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Lady bird [3.3K]
3 years ago
15

Factor 2x^3+3x^2-23x-12, given that x+4 is a factor

Mathematics
2 answers:
Paraphin [41]3 years ago
6 0
Hello,
2x^3+3x^2-23x-12\\
=2x^3+8x^2-5x^2-20x-3x-12\\
=2x^2(x+4)-5x(x+4)-3(x+4)\\
=(x+4)(2x^2-5x-3)\\
=(x+4)(2x^2-6x+x-3)\\
=(x+4)[2x(x-3)+1(x+3)]\\
=(x+4)(x-3)(2x+1)
Alex73 [517]3 years ago
5 0
<span>(2x + 1) • (x - 3) • (x + 4)</span>
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Calculus Problem
Roman55 [17]

The two parabolas intersect for

8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2

and so the base of each solid is the set

B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, |x^2-(8-x^2)| = 2|x^2-4|. But since -2 ≤ x ≤ 2, this reduces to 2(x^2-4).

a. Square cross sections will contribute a volume of

\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x

where ∆x is the thickness of the section. Then the volume would be

\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x

We end up with the same integral as before except for the leading constant:

\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx

Using the result of part (a), the volume is

\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x

and using the result of part (a) again, the volume is

\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}

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3 years ago
Let m = 7.
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32 is the answer to the problem


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Which represents a quadratic function y=2(x+2)(x-1) in standard form
ale4655 [162]

answer: y= 2x^2+2x-4
steps:
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3 years ago
How does a digit in the ten thousand place compere to a digit in the thousand place
Mademuasel [1]
Ten thousand - 10,000

thousand literally just haves 1,000


if the twenty is before the three zeros, it would be 20,000 which is twenty thousand. its the same with other numbers. its super simple!
7 0
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