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NikAS [45]
3 years ago
14

Two large and 1 small pumps can fill a swimming pool in 4 hours. One large and 3 small pumps can also fill the same swimming poo

l in 4 hours. How many hours will it take 4 large and 4 small pumps to fill the swimming pool.(We assume that all large pumps are similar and all small pumps are also similar.)
if u get this right i will mark brainliest
Mathematics
1 answer:
kolezko [41]3 years ago
6 0

Let <em>x</em> and <em>y</em> be the unit rates at which one large pump and one small pump works, respectively.

Two large/one small operate at a unit rate of

(1 pool)/(4 hours) = 0.25 pool/hour

so that

2<em>x</em> + <em>y</em> = 0.25

One large/three small operate at the same rate,

(1 pool)/(4 hours) = 0.25 pool/hour

<em>x</em> + 3<em>y</em> = 0.25

Solve for <em>x</em> and <em>y</em>. We have

<em>y</em> = 0.25 - 2<em>x</em>   ==>   <em>x</em> + 3 (0.25 - 2<em>x</em>) = 0.25

==>   <em>x</em> + 0.75 - 6<em>x</em> = 0.25

==>   5<em>x</em> = 0.5

==>   <em>x</em> = 0.1

==>   <em>y</em> = 0.25 - 2 (0.1) = 0.25 - 0.2 = 0.05

In other words, one large pump alone can fill a 1/10 of a pool in one hour, while one small pump alone can fill 1/20 of a pool in one hour.

Now, if you have four each of the large and small pumps, they will work at a rate of

4<em>x</em> + 4<em>y</em> = 4 (0.1) + 4 (0.05) = 0.6

meaning they can fill 3/5 of a pool in one hour. If it takes time <em>t</em> to fill one pool, we have

(3/5 pool/hour) (<em>t</em> hours) = 1 pool

==>   <em>t</em> = (1 pool) / (3/5 pool/hour) = 5/3 hours

So it would take 5/3 hours, or 100 minutes, for this arrangement of pumps to fill one pool.

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<u>252 square inches is the surface area of Josiah's cereal box.</u>

Step-by-step explanation:

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Width of the box = 2 inches

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Let's recall the surface area of a rectangular prism:

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Answer:

a(\frac{1}{5e})=5e

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we are given equation for position function as

s(t)=tln(5t)

Since, we have to find acceleration

For finding acceleration , we will find second derivative

s'(t)=\frac{d}{dt}\left(t\ln \left(5t\right)\right)

=\frac{d}{dt}\left(t\right)\ln \left(5t\right)+\frac{d}{dt}\left(\ln \left(5t\right)\right)t

=1\cdot \ln \left(5t\right)+\frac{1}{t}t

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now, we can find derivative again

s''(t)=\frac{d}{dt}\left(\ln \left(5t\right)+1\right)

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