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3 years ago

10

A group of students organized a car wash to raise money for a local charity. The students charged $8.00 for each car they washed

. In 4 hours, they washed 16 cars. At that rate, how much money could they earn from washing cars for ten hours?

2 answers:

faust18 [17]3 years ago

6 0

Answer:

$320

Step-by-step explanation:

They charged &8.00 for each car. In 4 hours they washed 16 cars. This averages to 4 cars every hour. If they did this for 10 hours then you would multiply 10 by 4. That would equal 40 cars in total. If you multiply 40 by 8 you get 320

Hope this helped everyone:)

3241004551 [841]3 years ago

3 0

Answer:

$320

Step-by-step explanation:

First, we can find how many cars they washed.

At 8 hours, they washed 32 cars, and since you still have two hours worth of cars to find, you can divide 16 by 2, and you get 8.

So in total, they washed 40 cars.

Now, you can multiply 40 by $8.00, and you get $320.

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find the volume of the solid formed by revolving the region bounded by the graphs of y = 4x - x^2 and f(x) = x^2 from [0,2] abou

Neko [114]

Answer:

$v = \frac{32\pi}{3}$

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$v=33.52$

Step-by-step explanation:

Given

$f(x) = 4x - x^2$

$g(x) = x^2$

$[a,b] = [0,2]$

Required

The volume of the solid formed

Rotating about the x-axis.

Using the washer method to calculate the volume, we have:

$\int dv = \int\limit^b_a \pi(f(x)^2 - g(x)^2) dx$

Integrate

$v = \int\limit^b_a \pi(f(x)^2 - g(x)^2)\ dx$

$v = \pi \int\limit^b_a (f(x)^2 - g(x)^2)\ dx$

Substitute values for a, b, f(x) and g(x)

$v = \pi \int\limit^2_0 ((4x - x^2)^2 - (x^2)^2)\ dx$

Evaluate the exponents

$v = \pi \int\limit^2_0 (16x^2 - 4x^3 - 4x^3 + x^4 - x^4)\ dx$

Simplify like terms

$v = \pi \int\limit^2_0 (16x^2 - 8x^3 )\ dx$

Factor out 8

$v = 8\pi \int\limit^2_0 (2x^2 - x^3 )\ dx$

Integrate

$v = 8\pi [ \frac{2x^{2+1}}{2+1} - \frac{x^{3+1}}{3+1} ]|\limit^2_0$

$v = 8\pi [ \frac{2x^{3}}{3} - \frac{x^{4}}{4} ]|\limit^2_0$

Substitute 2 and 0 for x, respectively

$v = 8\pi ([ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ] - [ \frac{2*0^{3}}{3} - \frac{0^{4}}{4} ])$

$v = 8\pi ([ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ] - [ 0 - 0])$

$v = 8\pi [ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ]$

$v = 8\pi [ \frac{16}{3} - \frac{16}{4} ]$

Take LCM

$v = 8\pi [ \frac{16*4- 16 * 3}{12}]$

$v = 8\pi [ \frac{64- 48}{12}]$

$v = 8\pi * \frac{16}{12}$

Simplify

$v = 8\pi * \frac{4}{3}$

$v = \frac{32\pi}{3}$

or

$v=\frac{32}{3} * \frac{22}{7}$

$v=\frac{32*22}{3*7}$

$v=\frac{704}{21}$

$v=33.52$

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