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2 years ago

What is the frequency of d=9cos(pi/2 t)

1 answer:

5 0

The period is equal to 2pi/n, where n is the coefficient of t. In this case pi/2. Therefore the period in this example is 4. Frequency is equal to 1/period. Hence the frequency for this problem is 1/4

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Looking at the histogram, where would the median fall?

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Answer:

The median would be found in the middle of the histogram

Step-by-step explanation:

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2 years ago

Keri pours 6 glasses of juice. Each glass holds 34 cup. How many cups of juice does Keri pour?

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4 1 /2 cups

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3 years ago

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2 years ago

Water drips into a circular puddle such that the radius of the puddle, in centimeters, at time t, in seconds, is given by the eq

bulgar [2K]

Part A

Given that the puddle is circular in shape and that the radius of the puddle, in centimeters, at time t, in seconds, is given by the equation $r(t)=\sqrt{t}$ .

Then the area of the puddle is given by the area of a circle = $Area=\pi r^2$
But, given that $r(t)=\sqrt{t}$ , then $A(t)=\pi (r(t))^2=\pi(\sqrt{t})^2=\pi t$

Therefore, the equation for the area of the puddle as a function of t is given by $A(t)=\pi t$

Part B

The average rate of change of a function f(x) between x = a and x = b is given by $\frac{f(b)-f(a)}{b-a}$ .

Thus, the average rate of change of the area of the puddle with respect to time between t = 0 and t = 16 is given by $\frac{A(16)-A(0)}{16-0} = \frac{16\pi-0}{16} = \frac{16\pi}{16} =\pi$

Therefore, the average rate of change of the area of the puddle with respect to time between t = 0 and t = 16 is π.

Part C

The area of the puddle with respect to the radius is given by $A(r)=\pi r^2$

Given that $r(t)=\sqrt{t}$ , thus when t = 0, $r(0)=\sqrt{0}=0$ and when t = 16, $r(16)=\sqrt{16}=4$

Thus, the average rate of change of the area of the puddle with respect to the radius between r = 0 and r = 4 is given by

$\frac{A(4)-A(0)}{4-0} = \frac{\pi(4)^2-\pi(0)^2}{4} = \frac{16\pi}{4} =4\pi$

Therefore, the average rate of change of the area of the puddle with respect to the radius between t = 0 and t = 16 is 4π.

Part D

The circumference of a circle is given by $C=2\pi r$

Thus, the radius of the puddle in terms of circumference is given by $r= \frac{C}{2\pi}$

Thus, the area of the puddle with respect to the circumference, C, of the puddle is given by $A(C)=\pi\left( \frac{C}{2\pi} \right)^2= \frac{1}{4\pi} C^2$

Since, $C=2\pi r$ and $r(t)= \sqrt{t}$ , thus when t = 0, r = 0 and C = 0; when t = 16, r = 4 and C = 8π.

Thus the area of the puddle with respect to the circumference, C, of the puddle between C = 0 and C = 8π is given by $\frac{A(8\pi)-A(0)}{8\pi-0} = \frac{ \frac{(8\pi)^2}{4\pi}- \frac{(0)^2}{4\pi} }{8\pi} = \frac{ \frac{64\pi^2}{4\pi} }{8\pi} = \frac{16\pi}{8\pi} =2$

Therefore, the average rate of change of the area of the puddle with respect to the circumference of the puddle between t = 0 and t = 16 is 2.

5 0

3 years ago