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melomori [17]
3 years ago
13

A hose with a larger diameter working alone can fill a swimming pool in 9 hours.A hose with a smaller diameter working alone can

fill a swimming pool in 18 hours.working together,how long would it take the two hoses to fill the swimming pool?
Mathematics
1 answer:
Arte-miy333 [17]3 years ago
8 0

The large hose fills 1 pool in 9 hours written as  1/9

The small hose fills 1 pool in 18 hours, written as 1/18


Now you have 1/9 + 1/18 = 1/x ( x is the unknown time for both hoses).

rewrite 1/9 with a common denominator as 1/18:

1/9 = 2/18


Now you have 2/18 + 1/18 = 1/x

Add the left side:

2/18 + 1/18 = 3/18


Now you have 3/18 = 1/x

Cross Multiply:

3x = 18

Divide both sides by 3"

x = 18/3

X = 6


It will take both hoses 6 hours.

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Janice had $125. She loved earrings so much that she spent $7.00 every week to buy a new pair. How much money will she have left
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she will have $90 left after 5 weeks

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2x - y = 6<br>4<br>x-y<br>13<br>anch​
yanalaym [24]

Answer:

x=-7, y= -20

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2x - y =6

x - y = 13

when I subract (x-y =13 ) from (2x-y =6)

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-x +y=-13

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x = -7

substitute x=-7 in the second equation

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A concert cleanup crew has 6 workers and one cleanup kit.
myrzilka [38]

Answer:

Option (D)

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And number of kits required for w workers = \frac{w}{6} = w ÷ 6

If w = 54

Number of kits required = \frac{54}{6}=9

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Therefore, table given in Option (D) is the correct table.

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