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3 years ago

Please solve:Common Factors Of Polynomials

Mathematics

2 answers:

EleoNora [17]3 years ago

8 0

The answer is D on e2020

GenaCL600 [577]3 years ago

4 0

1- $8x^2+12x \\ =4x*2x+4x*3 \\ =4x(2x+3) \\ =A$

2- $20mn-30m \\ =10m*2n-10m*3 \\ =10m(2n-3) \\ =B$

3- $15x^3-12x \\ =3x*5x^2-3x*4 \\ =3x(5x^2-4) \\ =B$

4- $12x^2y-3x^2y^2-15xy \\ =3xy*4x-3xy*xy-3xy*5 \\ =3xy(4x-xy-5) \\ =C$

5- $24w^2x^6-8wx^3 \\=8wx^3*3wx^3-8wx^3*1 \\ =8wx^3(3wx^3-1) \\ =D$

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Cans of regular soda have volumes with a mean of 13.3813.38 oz and a standard deviation of 0.110.11 oz. is it unusual for a can

dimulka [17.4K]

No 13.3813 + 0.1100 = 13.4913. (max. standard deviation)
13.4813 is less than max. standard deviation

8 0

3 years ago

For the function y = -2+5sin(pi/12(x-2)), what is the maximum value?

myrzilka [38]

To get the function y = -2+5sin(pi/12(x-2)), the maximum value can be determined by differentiating the function and equating it to zero. The value of x will give the maximum value of the function.
dy/dx = 5 cos (pi/12 (x-2)) (pi/12)
dy/dx = 5 pi/12 cos(pi/12 (x-2))

Equate to zero:
5 pi/12 cos(pi/12 (x-2)) =0
pi/12 (x-2) = 3pi/2
x = 8

Substituting,
y= -2 + 5sin( pi/12 (8-2)
y = -1.86

8 0

3 years ago

Can someone help me?

Pie

Answer:

x = 12 efg = 132

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

The Sugar Sweet Company is going to transport its sugar to market. It will cost $6500 to rent trucks, and it will cost an additi

inessss [21]

Answer:

$10000 Total cost to transport 14 tons

Step-by-step explanation:

The total cost C(S) = $6500 + ($250/ton)S. This is a linear function with initial value $6500 and slope $250/ton.

C(14) = $6500 + ($250/ton)(14) = $10000 = Total cost to transport 14 tons

5 0

3 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