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3 years ago

Can someone help me with this? its a relatively simple problem for extra credit in algebra 1.

Mathematics

1 answer:

suter [353]3 years ago

4 0

(2x^2+3x+4)(x^3+4x^2+3x)(x+4)

First use the distributive property for the first two:

(2x^5+8x^4+6x^3+3x^4+12x^3+9^2+4x^3+16x^2+12x)(x+4)

And then combine like terms:

(2x^5+11x^4+22x^3+25x^2+12x)(x+4)

Continue use the distributive property:

(2x^6+11x^5+22x^4+25x^3+12x^2+8x^5+44x^4+88x^3+100x^2+48x)

Continue combine like terms:

(2x^6+19x^5+66x^4+113x^3+112x^2+48x)
And that is your final answer.

Hope that help:)

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A diver descends to a depth of –25.6 feet relative to sea level. How many feet must the diver ascend to reach sea level?

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2 years ago

Read 2 more answers

"find the reduction formula for the integral" sin^n(18x)

dexar [7]

Let

$I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx$
For demonstration on how to tackle this sort of problem, I'll only work through the case where $n$ is odd. We can write

$\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx$
$\implies I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$

For the remaining integral, we can integrate by parts, taking

$u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx$ $\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax$

$\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx$

For this next integral, we rewrite the integrand

$\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax$
$\implies\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx=I(n-4,a)-2I(n-2,a)+I(n,a)$

So putting everything together, we found

$I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$
$I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)$
$I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$
$\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$

$\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax$

7 0

3 years ago

Factor completely 5ab(x + 6) − 4(x + 6).

konstantin123 [22]

It seems like you've begun to apply the grouping method.

Because the two binomials in the parentheses are the same, we can rewrite this as

(5ab-4)(x+6)

(x+6) cannot be further factored. (5ab-4) cannot be further factored either.

Final answer: (5ab-4)(x+6)

4 0

3 years ago

Convert 12,000+7*10^(4) to scientific notation

Nostrana [21]

Answer:

Step-by-step explanation:

Enter numbers or scientific notation.

For scientific notation, use a carat ^ for the powers of 10 or use e notation.

Scientific Notation: 3.45 x 10^5

E Notation: 3.45e5

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3 years ago