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2 years ago

Can someone help me out with this question plz

Mathematics

1 answer:

Genrish500 [490]2 years ago

6 0

Answer:

$\left(s\cdot t\right)\left(x\right)=2x^2+12x+16$

$\left(s-t\right)\left(x\right)=-x$

$\left(s+t\right)\left(4\right)=20$

Step-by-step explanation:

Given functions are:

$s\left(x\right)=x+4$

$t\left(x\right)=2x+4$

Then $\left(s\cdot t\right)\left(x\right)=s\left(x\right)\cdot t\left(x\right)$

or $\left(s\cdot t\right)\left(x\right)=\left(x+4\right)\left(2x+4\right)$

or $\left(s\cdot t\right)\left(x\right)=2x^2+4x+8x+16$

or $\left(s\cdot t\right)\left(x\right)=2x^2+12x+16$

---------

Similarly

$\left(s-t\right)\left(x\right)=s\left(x\right)-t\left(x\right)$

or $\left(s-t\right)\left(x\right)=\left(x+4\right)-\left(2x+4\right)=x+4-2x-4$

or $\left(s-t\right)\left(x\right)=-x$

---------

Similarly

$\left(s+t\right)\left(4\right)=s\left(4\right)+t\left(4\right)=\left(4+4\right)+\left(2\left(4\right)+4\right)=\left(8\right)+\left(12\right)=20$

$\left(s+t\right)\left(4\right)=20$

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Answer:

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3 years ago

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Answer:

$\frac{3}{28}$

Step-by-step explanation:

Probability (P) is calculated as

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3 years ago

Binomial Expansion/Pascal's triangle. Please help with all of number 5.

Mandarinka [93]

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}$

The rows add up to $1,2,4,8,16$ , respectively. (Notice they're all powers of 2)

The sum of the numbers in row $n$ is $2^{n-1}$ .

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

$(1+x)^1=1+x=\dbinom10+\dbinom11x$

so the base case holds. Assume the claim holds for $n=k$ , so that

$(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k$

Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}$

So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

and since $\dbinom{k+1}0=\dbinom{k+1}{k+1}=1$ , you have

$(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}$

and so the claim holds for $n=k+1$ , thus proving the claim overall that

$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

Setting $x=1$ gives

$(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n$

which agrees with the result obtained for part (c).

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2 years ago

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ella [17]

Answer:

Step-by-step explanation:

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2 years ago

I know y= 40 but what is x ?

puteri [66]

Answer:

x = 67; y = 40

Step-by-step explanation:

You are correct with y. Angles y + 23 and 63 are vertical, so they are congruent. That gives you y + 23 = 63, which gives y = 40.

To find x, look at angles 2x - 17 and 63.

They are adjacent angles whose outside rays lie on a line. Angles like these are called a linear pair, and they are supplementary. That means their measures add up to 180 deg.

2x - 17 + 63 = 180

2x + 46 = 180

2x = 134

x = 67

6 0

3 years ago