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MrRa [10]
3 years ago
5

5 dived by 1 over 10

Mathematics
2 answers:
d1i1m1o1n [39]3 years ago
8 0
2 in a decimal form 2/1
MakcuM [25]3 years ago
5 0
50. You’d multiply by the reciprocal, which is 10z
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Ann is adding water to a swimming pool at a constant rate. The table below shows the amount of water in the pool after different
N76 [4]

Answer:

(a)  As time increases, the amount of water in the pool increases.

     11 gallons per minute

(b)  65 gallons

Step-by-step explanation:

From inspection of the table, we can see that <u>as time increases, the amount of water in the pool increases</u>.

We are told that Ann adds water at a constant rate.  Therefore, this can be modeled as a linear function.  

The rate at which the water is increasing is the <em>rate of change</em> (which is also the <em>slope </em>of a linear function).

Choose 2 ordered pairs from the table:

\textsf{let}\:(x_1,y_1)=(8, 153)

\textsf{let}\:(x_2,y_2)=(12,197)

Input these into the slope formula:

\textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{197-153}{12-8}=\dfrac{44}{4}=11

Therefore, the rate at which the water in the pool is increasing is:

<u>11 gallons per minute</u>

To find the amount of water that was already in the pool when Ann started adding water, we need to create a linear equation using the found slope and one of the ordered pairs with the point-slope formula:

y-y_1=m(x-x_1)

\implies y-153=11(x-8)

\implies y-153=11x-88

\implies y=11x-88+153

\implies y=11x+65

When Ann had added no water, x = 0.  Therefore,

y=11(0)+65

y=65

So there was <u>65 gallons</u> of water in the pool before Ann starting adding water.

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2 years ago
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