3 years ago

How do you take the antiderivative of (sinxcosx)^2?

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aleksley [76]

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Step-by-step explanation:

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A circle has a radius of 7.5 centimeters and a central angle AOB that measures 90 degrees. What is the length of the intercepted

Julli [10]

Since there are $360^{\circ}$ in a circle, and central $\angle AOB = 90^{\circ}$ , you can see that the part "sliced out" by the central angle is $\frac{90^{\circ}}{360^{\circ}} = \frac{1}{4}$ of the whole circle.

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$\stackrel{\frown}{AB} = \frac{1}{4}(2 \pi r) = \frac{1}{4}(2 \pi (7.5 cm)) = \bf \frac{15 \pi}{4} cm$

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