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skad [1K]
3 years ago
15

The area of a playground is 204 yards squared the width of a playground at 5 yards longer than its length whats the length and t

he width of the playground
Mathematics
1 answer:
Ilya [14]3 years ago
6 0
W= width= L+5
L= length
Area= 204

Area= LW
204 square yards= LW
substitute w=L+5 for w
204= (L)(L+5)
204= L^2 + 5L
subtract 204 from both sides
0= L^2 + 5L - 204
factor
0= (L+17)(L-12)

0= L+17
-17= L

0= L-12
12 yards= L


WIDTH
Since the area cannot be a negative number, we'll use L=12 to find the width.

w=L+5
w=12+5
w=17 yards


CHECK:
204=(12)(17)
204=204


ANSWER: The length is 12 yards and the width is 17 yards.

Hope this helps! :)
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For more explanation please look at the image.
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V = (About) 22.2, Graph = First graph/Graph in the attachment

Step-by-step explanation:

Remember that in all these cases, we have a specified method to use, the washer method, disk method, and the cylindrical shell method. Keep in mind that the washer and disk method are one in the same, but I feel that the disk method is better as it avoids splitting the integral into two, and rewriting the curves. Here we will go with the disk method.

\mathrm{V\:=\:\pi \int _a^b\left(r\right)^2dy\:},\\\mathrm{V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy}

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V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=\pi \cdot \int _1^3\left(1+\frac{2}{y}\right)^2-1dy\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\= \pi \left(\int _1^3\left(1+\frac{2}{y}\right)^2dy-\int _1^31dy\right)\\\\

\int _1^3\left(1+\frac{2}{y}\right)^2dy=4\ln \left(3\right)+\frac{14}{3}, \int _1^31dy=2\\\\=> \pi \left(4\ln \left(3\right)+\frac{14}{3}-2\right)\\=> \pi \left(4\ln \left(3\right)+\frac{8}{3}\right)

Our exact solution will be V = π(4In(3) + 8/3). In decimal form it will be about 22.2 however. Try both solution if you like, but it would be better to use 22.2. Your graph will just be a plot under the curve y = 2/x, the first graph.

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Step-by-step explanation: all you have to do is work it out

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