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jonny [76]
2 years ago
10

When two pipes fill a pool together, they can finish in 5 hours. If one of the pipes fills half the pool then the other takes ov

er and finishes filling the pool, it will take them 18 hours. How long will it take each pipe to fill the pool if it were working alone?
Mathematics
1 answer:
Alexxandr [17]2 years ago
3 0

♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫

➷The answer is 18 hours.

Explanation: If you look carefully, you can see that the answer is the second condition as well. If one pipe did half, then the other pipe took over and that took 18 hours, that means that it equals one pipe filling the pool continuously by itself. :)

✽

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

DOGE

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Answer:

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Step-by-step explanation:

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1. Kim works as a book keeper and creates invoices for a variety of companies. Help Kim find the total
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This is about calculation of invoice.

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  • 1) 12 cameras for $200 each

Amount for cameras = 12 × 200 = $2400

25 cables with length of 10 ft each cost $1.25

Amount for 10 ft cables = 25 × 1.25 = $31.25

12 ft specialty cables cost $1.75 per foot

10 of these cables will have length =12 × 10 = 120 ft

Amount for 10 of these cables = 1.75 × 120 = $210

Total for the Invoice = $2400 + $31.25 + $210

Total for the invoice = $2641.25

  • 2) 250 yards of rope costs $500.

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For a customer that wants 15 yard;

Total for the invoice = 15 × $2 = $30

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On Tuesday; 3 hours will cost; 3 × $275 = $825

On wednesday; 2hours will cost; 2 × $275 = $550

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Read more at; brainly.com/question/17745127

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A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si
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Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

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We have that:

Area = 128

Let the dimension of the paper be x and y;

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Length = x

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So:

Area = x * y

Substitute 128 for Area

128 = x * y

Make x the subject

x = \frac{128}{y}

When 1 inch margin is at top and bottom

The length becomes:

Length = x + 1 + 1

Length = x + 2

When 2 inch margin is at both sides

The width becomes:

Width = y + 2 + 2

Width = y + 4

The New Area (A) is then calculated as:

A = (x + 2) * (y + 4)

Substitute \frac{128}{y} for x

A = (\frac{128}{y} + 2) * (y + 4)

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A = 128 + \frac{512}{y} + 2y + 8

Collect Like Terms

A = \frac{512}{y} + 2y + 8+128

A = \frac{512}{y} + 2y + 136

A= 512y^{-1} + 2y + 136

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Different A with respect to y

A' = -512y^{-2} + 2

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This gives:

0 = -512y^{-2} + 2

Collect Like Terms

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Multiply through by y^2

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Divide through by 2

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A" = 1024y^{-3}

A" = \frac{1024}{y^3}

The above value is:

A" = \frac{1024}{y^3} > 0

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

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