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3 years ago

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Water is being drained out of a swimming pool at a constant rate of 780 gallons per hour. The swimming pool initially contained

45,000 gallons of water. A chemical additive must be added to the pool when it has no more than 15,000 gallons of the water remaining in the pool.
Write an expression for the amount of water remaining in the pool after h (hours)

I need help writing the equation. I can solve it from there. Thanks!

2 answers:

german3 years ago

3 0

45,000-780h=y

It looses 780 gallons per hour so you do 780 time # of hours.

solong [7]3 years ago

3 0

Answer:

$y=45000-780\cdoth$

Step-by-step explanation:

If you had 45000 gallons of water before it is drained you have calculate how much long it takes to be drained at a rate of 780 gallons per hour. If the time is h hours and y is the amount of water remaining the expression is:

$y=45000-780\cdoth$

This is the initital amount minus the amount drained after h hours to give the total remaining

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The following are the solution to the given points:

Step-by-step explanation:

Given value:

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Solve point 1 that is $\sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\$ :

when,

$k= 1 \to s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\$

$= \frac{1}{2} - \frac{1}{3}\\\\$

$k= 2 \to s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\$

$= \frac{1}{3} - \frac{1}{4}\\\\$

$k= 3 \to s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\$

$= \frac{1}{4} - \frac{1}{5}\\\\$

$k= n^ \to s_n = \frac{1}{n+1} - \frac{1}{n+2}\\\\$

Calculate the sum $(S=s_1+s_2+s_3+......+s_n)$

$S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}\\\\$

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$=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}\\\\$

$\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}$

In point 2: $\sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

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$k= 1 \to s_1 = \frac{1}{(1+6)(1+7)}\\\\$

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$k= 2 \to s_1 = \frac{1}{(2+6)(2+7)}\\\\$

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$k= 3 \to s_1 = \frac{1}{(3+6)(3+7)}\\\\$

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$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\$

when $s_n \ \ dt_{n \to 0}$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066$

$\boxed{\text{In point 2:} \sum ^{\infty}_{k = 1} \frac{1}{(n+6)(n+7)} = 0.066}$

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