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3 years ago

Y= (x-9) (x+3)

Mathematics

1 answer:

juin [17]3 years ago

4 0

Answer:

y = x² − 6x − 27

Step-by-step explanation:

To distribute, you can use something called FOIL. It stands for First, Outer, Inner, Last.

First, multiply the First term in each factor.

x · x = x²

Now multiply the Outer terms in each factor.

x · 3 = 3x

Next multiply the Inner terms in each factor.

-9 · x = -9x

Finally, multiply the Last terms in each factor.

-9 · 3 = -27

Add them all up:

x² + 3x − 9x − 27

x² − 6x − 27

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9x8-29+30/15-15 I'm looking for the answer plz

olasank [31]

9×8-29+30/15-15 equals 30

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3 years ago

MATHMATECIS HELP. HELP APPRECIATED. I NEED AN EXPLANATION STEP BY STEP PLEASE

VARVARA [1.3K]

Answer:

i think B

Step-by-step explanation:

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2 years ago

Read 2 more answers

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).

4 0

3 years ago

What is 9x^2-18x-7÷(3x+1)

Phantasy [73]

81x - 126x ÷ 3x + 1
45x / 3x + 1
15x / x + 1

6 0

3 years ago

Assume that the weights of spawning Chinook salmon in the Columbia river are normally distributed. You randomly catch and weigh

s344n2d4d5 [400]

Answer: The point estimate for the mean weight of all spawning Chinook salmon in the Columbia River is the mean weight from your sample $(\overline{x})$ = 31.2 pounds .

Step-by-step explanation:

The weights of spawning Chinook salmon in the Columbia river are normally distributed.

Given : The mean weight from your sample is 31.2 pounds with a standard deviation of 4.4 pounds.

The point estimate for the mean $(\mu)$ weight of all spawning Chinook salmon in the Columbia River is the mean weight from your sample $(\overline{x})$ i.e. is 31.2 pounds because the best point estimate for population mean is the sample mean. [in normal distribution]

8 0

3 years ago