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puteri [66]
3 years ago
7

What is the best approximation for the perimeter of a semicircle with a diameter of 12 cm? Use 3.14 to approximate pi. A)18.84 c

m B)30.84 cm C)61.68 cm D)75.36 cm    hurry please
Mathematics
1 answer:
sp2606 [1]3 years ago
7 0
I say A but you probably arnt gonna trust my answers again...gd luck!! xx
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Answer:

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\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]

Now, we will utilize the product rule. The product rule is:

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\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1

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Then it follows that:

\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}

Therefore:

\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]

By simplification:

\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17

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