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3 years ago

Suppose the area of the square is 64 in.² what is the area of the circle

Mathematics

2 answers:

igor_vitrenko [27]3 years ago

8 0

Since I don’t know if pi should be left as pi or 3.14 or 22/7, my exact answer would be 16(3.14)

gregori [183]3 years ago

3 0

A=16 $\pi$

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steposvetlana [31]

Answer:

1) The velocity at this time is of $-9\frac{\sqrt{3}}{2}$ meters per second.

2) The velocity at this time is of $13\frac{\sqrt{2}}{2}$ meters per second.

Step-by-step explanation:

This question involves concepts of derivatives.

The velocity is the derivative of the position.

We use these following derivatives:

$(\sin{x})^{\prime} = \cos{x}$

$(\cos{x})^{\prime} = -\sin{x}$

1) s = 1 + 9(cos t)

Find the body's speed at time t=pi/3 sec.

We have to find the derivative at $t = \frac{\pi}{3}$ . So

$v = (1 + 9\cos{t})^{\prime} = -9\sin{t}$

$v(\frac{\pi}{3}) = -9\sin{\frac{\pi}{3}}$

$\frac{\pi}{3}$ is a common angle, which has a sine of $\frac{\sqrt{3}}{2}$ . So

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The velocity at this time is of $-9\frac{\sqrt{3}}{2}$ meters per second.

2) s = 12(sin t) - (cos t)

Find the body's velocity at time t=pi/4 sec.

We have to find the derivative at $t = \frac{\pi}{4}$ . So

$v = (12\sin{t} - \cos{t})^{\prime} = 12\cos{t} + \sin{t}$

$\frac{\pi}{4}$ is a common angle, which has both sine and cosine of $\frac{\sqrt{2}}{2}$ . So

$12\cos{\frac{\pi}{4}} + \sin{\frac{\pi}{4}} = 12\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 13\frac{\sqrt{2}}{2}$

The velocity at this time is of $13\frac{\sqrt{2}}{2}$ meters per second.

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