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-Dominant- [34]
3 years ago
9

ben is filling up his cylinder shaped pool up to 80% of its capacity . If his is 6 feet deep and has a diameter of 18 feet, how

much water will he put in the pool.
Mathematics
1 answer:
Nady [450]3 years ago
8 0

Answer:

You will put 1220.83ft³ of water in the pool

Step-by-step explanation:

First of all we have to calculate the volume of the pool and then calculate 80% of that volume

radius = half of diameter

d = 18ft

r = 18ft / 2 = 9ft

To calculate the volume of a cylinder we have to use the following formula:

v = volume

h = height = 6ft

π = 3.14

r = radius = 9ft

v = (π * r²) * h

we replace with the known values

v = (3.14 * (9ft)²) * 6ft

v = (3.14 * 81ft²) * 6ft

v = 254.34ft² * 6ft

v = 1526.04ft³

The volume of the cylinder is 1526.04ft³

If it is filled to 80% then we have to multiply the volume by 80/100

1526.04ft³  * 80/100 = 1220.83ft³

You will put 1220.83ft³ of water in the pool

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The sum of three numbers is 69. If the second number is equal to the first diminished by 8, and the third number is 5 times the
bonufazy [111]
X = first number
x - 8 = 2nd number
5x = 3rd number

x + x - 8 + 5x = 69
7x - 8 = 69 <=== this is the equation u would use

7x - 8 = 69
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3 0
3 years ago
Read 2 more answers
(a) Find the remainder when 15! is divided by 17. (b) Find the remainder when 2(26!) is divided by 29, Thatermine whether 17 is
dlinn [17]

Answer:

Step-by-step explanation:

a) We have 15! as the product of 1 to 15 natural numbers.  Since 17 is prime there will be no factor common to these

By actual division we find

15! (mod 17) =16

From this we deduce

even 16! mod 17 = 16 = -1

According to Wilson theorem

(17-1)! = -1 mod 17

Thus verified 17 is prime

Hence 15! (mod 17) =-1=16

-----------------------

b) 2(26!) is divided by 29

Since 29 is prime

(29-1)! = -1 mod 29

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4 years ago
Six friends attend a party. They form pairs for a game. How many different pairs are possible?
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Pairs, in this case, relates to a group of 2 or more. We have 6 friends. Let's call them A,B,C,D,E,F. This will allow us to make a [some sort of] combination tree:

1. ABC against DEF

2. ABD against CEF

3. ABE against CDF

4. ABF against CDE

5. ACD against BFE

6. ACE against BDF

7. ACF against BDE

8. ADE against BCF

9. ADF against BCE

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3 years ago
Find each function value<br><br>1. f(7) if (x)=5x<br><br>2. f(4) if f(x)=3x-1
Inessa [10]

<u>ANSWER</u>

1. \:  \: f(7) = 35

2. \: f(4) = 11

<u>EXPLANATION</u>

To find the value of a function, f(x) at a given value x=a, we plug in x=a into f(x) to obtain f(a) in simplified form.

The first function is

f(x) = 5x

To find f(7), we substitute x=7, into the function.

\implies f(7) =5(7) =  5 \times 7 = 35

The second function is

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To find f(4), we replace x with 4 to obtain,

f(4) = 3(4) - 1

f(4) = 12 - 1 = 11

6 0
3 years ago
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