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V125BC [204]
4 years ago
6

The diameter of a basketball rim is 18 inches. A standard basketball has a circumference of 30 inches. What is the distance betw

een the ball and the rim in a shot in which the ball goes exactly in the center of the rim? Show your work.
Please show the steps!! This unit is area, if that helps.
Mathematics
2 answers:
Ksivusya [100]4 years ago
6 0
Hello!

First you have to find the circumference of the rim

To find the circumference you use the equation

C = 2 \pi r

C is circumferece
r is radius which is half the diameter

Put in the values you know

C = 2 \pi (9)

C ≈ 56.55

You subtract the circumference of the rim by the circumference of the ball

56.55 - 30 = 26.55

The answer is 26.55 inches

Hope this helps!
Likurg_2 [28]4 years ago
5 0
Diameter of the basketball rim = 18 inchesCircumference of the standard basketball = 30 inches
Area of circle = π r² 
Basketball rim = 3.14 * (18/2)² = 3.14 * 9² = 3.14 * 81 = 254.34 in²
Circumference of a circle = 2 π r30 = 2 * 3.14 * rr = 30 / (2*3.14) = 30 / 6.28 = 4.78 inches
Standard basketball = 3.14 * (4.78)² = 3.14 * 22.85 = 71.75 in²
254.34 - 71.75 = 182.59 in²
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a) f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

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f'(t)=0.001155t(t-1980)^{0.5}

a) To find the function f(t) you integrate f(t):

\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt

To solve the integral you use:

\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}

Next, you replace in the integral:

\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C

Then, the function f(t) is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient​ visits.

Hence C' = 264,034,000

The function is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) For t = 2015 you have:

f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7

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