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3 years ago

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Working together, it takes two different sized hoses 30 minutes to fill a small swimming pool. If it takes 50 minutes for the la

rger hose to fill the swimming pool by itself, how long will it take the smaller hose to fill the pool on its own

2 answers:

katrin2010 [14]3 years ago

8 0

Answer:

75 minutes

Step-by-step explanation:

Let's call the volume of the pool by V.

If the larger hose takes 50 minutes to fill the pool, the amount of water this hose fills is V/50

Let's call the amount of time the small hole takes to fill the pool on its own by T, so its speed is V/T

The two holes together fill the pool in 30 minutes, so their speed summed is V/30:

V/50 + V/T = V/30

Dividing the equation by V, we have:

1/50 + 1/T = 1/30

(3T + 150)/150T = 5T/150T

3T + 150 = 5T

2T = 150

T = 75 minutes

Oduvanchick [21]3 years ago

7 0

Answer:

75 minutes

Step-by-step explanation:

In this question, we are asked to calculate the time it will take a smaller hose out of two hoses to fill a swimming pool by itself.

Firstly, let’s say the volume of the swimming pool to fill is Xm^3.

Hence for both, the rate of filling is x/30 m^3/minutes. For the larger hose, rate will be x/50 m^3/minutes. For the larger hose, let the time taken be z minutes. It’s rate will be x/z m^3/minute.

Mathematically, adding both rates together give the rate of both at the same time. This means;

X/30 = x/50 + x/z

X/z = x/30 - c/50

x/z = 20/1500

x/z = 1/75

z = x/75

This means the time that it will take the smaller hose will be 75 minutes

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As we know that for any polynomial let say<em> </em> $p(x)$ <em>, </em> $c$ is the root of the polynomial if $p(c)=0$ .

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$\left(-2\right)^4-9\left(-2\right)^2-4\left(-2\right)+12$

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$=\left(-2\right)^4-9\left(-2\right)^2+4\cdot \:2+12$

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