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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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A bus company has contracted with a local high school to carry 450 students on a field trip. The company has 18 large buses whic
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Answer:

The answer is below

Step-by-step explanation:

Let x represent the big buses and y represent small buses. The large buses can carry 30 students and the small buses can carry 15 students. The total number of students are 450, this can be represented by the inequality:

30x + 15y ≤ 450

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x + y ≤ 20

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After plotting the graph, the minimum solution to the graph are at:

A (15,0), B(18,0), C(10, 10), D(18, 2).

The cost function is given as:

The total cost of operating one large bus is $225 a day, and the total cost of operating one small bus is $100 per day.

​

F(x, y) = 225x + 100y

At point A:

F(x, y) = 225(15) + 100(0) = $3375

At point B:

F(x, y) = 225(18) + 100(0) = $4050

At point C:

F(x, y) = 225(10) + 100(10) = $3250

At point D:

F(x, y) = 225(18) + 100(2) = $4250

The minimum cost is at point C(10, 10) which is $3250

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