All categories

2 years ago

1 point

Mathematics

1 answer:

spayn [35]2 years ago

5 0

well, what we do is, simply get the area of the containing rectangle, is an 8x10, and then get the area of the circle, with a radius of half of 4.3 or r = 2.15, and then if we subtract the area of the circle from the rectangle's, what's leftover is the shaded part.

$\stackrel{\textit{\large Areas}}{\stackrel{rectangle's}{(8\cdot 10)}~~-~~\stackrel{circle's}{[\pi (2.15)^2]}}\implies 80-4.6225\pi ~~\approx~~65.5$

You might be interested in

Victor got 24 out of 30 correct on his math test. What percent did Victor get correct?

Snowcat [4.5K]

Answer:

$\frac{24}{30}$ x 100 = 80%

Step-by-step explanation:

8 0

2 years ago

What is the solution to the following equation?

Pachacha [2.7K]

Answer:

x = 4

Step-by-step explanation:

8x - 9 = 6x - 1

Bring variables to the left side, whereas numbers to the right.

8x - 6x = - 1 + 9

2x = 8

x = 8 / 2

x = 4

3 0

3 years ago

The speed of light in cold air is how many times the speed of sound in cold air?

Tju [1.3M]

Answer: B I think

Step-by-step explanation: hope it helps

5 0

2 years ago

What is the probability that a randomly selected camper likes swimming or having a cookout, but not both?

emmainna [20.7K]

I would think 0.10 would be the answer

3 0

2 years ago

The radius of a cone is decreasing at a constant rate of 7 inches per second, and the volume is decreasing at a rate of 948 cubi

inessss [21]

Answer:

The height of cone is decreasing at a rate of 0.085131 inch per second.

Step-by-step explanation:

We are given the following information in the question:

The radius of a cone is decreasing at a constant rate.

$\displaystyle\frac{dr}{dt} = -7\text{ inch per second}$

The volume is decreasing at a constant rate.

$\displaystyle\frac{dV}{dt} = -948\text{ cubic inch per second}$

Instant radius = 99 inch

Instant Volume = 525 cubic inches

We have to find the rate of change of height with respect to time.

Volume of cone =

$V = \displaystyle\frac{1}{3}\pi r^2 h$

Instant volume =

$525 = \displaystyle\frac{1}{3}\pi r^2h = \frac{1}{3}\pi (99)^2h\\\\\text{Instant heigth} = h = \frac{525\times 3}{\pi(99)^2}$

Differentiating with respect to t,

$\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)$

Putting all the values, we get,

$\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)\\\\-948 = \frac{1}{3}\pi\bigg(2(99)(-7)(\frac{525\times 3}{\pi(99)^2}) + (99)(99)\frac{dh}{dt}\bigg)\\\\\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi} = (99)^2\frac{dh}{dt}\\\\\frac{1}{(99)^2}\bigg(\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi}\bigg) = \frac{dh}{dt}\\\\\frac{dh}{dt} = -0.085131$

Thus, the height of cone is decreasing at a rate of 0.085131 inch per second.

3 0

2 years ago