that depicts the hours.
Let us now summarize the key points related to the ratios:
• Ratios only exist between similar quantities
• When we use the ratio to compare two quantities, then their units must be the same
• The order of terms in the ratio is also important
• The ratio remains unaffected if both the numbers are multiplied or divided by the same number.
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## Ratio and Proportion Formulas
Now, let us see what are some of the formulas of ratio and proportion.
### Ratio Formula
Suppose we have two quantities (or numbers) x and y and we have to find the ratio between these two quantities. The formula for finding the ratio between these quantities will be:
, where x and y are any two quantities.
Here, x is the first term, also known as an antecedent, and y is the second term also known as a consequent.
For example, what is the ratio between the numbers 6 and 24.
We can write the numbers in ratio form like this:
To find the ratio, simply divide 6 by 24:
Hence, the ratio between the numbers 6 and 24 in simplified form is 1 : 4 or .
Consider another example below:
Two numbers exist in the ratio of 4:5 and their sum is 90. Find the numbers.
Suppose the numbers are 4x and 5x respectively. Given the information in the example, we can construct the following equation:
Now, we will substitute this value of x in 4x and 5x respectively to get the numbers.
First number = 4 (10) = 40
Second number = 5 (10) = 50
Hence, the two numbers are 40 and 50.
### Proportion Formula
We know that the proportion is an equality between two ratios a:b and c:d. The terms a and d are known as extreme or extreme terms, whereas, the terms b and c are known as means or mean terms. Mathematically, we can write proportion as:
In fractional form, we can write it as:
For instance, consider the following proportion:
In the above proportion, 8 and 18 are the extreme values and 9 and 16 are mean values.
## Properties of Proportion
Some of the properties of proportion are given below:
If a : b = c : d, then a + c : b + d.
### Subtrahendo
If a : b = c : d, then a - b : b = c - d : d.
### Componendo
If a : b = c : d, then a + b : b = c + d : d.
### Alternendo
If a : b = c : d, then a : c = b : d.
### Invertendo
If a : b = c : d, then b : a : d : c.
### Componendo and dividendo
If a : b = c : d, then a + b : a - b = c + d : c - d.
## Third, Fourth and Mean Proportional
• Third proportional : If , then c is known as a third proportional to a, b, and d.
• Fourth proportional: If , then d is known as the fourth proportional to a, b and c.
• Mean proportional: The mean proportional of the ratio a:b is equal to the square root of the product of the terms a and b. It is represented by the following formula:
Mean proportional =
## Example 1
Three friends invest $2,000,$2,500, and $3,500 in a business. After a year, the business made a profit of$4000. If the money is withdrawn, how much will each individual receive if their earnings are directly proportional to the money they originally invested?
### Solution
The amount that is invested by three friends is directly proportional to the profit earned.
Amount that will be received by the first individual =
Hence, the first individual will receive 1000 dollars.
Amount received the second individual =
Hence, the second individual will receive 1250 dollars
Amount received by the third individual =
Hence, the third individual will receive 1750 dollars.
## Example 2
The sum of two numbers is 280. These numbers exist in the ratio of 6: 8. Find the numbers.
### Solution
Suppose the numbers are x and y.
The ratio between these numbers is 6:8, so we can say that the numbers are 6x and 8x respectively.
Divide both sides by 14 to get the value of x:
Now, substitute the value of x = 20 in 6x and 8x respectively:
First number = 6x = 6(20) = 120
Second number = 8x = 8 (20) = 160
Hence, the two numbers are 120 and 160.
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Review question
# The liar, the switcher and the truth-teller; who's who? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R8279
## Solution
1. A, B and C are three people. One of them is a liar who always tells lies, another is a saint who always tells the truth, and the third is a switcher who sometimes tells the truth and sometimes lies. They make the following statements:
A: I am the liar.
B: A is the liar.
C: I am not the liar.
Who is the liar among A, B and C? Briefly explain your answer.
A cannot be the liar, since then he would have told the truth, and A cannot be the saint because then he would have lied. Therefore A must be the switcher.
B is therefore lying, so is the liar. C is telling the truth and is the saint.
We can show this reasoning in a table. The top half of the table shows the six possible combinations of T (truth-telling saint), L (liar) and S (switcher). The bottom half shows which combinations become impossible at each step in the conversation.
A is B is C is T L S T S L L T S L S T S T L S L T A says A=L x x x x $\checkmark$ $\checkmark$ B says A=L – – – – x $\checkmark$ C says C$\neq$L – – – – – $\checkmark$
1. P, Q and R are three more people, one a liar, one a saint, and the third a contrarian who tells a lie if he is the first to speak or if the preceding speaker told the truth, but otherwise tells the truth. They make the following statements:
P: Q is the liar.
Q: R is the liar.
R: P is the liar.
Who is the liar among P, Q and R? Briefly explain your answer.
Again, we draw up a table, this time using C to mean contrarian.
P is Q is R is T L C T C L L T C L C T C T L C L T P says Q=L $\checkmark$ x $\checkmark$ $\checkmark$ $\checkmark$ x Q says R=L $\checkmark$ – x x $\checkmark$ – R says P=L x – – – $\checkmark$ –
The only possible combination is PQR=CTL. Let’s check that this is consistent with the statements.
P: Q is the liar. (FALSE. Consistent with P=contrarian, so lies if speaks first.)
Q: R is the liar. (TRUE. Consistent with Q=saint.)
R: P is the liar. (FALSE. Consistent with R=liar.)
So R is the liar.
1. X, Y and Z are three more people, one a liar, one a switcher and one a contrarian. They make the following statements:
X: Y is the liar.
Y: Z is the liar.
Z: X is the liar.
X: Y is the liar.
Y: X is the liar.
Who is the liar among X, Y and Z? Briefly explain your answer.
X is Y is Z is L S C L C S S L C S C L C S L C L S X says Y=L $\checkmark$ $\checkmark$ $\checkmark$ (truth) $\checkmark$ (lie) $\checkmark$ x Y says Z=L $\checkmark$ (lie) x $\checkmark$ $\checkmark$ $\checkmark$ (truth) – Z says X=L $\checkmark$ – x $\checkmark$ $\checkmark$ – X says Y=L $\checkmark$ – – $\checkmark$ (lie) x – Y says X=L $\checkmark$ – – x – –
Let’s check LSC is consistent.
X: Y is the liar. (FALSE, consistent with X=liar).
Y: Z is the liar. (FALSE, consistent with Y=switcher).
Z: X is the liar. (TRUE, consistent with Z=contrarian, previous statement was false).
X: Y is the liar. (FALSE, consistent with X=liar).
Y: X is the liar. (TRUE, consistent with Y=switcher).
So X is the liar.
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Nov10
# Learn the Signs of the Power: Positive or Negative
In this post, we’re going to decipher an answer’s sign by looking at the sign of a number’s power.
#### To start off, let’s look at an example of what happens when the factor’s signs mingle.
As we already know, we multiply the factors normally and later, we add the appropriate sign to the answer.
#### Let’s see what happens with a power.
A power is a succession of the same product, as shown in the example below.
So then, if the power has a positive sign and the exponent is even and positive as well, like in the example, the answer will be positive, too.
If the even exponent and sign are negative, the answer would also be positive. This happens because each negative sign cancels the other out (- x – = +) and they form a positive answer.
We can also see cases where the exponent is odd and positive. In such cases, we’d get a positive answer yet again.
If we take all these situations into account, we could say that the last case, an odd exponent with a negative sign, is the only exception that gives a negative answer.
This happens because the odd number of factors means that we have a negative sign that’s left out. So despite all the other pairs having a positive (+) result, the last negative number (-) makes the final answer negative.
I hope that this post helps you quickly determine the signs of all the powers that you encounter in problems.
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# 9.2: Multiplication Properties of Radicals
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Recall that the equation $$x^2 = a$$, where a is a positive real number, has two solutions, as indicated in Figure 1.
Here are the key facts.
Solution
Solutions of $$x^2 = a$$. If a is a positive real number, then:
1. The equation $$x^2 = a$$ has two real solutions.
2. The notation $$\sqrt{a}$$ denotes the unique positive real solution.
3. The notation $$−\sqrt{a}$$ denotes the unique negative real solution.
Note the use of the word unique. When we say that solution $$\sqrt{a}$$ is the unique positive real solution, we mean that it is the only one. There are no other positive real numbers that are solutions of $$x^2 = a$$. A similar statement holds for the unique negative solution.
Thus, the equations $$x^2 = a$$ and $$x^2 = b$$ have unique positive solutions $$x = \sqrt{a}$$ and $$x = \sqrt{b}$$, respectively, provide that a and b are positive real numbers. Furthermore, because they are solutions, they can be substituted into the equations $$x^2 = a$$ and $$x^2 = b$$ to produce the results
$$(\sqrt{a})^2 = a$$ and $$(\sqrt{b})^2 = b$$
respectively. Again, these results are dependent upon the fact that a and b are positive real numbers.
Similarly, the equation
$$x^2 = ab$$
has unique positive solution $$x = \sqrt{ab}$$, provided a and b are positive numbers. However, note that
$$(\sqrt{a}\sqrt{b})^2 = (\sqrt{a})^{2}(\sqrt{b})^{2} = ab$$,
making $$\sqrt{a}\sqrt{b}$$ a second positive solution of $$x^2 = ab$$. However, because $$\sqrt{ab}$$ is the unique positive solution of $$x^2 = ab$$, this forces
$$\sqrt{ab} = \sqrt{a}\sqrt{b}$$
This discussion leads to the following property of radicals.
Property 1
Let $$a$$ and $$b$$ be positive real numbers. Then,
$\sqrt{ab} = \sqrt{a}\sqrt{b}$
This result can be used in two distinctly different ways.
• You can use the result to multiply two square roots, as in
$$\sqrt{7}\sqrt{5} = \sqrt{35}$$.
• $$\sqrt{35} = \sqrt{7}\sqrt{5}$$
It is interesting to check this result on the calculator, as shown in Figure 2.
## Simple Radical Form
In this section we introduce the concept of simple radical form, but let’s first start with a little story. Martha and David are studying together, working a homework problem from their textbook. Martha arrived at an answer of $$\sqrt{32}$$, while David gets the result $$2\sqrt{8}$$. At first, David and Martha believe that their solutions are different numbers, but they’ve been mistaken before so they decide to compare decimal approximations of their results on their calculators. Martha’s result is shown in Figure 3(a), while David’s is shown Figure 3(b).
Martha finds that $$\sqrt{32} \approx 5.656854249$$ and David finds that his solution $$2\sqrt{8} \approx 5.656854249$$. David and Martha conclude that their solutions match, but they want to know why the two very different looking radical expressions are identical.
The following calculation, using Property 1, shows why David’s result is identical to Martha’s.
$$\sqrt{32}=\sqrt{4}\sqrt{8}=2\sqrt{8}$$
Indeed, there is even a third possibility, one that is much different from the results found by David and Martha. Consider the following calculation, which again uses Property 1.
$$\sqrt{32}=\sqrt{16}\sqrt{2}=4\sqrt{2}$$
In Figure 4, note that the decimal approximation of $$4\sqrt{2}$$ approximations for $$\sqrt{32}$$ (Martha’s result in Figure 3(a)) and $$2\sqrt{8}$$ (David’s result inFigure 3(b)).
While all three of these radical expressions ($$\sqrt{32}$$, $$2\sqrt{8}$$, and $$4\sqrt{2}$$) are identical, it is somewhat frustrating to have so many different forms, particularly when we want to compare solutions. Therefore, we offer a set of guidelines for a special form of the answer which we will call simple radical form.
The First Guideline for Simple Radical Form.
When possible, factor out a perfect square.
Thus, $$\sqrt{32}$$ is not in simple radical form, as it is possible to factor out a perfect square, as in
$$\sqrt{32}=\sqrt{16}\sqrt{2}=4\sqrt{2}$$.
Similarly, David’s result $$(2\sqrt{8})$$ is not in simple radical form, because he too can factor out a perfect square as follows.
$$2\sqrt{8}=2(\sqrt{4}\sqrt{2})=2(2\sqrt{2})=(2 \cdot 2)\sqrt{2}=4\sqrt{2}$$.
If both Martha and David follow the “first guideline for simple radical form,” their answer will look identical (both equal $$4\sqrt{2}$$). This is one of the primary advantages of simple radical form: the ability to compare solutions.
In the examples that follow (and in the exercises), it is helpful if you know the squares of the first 25 positive integers. We’ve listed them in the margin for you in Table 1 for future reference.
Let’s place a few more radical expressions in simple radical form.
Example $$\PageIndex{3}$$
Place $$\sqrt{50}$$ in simple radical form.
In Table 1, 25 is a square. Because $$50 = 25 \cdot 2$$, we can use Property 1 to write
$$\sqrt{50}=\sqrt{25}\sqrt{2}=5\sqrt{2}$$.
Example $$\PageIndex{4}$$
Place $$\sqrt{98}$$ in simple radical form.
In Table 1, 49 is a square. Because $$98 = 49 \cdot 2$$, we can use Property 1 to write
$$\sqrt{98}=\sqrt{49}\sqrt{2}=7\sqrt{2}$$.
Example $$\PageIndex{5}$$
Place $$\sqrt{288}$$ in simple radical form.
Some students seem able to pluck the optimal “perfect square” out of thin air. If you consult Table 1, you’ll note that 144 is a square. Because $$288 = 144 \cdot 2$$, we can write
$$\sqrt{288}=\sqrt{144}\sqrt{2}=12\sqrt{2}$$.
However, what if you miss that higher perfect square, think $$288 = 4 \cdot 72$$, and write
$$\sqrt{288}=\sqrt{4}\sqrt{72}=2\sqrt{72}$$.
This approach is not incorrect, provided you realize that you’re not finished. You can still factor a perfect square out of 72. Because $$72 = 36 \cdot 2$$, you can continue and write
$$2\sqrt{72}=2(\sqrt{36}\sqrt{2})=2(6\sqrt{2})=(2\cdot 6)\sqrt{2}=12\sqrt{2}$$.
Note that we arrived at the same simple radical form, namely $$12\sqrt{2}$$. It just took us a little longer. As long as we realize that we must continue until we can no longer factor out a perfect square, we’ll arrive at the same simple radical form as the student who seems to magically pull the higher square out of thin air.
Indeed, here is another approach that is equally valid.
$$\sqrt{288}=\sqrt{4}\sqrt{72}=2(\sqrt{4}\sqrt{18})=2(2\sqrt{18})=(2\cdot 2)\sqrt{18}=4\sqrt{18}$$
We need to recognize that we are still not finished because we can extract another perfect square as follows.
$$4\sqrt{18}=4(\sqrt{9}\sqrt{2})=4(3\sqrt{2})=(4\cdot 3)\sqrt{2}=12\sqrt{2}$$
Once again, same result. However, note that it behooves us to extract the largest square possible, as it minimizes the number of steps required to attain simple radical form.
Checking Results with the Graphing Calculator. Once you’ve placed a radical expression in simple radical form, you can use your graphing calculator to check your result. In this example, we found that
$$\sqrt{288}=12\sqrt{2}$$. (6)
Enter the left- and right-hand sides of this result as shown in Figure 5. Note that each
side produces the same decimal approximation, verifying the result in equation (6).
Recall that raising a power of a base to another power requires that we multiply exponents.
Raising a Power of a Base to another Power.
$$(a^m)^n = a^{mn}$$
In particular, when you square a power of a base, you must multiply the exponent by 2. For example,
$$(2^5)^2 = 2^{10}$$.
Conversely, because taking a square root is the “inverse” of squaring, when taking a square root we must divide the existing exponent by 2, as in
$$\sqrt{2^{10}} = 2^5$$.
Note that squaring $$2^5$$ gives $$2^{10}$$, so taking the square root of $$2^{10}$$ must return you to $$2^5$$. When you square, you double the exponent. Therefore, when you take the square root, you must halve the exponent.
Similarly,
• $$(2^6)^2 = 2^{12}$$ so $$\sqrt{2^{12}} = 2^6$$.
• $$(2^7)^2 = 2^{14}$$ so $$\sqrt{2^{14}} = 2^7$$.
• $$(2^8)^2 = 2^{16}$$ so $$\sqrt{2^{16}} = 2^8$$.
This leads to the following result.
Taking the Square Root of an Even Power.
When taking a square root of $$x^n$$, when x is a positive real number and n is an even natural number, divide the exponent by two. In symbols,
$$\sqrt{x^n} = x^{\frac{n}{2}}$$.
Note that this agrees with the definition of rational exponents presented in Chapter 8, as in
$$\sqrt{x^n} = (x^n)^{\frac{1}{2}} = x^{\frac{n}{2}}$$.
On another note, recall that raising a product to a power requires that we raise each factor to that power.
Raising a Product to a Power.
$$(ab)^n = a^{n}b^{n}$$.
In particular, if you square a product, you must square each factor. For example,
$$(5^{3}7^{4})^2 = (5^3)^{2}(7^4)^2 = 5^{6}7^{8}$$.
Note that we multiplied each existing exponent in this product by 2.
Property 1 is similar, in that when we take the square root of a product, we take the square root of each factor. Because taking a square root is the inverse of squaring, we must divide each existing exponent by 2, as in
$$\sqrt{5^{6}7^{8}} = \sqrt{5^{6}}\sqrt{7^{8}} = 5^{3}7^{4}$$
Let’s look at some examples that employ this technique.
Example $$\PageIndex{7}$$
Simplify $$\sqrt{2^{4}3^{6}5^{10}}$$
When taking the square root of a product of exponential factors, divide each exponent by 2.
$$\sqrt{2^{4}3^{6}5^{10}} = 2^{2}3^{3}5^{5}$$
If needed, you can expand the exponential factors and multiply to provide a single numerical answer.
$$2^{2}3^{3}5^{5} = 4 \cdot 27 \cdot 3125 = 337500$$
A calculator was used to obtain the final solution.
Example $$\PageIndex{8}$$
Simplify $$\sqrt{2^{5}3^{3}}$$
In this example, the difficulty is the fact that the exponents are not divisible by 2. However, if possible, the “first guideline of simple radical form” requires that we factor out a perfect square. So, extract each factor raised to the highest possible power that is divisible by 2, as in
$$\sqrt{2^{5}3^{3}} = \sqrt{2^{4}3^{2}}\sqrt{2 \cdot 3}$$
Now, divide each exponent by 2.
$$\sqrt{2^{4}3^{2}}\sqrt{2\cdot 3} = 2^{2}3^{1}\sqrt{2 \cdot 3}$$
Finally, simplify by expanding each exponential factor and multiplying.
$$2^{2}3^{1}\sqrt{2 \cdot 3} = 4 \cdot 3\sqrt{2 \cdot 3} = 12\sqrt{6}$$
Example $$\PageIndex{9}$$
Simplify $$\sqrt{3^{7}5^{2}7^{5}}$$.
Extract each factor to the highest possible power that is divisible by 2.
$$\sqrt{3^{7}5^{2}7^{5}} = \sqrt{3^{6}5^{2}7^{4}}\sqrt{3 \cdot 7}$$
Divide each exponent by 2.
$$\sqrt{3^{6}5^{2}7^{4}}\sqrt{3 \cdot 7} = 3^{3}5^{1}7^{2}\sqrt{3 \cdot 7}$$
Expand each exponential factor and multiply.
$$3^{3}5^{1}7^{2}\sqrt{3 \cdot 7} = 27 \cdot 5 \cdot 49 \sqrt{3 \cdot 7} =6615\sqrt{21}$$
Example $$\PageIndex{10}$$
Place $$\sqrt{216}$$ in simple radical form.
If we prime factor 216, we can attack this problem with the same technique used in the previous examples. Before we prime factor 216, here are a few divisibility tests that you might find useful.
Divisibility Test
• If a number ends in 0, 2, 4, 6, or 8, it is an even number and is divisible by 2.
• If the last two digits of a number form a number that is divisible by 4, then the entire number is divisible by 4.
• If a number ends in 0 or 5, it is divisible by 5.
• If the sum of the digits of a number is divisible by 3, then the entire number is divisible by 3.
• If the sum of the digits of a number is divisible by 9, then the entire number is divisible by 9.
For example, in order:
• The number 226 ends in a 6, so it is even and divisible by 2. Indeed, $$226 = 2 \cdot 113$$.
• The last two digits of 224 are 24, which is divisible by 4, so the entire number is divisible by 4. Indeed, $$224 = 4 \cdot 56$$.
• The last digit of 225 is a 5. Therefore 225 is divisible by 5. Indeed, $$225 = 5 \cdot 45$$.
• The sum of the digits of 222 is 2 + 2 + 2 = 6, which is divisible by 3. Therefore, 222 is divisible by 3. Indeed, $$222 = 3 \cdot 74$$.
• The sum of the digits of 684 is 6 + 8 + 4 = 18, which is divisible by 9. Therefore, 684 is divisible by 9. Indeed, $$684 = 9 \cdot 76$$.
Now, let’s prime factor 216. Note that 2+1+6 = 9, so 216 is divisible by 9. Indeed, $$216 = 9 \cdot 24$$. In Figure 6, we use a “factor tree” to continue factoring until all of the “leaves” are prime numbers.
Thus,
$$216 = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3$$,
or in exponential form,
$$216 = 2^{3} \cdot 3^{3}$$.
Thus,
$$\sqrt{216} = \sqrt{2^{3}3^{3}} = \sqrt{2^{2}3^{2}}\sqrt{2\cdot 3}=2 \cdot 3\sqrt{2 \cdot 3}=6\sqrt{6}$$.
Prime factorization is an unbelievably useful tool!
Let’s look at another example.
Example $$\PageIndex{11}$$
Place $$\sqrt{2592}$$ in simple radical form.
If we find the prime factorization for 2592, we can attack this example using the same technique we used in the previous example. We note that the sum of the digits of 2592 is 2 + 5 + 9 + 2 = 18, which is divisible by 9. Therefore, 2592 is also divisible by 9.
$$2592 = 9 \cdot 288$$
The sum of the digits of 288 is 2+8+8 = 18, which is divisible by 9, so 288 is also divisible by 9.
$$2592 = 9 \cdot (9 \cdot 32)$$
Continue in this manner until the leaves of your “factor tree” are all primes. Then, you should get
$$2592 = 2^{5}3^{4}$$.
Thus,
$$\sqrt{2592} = \sqrt{2^{5}3^{4}} = \sqrt{2^{4}3^{4}}\sqrt{2} = 2^{2}3^{2}\sqrt{2} = 4 \cdot 9\sqrt{2}=36\sqrt{2}$$.
Let’s use the graphing calculator to check this result. Enter each side of $$\sqrt{2592} = 36\sqrt{2}$$ separately and compare approximations, as shown in Figure 7.
An Important Property of Square Roots
One of the most common mistakes in algebra occurs when practitioners are asked to simplify the expression $$\sqrt{x^2}$$, where x is any arbitrary real number. Let’s examine two of the most common errors.
• Some will claim that the following statement is true for any arbitrary real number x.
$$\sqrt{x^2} = \pm x$$.
This is easily seen to be incorrect. Simply substitute any real number for x to check this claim. We will choose x = 3 and substitute it into each side of the proposed statement.
$$\sqrt{3^2} = \pm 3$$.
If we simplify the left-hand side, we produce the following result.
$$\sqrt{3^2} = \pm 3$$.
$$3 = \pm 3$$
It is not correct to state that 3 and $$\pm 3$$ are equal.
• A second error is to claim that
$$\sqrt{x^2} = x$$
for any arbitrary real number x. Although this is certainly true if you substitute nonnegative numbers for x, look what happens when you substitute 3 for x.
$$\sqrt{(−3)^2} = 3$$
If we simplify the left-hand side, we produce the following result.
$$\sqrt{9} = −3$$
3 = −3
Clearly, 3 and 3 are not equal.
In both cases, what has been forgotten is the fact that $$\sqrt{}$$ calls for a positive (nonnegative if you want to include the case $$\sqrt{0}$$) square root. In both of the errors above, namely $$\sqrt{x^2} = \pm x$$ and $$\sqrt{x^2} = x$$, the left-hand side is calling for a nonnegative response, but nothing has been done to insure that the right-hand side is also nonnegative. Does anything come to mind?
Sure, if we wrap the right-hand side in absolute values, as in
$$\sqrt{x^2} = |x|$$,
then both sides are calling for a nonnegative response. Indeed, note that
$$\sqrt{(−3)^2} =|−3|$$, $$\sqrt{0^2} =|0|$$, and $$\sqrt{3^2} =|3|$$
are all valid statements.
This discussion leads to the following result.
The positive Square Root of the square of x
If x is any real number, then
$$\sqrt{x^2} = |x|$$,
The next task is to use this new property to produce a extremely useful property of absolute value.
## A Multiplication Property of Absolute Value
If we combine the law of exponents for squaring a product with our property for taking the square root of a product, we can write
$$\sqrt{(ab)^2} = \sqrt{a^{2}b^{2}} = \sqrt{a^2}\sqrt{b^2}$$.
However, $$\sqrt{(ab)^2} = |ab|$$, while $$\sqrt{a^2}{b^2} = |a||b|$$. This discussion leads to the following result.
Product rule for absolute value
If a and b are any real numbers,
|ab| = |a||b|.
In words, the absolute value of a product is equal to the product of the absolute values.
We saw this property previously in the chapter on the absolute value function, where we provided a different approach to the proof of the property. It’s interesting that we can prove this property in a completely new way using the properties of square root. We’ll see we have need for the Product Rule for Absolute Value in the examples that follow.
For example, using the product rule, if x is any real number, we could write
|3x| = |3||x| = 3|x|
However, there is no way we can remove the absolute value bars that surround x unless we know the sign of x. If $$x \ge 0$$, then |x| = x and the expression becomes
3|x| = 3x.
On the other hand, if x < 0, then |x| = −x and the expression becomes
3|x| = 3(−x) = −3x.
Let’s look at another example. Using the product rule, if x is any real number, the expression $$| − 4x^3|$$ can be manipulated as follows.
$$|−4x^3| = |−4||x^2||x|$$
However,|−4|=4 and since $$x^2 \ge 0$$ for any value of x, $$|x^2|=x^2$$. Thus,
$$|−4||x^2||x| = 4x^{2}|x|$$.
Again, there is no way we can remove the absolute value bars around x unless we know the sign of x. If $$x \ge 0$$, then|x|=x and
$$4x^{2}|x| = 4x^{2}(x) = 4x^3$$.
On the other hand, if x < 0, then |x| = −x and
$$4x^{2}|x| = 4x^{2}(−x) = −4x^3$$.
Let’s use these ideas to simplify some radical expressions that contain variables.
## Variable Expressions
Example $$\PageIndex{13}$$
Given that the x represents any real numbers, place the radical expression
$$\sqrt{48x^6}$$
in simple radical form.
Simple radical form demands that we factor out a perfect square, if possible. In this case, $$48 = 16 \cdot 3$$ and we factor out the highest power of x that is divisible by 2.
$$\sqrt{48x^6} = \sqrt{16x^6}\sqrt{3}$$
We can now use Property 1 to take the square root of each factor.
$$\sqrt{16x^6}\sqrt{3} = \sqrt{16}\sqrt{x^6}\sqrt{3}$$
Now, remember the notation $$\sqrt{}$$ calls for a nonnegative square root, so we must insure that each response in the equation above is nonnegative. Thus,
$$\sqrt{16}\sqrt{x^6}\sqrt{3} = 4|x^3|\sqrt{3}$$
Some comments are in order.
• The nonnegative square root of 16 is 4. That is, $$\sqrt{16} = 4$$
• The nonnegative square root of x6 is trickier. It is incorrect to say $$\sqrt{x^6} = x^3$$, because $$x^3$$ could be negative (if x is negative). To insure a nonnegative square root, in this case we need to wrap our answer in absolute value bars. That is, $$\sqrt{x^6} = |x^3|$$.
We can use the Product Rule for Absolute Value to write $$|x^3| = |x^2||x|$$. Because $$x^2$$ is nonnegative, absolute value bars are redundant and not needed. That is, $$|x^2||x| = x^{2}|x|$$. Thus, we can simplify our solution a bit further and write
$$4|x^3|\sqrt{3} = 4x^{2}|x|\sqrt{3}$$
Thus,
$$\sqrt{48x^6} = 4x^{2}|x|\sqrt{3}$$.
Alternate Solution. There is a variety of ways that we can place a radical expression in simple radical form. Here is another approach. Starting at the step above, where we first factored out a perfect square,
$$\sqrt{48x^6} = \sqrt{16x^6}\sqrt{3}$$
we could write
$$\sqrt{16x^6}\sqrt{3} = \sqrt{(4x^3)^2}\sqrt{3}$$.
Now, remember that the nonnegative square root of the square of an expression is the absolute value of that expression (we have to guarantee a nonnegative answer), so
$$\sqrt{(4x^3)^2}\sqrt{3} = |4x^3|\sqrt{3}$$.
However, $$|4x^3| = |4||x^3|$$ by our product rule and $$|4||x^3| = 4|x^3|$$. Thus,
$$|4x^3|\sqrt{3} = 4|x^3|\sqrt{3}$$.
Finally, $$|x^3| = |x^2||x| = x^2|x|$$ because $$x^2 \ge 0$$, so we can write
$$4|x^3|\sqrt{3} = 4x^{2}|x|\sqrt{3}$$
We cannot remove the absolute value bar that surrounds x unless we know the sign of x.
Note that the simple radical form in the alternate solution is identical to the simple radical form found with the previous solution technique.
Let’s look at another example.
Example $$\PageIndex{14}$$
Given that x < 0, place $$\sqrt{24x^6}$$ in simple radical form.
First, factor out a perfect square and write
$$\sqrt{24x^6} = \sqrt{4x^6}\sqrt{6}$$
Now, use Property 1 and take the square root of each factor.
$$\sqrt{4x^6}\sqrt{6}= \sqrt{4}\sqrt{x^6}\sqrt{6}$$
To insure a nonnegative response to $$\sqrt{x^6}$$, wrap your response in absolute values.
$$\sqrt{4}\sqrt{x^6}\sqrt{6}=2|x^3|\sqrt{6}$$
However, as in the previous problem, $$|x^3| = |x^2||x| = x^{2}|x|$$, since $$x^2 \ge 0$$. Thus,
$$2|x^3|\sqrt{6} = 2x^{2}|x|\sqrt{6}$$.
In this example, we were given the extra fact that x < 0, so |x| = −x and we can write
$$2x^{2}|x|\sqrt{6} = 2x^{2}(−x)\sqrt{6} = −2x^{3}\sqrt{6}$$.
It is instructive to test the validity of the answer
$$\sqrt{24x^6} = −2x^{3}\sqrt{6}$$, x<0.
using a calculator.
It is somewhat counterintuitive that the result
$$\sqrt{24x^6} = −2x^{3}\sqrt{6}$$, x<0.
contains a negative sign. After all, the expression $$\sqrt{24x^6}$$ call for a nonnegative result, but we have a negative sign. However, on closer inspection, if x < 0, then x is a negative number and the right-hand side $$−2x^{3}\sqrt{6}$$ is a positive number (−2 is negative, $$x^3$$ is negative because x is negative, and the product of two negatives is a positive).
Let’s look at another example.
Example $$\PageIndex{15}$$
If x < 3,simplify $$\sqrt{x^2−6x+9}$$.
The expression under the radical is a perfect square trinomial and factors.
$$\sqrt{x^2−6x+9}=\sqrt{(x−3)^2}$$
However, the nonnegative square root of the square of an expression is the absolute value of that expression, so
$$\sqrt{(x−3)^2} =|x−3|$$.
Finally, because we are told that x < 3, this makes x 3 a negative number, so
|x−3| = −(x−3).
Again, the result $$\sqrt{x^2−6x+9} = −(x−3)$$, provided x < 3, is somewhat counterintuitive as we are expecting a positive result. However, if x < 3, the result −(x−3) is positive. You can test this by substituting several values of x that are less than 3 into the expression −(x−3) and noting that the result is positive. For example, if x = 2, then x is less than 3 and
−(x−3) = −(2−3) = −(−1) = 1,
which, of course, is a positive result.
It is even more informative to note that our result is equivalent to
$$\sqrt{x^2−6x+9}=−x+3$$, x<3.
This is easily seen by distributing the minus sign in the result −(x−3).
We’ve drawn the graph of $$y = \sqrt{x^2−6x+9}$$ on our calculator on Figure 9(a). In Figure 9(b), we’ve drawn the graph of y = −x + 3. Note that the graphs agree when x < 3. Indeed, when you consider the left-hand branch of the “V” in Figure 9(a), you can see that the slope of this branch is −1 and the y-intercept is 3. The equation of this branch is y = −x+3, so it agrees with the graph of y = −x+3 in Figure 9(b) when x is less than 3.
Exercise $$\PageIndex{1}$$
Use a calculator to first approximate $$\sqrt{2}\sqrt{5}$$. One the same screen, approximate $$\sqrt{10}$$. Report the results on your homework paper.
Note that $$\sqrt{5}\sqrt{2} = \sqrt{10} \approx 3.16227766$$.
Exercise $$\PageIndex{2}$$
Use a calculator to first approximate $$\sqrt{7}\sqrt{10}$$. One the same screen, approximate $$\sqrt{70}$$. Report the results on your homework paper.
Exercise $$\PageIndex{3}$$
Use a calculator to first approximate $$\sqrt{3}\sqrt{11}$$. One the same screen, approximate $$\sqrt{33}$$. Report the results on your homework paper.
Note that $$\sqrt{3}\sqrt{11} = \sqrt{33} \approx 5.744562647$$
Exercise $$\PageIndex{4}$$
Use a calculator to first approximate $$\sqrt{5}\sqrt{13}$$. One the same screen, approximate $$\sqrt{65}$$. Report the results on your homework paper.
In Exercises 5-20, place each of the radical expressions in simple radical form. As in Example 3 in the narrative, check your result with your calculator.
Exercise $$\PageIndex{5}$$
$$\sqrt{18}$$
$$\sqrt{18} = \sqrt{3^{2} \cdot 2}= \sqrt{3^{2}}\sqrt{2}=3\sqrt{2}$$
Exercise $$\PageIndex{6}$$
$$\sqrt{80}$$
Exercise $$\PageIndex{7}$$
$$\sqrt{112}$$
$$\sqrt{112} = \sqrt{4^{2} \cdot 7}= \sqrt{4^{2}}\sqrt{7}=4\sqrt{7}$$
Exercise $$\PageIndex{8}$$
$$\sqrt{72}$$
Exercise $$\PageIndex{9}$$
$$\sqrt{108}$$
$$\sqrt{108} = \sqrt{6^{2} \cdot 3}= \sqrt{6^{2}}\sqrt{3}=6\sqrt{3}$$
Exercise $$\PageIndex{10}$$
$$\sqrt{54}$$
Exercise $$\PageIndex{11}$$
$$\sqrt{50}$$
$$\sqrt{50} = \sqrt{5^{2} \cdot 2}= \sqrt{5^{2}}\sqrt{2}=5\sqrt{2}$$
Exercise $$\PageIndex{12}$$
$$\sqrt{48}$$
Exercise $$\PageIndex{13}$$
$$\sqrt{245}$$
$$\sqrt{245} = \sqrt{7^{2} \cdot 5}= \sqrt{7^{2}}\sqrt{5}=7\sqrt{5}$$
Exercise $$\PageIndex{14}$$
$$\sqrt{150}$$
Exercise $$\PageIndex{15}$$
$$\sqrt{98}$$
$$\sqrt{98} = \sqrt{7^{2} \cdot 2}= \sqrt{7^{2}}\sqrt{2}=7\sqrt{2}$$
Exercise $$\PageIndex{16}$$
$$\sqrt{252}$$
Exercise $$\PageIndex{17}$$
$$\sqrt{45}$$
$$\sqrt{45} = \sqrt{3^{2} \cdot 5}= \sqrt{3^{2}}\sqrt{5}=3\sqrt{5}$$
Exercise $$\PageIndex{18}$$
$$\sqrt{294}$$
Exercise $$\PageIndex{19}$$
$$\sqrt{24}$$
$$\sqrt{24} = \sqrt{2^{2} \cdot 6}= \sqrt{2^{2}}\sqrt{6}=2\sqrt{6}$$
Exercise $$\PageIndex{20}$$
$$\sqrt{32}$$
In Exercises 21-26, use prime factorization (as in Example 10 and 11 in the narrative) to assist you in placing the given radical expression in simple radical form. Check your result with your calculator.
Exercise $$\PageIndex{21}$$
$$\sqrt{2016}$$
Note that 2+0+1+6 = 9, which is divisible by 9. Thus, 2016 is divisible by 9. Indeed,
$$2019 = 9 \cdot 224$$
The last two digits of 224 are 24, which is divisible by 4. Thus, 224 is divisible by 4. Indeed, $$224 = 4 \cdot 56$$.
$$2016 = 9 \cdot 224 = (3 \cdot 3) \cdot (4 \cdot 56)$$.
Continue to primes.
$$2016=3 \cdot 3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 7 = 2^{5} \cdot 3^{2} \cdot 7$$.
Checking,
Exercise $$\PageIndex{22}$$
$$\sqrt{2700}$$
Exercise $$\PageIndex{23}$$
$$\sqrt{14175}$$
Money! Anything that ends in 00, 25, 50, or 75 is divisible by 25. Indeed, $$14175 = 25 \cdot 567$$. Further, 5+6+7 = 18, so 567 is divisible by 9; i.e., $$567 = 9 \cdot 63$$. Continuing to primes,
$$14175 = 25 \cdot 567 = 5 \cdot 5 \cdot 9 \cdot 63 = 5 \cdot 5 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 7$$.
Factor our a perfect square (exponents divisible by 2).
$$\sqrt{14175} = \sqrt{3^{4} \cdot 5^{2}} \cdot \sqrt{7} = 3^{2} \cdot 5\sqrt{7} = 45\sqrt{7}$$
Checking,
Exercise $$\PageIndex{24}$$
$$\sqrt{44000}$$
Exercise $$\PageIndex{25}$$
$$\sqrt{20250}$$
Money! Anything that ends in 00, 25, 50, or 75 is divisible by 25. Indeed, $$20250 = 25 \cdot 810$$. Continuing to primes,
$$20250 = 5 \cdot 5 \cdot 9 \cdot 9 \cdot 10 = 5 \cdot 5 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 2 \cdot 5$$.
Factor out a perfect square.
$$\sqrt{20250} = \sqrt{2 \cdot 3^{4} \cdot 5^{3}} = \sqrt{3^{4} \cdot 5^{2}} \cdot \sqrt{2 \cdot 5} = 3^{2} \cdot 5 \sqrt{2 \cdot 5}=45\sqrt{10}$$
Checking,
Exercise $$\PageIndex{26}$$
$$\sqrt{3564}$$
In Exercises 27-46, place each of the given radical expressions in simple radical form. Make no assumptions about the sign of the variables. Variables can either represent positive or negative numbers.
Exercise $$\PageIndex{27}$$
$$\sqrt{(6x−11)^4}$$
$$\sqrt{(6x−11)^4} = \sqrt{((6x−11)^2)^2} = |(6x−11)^2|$$
However, $$(6x − 11)^2$$ is already nonnegative, so the absolute value bars are unnecessary. Hence,
$$\sqrt{(6x−11)^4} = (6x−11)^2$$
Exercise $$\PageIndex{28}$$
$$\sqrt{16h^8}$$
Exercise $$\PageIndex{29}$$
$$\sqrt{25f^2}$$
$$\sqrt{25f^2} = \sqrt{25}\sqrt{f^2} = 5|f|$$
Because f can be any real number, we cannot remove the absolute value bars without more information.
Exercise $$\PageIndex{30}$$
$$\sqrt{25j^8}$$
Exercise $$\PageIndex{31}$$
$$\sqrt{16m^2}$$
$$\sqrt{16m^2} = \sqrt{4^{2}m^2} = \sqrt{4^2}\sqrt{m^2} = 4|m|$$
Since the index on the radical is even and, after simplification, the variable is raised to an odd power, absolute value signs around the simplified variable are necessary.
Exercise $$\PageIndex{32}$$
$$\sqrt{25a^2}$$
Exercise $$\PageIndex{33}$$
$$\sqrt{(7x+5)^{12}}$$
$$\sqrt{(7x+5)^{12}} = \sqrt{((7x+5)^6)^2} = |(7x+5)^6|$$
However, $$(7x+5)^6$$ is already nonnegative, so absolute value signs are unnecessary.
$$\sqrt{(7x+5)^{12}} = (7x+5)^6$$
Exercise $$\PageIndex{34}$$
$$\sqrt{9w^{10}}$$
Exercise $$\PageIndex{35}$$
$$\sqrt{25x^2−50x+25}$$
$$\sqrt{25x^2−50x+25}=\sqrt{(5x−5)^2} =|5x−5|$$
Because x can be any real number, the absolute value signs around the simplified binomial are necessary.
Exercise $$\PageIndex{36}$$
$$\sqrt{49x^2−42x+9}$$
Exercise $$\PageIndex{37}$$
$$\sqrt{25x^2+90x+81}$$
$$\sqrt{25x^2+90x+81} = \sqrt{(5x+9)^2} = |5x+9|$$
Because x can be any real number, the absolute value signs around the simplified binomial are necessary.
Exercise $$\PageIndex{38}$$
$$\sqrt{25f^{14}}$$
Exercise $$\PageIndex{39}$$
$$\sqrt{(3x+6)^{12}}$$
$$\sqrt{(3x+6)^{12}} = \sqrt{((3x+6)^6)^2} = |(3x+6)^6|$$
However, the expression $$(3x+6)^6$$ is already nonnegative, so the absolute value bars are unnecessary.
$$\sqrt{(3x+6)^{12}} = (3x+6)^6$$
Exercise $$\PageIndex{40}$$
$$\sqrt{(9x−8)^{12}}$$
Exercise $$\PageIndex{41}$$
$$\sqrt{36x^2+36x+9}$$
$$\sqrt{36x^2+36x+9} = \sqrt{(6x+3)^2} =|6x+3|$$
Because x can be any real number, the absolute value signs around the simplified binomial are necessary.
Exercise $$\PageIndex{42}$$
$$\sqrt{4e^2}$$
Exercise $$\PageIndex{43}$$
$$\sqrt{4p^{10}}$$
$$\sqrt{4p^{10}} = \sqrt{4}\sqrt{(p^5)^2}=2|p^5|$$
Now, we can use the multiplicative property of absolute values and write
$$2|p^5| = 2|p^4||p| = 2p^{4}|p|$$.
Since p can be any real number, absolute value signs around the simplified variable are necessary.
Exercise $$\PageIndex{44}$$
$$\sqrt{25x^{12}}$$
Exercise $$\PageIndex{45}$$
$$\sqrt{25q^6}$$
$$\sqrt{25q^6} = \sqrt{25}\sqrt{(q^3)^2} = 5|q^3|$$
Now, we can use the multiplicative property of absolute values and write
$$5|q^3| = 5|q^2||q| = 5q^{2}|q|$$.
Because q can be any real number, absolute value signs around the simplified variable are necessary.
Exercise $$\PageIndex{46}$$
$$\sqrt{16h^{12}}$$
Exercise $$\PageIndex{47}$$
Given that x < 0, place the radical expression $$\sqrt{32x^6}$$ in simple radical form. Check your solution on your calculator for x = −2.
Factor out a perfect square.
$$\sqrt{32x^6} = \sqrt{16x^6}\sqrt{2} = \sqrt{16}\sqrt{x^6}\sqrt{2}=4|x^3|\sqrt{2}$$
However, $$|x^3| = |x^2||x| = x^2|x|$$, since $$x^2 \ge 0$$. Thus
$$\sqrt{32x^6} = 4x^{2}|x|\sqrt{2}$$.
If x < 0, then |x| = −x and
$$\sqrt{32x^6} = 4x^{2}(−x)\sqrt{2} =−4x^{3}\sqrt{2}$$.
Checking with x = 2.
Exercise $$\PageIndex{48}$$
Given that x < 0, place the radical expression $$\sqrt{54x^8}$$ in simple radical form. Check your solution on your calculator for x = −2.
Exercise $$\PageIndex{49}$$
Given that x < 0, place the radical expression $$\sqrt{27x^{12}}$$ in simple radical form. Check your solution on your calculator for x = −2.
Factor out a perfect square.
$$\sqrt{27x^{12}} = \sqrt{9x^{12}}\sqrt{3} = \sqrt{9}\sqrt{x^{12}}\sqrt{3}=3|x^6|\sqrt{3}$$
However, $$|x^6| = x^6$$ since $$x^6 \ge 0$$. Thus,
$$\sqrt{27x^{12}}=3x^{6}\sqrt{3}$$
Checking with x = 2.
Exercise $$\PageIndex{50}$$
Given that x < 0, place the radical expression $$\sqrt{44x^{10}}$$ in simple radical form. Check your solution on your calculator for x = −2.
In Exercises 51-54, follow the lead of Example 17 in the narrative to simplify the given radical expression and check your result with your graphing calculator.
Exercise $$\PageIndex{51}$$
Given that x < 4, place the radical expression $$\sqrt{x^2−8x+16}$$ in simple radical form. Use a graphing calculator to show that the graphs of the original expression and your simple radical form agree for all values of x such that x < 4.
Factor the perfect square trinomial.
$$\sqrt{x^2−8x+16} = \sqrt{(x−4)^2} =|x−4|$$
If x < 4, or equivalently, if x−4 < 0, then |x−4| = −(x−4). Thus,
$$\sqrt{x^2−8x+16} =−x+4$$.
In (b), we've drawn the graph of $$y = \sqrt{x^2−8x+16}$$. In (d), we've drawn the graph of y = −x+4. Note that the graphs in (b) and (d) agree when x < 4, leading credence to the fact that $$\sqrt{x^2−8x+16} =−x+4$$ when x < 4.
Exercise $$\PageIndex{52}$$
Given that $$x \ge −2$$, place the radical expression $$\sqrt{x^2+4x+4}$$ in simple radical form. Use a graphing calculator to show that the graphs of the original expression and your simple radical form agree for all values of x such that $$x \ge −2$$.
Exercise $$\PageIndex{53}$$
Given that $$x \ge 5$$, place the radical expression $$\sqrt{x^2−10x+25}$$ in simple radical form. Use a graphing calculator to show that the graphs of the original expression and your simple radical form agree for all values of x such that $$x \ge 5$$.
Factor the perfect square trinomial.
$$\sqrt{x^2−10x+25} = \sqrt{(x−5)^2} =|x−5|$$
If $$x \ge 5$$, or equivalently, if $$x−5 \ge 0$$, then |x−5| = x−5. Thus,
$$\sqrt{x^2−8x+16} =x−5$$.
In (b), we've drawn the graph of $$y = \sqrt{x^2−10x+25}$$. In (d), we've drawn the graph of y = x−5. Note that the graphs in (b) and (d) agree when $$x \ge 5$$, leading credence to the fact that $$\sqrt{x^2−10x+25} = x−5$$ when $$x \ge 5$$.
Exercise $$\PageIndex{54}$$
Given that x < −1, place the radical expression $$\sqrt{x^2+2x+1}$$ in simple radical form. Use a graphing calculator to show that the graphs of the original expression and your simple radical form agree for all values of x such that x < −1.
In Exercises 55-72, place each radical expression in simple radical form. Assume that all variables represent positive numbers.
Exercise $$\PageIndex{55}$$
$$\sqrt{9d^{13}}$$
$$\sqrt{9d^{13}} = \sqrt{9}\sqrt{d^{12}}\sqrt{d}=3d^6\sqrt{d}$$
Exercise $$\PageIndex{56}$$
$$\sqrt{4k^2}$$
Exercise $$\PageIndex{57}$$
$$\sqrt{25x^2+40x+16}$$
$$\sqrt{25x^2+40x+16} = \sqrt{(5x+4)^2} = 5x+4$$
Exercise $$\PageIndex{58}$$
$$\sqrt{9x^2−30x+25}$$
Exercise $$\PageIndex{59}$$
$$\sqrt{4j^{11}}$$
$$\sqrt{4j^{11}} = \sqrt{4}\sqrt{j^{10}}\sqrt{j}=3j^5\sqrt{j}$$
Exercise $$\PageIndex{60}$$
$$\sqrt{16j^6}$$
Exercise $$\PageIndex{61}$$
$$\sqrt{25m^2}$$
$$\sqrt{25m^2} = \sqrt{25}\sqrt{m^2} = 5m$$
Exercise $$\PageIndex{62}$$
$$\sqrt{9e^9}$$
Exercise $$\PageIndex{63}$$
$$\sqrt{4c^5}$$
$$\sqrt{4c^5} = \sqrt{4c^4}\sqrt{c} = 2c^{2}\sqrt{c}$$
Exercise $$\PageIndex{64}$$
$$\sqrt{25z^2}$$
Exercise $$\PageIndex{65}$$
$$\sqrt{25h^{10}}$$
$$\sqrt{25h^{10}} = \sqrt{25}\sqrt{h^{10}} = 5h^5$$
Exercise $$\PageIndex{66}$$
$$\sqrt{25b^2}$$
Exercise $$\PageIndex{67}$$
$$\sqrt{9s^7}$$
$$\sqrt{9s^7} = \sqrt{9s^6}\sqrt{s} = 3s^{3}\sqrt{s}$$
Exercise $$\PageIndex{68}$$
$$\sqrt{9e^7}$$
Exercise $$\PageIndex{69}$$
$$\sqrt{4p^8}$$
$$\sqrt{4p^8} = \sqrt{4}\sqrt{p^8} = 2\sqrt{p^4}$$
Exercise $$\PageIndex{70}$$
$$\sqrt{9d^{15}}$$
Exercise $$\PageIndex{71}$$
$$\sqrt{9q^{10}}$$
$$\sqrt{9q^{10}} = \sqrt{9}\sqrt{q^{10}} = 3\sqrt{q^5}$$
Exercise $$\PageIndex{72}$$
$$\sqrt{4w^7}$$
In Exercises 73-80, place each given radical expression in simple radical form. Assume that all variables represent positive numbers.
Exercise $$\PageIndex{73}$$
$$\sqrt{2f^5}\sqrt{8f^3}$$
$$\sqrt{2f^5}\sqrt{8f^3} =\sqrt{2 \cdot 8 \cdot f^5 \cdot f^3} = \sqrt{16f^8} = \sqrt{16}\sqrt{(f^4)^2} =4f^4$$
Exercise $$\PageIndex{74}$$
$$\sqrt{3s^3}\sqrt{243s^3}$$
Exercise $$\PageIndex{75}$$
$$\sqrt{2k^7}\sqrt{32k^3}$$
$$\sqrt{2k^7}\sqrt{32k^3} = \sqrt{2 \cdot 32 \cdot k^7 \cdot k^3} = \sqrt{64k^{10}} = \sqrt{64}\sqrt{(k^5)^2}=8k^5$$
Exercise $$\PageIndex{76}$$
$$\sqrt{2n^9}\sqrt{8n^3}$$
Exercise $$\PageIndex{77}$$
$$\sqrt{2e^9}\sqrt{8e^3}$$
$$\sqrt{2e^9}\sqrt{8e^3} = \sqrt{2 \cdot 8 \cdot e^9 \cdot e^3} = \sqrt{16e^{12}} = \sqrt{16}\sqrt{(e^6)^2}=4e^6$$
Exercise $$\PageIndex{78}$$
$$\sqrt{5n^9}\sqrt{125n^3}$$
Exercise $$\PageIndex{79}$$
$$\sqrt{3z^5}\sqrt{27z^3}$$
$$\sqrt{3z^5}\sqrt{27z^3} = \sqrt{3 \cdot 27 \cdot z^5 \cdot z^3} = \sqrt{81z^8} = \sqrt{81}\sqrt{(z^4)^2}=9z^4$$
Exercise $$\PageIndex{80}$$
$$\sqrt{3t^7}\sqrt{27t^3}$$
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# 8.2: One-Sided Limits
Difficulty Level: At Grade Created by: CK-12
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Practice One-Sided Limits
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Tuscany and Sophia were out hiking. As they followed the path up the side of a hill, they discovered that there had been a washout from the recent storm. The path had been obliterated for a space of about 15 feet in front of them, whereafter it continued on up the mountain from about 10 feet higher.
How could this situation be explained with one-sided limits?
Embedded Video:
### Guidance
Unlike the functions from prior lessons where the limit was the same from both directions, the functions we explore in this lesson may have a different limit for each "side." To evaluate these functions, we deal with each "side" separately, first evaluating what happens as the limit is approached from either the positive direction (the values of x are bigger than at the limit) or the negative direction (the values of x are smaller than at the limit) and then evaluating the other direction afterward as if it were effectively a separate function.
Sometimes the same value is approached from either "side" of the limit value, and some functions have different limits on the two sides of x = x0.
When the value of f(x) does not get closer and closer to some single value L as \begin{align*}x \rightarrow x_0\end{align*} we say that the limit of the function as x approaches x0 does not exist. We may separately state the limits for each "side" of x0, but the complete function will only have a specified limit if it is the same for both sides.
For example, the two-sided limit of the complete function \begin{align*}lim_{x \rightarrow 0} \frac{ |x|}{x}\end{align*} does not exist because the values of f(x) do not approach a single number as x approaches 0. We can state the one-sided limits for each side, since the values approach -1 from the left and 1 from the right.
#### Example A
Identify the limit of the function: \begin{align*}f(x) = \frac{|x|} {x} = \begin{cases} 1, x > 0\\ -1, x < 0\\ \end{cases} \end{align*}
which is shown in the graph below:
Solution'
As x approaches 0 from the right, f(x) approaches 1.
On the other hand, as x approaches 0 from the left, the function f(x) approaches -1.
Since the limit is not the same from both sides, the limit of the function does not exist. However we can say that:
\begin{align*}\lim_{x \rightarrow 0^+} \frac{|x|} {x} = 1\end{align*}
\begin{align*}\lim_{x \rightarrow 0^-} \frac{|x|} {x} = -1\end{align*}
Where the superscript “+” indicates a limit from the right and the superscript “-” indicates a limit from the left.
#### Example B
Consider the function f graphed in the accompanying figure and find
a) \begin{align*} \lim_{x \rightarrow 2^-}f(x)\end{align*} b) \begin{align*} \lim_{x \rightarrow 2^+}f(x)\end{align*} c) \begin{align*}\lim_{x \rightarrow 2}f(x)\end{align*} d) \begin{align*}f(2)\end{align*}
Solution:
a) From graph, we can see that, \begin{align*}\lim_{x \rightarrow 2^-}f(x) = -2\end{align*}.
b) We can also see from the graph that \begin{align*}\lim_{x \rightarrow 2^+}f(x) = 4\end{align*}.
c) Since the limits from the right and the left are not equal (they do not approach a single value L), the limit does not exist. That is, \begin{align*} \lim_{x \rightarrow 2}f(x) \end{align*} does not exist.
d) \begin{align*}f(2)=1\end{align*}.
#### Example C
Consider \begin{align*}g(x) = \frac{|x - 2|}{x - 2}|3|\end{align*} in the image below:
Find: a) \begin{align*} \lim_{x \rightarrow 2^-}f(x)\end{align*} b) \begin{align*} \lim_{x \rightarrow 2^+}f(x)\end{align*} c) \begin{align*}\lim_{x \rightarrow 2}f(x)\end{align*} d) \begin{align*}f(2)\end{align*}
Solution:
a) From graph, we can see that, \begin{align*}\lim_{x \rightarrow 2^-}f(x) = -3\end{align*}.
b) We can also see from the graph that \begin{align*}\lim_{x \rightarrow 2^+}f(x) = 3\end{align*}.
c) Since the limits from the right and the left are not equal (they do not approach a single value L), the limit does not exist. That is, \begin{align*} \lim_{x \rightarrow 2}f(x) \end{align*} does not exist.
d) \begin{align*}f(2) = \not0\end{align*} undefined
Concept question wrap-up Tuscany and Sophia's path could be examined as a discontinuous function of elevation based on distance traveled along the path. For instance, if Tuscany and Sophia had traveled for 500 yards along the path before encountering the washout, then the limit of the function from the "trailhead side" would be the elevation at the edge they encountered. The function would then be undefined for the next 5 yards or so (since the path does not exist), and would pick up at 506 yards, where the elevation would be 10ft higher. If Sayber were coming down the path toward Tuscany and Sophia, from his point of view the "limit" of the elevation would be 10ft greater, and would be the lowest elevation that "his side" of the function could attain.
### Vocabulary
A one-sided-limit is a limit that only applies as the limit is approached from a specific side.
A two-sided-limit is the same value from both directions.
### Guided Practice
Questions
1) Identify the specified limit:
\begin{align*}\lim{x\to 0^+} \frac{-5x^2 - 4x}{x}\end{align*}
2) Use the image to identify the specified limits:
a) \begin{align*}\lim_{x\to2^+}\end{align*}
b) \begin{align*}\lim_{x\to2^-}\end{align*}
c) \begin{align*}f(2)\end{align*}
3) Identify the limit based on the equation:
\begin{align*}g(x)= \begin{cases} 7; x = - 5\\ 2; x \not= -5\\ \end{cases} \end{align*}
4) Identify the limit based on the equation, use a graphing tool:
\begin{align*}\lim_{x\to2^+}\frac{-x^2 - 2x + 8}{x - 2}= \end{align*}
Solutions
1) As you can see from the graph,
a) \begin{align*} \lim_{x \rightarrow + \infty} f(x) = 4\end{align*}, in other words, as x becomes very large, f(x) stays at the same value, 4.
b)\begin{align*} \lim_{x \rightarrow + \infty} f(x) = -2\end{align*}, which says that as x becomes smaller and smaller, f(x) stays at the same value, -2.
2) From the image, we can see that:
a) \begin{align*}\lim_{x\to2^+} = 5\end{align*}
b) \begin{align*}\lim_{x\to2^-} = -1\end{align*}
c) \begin{align*}f(x) = 2\end{align*} as there is a specified value for \begin{align*}f(x)\end{align*} when \begin{align*}x = 2\end{align*}
3) The cases specify that if \begin{align*}x = -5\end{align*} then \begin{align*}g(x) =7\end{align*} and if \begin{align*}x\end{align*} is anything else, then \begin{align*}g(x) = 2\end{align*}
\begin{align*}\therefore\end{align*} the limit as x approaches -5 from either direction is 2.
4) To find the limit of \begin{align*}\lim_{x\to3^+}\frac{x^2 - 5x + 6}{x - 3}\end{align*}
Factor the numerator: \begin{align*}(x - 2)(x - 3)\end{align*}
Now that you now have \begin{align*}(x - 3)\end{align*} in the numerator and in the denominator
Substitute 3 in for x in \begin{align*}x - 2\end{align*} since 3 is the number we want to evaluate
\begin{align*}3 - 2 = 1\end{align*}
\begin{align*}\therefore \lim_{x\to3^+}\frac{x^2 - 5x + 6}{x - 3} = 1\end{align*}
### Practice
Identify the limit, based on each graph.
1. \begin{align*}\lim_{x\to-3^-}\end{align*}
2. \begin{align*}\lim_{x\to2^+}\end{align*}
3. \begin{align*}\lim_{x\to-1^+}\end{align*} and \begin{align*}\lim_{x\to-1^-}\end{align*}
4. \begin{align*}\lim_{x\to-1}\end{align*}
5. \begin{align*}\lim_{x\to-2^-}\end{align*} and \begin{align*}\lim_{x\to5^+}\end{align*}
Identify the limit based on the equation:
1. \begin{align*}\lim_{x\to2^+}\frac{-x^2 - 2x + 8}{x - 2}= \end{align*}
2. \begin{align*}g(x)= \begin{cases} 4; x = - 3\\ 1; x \not= -3\\ \end{cases}\end{align*}
3. \begin{align*}\lim_{x\to0^+}\frac{-x^2 + 4x}{x}= \end{align*}
4. \begin{align*}g(x)= \begin{cases} -5; x \not= -1\\ -1; x = -1\\ \end{cases} \end{align*}
5. \begin{align*}\lim_{x\to1^+}\frac{4x^2 - x - 3}{x - 1}= \end{align*}
6. \begin{align*}f(x)= \begin{cases} 4 ; x \geq 3\\ x + 1; x < 3\\ \end{cases} \end{align*}
7. \begin{align*}\lim_{x\to0^+}\frac{x^2 - 4x}{x}= \end{align*}
8. \begin{align*}h(x)= \begin{cases} 4x + 4 ; x \not= 2\\ 1 ; x = 2\\ \end{cases} \end{align*}
9. \begin{align*}\lim_{x\to2^-}\frac{4x^2 - 7x - 2}{x - 2}= \end{align*}
10. \begin{align*}g(x)= \begin{cases} x - 5 ; x = -2\\ 4x + 1 ; x \not= -2\\ \end{cases} \end{align*}
11. \begin{align*}g(x)= \begin{cases} -3x ; \not= 3\\ -9 ; x = 3\\ \end{cases} \end{align*}
12. \begin{align*}\lim_{x \to -5^-}\frac{-3x^2 - 13x + 10}{x + 5}= \end{align*}
13. \begin{align*}f(x)= \begin{cases} x ; x = 2\\ 3x - 3 ; x \not= 2\\ \end{cases} \end{align*}
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
TermDefinition
limit A limit is the value that the output of a function approaches as the input of the function approaches a given value.
one-sided limit A one-sided limit is the value that a function approaches from either the left side or the right side.
two-sided limit A two-sided limit is the value that a function approaches from both the left side and the right side.
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# Fractions as Percents
## Convert fractions to percents.
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MCC6.RP.3c - Fractions as Percents
Have you ever met anyone who liked to clean up?
Sam’s least favorite part of her after school job at the supermarket is closing. When closing the store, the break room has to be swept and mopped. Each of the students who work at the store part-time take turns closing, and every Friday night is Sam’s turn. On Friday, Sam got her mop and broom and headed up to the break room. It seemed to be even messier than usual. “Oh no, I will never get done,” Sam sighed, but she picked up the broom and began to sweep.
In just fifteen minutes, Sam had swept four-fifths of the room. She was amazed at how quickly the task was getting done with a little focus and effort.
What fraction of the room does Sam still have to sweep? What percent of the room is this?
In this Concept, you will learn how to identify fractions and convert them to percents. This is exactly what Sam will need to accomplish the task.
### Guidance
In our last Concept, you learned how to understand percents. Remember that a percent means a part of a whole, and the whole is 100. You also learned how to take a percent and write it as a decimal and as a fraction. In this Concept, you will learn to work the other way around. Let’s begin by learning how to write fractions as a percent.
When writing a percent as a fraction, we can drop the % sign and make that the denominator of 100. Then we take the quantity and write it in the numerator above the 100. This is our fraction.
Write \begin{align*}14%\end{align*} as a fraction.
First, we drop the % sign. % means out of 100, so 100 becomes our new denominator.
\begin{align*}14%\end{align*} becomes \begin{align*}\frac{14}{100}\end{align*}.
Yes! We can change the denominator of 100 to a % sign and add that to the quantity in the numerator. Here is what that looks like.
Write \begin{align*}\frac{47}{100}\end{align*} as a percent.
First, drop the 100. Then write 47 with a % sign.
\begin{align*}\frac{47}{100}\end{align*} becomes \begin{align*}47%\end{align*}
Our answer is \begin{align*}47%\end{align*}.
Not all fractions have a denominator of 100. How do we write a fraction as a percent when the denominator is not 100?
This is the next thing that we need to learn. If a fraction does not have a denominator of 100, to write it as a percent we need to rewrite it as an equal fraction with a denominator of 100.
Write \begin{align*}\frac{2}{5}\end{align*} as a percent.
To write \begin{align*}\frac{2}{5}\end{align*} as a percent, we must first rewrite it as a fraction with a denominator of 100. Write a proportion to show this comparison.
\begin{align*}\frac{2}{5}=\frac{}{100}\end{align*}
What number was multiplied by 5 to get 100 as a product? 20! So we multiply the numerator by 20 and we will have an equivalent fraction with a denominator of 100.
\begin{align*}2 \times 20 &= 40\\ \frac{2}{5}=\frac{40}{100}&=40\%\end{align*}
Our answer is \begin{align*}40%\end{align*}.
One special fraction to work with is one-third. To convert one-third to a percent is a little tricky because 3 does not divide evenly into 100. Take a look.
\begin{align*}\frac{1}{3}=\frac{}{100}\end{align*}
When completing this problem, we end up with a repeating decimal; the 3’s just continue on and on and on.
\begin{align*}.33333333\end{align*} etc.
To work with this fraction, we call it 33 \begin{align*}\frac{1}{3}\end{align*} %.
Write each fraction as a percent.
#### Example A
\begin{align*}\frac{48}{100}\end{align*}
Solution:\begin{align*}48%\end{align*}
#### Example B
\begin{align*}\frac{82}{100}\end{align*}
Solution:\begin{align*}82%\end{align*}
#### Example C
\begin{align*}\frac{91}{100}\end{align*}
Solution:\begin{align*}91%\end{align*}
Remember Sam? Let's use what we have learned to help Sam with her sweeping dilemma.
In just fifteen minutes, Sam had swept four-fifths of the room. She was amazed at how quickly the task was getting done with a little focus and effort.
What fraction of the room does Sam still have to sweep?
What percent of the room has she finished? What percent of the room is still left?
Let’s work through this solution. If Sam has completed four-fifths of the room, then she has one-fifth left to complete.
What percent of the room has she completed?
To figure this out, we have to figure out what four–fifths is as a percent. To do this, we can figure it out by using an equal ratio out of 100.
\begin{align*}\frac{4}{5}=\frac{80}{100}=80\%\end{align*}
Sam has completed 80% of the room.
What percent of the room does she have left?
You can figure this out two different ways. The first way is to simply subtract 80% from 100%. 100% - 80% (what Sam has completed) \begin{align*}= 20\%\end{align*}
The other way is to convert one-fifth (the amount left) to a percent. We can do this by creating an equal ratio out of 100.
\begin{align*}\frac{1}{5}=\frac{20}{100}=20\%\end{align*}
Since it only took Sam 15 minutes to complete 80% of the room, if she continues with her great effort she will be finished in no time at all.
### Vocabulary
Here are the vocabulary words in this Concept.
Percent
is "per-cent" or "per-hundred", it is a quantity written with a % sign, a part of a whole (100)
Fraction
a part of a whole, related to decimals and percents.
Decimal
a part of a whole shown by a decimal point, hundredths means two decimal places.
Equivalent
means equal
### Guided Practice
Here is one for you to try on your own.
Write \begin{align*} \frac{23}{50}\end{align*} as a percent.
To write this fraction as a percent, we have to rewrite the fraction with a denominator of 100. We can set up a proportion to do this.
\begin{align*}\frac{23}{50} = \frac{x}{100}\end{align*}
Now we can solve the proportion.
The answer is \begin{align*}46%\end{align*}.
### Video Review
Here are videos for review.
### Practice
Directions: Write each fraction as a percent.
1. \begin{align*}\frac{4}{100}\end{align*}
2. \begin{align*}\frac{24}{100}\end{align*}
3. \begin{align*}\frac{20}{100}\end{align*}
4. \begin{align*}\frac{76}{100}\end{align*}
5. \begin{align*}\frac{61}{100}\end{align*}
6. \begin{align*}\frac{1}{4}\end{align*}
7. \begin{align*}\frac{3}{4}\end{align*}
8. \begin{align*}\frac{3}{6}\end{align*}
9. \begin{align*}\frac{2}{5}\end{align*}
10. \begin{align*}\frac{4}{5}\end{align*}
11. \begin{align*}\frac{8}{10}\end{align*}
12. \begin{align*}\frac{6}{10}\end{align*}
13. \begin{align*}\frac{6}{50}\end{align*}
14. \begin{align*}\frac{3}{25}\end{align*}
15. \begin{align*}\frac{20}{50}\end{align*}
### Notes/Highlights Having trouble? Report an issue.
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# Square
## Definition
It is a geometric shape having four equal line segments that intersect at four points to create four internal angles of 90° each.
## Properties
1. All four interior angles are right angles; in □ ABCD, ∠ABC = ∠BCD = ∠CDA =∠DAB = 90°
2. Has four vertices and four sides; A, B, C and D are vertices and AB, BC, CD and DA are sides
3. All four sides are congruent; so AB ≅ BC ≅ CD ≅DA
4. Opposite sides are parallel to each other; AB ∥ CD and BC ∥ DA
5. Diagonals of the square bisect each other at 90°
6. Two diagonals of the square are perpendicular to each other; AC = BD
## Formulas
### Diagonal
The segment that connects two opposite vertices of the square at right-angle to each other. The formula is given below:
Diagonal (d) = a × √2, here a = side length
Problem: Finding the diagonal of a square when only the SIDES are known
Find the diagonal of a square whose sides measure 4 cm.
Solution:
As we know,
Diagonal (d) = a × √2, where a = 4 cm
= 4 x √2 cm
= 5.656 cm
### Area
The total space enclosed by the square. The formula is given below:
Area (A) = a2, here a = side length
Problem: Finding the area of a square when only the SIDES are known
Find the area of a square whose sides measure 6 cm.
Solution:
As we know,
Area (A) = a2, where a = 6 cm
= 6 x 6 cm2
= 36 cm2
### Perimeter
The total distance covered around the edge of the square. The formula is given below:
Perimeter (P ) = 4a ,here a = side length
Problem: Finding the perimeter of a square when only the SIDES are known
Find the perimeter of a square whose sides measure 10 cm.
Solution:
As we know,
Perimeter (P ) = 4a,where a = 10 cm
= 4 x 10 cm
= 40 cm
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# What is 16 Out of 25 as a Percentage?
If you’re wondering what 16 out of 25 represents as a percentage, you’re not alone. This is a common question, and the answer is that it equals 64%.
Checkout this video:
## Introduction
16 out of 25 is equal to 64 percent. To find the percentage, divide the numerator (16) by the denominator (25), and then multiply by 100.
## What is 16 Out of 25 as a Percentage?
Understanding what16 out of 25 as a percentage means can be difficult for some people. After all, when you are dealing with percentages, you are really dealing with a fraction of a whole. In this case, 16 out of 25 is equal to 64%.
## Simple Method
To work out 16% of 25, you can use mental math.
Recall that percent means out of 100, so 16% means 16 out of 100, or equivalently, 16/100.
Since 25 is 4 times as large as 100 (25 = 4 × 100), it follows that 16 is 4 times as large as 16 (16 = 4 × 16). Therefore:
16% of 25
= (16/100) × 25
= (4/4) × (16/100) × 25
= ((16/100) × (4/4)) × 25
= (16/100) × 25
= 16/100 × 25
= 400/10000
= 0.400
≈ 0.40
## Long Method
To convert a fraction to a percent, multiply the top number of the fraction by 100 and then divide it by the bottom number of the fraction.
Here is the calculation for 16 out of 25 as a percent. To calculate 16 out of 25 as a percent, we multiply 16 by 100 and then divide by 25. So, 16 out of 25 as a percent is calculated like this:
16/25 × 100 = 64%.
Here is the calculation for 16 out of 25 as a decimal. To calculate 16 out of 25 as a decimal, we divide 16 by 25. So, 16 out of 25 as a decimal is calculated like this:
16/25 = 0.64
## Conclusion
16 out of 25 is equal to 64 percent.
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## 2.5.3 Quadratic Equations, SPM Paper 2 (Long Questions)
Question 9:Solve the following quadratic equation:4x (x + 4) = 9 + 16xSolution: 4x( x+4 )=9+16x 4 x 2 +16x=9+16x 4 x 2 −9=0 ( 2x ) 2 − 3 2 =0 ( 2x+3 )( 2x−3 )=0 2x+3=0 or 2x−3=0 2x=−3 2x=3 x=− 3 2 x= 3 2 Question 10:Solve the following quadratic equation:(x + 2)2 = … Read more2.5.3 Quadratic Equations, SPM Paper 2 (Long Questions)
## 2.4 Roots of Quadratic Equations
2.4 Roots of Quadratic Equations 1. A root of quadratic equation is the value of the unknown which satisfies the quadratic equation. 2. Roots of an equation are also called the solution of an equation. 3. To solve a quadratic equation by the factorisation method, follow the steps below: Step 1: Express the quadratic equation in general form ax2 … Read more2.4 Roots of Quadratic Equations
2.3 Quadratic Equations 1. Quadratic equations are equations which fulfill the following characteristics: (a) Have an equal ‘=’ sign (b) Contain only one unknown (c) Highest power of the unknown is 2. For example, 2. The general form of a quadratic equation is written as: (a) ax2 + bx + c = 0, where a ≠ 0, b ≠ … Read more2.3 Quadratic Equations
## 2.2 Factorisation of Quadratic Expression
2.2 Factorisation of Quadratic Expression (A) Factorisation quadratic expressions of the form ax2 + bx + c, b = 0 or c = 0 1. Factorisation of quadratic expressions is a process of finding two linear expressions whose product is the same as the quadratic expression. 2. Quadratic expressions ax2 + c and ax2 … Read more2.2 Factorisation of Quadratic Expression
## 2.1.1 Quadratic Expression (Sample Questions)
Example 1: Form a quadratic expression by multiplying each of the following. (a) (6p – 2)(2p – 1) (b) (m + 5)(4 – 7m) (c) (x + 2) (2x – 3) Solution: (a) (6p – 2)(2p – 1) = (6p)(2p) + (6p)(-1) + (-2)(2p) +(-2)(-1) = 12p2 – 6p – 4p + 2 = 12p2 – … Read more2.1.1 Quadratic Expression (Sample Questions)
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RD Sharma XI 2020 2021 Volume 1 Solutions for Class 11 Commerce Maths Chapter 10 Sine And Cosine Formulae And Their Applications are provided here with simple step-by-step explanations. These solutions for Sine And Cosine Formulae And Their Applications are extremely popular among class 11 Commerce students for Maths Sine And Cosine Formulae And Their Applications Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XI 2020 2021 Volume 1 Book of class 11 Commerce Maths Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XI 2020 2021 Volume 1 Solutions. All RD Sharma XI 2020 2021 Volume 1 Solutions for class 11 Commerce Maths are prepared by experts and are 100% accurate.
#### Question 1:
If in ∆ABC, ∠A = 45°, ∠B = 60° and ∠C = 75°, find the ratio of its sides.
Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Then,
On multiplying by , we get:
Hence, the ratio of the sides is .
#### Question 2:
If in ∆ABC, ∠C = 105°, ∠B = 45° and a = 2, then find b.
#### Question 3:
In ∆ABC, if a = 18, b = 24 and c = 30 and ∠c = 90°, find sin A, sin B and sin C.
Given,∠C = 90°, a = 18, b = 24 and c = 30
According to sine rule, .
#### Question 4:
In triangle ABC, prove the following:
$\mathrm{Assume}\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k\phantom{\rule{0ex}{0ex}}$
Consider the LHS of the equation .
.
#### Question 5:
In triangle ABC, prove the following:
Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$ ...(1)
Consider the LHS of the equation
Hence proved.
#### Question 6:
In triangle ABC, prove the following:
Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ ...(1)
We need to prove:
Consider
#### Question 7:
In triangle ABC, prove the following:
Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ ...(1)
We need to prove:
Consider
#### Question 8:
In triangle ABC, prove the following:
Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ ...(1)
Then,
Consider the LHS of the equation .
#### Question 9:
In any triangle ABC, prove the following:
Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Then,
Consider the RHS of the equation
#### Question 10:
In triangle ABC, prove the following:
Let
Then,
Consider the LHS of the equation .
#### Question 11:
In triangle ABC, prove the following:
Let
Then,
Consider the LHS of he equation .
LHS
#### Question 12:
In triangle ABC, prove the following:
Let
Then,
Consider the RHS of the equation .
#### Question 13:
In triangle ABC, prove the following:
Consider the LHS of the equation .
Let
Then,
#### Question 14:
In triangle ABC, prove the following:
Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$
Then,
Consider the LHS of the equation .
#### Question 15:
In triangle ABC, prove the following:
Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$
Then,
Consider the LHS of the equation .
#### Question 16:
In triangle ABC, prove the following:
Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$
Then,
Consider the LHS of the equation .
#### Question 17:
In triangle ABC, prove the following:
Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$
Then,
Consider the LHS of the equation .
#### Question 18:
In triangle ABC, prove the following:
Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$
Then,
Consider the LHS of the equation .
#### Question 19:
In triangle ABC, prove the following:
Let $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$
Then,
Consider the LHS of the equation .
Also,
Similarly,
$\frac{{\mathrm{cos}}^{2}A-{\mathrm{cos}}^{2}B}{a+b}=\frac{\mathrm{sin}B-\mathrm{sin}A}{k}$
Thus,
Hence, in any triangle ABC, .
#### Question 20:
In ∆ABC, prove that:
.
Consider
$a\mathrm{sin}\frac{A}{2}\mathrm{sin}\left(\frac{B-C}{2}\right)+b\mathrm{sin}\frac{B}{2}\mathrm{sin}\left(\frac{C-A}{2}\right)+c\mathrm{sin}\frac{C}{2}\mathrm{sin}\left(\frac{A-B}{2}\right)$
Hence proved.
#### Question 21:
In ∆ABC, prove that:
Let ABC be any triangle.
Suppose $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$
Now,
Also,
From (1), (2) and (3), we get:
Hence proved.
#### Question 22:
In triangle ABC, prove the following:
So, from (1), we have
.
Hence proved.
#### Question 23:
Suppose $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$
Consider:
From (1), (2) and (3), we get:
Hence proved.
#### Question 24:
In ∆ABC, prove that
#### Question 25:
In ∆ABC, prove that if θ be any angle, then b cosθ = c cos (A − θ) + a cos (C + θ).
Suppose $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$. ...(1)
Consider the RHS of the equation b cosθ = c cos (A − θ) + a cos (C + θ).
#### Question 26:
In ∆ABC, if sin2A + sin2B = sin2C. show that the triangle is right-angled.
In ∆ ABC,
Given,
Suppose $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$ .
On putting these values in equation (1), we get:
Thus, ∆ ABC is right-angled.
#### Question 27:
In ∆ABC, if a2, b2 and c2 are in A.P., prove that cot A, cot B and cot C are also in A.P.
Then,
a2, b2 and c2 are in A.P.
#### Question 28:
The upper part of a tree broken by the wind makes an angle of 30° with the ground and the distance from the root to the point where the top of the tree touches the ground is 15 m. Using sine rule, find the height of the tree.
Suppose BD be the tree and the upper part of the tree is broken over by the wind at point A.
#### Question 29:
At the foot of a mountain, the elevation of it summit is 45°; after ascending 1000 m towards the mountain up a slope of 30° inclination, the elevation is found to be 60°. Find the height of the mountain.
Suppose, AB is a mountain of height t + x.
Hence, height of the mountain = .
#### Question 30:
A person observes the angle of elevation of the peak of a hill from a station to be α. He walks c metres along a slope inclined at an angle β and finds the angle of elevation of the peak of the hill to be ϒ. Show that the height of the peak above the ground is .
Suppose, AB is a peak whose height above the ground is t+x.
#### Question 31:
If the sides a, b and c of ∆ABC are in H.P., prove that are in H.P.
#### Question 1:
In , show that its area is . units.
#### Question 2:
In , show that its area is units.
#### Question 3:
The sides of a triangle are a = 4, b = 6 and c = 8. Show that .
Given:
Then,
Hence proved.
#### Question 4:
In ∆ ABC, if a = 18, b = 24 and c = 30, find cos A, cos B and cos C.
Hence,
#### Question 5:
In ∆ABC, prove the following:
Let ABC be any triangle.
Hence proved.
#### Question 6:
In ∆ABC, prove the following:
Consider
Hence proved.
#### Question 7:
In ∆ABC, prove the following:
LHS =
On using the cosine law, we get:
$\mathrm{LHS}=2\left[bc\left(\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\right)+ca\left(\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}\right)+ab\left(\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}\right)\right]$
$={b}^{2}+{c}^{2}-{a}^{2}+{a}^{2}+{c}^{2}-{b}^{2}+{a}^{2}+{b}^{2}-{c}^{2}\phantom{\rule{0ex}{0ex}}={a}^{2}+{b}^{2}+{c}^{2}=\mathrm{RHS}$
Hence proved.
#### Question 8:
In ∆ABC, prove the following:
From (1), (2) and (3), we get:
#### Question 9:
In ∆ABC, prove the following:
Let ABC be any triangle.
Hence proved.
#### Question 10:
In ∆ABC, prove that:
#### Question 11:
a cos A + b cos B + c cos C = 2b sin A sin C
Hence, a cos A + b cos B + c cos C = 2b sin A sin C.
#### Question 12:
In ∆ABC, prove the following:
Hence proved.
#### Question 13:
In ∆ABC, prove the following:
$=2\left(ab+bc+ac\right)+{b}^{2}+{c}^{2}-{a}^{2}+{c}^{2}+{a}^{2}-{b}^{2}+{a}^{2}+{b}^{2}-{c}^{2}\phantom{\rule{0ex}{0ex}}={a}^{2}+{b}^{2}+{c}^{2}+2ab+2bc+2ac\phantom{\rule{0ex}{0ex}}={\left(a+b+c\right)}^{2}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}$
Hence proved.
#### Question 14:
In ∆ABC, prove the following:
#### Question 15:
In . Prove that .
Hence proved.
In prove that .
Hence proved.
#### Question 17:
If in , prove that the triangle is right-angled.
Let ABC be any triangle.
In $∆\mathrm{ABC}$,
Hence, $∆$ABC is right angled.
#### Question 18:
In , prove that the triangle is isosceles.
Let $∆ABC$ be any triangle.
Suppose $\frac{\mathrm{sin}A}{a}=\frac{\mathrm{sin}B}{b}=\frac{\mathrm{sin}C}{c}=k$
If $\mathrm{cos}C=\frac{\mathrm{sin}A}{2\mathrm{sin}B}$, then
Thus, the lengths of two sides of the $∆ABC$ are equal.
Hence, $∆\mathrm{ABC}$ is an isosceles triangle.
#### Question 19:
Two ships leave a port at the same time. One goes 24 km/hr in the direction N 38° E and other travels 32 km/hr in the direction S 52° E. Find the distance between the ships at the end of 3 hrs.
#### Question 1:
Mark the correct alternative in each of the following:
In any ∆ABC, $\sum _{}^{}{a}^{2}\left(\mathrm{sin}B-\mathrm{sin}C\right)=$
(a) ${a}^{2}+{b}^{2}+{c}^{2}$ (b) ${a}^{2}$ (c) ${b}^{2}$ (d) 0
Using sine rule, we have
$\sum _{}^{}{a}^{2}\left(\mathrm{sin}B-\mathrm{sin}C\right)$
$={a}^{2}\left(\frac{b}{k}-\frac{c}{k}\right)+{b}^{2}\left(\frac{c}{k}-\frac{a}{k}\right)+{c}^{2}\left(\frac{a}{k}-\frac{b}{k}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{k}\left({a}^{2}b-{a}^{2}c+{b}^{2}c-{b}^{2}a+{c}^{2}a-{c}^{2}b\right)$
This expression cannot be simplified to match with any of the given options.
However, if the quesion is "In any ∆ABC, $\sum _{}^{}{a}^{2}\left({\mathrm{sin}}^{2}B-{\mathrm{sin}}^{2}C\right)=$", then the solution is as follows.
Using sine rule, we have
$\sum _{}^{}{a}^{2}\left({\mathrm{sin}}^{2}B-{\mathrm{sin}}^{2}C\right)$
$={a}^{2}\left(\frac{{b}^{2}}{{k}^{2}}-\frac{{c}^{2}}{{k}^{2}}\right)+{b}^{2}\left(\frac{{c}^{2}}{{k}^{2}}-\frac{{a}^{2}}{{k}^{2}}\right)+{c}^{2}\left(\frac{{a}^{2}}{{k}^{2}}-\frac{{b}^{2}}{{k}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{{k}^{2}}\left({a}^{2}{b}^{2}-{a}^{2}{c}^{2}+{b}^{2}{c}^{2}-{b}^{2}{a}^{2}+{c}^{2}{a}^{2}-{c}^{2}{b}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{{k}^{2}}×0\phantom{\rule{0ex}{0ex}}=0$
Hence, the correct answer is option (d).
Disclaimer: The question given in the book in incorrect or there is some printing mistake in the question.
#### Question 2:
Mark the correct alternative in each of the following:
In a ∆ABC, if a = 2, $\angle B=60°$ and $\angle C=75°$, then b =
(a) $\sqrt{3}$ (b) $\sqrt{6}$ (c) $\sqrt{9}$ (d) $1+\sqrt{2}$
It is given that a = 2, $\angle B=60°$ and $\angle C=75°$.
In ∆ABC,
Using sine rule, we get
Hence, the correct answer is option (b).
#### Question 3:
Mark the correct alternative in each of the following:
If the sides of a triangle are in the ratio $1:\sqrt{3}:2$, then the measure of its greatest angle is
(a) $\frac{\mathrm{\pi }}{6}$ (b) $\frac{\mathrm{\pi }}{3}$ (c) $\frac{\mathrm{\pi }}{2}$ (d) $\frac{2\mathrm{\pi }}{3}$
Let ∆ABC be the given triangle such that its sides are in the ratio $1:\sqrt{3}:2$.
$\therefore a=k,b=\sqrt{3}k,c=2k$
Now, ${a}^{2}+{b}^{2}={k}^{2}+3{k}^{2}=4{k}^{2}={c}^{2}$
So, ∆ABC is a right triangle right angled at C.
$\therefore C=90°$
Using sine rule, we have
Thus, the measure of its greatest angle is $\frac{\mathrm{\pi }}{2}$.
Hence, the correct answer is option (c).
#### Question 4:
Mark the correct alternative in each of the following:
In any ∆ABC, 2(bc cosA + ca cosB + ab cosC) =
(a) $abc$ (b) $a+b+c$ (c) ${a}^{2}+{b}^{2}+{c}^{2}$ (d) $\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}+\frac{1}{{c}^{2}}$
Using cosine rule, we have
$2\left(bc\mathrm{cos}A+ca\mathrm{cos}B+ab\mathrm{cos}C\right)\phantom{\rule{0ex}{0ex}}=2bc\left(\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\right)+2ca\left(\frac{{c}^{2}+{a}^{2}-{b}^{2}}{2ca}\right)+2ab\left(\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}\right)\phantom{\rule{0ex}{0ex}}={b}^{2}+{c}^{2}-{a}^{2}+{c}^{2}+{a}^{2}-{b}^{2}+{a}^{2}+{b}^{2}-{c}^{2}\phantom{\rule{0ex}{0ex}}={a}^{2}+{b}^{2}+{c}^{2}$
Hence, the correct answer is option (c).
#### Question 5:
Mark the correct alternative in each of the following:
In a triangle ABC, a = 4, b = 3, $\angle A=60°$ then c is a root of the equation
(a) ${c}^{2}-3c-7=0$ (b) ${c}^{2}+3c+7=0$ (c) ${c}^{2}-3c+7=0$ (d) ${c}^{2}+3c-7=0$
It is given that a = 4, b = 3 and $\angle A=60°$.
Using cosine rule, we have
$\mathrm{cos}A=\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}60°=\frac{9+{c}^{2}-16}{2×3×c}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=\frac{{c}^{2}-7}{6c}\phantom{\rule{0ex}{0ex}}⇒{c}^{2}-7=3c\phantom{\rule{0ex}{0ex}}⇒{c}^{2}-3c-7=0$
Thus, c is the root of ${c}^{2}-3c-7=0$.
Hence, the correct answer is option (a).
#### Question 6:
Mark the correct alternative in each of the following:
In a ∆ABC, if $\left(c+a+b\right)\left(a+b-c\right)=ab$, then the measure of angle C is
(a) $\frac{\mathrm{\pi }}{3}$ (b) $\frac{\mathrm{\pi }}{6}$ (c) $\frac{2\mathrm{\pi }}{3}$ (d) $\frac{\mathrm{\pi }}{2}$
Given: $\left(c+a+b\right)\left(a+b-c\right)=ab$
$⇒{\left(a+b\right)}^{2}-{c}^{2}=ab\phantom{\rule{0ex}{0ex}}⇒{a}^{2}+{b}^{2}+2ab-{c}^{2}=ab\phantom{\rule{0ex}{0ex}}⇒{a}^{2}+{b}^{2}-{c}^{2}=-ab\phantom{\rule{0ex}{0ex}}⇒\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}=-\frac{1}{2}$
Thus, the measure of angle C is $\frac{2\mathrm{\pi }}{3}$.
Hence, the correct answer is option (c).
#### Question 7:
Mark the correct alternative in each of the following:
In any ∆ABC, the value of $2ac\mathrm{sin}\left(\frac{A-B+C}{2}\right)$ is
(a) ${a}^{2}+{b}^{2}-{c}^{2}$ (b) ${c}^{2}+{a}^{2}-{b}^{2}$ (c) ${b}^{2}-{c}^{2}-{a}^{2}$ (d) ${c}^{2}-{a}^{2}-{b}^{2}$
In ∆ABC,
$\therefore 2ac\mathrm{sin}\left(\frac{A-B+C}{2}\right)\phantom{\rule{0ex}{0ex}}=2ac\mathrm{sin}\left(\frac{\mathrm{\pi }-2B}{2}\right)\phantom{\rule{0ex}{0ex}}=2ac\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}-B\right)\phantom{\rule{0ex}{0ex}}=2ac\mathrm{cos}B$
Hence, the correct answer is option (b).
#### Question 8:
Mark the correct alternative in each of the following:
In any ∆ABC, $a\left(b\mathrm{cos}C-c\mathrm{cos}B\right)=$
(a) ${a}^{2}$ (b) ${b}^{2}-{c}^{2}$ (c) 0 (d) ${b}^{2}+{c}^{2}$
Using cosine rule, we have
$a\left(b\mathrm{cos}C-c\mathrm{cos}B\right)\phantom{\rule{0ex}{0ex}}=ab\left(\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}\right)-ca\left(\frac{{c}^{2}+{a}^{2}-{b}^{2}}{2ca}\right)\phantom{\rule{0ex}{0ex}}=\frac{{a}^{2}+{b}^{2}-{c}^{2}-{c}^{2}-{a}^{2}+{b}^{2}}{2}\phantom{\rule{0ex}{0ex}}=\frac{2{b}^{2}-2{c}^{2}}{2}\phantom{\rule{0ex}{0ex}}={b}^{2}-{c}^{2}$
Hence, the correct answer is option (b).
#### Question 1:
In a ∆ABC, if $\frac{\mathrm{sin}\left(A–B\right)}{\mathrm{sin}\left(A+B\right)}=\frac{{a}^{2}-{b}^{2}}{k},$ then k = ___________.
In ∆ABC,
Given
#### Question 2:
In a ∆ABC, if c2 + a2b2 = ac, then the measure of angle B is ____________.
In a ∆ABC,
if c2 + a2 − b2 = ac
i.e cosB$\frac{1}{2}$
i.e B =$\frac{\mathrm{\pi }}{3}$
#### Question 3:
In a ∆ABC, if and a = 2, then area of ∆ABC is equal to __________.
In a ∆ABC
Given
Using sine formula
using above two, (1) and (2)
we get,
⇒ cotA = cotB = cotC
⇒ angle A = B = C = 60°
⇒ ∆ABC is an equilateral triangle
#### Question 4:
In a triangle ABC, if a = 2, b = 4 and $A+B=\frac{2\mathrm{\pi }}{3},$ then area of ∆ABC is __________.
Since A + B$\frac{2\mathrm{\pi }}{3}$ given
and A + B + C$\mathrm{\pi }$ (Angle sum property)
$\begin{array}{rcl}& ⇒& C=\mathrm{\pi }-\frac{2\mathrm{\pi }}{3}\\ C& =& \frac{\mathrm{\pi }}{3}\end{array}$
Since area of triangle ABC = $\frac{1}{2}$ ab sinC
#### Question 5:
The angles A, B, C of a ∆ABC are in AP and the sides a, b, c are in G.P. If a2 + c2 = λb2, then λ = ____________.
Since A, B, C are A.P
⇒ 2B = A + C
Since A + B + C$\mathrm{\pi }$ (By angle sum property)
⇒ 3B$\mathrm{\pi }$
also,
Since a, b, c are in g.p
b2 = ac ....(2)
Using cosB$\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}$
i.e from (1) and (2).
⇒ ac = a2 + c2 - ac
a2 + c2 = 2ac
a2 + c2 = 2b2 from (2)
Hence $\lambda$ = 2
#### Question 6:
In a ∆ABC, if ∠C = 60°, a = 47 cm and b = 94 cm, then c2 = ____________.
Given ∠C = 60∘ , a = 47 , b = 94 In ∆ABC
Using cosC
i.e cos 60∘
i.e $2×\frac{1}{2}×47×94={\left(47\right)}^{2}+{\left(94\right)}^{2}-{C}^{2}$
i.e C2 = (47)2 + (94)2 − 47 × 47 × 2
= − (47)2 + (94)2
C2 = 8836 − 22049 = 6627
#### Question 7:
In a ∆ABC, if c = 20, then a = ____________.
In ∆ABC
∠C = $\frac{\mathrm{\pi }}{2}$, ∠ A = $\frac{\mathrm{\pi }}{6}$
∠B = $\frac{\mathrm{\pi }}{3}$ (By angle sum property A + B + C = $\mathrm{\pi }$)
using sina formula,
$\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}$
⇒ $\frac{a}{\mathrm{sin}\frac{\mathrm{\pi }}{6}}=\frac{20}{s\mathrm{in}\frac{\mathrm{\pi }}{2}}$ ($\because$c = 20 given)
a = sin ($\because$ sin π/2 = 1)
i.e a = 10
#### Question 8:
In a ∆ABC, if then k = ___________________.
Using cosine formula
#### Question 9:
In a ∆ABC, if c2 sin A sin B = ab, then A + B = ______________.
Since c2 sinA sinB = ab
#### Question 10:
In a ∆ABC, if a = 8, b = 9 and 3 cos C = 2, then C = ____________.
a = 8 , b = 9
3 cosC = 2
⇒ cosC$\frac{2}{3}$
also using, cosine formula
#### Question 11:
In a ∆ABC, if and $B–C=\frac{\mathrm{\pi }}{2},$ then A = ______________.
Given b $\sqrt{3}$ , c = 1 and BC$\frac{\mathrm{\pi }}{2}$ In a triangle ABC
By angle sum property,
Since A + B + C = $\frac{\mathrm{\pi }}{2}$ ....(1)
and by Sine formula,
$\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}\phantom{\rule{0ex}{0ex}}⇒\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}\phantom{\rule{0ex}{0ex}}⇒\frac{\sqrt{3}}{\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}+c\right)}=\frac{c=1}{\mathrm{sin}C}$
#### Question 12:
If angles of a triangle are in A.P. and then C = ______________.
If angle of a triangle ABC are in A.P
⇒ 2∠B = ∠A + ∠C
and $\frac{b}{c}=\frac{\sqrt{3}}{\sqrt{2}}$
By angle sum property
∠A + ∠B + ∠C$\mathrm{\pi }$
⇒ 2∠B + ∠B$\mathrm{\pi }$
⇒ ∠B$\frac{\mathrm{\pi }}{3}$
also,
Using Sine formula
#### Question 13:
If the sides of a ∆ABC are then the measure of the largest angle is ____________.
Let us suppose the greatest angle is c
Using cosine formula,
#### Question 14:
In a ∆ABC, if a4 + b4 + c4 = 2a2b2 + 2b2c2, then B = __________.
In a ∆ABC,
Given a4 + b4 + c4 = 2a2b2 + 2b2c2
i.e a4 + b4 − 2a2b2 + c2 − 2b2c2 = 0
i.e (a2 + c2)2 + b4− 2a2b2 − 2b2c2 − 2a2c2 = 0
i.e (a2 + c2)2 + b4 − 2b2 (a2 + c2) − 2a2c2 = 0
i.e (a2 + c2 − b2)2 = 2a2c2
i.e a2 + c2 − b2
Using cosine formula,
i.e cosB = $\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}=\frac{\sqrt{2}ac}{2ac}$
cosB = $\frac{1}{\sqrt{2}}$
i.e
#### Question 15:
In a ∆ABC, if a = 4, b = 3, $A=\frac{\mathrm{\pi }}{3}.$ Then side C is given by ____________.
In ∆ABC
if a = 4, b = 3, A = $\frac{\mathrm{\pi }}{3}$
#### Question 1:
Answer each of the following questions in one word or one sentence or as per exact requirement of the question.
Find the area of the triangle ∆ABC in which a = 1, b = 2 and $\angle C=60°$.
In ∆ABC, a = 1, b = 2 and $\angle C=60°$.
∴ Area of the ∆ABC
#### Question 2:
Answer each of the following questions in one word or one sentence or as per exact requirement of the question.
In a ∆ABC, if b = $\sqrt{3}$, c = 1 and $\angle A=30°$, find a.
In ∆ABC, b = $\sqrt{3}$, c = 1 and $\angle A=30°$.
Using cosine formula, we have
$\mathrm{cos}A=\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}30°=\frac{{\left(\sqrt{3}\right)}^{2}+{\left(1\right)}^{2}-{a}^{2}}{2×\sqrt{3}×1}\phantom{\rule{0ex}{0ex}}⇒\frac{\sqrt{3}}{2}=\frac{4-{a}^{2}}{2\sqrt{3}}\phantom{\rule{0ex}{0ex}}⇒3=4-{a}^{2}\phantom{\rule{0ex}{0ex}}⇒{a}^{2}=4-3=1\phantom{\rule{0ex}{0ex}}⇒a=1$
#### Question 3:
Answer each of the following questions in one word or one sentence or as per exact requirement of the question.
In a ∆ABC, if $\mathrm{cos}A=\frac{\mathrm{sin}B}{2\mathrm{sin}C}$, then show that c = a.
Given: $\mathrm{cos}A=\frac{\mathrm{sin}B}{2\mathrm{sin}C}$
$⇒\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}=\frac{b}{2c}$ (Using sine rule and cosine rule)
$⇒{b}^{2}+{c}^{2}-{a}^{2}={b}^{2}$
$⇒{c}^{2}={a}^{2}$
$⇒c=a$
#### Question 4:
Answer each of the following questions in one word or one sentence or as per exact requirement of the question.
In a ∆ABC, if b = 20, c = 21 and $\mathrm{sin}A=\frac{3}{5}$, find a.
In ∆ABC, b = 20, c = 21 and $\mathrm{sin}A=\frac{3}{5}$.
Using cosine rule, we have
$⇒{a}^{2}=841-672=169\phantom{\rule{0ex}{0ex}}⇒a=13$
#### Question 5:
Answer each of the following questions in one word or one sentence or as per exact requirement of the question.
In a ∆ABC, if sinA and sinB are the roots of the equation ${c}^{2}{x}^{2}-c\left(a+b\right)x+ab=0$, then find $\angle C$.
It is given that sinA and sinB are the roots of the equation ${c}^{2}{x}^{2}-c\left(a+b\right)x+ab=0$.
$⇒\mathrm{sin}A+\mathrm{sin}B=\frac{\mathrm{sin}A+\mathrm{sin}B}{\mathrm{sin}C}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}C=1=\mathrm{sin}90°\phantom{\rule{0ex}{0ex}}⇒C=90°$
#### Question 6:
Answer each of the following questions in one word or one sentence or as per exact requirement of the question.
In ∆ABC, if a = 8, b = 10, c = 12 and C = λA, find the value of λ.
Using cosine rule, we have
Now, using sine rule, we have
$\frac{a}{\mathrm{sin}A}=\frac{c}{\mathrm{sin}C}\phantom{\rule{0ex}{0ex}}⇒\frac{8}{\mathrm{sin}A}=\frac{12}{\mathrm{sin}\lambda A}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\lambda A=\frac{3}{2}\mathrm{sin}A\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\lambda A=2×\frac{3}{4}\mathrm{sin}A$
#### Question 7:
Answer each of the following questions in one word or one sentence or as per exact requirement of the question.
If the sides of a triangle are proportional to 2, $\sqrt{6}$ and $\sqrt{3}-1$, find the measure of its greatest angle.
Let ∆ABC be the triangle such that a = 2, b = $\sqrt{6}$ and c = $\sqrt{3}-1$.
Clearly, b > a > c. Then,
$\angle$B is the greatest angle of ∆ABC. (Greatest side has greatest angle opposite to it)
Using cosine formula, we have
$\mathrm{cos}B=\frac{{c}^{2}+{a}^{2}-{b}^{2}}{2ca}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}B=\frac{{\left(\sqrt{3}-1\right)}^{2}+{2}^{2}-{\left(\sqrt{6}\right)}^{2}}{2×\left(\sqrt{3}-1\right)×2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}B=\frac{3+1-2\sqrt{3}+4-6}{4\left(\sqrt{3}-1\right)}$
$⇒\mathrm{cos}B=\frac{2-2\sqrt{3}}{4\left(\sqrt{3}-1\right)}=\frac{-2\left(\sqrt{3}-1\right)}{4\left(\sqrt{3}-1\right)}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}B=-\frac{1}{2}=\mathrm{cos}120°\phantom{\rule{0ex}{0ex}}⇒B=120°$
Hence, the measure of its greatest angle is 120º.
#### Question 8:
Answer each of the following questions in one word or one sentence or as per exact requirement of the question.
If in a ∆ABC, $\frac{\mathrm{cos}A}{a}=\frac{\mathrm{cos}B}{b}=\frac{\mathrm{cos}C}{c}$, then find the measures of angles A, B, C.
In ∆ABC,
⇒ ∆ABC is an equilateral triangle.
A = B = C = 60º
#### Question 9:
Answer each of the following questions in one word or one sentence or as per exact requirement of the question.
In any triangle ABC, find the value of $a\mathrm{sin}\left(B-C\right)+b\mathrm{sin}\left(C-A\right)+c\mathrm{sin}\left(A-B\right)$.
Using sine rule, we have
$=k\left[\mathrm{sin}\left(B+C\right)\mathrm{sin}\left(B-C\right)+\mathrm{sin}\left(C+A\right)\mathrm{sin}\left(C-A\right)+\mathrm{sin}\left(A+B\right)\mathrm{sin}\left(A-B\right)\right]\phantom{\rule{0ex}{0ex}}=k\left({\mathrm{sin}}^{2}B-{\mathrm{sin}}^{2}C+{\mathrm{sin}}^{2}C-{\mathrm{sin}}^{2}A+{\mathrm{sin}}^{2}A-{\mathrm{sin}}^{2}B\right)\phantom{\rule{0ex}{0ex}}=k×0\phantom{\rule{0ex}{0ex}}=0$
Hence, the required value is 0.
#### Question 10:
Answer each of the following questions in one word or one sentence or as per exact requirement of the question.
In any ∆ABC, find the value of $\sum _{}^{}a\left(\mathrm{sin}B-\mathrm{sin}C\right)$.
Using sine rule, we have
$\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k\phantom{\rule{0ex}{0ex}}⇒a=k\mathrm{sin}A,b=k\mathrm{sin}B,c=k\mathrm{sin}C$
$\therefore \sum _{}^{}a\left(\mathrm{sin}B-\mathrm{sin}C\right)\phantom{\rule{0ex}{0ex}}=\sum _{}^{}k\mathrm{sin}A\left(\mathrm{sin}B-\mathrm{sin}C\right)$
$=k\sum _{}^{}\mathrm{sin}A\left(\mathrm{sin}B-\mathrm{sin}C\right)$
$=k\left[\mathrm{sin}A\left(\mathrm{sin}B-\mathrm{sin}C\right)+\mathrm{sin}B\left(\mathrm{sin}C-\mathrm{sin}A\right)+\mathrm{sin}C\left(\mathrm{sin}A-\mathrm{sin}B\right)\right]$
$=k\left(\mathrm{sin}A\mathrm{sin}B-\mathrm{sin}A\mathrm{sin}C+\mathrm{sin}B\mathrm{sin}C-\mathrm{sin}B\mathrm{sin}A+\mathrm{sin}C\mathrm{sin}A-\mathrm{sin}C\mathrm{sin}B\right)$
$=k×0=0$
View NCERT Solutions for all chapters of Class 13
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# Solve Geometric Limits
In math, the model of a "limit" is used to illustrate the value that a function or order "approaches" as the input or key approaches a number of values. The model of limit allows one to, in a total space; define a new point since a Cauchy sequence of previously clear points. In this article we study about geometric limits and develop the knowledge of the geometry.
Proof for Solve Geometric Limits:
Prove and solve that `lim_(A->0)` ` (sin A)/ A` = 1
proof:
geometric limit
We take y =` (sin A)/A` . This function is defined for all A, other than A = 0, for which both numerator and denominator become zero. When A is replaced by − A, the magnitude of the fraction`(sin A)/A` does not change since `(sin (-A))/(- A)` =`(sin A)/A .`
Therefore it is enough to find the limit of the fraction as θ tends to 0 through positive values. i.e. in the first quadrant. We consider a circle with centre at O radius unity. P, Q are two points on this circle so
OP = OQ = 1. Let A be the angle subtended at the centre by the arc PT. Measuring angle in radians, we have sin A = PR, R being a point on PR such that OS passes through R. cosA = OR,P =`1/2` arc PQ, OPS = 90°
In triangle OPS, PS = tanA. Now length of arc AB = 2A and length of the chord AB = 2 sinA sum of the tangents = PS + QS = 2 tan A
Since the length of the arc is intermediate between the length of chord and the sum of the tangents we can write 2 sin A < 2A < 2 tan A.
Dividing by 2 sin A , we have 1 < A sinA <`1/(cos A)` or 1 >`(sin A)/A ` > cos A . But as A → 0, cos A, given by the distance OR, tends to 1
That is, `lim_(A->0)` cos A = 1 . Therefore 1 >`lim_(A->0)``( sin A)/A` > 1, by 3 .
That is, the variable y =`(sin A)/A` always lies between unity and a magnitude tending to unity, and hence `lim_(A->0)` `(sin A)/A` = 1.
Examples for Solve Geometric Limits:
Solve that using geometric limits `lim_(A->0)` `(1 - cos A)/A^2` .
Solution:
`(1 - cosA)/A^2` = `{2(sin^2 A/2)}/(A^2)`
`(1 - cosA)/A^2` = `{(1/2){ sin^2(A/2)}/(A/2)^2}`
=`1/2` `{ sin(A/2)/(A/2)}^2`
If A→ 0, α =`A/2` also tends to 0 and `lim_(A->0)` `sin(A/2)/(A/2)` =limα → 0 ( sin α)/α = 1 and
hence`lim_(A->0)` ` (1- cosA)/A^2` =`lim_(A->0)` ` ((1/2)(sin(A/2))/(A/2))^2`
=`1/2` `lim_(A->0)` `{ sin(A/2)/(A/2)}^2`
=`1/2` × `1^2 ` =`1/2`
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Chapter 3 Hm. 2
# Chapter 3 Hm. 2 - Solutions to Homework#2 Chapter 3 12...
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Unformatted text preview: Solutions to Homework #2 Chapter 3 12. Picture the Problem : The four possible locations of the treasure are labeled A , B , C , and D in the figure at right. The position vector for location A is also drawn. North is up and east is to the right. Strategy: Use the vector components to find the magnitude and direction of each vector. Solution: 1. Find the magnitude of A r : ( 29 ( 29 2 2 25.0 m 12.0 5.00 m 30.2 m A = + + = 2. Find the direction from north of A r : 1 25.0 m tan 55.8 west of north 12.0 5.00 m A θ- = = ° ÷ + 3. Find the magnitude of B r : ( 29 ( 29 2 2 25.0 m 12.0 5.00 m 26.0 m B = +- = 4. Find the direction from north of B r : 1 25.0 m tan 74.4 west of north 12.0 5.00 m B θ- = = ° ÷- 5. Find the magnitude of C r : ( 29 ( 29 2 2 25.0 m 12.0 5.00 m 30.2 m C = + + = 6. Find the direction from north of C r : 1 25.0 m tan 55.8 east of north 12.0 5.00 m C θ- = = ° ÷ + 7. Find the magnitude of D r : ( 29 ( 29 2 2 25.0 m 12.0 5.00 m 26.0 m D = +- = 8. Find the direction from north of D r : 1 25.0 m tan 74.4 east of north 12.0 5.00 m D θ- = = ° ÷- Insight: If you ever find a treasure map like this one, you’ll be glad you mastered vectors in physics! 18. Picture the Problem : The vectors involved in the problem are depicted at right. Strategy: Since A r points entirely in the x direction, and B r points entirely in the y direction, A r and B r are the x and y components of their sum + A B r r . Use the known lengths of + A B r r and A r to find B . Solution: 1. Set the length of + A B r r equal to 37 units: 2. Solve for B : Insight: Here the length of B r is determined by the lengths of the other two vectors because the directions of A r and B r are stipulated. Learning to manipulate vector components will be a useful skill when tackling many physics problems. 12.0 m 25.0 m 25.0 m 5.00 m A B C D palm tree A r θ A x y B r A r + A B r r −22 O 30 26. Picture the Problem : The vectors involved in the problem are depicted at right....
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Chapter 3 Hm. 2 - Solutions to Homework#2 Chapter 3 12...
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