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Roman55 [17]
3 years ago
14

The sum of two polynomials is 8d5 – 3c3d2 + 5c2d3 – 4cd4 + 9. If one addend is 2d5 – c3d2 + 8cd4 + 1, what is the other addend?

Mathematics
2 answers:
MatroZZZ [7]3 years ago
8 0

Answer:

a. x = 6d^5 - 2c^3d^2 + 5c^2d^3 - 12cd^4 + 8

Step-by-step explanation:

Given that

8d^5 + 3c^3d62 + 5c^2d^3 - 4cd^4 + 9

Now

if one is added i.e.

2d^5 - c^3d^2 + 8cd^4 + 1

Now let us assume the other polynomial be x

So,

8d^5 + 3c^3d62 + 5c^2d^3 - 4cd^4 + 9 = x  + (2d^5 - c^3d^2 + 8cd^4 + 1)\\\\x = 8d^5 + 3c^3d62 + 5c^2d^3 - 4cd^4 + 9 - (2d^5 - c^3d^2 + 8cd^4 + 1)\\\\

x = (8d^5 - 2d^5) + (-3c^3d^2 + c^3d^2) + 5c^2d^3+ (-4cd^4-8cd^4) + (9-1)\\\\x = 6d^5 - 2c^3d^2 + 5c^2d^3 - 12cd^4 + 8

My name is Ann [436]3 years ago
7 0

Answer:

a

Step-by-step explanation:

just use it mannnn

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3 years ago
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LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

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so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill

\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}

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