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yarga [219]
2 years ago
8

Find the quotient. Simplify your answer.

Mathematics
1 answer:
brilliants [131]2 years ago
8 0

Let's see

\\ \rm\Rrightarrow \dfrac{s+1}{s^2+1}\div \dfrac{s^2}{6s^2}

\\ \rm\Rrightarrow \dfrac{s+1}{s(s+1)}\div\dfrac{1}{5s^2}

\\ \rm\Rrightarrow \dfrac{1}{s}\times{5s^2}

\\ \rm\Rrightarrow \dfrac{5s^2}{s}

\\ \rm\Rrightarrow 5s

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Using derivatives, it is found that the x-values in which the slope belong to the interval (-1,1) are in the following interval:

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<h3>What is the slope of the tangent line to a function f(x) at point x = x0?</h3>

It is given by the derivative at x = x0, that is:

m = f^{\prime}(x_0).

In this problem, the function is:

f(x) = 0.2x^2 + 5x - 12

Hence the derivative is:

f^{\prime}(x) = 0.4x + 5

For a slope of -1, we have that:

0.4x + 5 = -1

0.4x = -6

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For a slope of 1, we have that:

0.4x + 5 = 1.

0.4x = -4

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Hence the interval is:

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More can be learned about derivatives and tangent lines at brainly.com/question/8174665

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