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Sergio [31]
3 years ago
8

Parker was able to pay for 44 percent of his college tuition with his scholarship. The remaining $10,054.52 he paid for with a s

tudent loan. What was the cost of parkers tuition?
Mathematics
2 answers:
NARA [144]3 years ago
6 0
The answer is.............
22851.18182
Darina [25.2K]3 years ago
6 0
We know that 100-44=56 which is the percent NOT covered by the scholarship.  So 56% was paid by student loans.  Cross multiply.  56/100 (which is 56%) equals $10,054.52.  56/100=10054.52/x (letting x be the total cost)

100x10,054.52=1005452
divide by 56=$17,954.50 which is the total cost of tuition.
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Step-by-step explanation:

Given that:

the inverted conical water tank with a height of 20 ft and a radius of 8 ft  is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.

We are meant to find the  rate of change of the water depth when the water depth is 10 ft.

The diagrammatic expression below clearly interprets the question.

From the image below, assuming h = the depth of the tank at  a time t and r = radius of the cone shaped at a time t

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\dfrac{h}{r}= \dfrac{5}{2}

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V = \dfrac{1}{3} \pi r^2 h

V = \dfrac{1}{3} \pi (\dfrac{h^2}{6.25}) h

V = \dfrac{1}{18.75} \pi \ h^3

The rate of change of the water depth  is :

\dfrac{dv}{dt}= \dfrac{\pi r^2}{6.25}\  \dfrac{dh}{dt}

Since the water is drained  through a hole in the vertex (bottom) at a rate of 4 ft^3/sec

Then,

\dfrac{dv}{dt}= - 4  \ ft^3/sec

Therefore,

-4 = \dfrac{\pi r^2}{6.25}\  \dfrac{dh}{dt}

the rate of change of the water at depth h = 10 ft is:

-4 = \dfrac{ 100 \ \pi }{6.25}\  \dfrac{dh}{dt}

100 \pi \dfrac{dh}{dt}  = -4 \times 6.25

100  \pi \dfrac{dh}{dt}  = -25

\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi}

Thus, the rate of change of the water depth when the water depth is 10 ft is;  \mathtt{\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi} \  \ ft/s}

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