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2

## How do use the binomial theorem to calculate ""^8C_5?

George C.
Featured 3 weeks ago

""^8C_5 = 56

#### Explanation:

This seems a curious question to me, since normally you would use ""^8C_5 = (8!)/(3!5!) to find a binomial coefficient, but let's give it a try.

Applying the Binomial theorem to $f \left(x\right) = {\left(x + 1\right)}^{8}$ we have:

${\left(x + 1\right)}^{8} = {\text{^8C_0 x^8 + ""^8C_1 x^7 + ""^8C_2 x^6 + ""^8C_3 x^5 + ""^8C_4 x^4 + ""^8C_5 x^3 + ""^8C_6 x^2 + ""^8C_7 x + }}^{8} {C}_{8}$

Taking the derivative $3$ times, we find:

$8 \cdot 7 \cdot 6 {\left(x + 1\right)}^{5}$

$= {f}^{\left(3\right)} \left(x\right)$

$= 8 \cdot 7 \cdot 6 \cdot {\text{^8C_0 x^5 + 7 * 6 * 5 * ""^8C_1 x^4 + 6 * 5 * 4 * ""^8C_2 x^3 + 5 * 4 * 3 * ""^8C_3 x^2 + 4 * 3 * 2 * ""^8C_4 x + 3 * 2 * 1 * }}^{8} {C}_{5}$

So:

$8 \cdot 7 \cdot 6$

$= 8 \cdot 7 \cdot 6 \cdot {\left(\left(\textcolor{b l u e}{0}\right) + 1\right)}^{5}$

$= {f}^{\left(3\right)} \left(0\right)$

$= 8 \cdot 7 \cdot 6 \cdot {\text{^8C_0 (color(blue)(0))^5 + 7 * 6 * 5 * ""^8C_1 (color(blue)(0))^4 + 6 * 5 * 4 * ""^8C_2 (color(blue)(0))^3 + 5 * 4 * 3 * ""^8C_3 (color(blue)(0))^2 + 4 * 3 * 2 * ""^8C_4 (color(blue)(0)) + 3 * 2 * 1 * }}^{8} {C}_{5}$

= 3 * 2 * 1 * ""^8C_5

So:

""^8C_5 = (8 * 7 * 6) / (3 * 2 * 1) = 56

Alternatively, you might recognise that the method of constructing Pascal's triangle relates precisely to the values of the coefficients of ${\left(a + b\right)}^{n}$ and hence deduce that the appropriate entry in Pascal's triangle must be the binomial coefficient.

The row of Pascal's triangle starting $1 , 8 , \ldots$ contains the coefficients for ${\left(a + b\right)}^{8}$, with the coefficient of ${a}^{3} {b}^{5}$ being $56$, a.k.a. ""^8C_5 ...

1

## How do you write an equation for the hyperbola with asymptote y=+-(4x)/3, contains (3sqrt(2,)4)?

Ratnaker Mehta
Featured 3 weeks ago

${x}^{2} / 9 - {y}^{2} / 16 = 1$.

#### Explanation:

Recall that, for the Hyperbola $S : {x}^{2} / {a}^{2} - {y}^{2} / {b}^{2} = 1$, the

asymptotes are given by, y=+-b/a*x;" where, "a,b gt 0.

In our Case, the asymptotes are, $y = \pm \frac{4}{3} \cdot x$.

$\therefore \frac{b}{a} = \frac{4}{3.} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$.

Next, $\left(3 \sqrt{2} , 4\right) \in S \Rightarrow {\left(3 \sqrt{2}\right)}^{2} / {a}^{2} - {4}^{2} / {b}^{2} = 1$.

$\therefore \frac{18}{a} ^ 2 - \frac{16}{b} ^ 2 = 1$.

Multiplying by ${b}^{2}$, we get, $18 \cdot {b}^{2} / {a}^{2} - 16 = {b}^{2}$.

By $\left(1\right)$, then, $18 \cdot {\left(\frac{4}{3}\right)}^{2} - 16 = {b}^{2} , i . e . , {b}^{2} = 16$.

$\therefore b = 4 \left(\because , b > 0\right) , \mathmr{and} , \left(1\right) \Rightarrow a = 3 , \mathmr{and} , {a}^{2} = 9$.

With ${a}^{2} = 9 , \mathmr{and} {b}^{2} = 16$, the reqd. eqn. of the Hyperbola is,

${x}^{2} / 9 - {y}^{2} / 16 = 1$.

1

## How do you solve x^2-4x>=5?

Nimo N.
Featured 2 weeks ago

Solution set
 color(red) (-oo <= x <= -1"," or 5 <= x <= oo
Alternate notation:
 color(red) ( x in (-oo, -1] uu [5, oo)

#### Explanation:

Solve:
 x^2 − 4x ≥ 5

1) Examine the related equation to find the boundaries.
 x^2 − 4x ≥ 5
 x^2 − 4x - 5 = 0
$\left(x - 5\right) \left(x + 1\right) = 0$
$\left(x - 5\right) = 0$
$x = 5$
or
$\left(x + 1\right) = 0$
$x = - 1$

The solutions to the related equation define boundary points that divide the number line into 5 parts, 3 regions and the 2 boundary points.

• Since the inequality is inclusive ("non-strict") the boundary points are in the solution set.

A graph of the number line, with solid dots on (-1), and (5) will help in thinking about the problem.

To check for the solution, use test points, one from each region, in the inequality to see if they qualify:

Try (-2), (0), and (6). It is best not to pick numbers that make the arithmetic difficult.

In region, $- \infty < x < - 1$
 ( -2 )^2 − 4( -2) ≥ 5 " ?"
 4 + 8 ≥ 5 " ?"
 12 ≥ 5 " YES"

In region, $- 1 < x < 5$
 ( 0 )^2 − 4( 0 ) ≥ 5 " ?"
 0 ≥ 5 " NO"

In region, $5 < x < \infty$
 ( 6 )^2 − 4( 6 ) ≥ 5 " ?"
 36 − 24 ≥ 5 " ?"
 12 ≥ 5 " YES"

We must be careful in stating the solution inequalities. x can not be in both regions at once, since they are disconnected. There is a full region between the two regions in the solution, so we can't use "and", we must use "or".

Solution set
 color(red) (-oo <= x <= -1"," or 5 <= x <= oo
Alternate notation:
 color(red) ( x in (-oo, -1] uu [5, oo)

2-D Graph of the inequality:

1

## How do you solve the system r+s+t=15, r+t=12, s+t=10 using matrices?

Cem Sentin
Featured 2 weeks ago

$r = 5$, $s = 3$ and $t = 7$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

$A = \left(\begin{matrix}1 & 1 & 1 & | & 15 \\ 1 & 0 & 1 & | & 12 \\ 0 & 1 & 1 & | & 10\end{matrix}\right)$

I have written the equations not in the sequence as in the question in order to get $1$ as pivot.

Perform the folowing operations on the rows of the matrix

$R 2 \leftarrow R 2 - R 1$

$A = \left(\begin{matrix}1 & 1 & 1 & | & 15 \\ 0 & - 1 & - 0 & | & - 3 \\ 0 & 1 & 1 & | & 10\end{matrix}\right)$

$R 1 \leftarrow R 1 + R 2$; $R 3 \leftarrow R 3 + R 2$

$A = \left(\begin{matrix}1 & 0 & 1 & | & 12 \\ 0 & - 1 & - 0 & | & - 3 \\ 0 & 0 & 1 & | & 7\end{matrix}\right)$

$R 1 \leftarrow R 1 - R 3$

$A = \left(\begin{matrix}1 & 0 & 0 & | & 5 \\ 0 & - 1 & - 0 & | & - 3 \\ 0 & 0 & 1 & | & 7\end{matrix}\right)$

$R 2 \leftarrow \left(R 2\right) \cdot \left(- 1\right)$

$A = \left(\begin{matrix}1 & 0 & 0 & | & 5 \\ 0 & 1 & 0 & | & 3 \\ 0 & 0 & 1 & | & 7\end{matrix}\right)$

Thus $r = 5$, $s = 3$ and $t = 7$

1

## If ( 1 + x )^n = C_0 + C_1 x_1 + C_2 x_2 + ⋯ + C_n x_n then show that C_0C_r+C_1C_(r+1)+C_2C_(r+2)+....C_nC_(r+n)=((2n)!)/((n+r)!(n-r)!) ?

Shwetank Mauria
Featured 2 weeks ago

#### Explanation:

We know that

(a+b)^n=C_0^na^n+C_1^na^(n-1)b+C_2^na^(n-2)b^2+ ⋯ + C_n^nb^n

and ${r}^{t h}$ term is ${C}_{r}^{n} {a}^{r} {b}^{n - r}$ and C_r^n=(n!)/(r!(n-r)!)

and hence (1+x)^n=C_0 1^n+C_1 1^(n-1)*x+C_2 1^(n-2)x^2+ ⋯ + C_nx^n

= (1+x)^n=C_0 1^n+C_1x+C_2x^2+ ⋯ + C_nx^n (A)

and similarly (x+1)^n=C_0 x^n+C_1 x^(n-1)*1+C_2 x^(n-2)1^2+ ⋯ + C_n1^n

= (x+1)^n=C_0x^n+C_1x^(n-1)+C_2x^(n-2)+ ⋯ +C_n1 (B)

Multipllying (A) and (B), we get ${\left(1 + x\right)}^{n} {\left(x + 1\right)}^{n}$ on the LHS and multiplication of expansion on the RHS.

Now what is coefficient of ${x}^{n + r}$ in this.

While on LHS we have ${\left(1 + x\right)}^{2 n}$ and coefficient of ${x}^{n + r}$ is (2n!)/((n+r)!(2n-n-r)!)=(2n!)/((n+r)!(n-r)!)

On the RHS we get ${x}^{n + r}$, when ${C}_{0} {x}^{n}$ is multiplied by ${C}_{r} {x}^{r}$; ${C}_{1} {x}^{n - 1}$ is multiplied by ${C}_{r + 1} {x}^{r + 1}$; ${C}_{2} {x}^{n - 2}$ is multiplied by ${C}_{r + 2} {x}^{r + 2}$ and so on till we get ${C}_{n} {x}^{n}$ multipled by C_(r+n)x^(n-(r+n).

Hence coefficient of ${x}^{n + r}$ is

${C}_{0} {C}_{r} + {C}_{1} {C}_{r + 1} + {C}_{2} {C}_{r + 2} + \ldots . {C}_{n} {C}_{r + n}$

and hence C_0C_r+C_1C_(r+1)+C_2C_(r+2)+....C_nC_(r+n)=((2n)!)/((n+r)!(n-r)!)

1

## How do you find the horizontal and vertical asymptote of the following: f(x) = (2x-3)/(x^2+2)?

Jim G.
Featured 2 weeks ago

$\text{horizontal asymptote at } y = 0$

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

$\text{solve } {x}^{2} + 2 = 0 \Rightarrow {x}^{2} = - 2$

$\text{this has no real solutions hence there are no vertical}$
$\text{asymptotes}$

$\text{horizontal asymptotes occur as}$

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ ( a constant)}$

$\text{divide terms on numerator/denominator by the highest}$
$\text{power of x that is } {x}^{2}$

$f \left(x\right) = \frac{\frac{2 x}{x} ^ 2 - \frac{3}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{2}{x} ^ 2} = \frac{\frac{2}{x} - \frac{3}{x} ^ 2}{1 + \frac{2}{x} ^ 2}$

$\text{as } x \to \pm \infty , f \left(x\right) \to \frac{0 - 0}{1 + 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$
graph{(2x-3)/(x^2+2) [-10, 10, -5, 5]}

2

## How to do questions b?

Somebody N.
Featured 2 weeks ago

$\boldsymbol{Y} = \left[\begin{matrix}- \frac{11}{16} & \frac{17}{16} \\ - \frac{1}{4} & \frac{3}{4}\end{matrix}\right]$

#### Explanation:

First find $\boldsymbol{{A}^{-} 1}$

The easiest way to find the inverse of $\boldsymbol{A}$ is to find the determinant of $\boldsymbol{A}$:

This is just:

$\left(3 \times 6\right) - \left(1 \times 2\right) = 16$

Next switch the elements on the leading diagonal of $\boldsymbol{A}$ and change the signs of the elements on the non-leading diagonal of $\boldsymbol{A}$

So you should have:

$\left[\begin{matrix}6 & - 2 \\ - 1 & 3\end{matrix}\right]$

Divide each element by the determinant $\boldsymbol{16}$:

$\left[\begin{matrix}\frac{6}{16} & - \frac{2}{16} \\ - \frac{1}{16} & \frac{3}{16}\end{matrix}\right] = \left[\begin{matrix}\frac{3}{8} & - \frac{1}{8} \\ - \frac{1}{16} & \frac{3}{16}\end{matrix}\right]$

$\boldsymbol{{A}^{-} 1} = \left[\begin{matrix}\frac{3}{8} & - \frac{1}{8} \\ - \frac{1}{16} & \frac{3}{16}\end{matrix}\right]$

Now:

$\boldsymbol{Y A} + \boldsymbol{B} = \boldsymbol{C}$

$\boldsymbol{Y A} = \boldsymbol{C - B}$

Using $\boldsymbol{{A}^{-} 1}$

$\boldsymbol{Y A {A}^{-} 1} = \boldsymbol{\left(C - B\right) {A}^{-} 1}$

$\boldsymbol{Y} = \boldsymbol{\left(C - B\right) {A}^{-} 1}$

Note we are post multiplying on both sides. This is important as Matrix multiplication is non-commutative.

i.e.

$\boldsymbol{A B} \ne \boldsymbol{B A}$ ( In general )

$\boldsymbol{C - B} = \left[\begin{matrix}3 & 4 \\ 2 & 6\end{matrix}\right] - \left[\begin{matrix}4 & - 1 \\ 2 & 2\end{matrix}\right] = \left[\begin{matrix}- 1 & 5 \\ 0 & 4\end{matrix}\right]$

$\therefore$

$\boldsymbol{Y} = \left[\begin{matrix}- 1 & 5 \\ 0 & 4\end{matrix}\right] \left[\begin{matrix}\frac{3}{8} & - \frac{1}{8} \\ - \frac{1}{16} & \frac{3}{16}\end{matrix}\right] = \left[\begin{matrix}- \frac{11}{16} & \frac{17}{16} \\ - \frac{1}{4} & \frac{3}{4}\end{matrix}\right]$

$\boldsymbol{Y} = \left[\begin{matrix}- \frac{11}{16} & \frac{17}{16} \\ - \frac{1}{4} & \frac{3}{4}\end{matrix}\right]$

1

## Are there polynomial functions whose graphs have: 11 points of inflection, but no max or min ?

Cesareo R.
Featured 1 week ago

See below.

#### Explanation:

You can build a polynomial with as many inflexion points as needed using the truncated series expansion for sinx = sum_(k=0)^n (-1)^kx^(2k-1)/((2k-1)!) added to a line with convenient gradient or as

p_n(x,m) = sum_(k=0)^n (-1)^kx^(2k-1)/((2k-1)!)+ m x

For instance, an example for $m = - 2$ and $n = 21$ with exactly $11$ inflexion points.

The next plot shows $\frac{d}{\mathrm{dx}} {p}_{21} \left(x , - 2\right)$. As we can observe $\frac{d}{\mathrm{dx}} {p}_{21} \left(x , - 2\right) = 0$ does not have real roots, then ${p}_{21} \left(x , - 2\right)$ has not relative maxima/minima.

And finally the plot for ${d}^{2} / \left({\mathrm{dx}}^{2}\right) {p}_{21} \left(x , - 2\right)$ showing the inflexion points location. as the roots of ${d}^{2} / \left({\mathrm{dx}}^{2}\right) {p}_{21} \left(x , - 2\right) = 0$ . Here we can count exactly $11$ inflexion points,

NOTE

Depending on $n$ the sign for $m$ can be positive or negative, and $n > 4$. It is left as an exercise to determine the connection for the $m$ sign with the $n$ value as well as the dependency between the sought inflection points number, with the $n$ value.

4

## Can you calculate \qquad \qquad e^{ ( ( ln(2), 1, 1, 1 ), ( 0, ln(2), 1, 1), ( 0, 0, ln(2), 1 ), ( 0, 0, 0, ln(2) ) ) } \qquad  ?

Ananda Dasgupta
Featured 1 week ago

$\left(\begin{matrix}2 & 2 & 3 & \frac{13}{3} \\ 0 & 2 & 2 & 3 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 2\end{matrix}\right)$

#### Explanation:

Lets start by writing
$\left(\begin{matrix}\ln \left(2\right) & 1 & 1 & 1 \\ 0 & \ln \left(2\right) & 1 & 1 \\ 0 & 0 & \ln \left(2\right) & 1 \\ 0 & 0 & 0 & \ln \left(2\right)\end{matrix}\right)$

in the form $\ln \left(2\right) {I}_{4} + A$, where ${I}_{4}$ is the $4 \setminus \times 4$ identity matrix and $A$ is the upper triangular matrix

$A = \left(\begin{matrix}0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{matrix}\right)$

Since ${I}_{4}$ commutes with all $4 \times 4$ matrices in general, and $A$ in particular, the exponential matrix is simply

$\exp \left(\ln \left(2\right) {I}_{4} + A\right) = \exp \left(\ln \left(2\right) {I}_{4}\right) \exp \left(A\right) = {e}^{\ln \left(2\right)} {I}_{4} \exp \left(A\right) = 2 \exp \left(A\right)$

So the calculation essentially reduces to the determination of ${e}^{A}$. Note that the standard approach using diagonalization does not work here, since $A$ is not diagonalizable. However, the characteristic equation for $A$ is easily seen to be
${\lambda}^{4} = 0$
and thus the Cayley-Hamilton theorem says that
${A}^{4} = 0$
so that the infinite series for ${e}^{A}$ truncates :

e^A = I+A+1/{2!} A^2 + 1/{3!}A^3

It is easy to calculate

${A}^{2} = \left(\begin{matrix}0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{matrix}\right) \mathmr{and} {A}^{3} = \left(\begin{matrix}0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{matrix}\right)$

Thus
${e}^{A} = \left(\begin{matrix}1 & 1 & \frac{3}{2} & \frac{13}{6} \\ 0 & 1 & 1 & \frac{3}{2} \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1\end{matrix}\right)$

Our solution is twice this matrix!

2

## Sum of the series?? 1/2+3/4+5/8+....n

Shwetank Mauria
Featured yesterday

${S}_{n} = 3 - \frac{2 n + 3}{2} ^ n$

#### Explanation:

Let ${S}_{n} = \frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \ldots \ldots \ldots \ldots \ldots \ldots \text{n terms}$

What is ${n}^{t h}$ term of the series? Well numerators are in arithmatic sequence and numerator of ${n}^{t h}$ term is $2 n - 1$ and denominator is in geometric series with ${n}^{t h}$ term being ${2}^{n}$. Hence ${n}^{t h}$ term of given series is $\frac{2 n - 1}{2} ^ n$ and series is

${S}_{n} = \frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \frac{7}{16} + \ldots \ldots \ldots \ldots \ldots \ldots + \frac{2 n - 1}{2} ^ n$ and then

$\frac{1}{2} {S}_{n} = \frac{1}{4} + \frac{3}{8} + \frac{5}{16} + \ldots \ldots \ldots \ldots \ldots \ldots + \frac{2 n - 1}{2} ^ \left(n + 1\right)$

Subtracting latter from former, we get

${S}_{n} - \frac{1}{2} {S}_{n} = \frac{1}{2} + \frac{2}{4} + \frac{2}{8} + \ldots . + \frac{2}{2} ^ n - \frac{2 n - 1}{2} ^ \left(n + 1\right)$

or $\frac{1}{2} {S}_{n} = \frac{1}{2} + \frac{2}{4} \left[1 + \frac{1}{2} + \ldots . + \frac{1}{2} ^ \left(n - 2\right)\right] - \frac{2 n - 1}{2} ^ \left(n + 1\right)$

= $\frac{1}{2} + \frac{1}{2} \frac{1 - \frac{1}{2} ^ \left(n - 1\right)}{1 - \frac{1}{2}} - \frac{2 n - 1}{2} ^ \left(n + 1\right)$

= $\frac{1}{2} + 1 - \frac{1}{2} ^ \left(n - 1\right) - \frac{2 n - 1}{2} ^ \left(n + 1\right)$

= $\frac{3}{2} - \frac{4 + 2 n - 1}{2} ^ \left(n + 1\right)$

= $\frac{3}{2} - \frac{2 n + 3}{2} ^ \left(n + 1\right)$

and ${S}_{n} = 3 - \frac{2 n + 3}{2} ^ n$