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arlik [135]
3 years ago
8

A cotton candy holder is shaped like a cone. The height is 12 in., and the diameter is 12 in. What is the volume of the holder?

Use 3.14 to approximate pi, and express your final answer to the nearest tenth.. ______in3
Mathematics
2 answers:
babymother [125]3 years ago
6 0
The formula for the volume of a cone is
V=\frac{1}3\pi r^2h
(you can remember it like this: it's literally just 1/3 the volume of a cylinder, just as a pyramid is 1/3 the volume of a prism)

We know that the diameter is 12...the radius is always half the diameter, so 6.
The height is 12.
Let's plug these into our volume formula.

V=\frac{1}3\pi \times6^2\times12
V=\frac{1}3\pi \times36\times12
V=\frac{1}3\pi \times432
V=144\times \pi
V\approx144\times3.14
\boxed{V\approx452.16\ in^3}
alexgriva [62]3 years ago
6 0

that answer was close but I just took the test and its 452.2

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Answer:

115 3/4

Step-by-step explanation:

\frac{1}{2} * \frac{1}{2} Because d=1/2. 1/2*1/2=1/4

C=5 so, C^3=5x5x5

C=125

Next, b=9

Then, you put everything together.

1/4+125-9=125 1/4-9

115 3/4

115 3/4+9 1/4

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2 years ago
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2 years ago
The ratio two numbers is 3 and 5 their sum is 135find two number
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Let, the numbers = x,y

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value of 1 unit = 135/8 = 16.875

So, numbers would be 3(16.875) & 5(16.875) = 50.625 & 84.375

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3 years ago
The sum of two numbers is 100, the difference between the two numbers is 26. What is the number that is less?
Zepler [3.9K]
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3 years ago
One pipe can empty a tank 5 times faster than another pipe can fill the tank. Starting with a full tank, if both pipes are turne
xxTIMURxx [149]

Answer:

40 hours

Step-by-step explanation:

Let's say that the slower pipe can fill x amount of the tank in one hour. It adds x amount to the tank every hour. Therefore, we can say, for the slower pipe,

1 hour = x amount

Then, the faster pipe empties the tank 5 times faster than the slower pipe fills it, so it removes 5 times the amount that the smaller pipe puts in, so for the faster pipe,

1 hour = -5x amount (negative to symbolize removing).

For the problem at hand, the tank starts at full, or 100%=1. It ends empty, or at 0% = 0. After 10 hours, if we only account for the slower pipe, we add x amount to the tank every hour, so we add 10 times that total, resulting in

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1 + 10 * x - 50 * x = 0

1 - 40 * x = 0

add 40*x to both sides to isolate the x and its coefficient

1 = 40 * x

divide both sides by 40 to isolate x

x= 0.025

Therefore, the slower pipe adds 0.025, or 1/40 = 2.5% to the tank every hour. We want to figure out how long it would take for the slower pipe to fill up an empty tank, or turn it from 0% to 100% full.

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2.5% * y = 100%

divide both sides by 2.5% to isolate y

100%/2.5% = y = 40

5 0
3 years ago
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