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2 years ago

HELP FAST PLEASE!!!!!

Mathematics

1 answer:

lidiya [134]2 years ago

3 0

Alright, lets get started.

The answer Sandra got is correct that is zero but there are some mistakes she made while writing steps.

$2 cot(\frac{-9\pi}{2}) = 2 cot(\frac{9\pi}{2})$ Incorrect step

Because as per even/odd identity,

Cot (-Θ) = - cot (Θ)

Hence

$2 cot (\frac{-9\pi}{2}) = - 2 cot (\frac{9\pi}{2})$

Now,

It could be written as

$-2 cot (\frac{9\pi}{2}) = -2 cot (\frac{\pi}{2})$

$-2 * 0$

Hence the answer is 0. : Answer

Hope it will help :)

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Let $f(x)=4x^3+20x^2+2x-3\\\\\text{when x=-5}\\\\f(-5)=4(-5)^3+20(-5)^2+2(-5)-3=-13\\\\\text{when x=}\frac{1}{2}\\$

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Hence we do not find any relation between the coefficients and zeros.

Anyway if the given values doesn't represents the zeros then those given values will not have any relation with the coefficients of the p[polynomial.

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