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s2008m [1.1K]
3 years ago
7

Rosa just finished installing a small koi Pond in her backyard and now needs to re-sod. One side of the koi pond is 2 feet short

er than the other side. One side of the garden is three times the longer side of the koi pond The other side of the garden is 5 feet longer than the first side. How much psi does Rosa need to buy? K

Mathematics
1 answer:
hram777 [196]3 years ago
6 0
The sketch of the garden and the pond is shown below

Area of garden = length × width
Area of garden = [3x + 5] × 3x
Area of garden = 3x[3x+5]
Area of garden = 9x² + 15x

Area of pond = [x-2] × x
Area of pond = x[x-2]
Area of pond = x² - 2x

Area to be re-sod = Area of garden - Area of pond
Area to be re-sod = 9x² + 15x - [x² - 2x]
Area to be re-sod = 9x² - x² + 15x + 2x
Area to be re-sod = 8x² + 17x

The amount of sod needed to be bought would depend on how much in each packaging they are being sold.

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well, let's say the the rate of speed of the cold water faucet is "c", so in 1minute it has done 1/c of the whole work

now, we know the hot water faucet is slower, it takes it 3 times as long, that means is 1/3 the speed of the cold water faucet, so, how much does the hot water faucet do in 1min?   well, we know is 1/3 fast as the cold one, the cold one is 1/c, what's 1/3 of 1/c? well, 1/3 * 1/c

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thus

\bf \begin{array}{cccccclllllll}
\cfrac{1}{c}&+&\left( \cfrac{1}{3}\cdot \cfrac{1}{c} \right)&=&\cfrac{1}{6}\\\\
\cfrac{1}{c}&+&\cfrac{1}{3c}&=&\cfrac{1}{6}\\\\
\uparrow &&\uparrow &&\uparrow \\
\textit{cold water rate}&&\textit{hot water rate}&&\textit{total done by both}
\end{array}\\\\
-----------------------------\\\\

\bf \cfrac{3+1}{3c}=\cfrac{1}{6}\implies 24=3c\implies \cfrac{24}{3}=c\implies \boxed{8=c}
\\\\\\
\textit{now, if the cold water faucet can do it in 8 minutes by itself}\\\\
\textit{then the hot water faucet can do it in 3 times as long}
\\\\\\
3\cdot 8\implies \boxed{24}
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