If x-y doesn't equal 0 then x^2-y^2 / x-y =

1 answer:

3 0

Recognize the differnce of 2 perfect squares

a²-b²=(a-b)(a+b)

so
$\frac{x^2-y^2}{x-y}=\frac{(x-y)(x+y)}{x-y}=(\frac{x+y}{1})(\frac{x-y}{x-y})=(x+y)(1)=x+y$

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myrzilka [38]

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The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s

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The positions when the particle reverses direction are:

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Why?

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Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

$s(t)=2t^{3}-21t^{2}+60t+3$