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ivann1987 [24]
3 years ago
13

Hoses A and B spout water at different constant rates, and hose A can fill a certain pool in 6 hours. Hose A filled the pool alo

ne for the first 2 hours and the two hoses, working together, then finished filling the pool in another 3 hours. How many hours would it have taken hose B, working alone, to fill the entire pool
Mathematics
1 answer:
Naddik [55]3 years ago
8 0

Answer:

18 hours

Step-by-step explanation:

Let the volume of the pool be x. Since pipe A filled the  pool in 6 hours, the rate of pipe A = x / 6.

Let the rate of pipe b be y, Hose A filled the pool alone for the first 2 hours, this means that the volume filled in the 2 hours is x/6(2 hours) and the two hoses, working together, then finished filling the pool in another 3 hours for the 3 hours the volume filled is x/6(3) + y(3).  hence the total time is:

x/6(2) + x/6(3) + y(3) = x

x/3 + x/2 + 3y = x

Multiply through by 6:

2x + 3x + 18y = 6x

5x + 18y = 6x

18y = x

y = x/18

The rate of pipe B is x/18, this means it would take pipe B 18 hours to full the pool alone

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1.Assume that 20 squirrels are put in an enclosed wildlife ranch and the squirrel population grows for the next 5 years, as show
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Answer:

For the first group we have the pairs:

year           population

0                20

1                 80

2                320

3                 1,280

4                 5,120

5                 20,480

Here we can see that the population quadruples each year

(4*20 = 80, 80*4 = 320, 320*4 = 1,280, etc...)

then the population equation is:

P(0) = 20

P(1) = 20*4

P(2) = (20*4)*4 = 20*4^2

We already can see the pattern, then we can write this relationship as:

P(t) = A*(4)^(t)

Where:

t represents time in years, and A is the initial population, that we know it is 20, then:

P(t) = 20*(4)^t

This is the function that represents the table.

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Q(t) = 20*(3)^t

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Q(t) = 40*(3)^t

The population by year 3 is given by replacing t by 3, then:

Q(3) = 40*(3)^3 = 1080

And the population of the other group in year 3 is seen in the table, it is 1,280, then the population of the first group is bigger by year 3, and it is greater by:

1,280 - 1,080 = 200

So the first group is larger by 200 squirrels.

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