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4 years ago

Plz hurry and answer this correctly!!!

Mathematics

1 answer:

In-s [12.5K]4 years ago

7 0

Answer:

$\large\boxed{C)\ \dfrac{8}{5}\pi\ cm}$

Step-by-step explanation:

The formula of an area of a circle:

$A_O=\pi r^2$

The formula of a perimeter of a circle:

$P_O=2\pi r$

r - radius

We have the area of a circle

$A_O=\dfrac{16}{25}\pi\ cm^2$

Substitute:

$\pi r^2=\dfrac{16}{25}\pi$ <em>divdie both sides by π</em>

$r^2=\dfrac{16}{25}\to r=\sqrt{\dfrac{16}{25}}\\\\r=\dfrac{\sqrt{16}}{\sqrt{25}}\\\\r=\dfrac{4}{5}\ cm$

Calculate the perimeter:

$P_O=2\pi\left(\dfrac{4}{5}\right)=\dfrac{8}{5}\pi$

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Using the data in the table, which statement is true?

Igoryamba

The true statement is mean > median

<h3>How to determine the true statement?</h3>

The mean is the average value.

So, we have

Mean = Sum/Count

This gives

Mean = (8 * 1 + 10 * 3 + 14 * 2)/(1 + 3 + 2)

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At 2:00 PM a car's speedometer reads 30 mi/h. At 2:20 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:20 the acce

Bad White [126]

Answer:

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$ . By the Mean Value Theorem, there is a number c such that $0 < c$ with $v'(c)=60 \:{\frac{mi}{h^2}}$ . Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly $60 \:{\frac{mi}{h^2}}$ .

Step-by-step explanation:

The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

f is continuous on the closed interval [a, b].
f is differentiable on the open interval (a, b).

Then there is a number c in (a, b) such that

$f'(c)=\frac{f(b)-f(a)}{b-a}$

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $v(0 \:h) = 30 \:{\frac{mi}{h} }$ and $v( \frac{1}{3} \:h) = 50 \:{\frac{mi}{h} }$ (note that 20 minutes is $20/60=1/3$ of an hour), so the average rate of change of v on the interval $[0 \:h, \frac{1}{3} \:h]$ is

$\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in $(0 \:h, \frac{1}{3} \:h)$ at which $v'(c)=60 \:{\frac{mi}{h^2}}$ .