H(x)=x^2-1 what is the average rate of change of h over the interval −3≤x≤−1

1 answer:

7 0

$slope = m = \cfrac{rise}{run} \implies \cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby \begin{array}{llll} average~rate\\ of~change \end{array}\\\\[-0.35em] \rule{34em}{0.25pt}\\\\ h(x)=x^2-1 \qquad \stackrel{-3\leqslant ~~x~~\leqslant -1}{\begin{cases} x_1=-1\\ x_2=-3 \end{cases}}\implies \cfrac{h(-3)-h(-1)}{-3-(-1)} \\\\\\ \cfrac{[(-3)^2-1]~~ -~~[(-1)^2-1]}{-3+1}\implies \cfrac{8~~-~~0}{-2}\implies -4$

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The area of the base of a prism is 55 mm2. The perimeter of the base is 30 mm. The height of the prism is 7 mm. . . What is the

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The solution would be like this for this specific problem:

7 x 30 = 210

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The surface area of the prism is 320 mm2.

The correct answer between all the choices given is the second choice or letter B. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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4 years ago

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Sveta_85 [38]

Answer:

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Step-by-step explanation:

We can use the relation between tan and sec:

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