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BigorU [14]
4 years ago
8

A town is designing a rectangular park that will be 600 feet by 1000 feet. A rectangular area of the park for swing sets will be

25 feet by 100 feet. On a scale drawing of the park, the swing set area is 0.5 inch by 2 inches.
What are the dimensions of the park on the scale drawing?
Mathematics
2 answers:
Cerrena [4.2K]4 years ago
7 0
Do length*hight find width and divided by 2
Kay [80]4 years ago
5 0
12by20


is the correct
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Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck
Nimfa-mama [501]
\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}
\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}

Now, isolate the variables, so you can integrate.
(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}
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4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C
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(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C


y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}
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At x = 1, y = 0.
0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}
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4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0
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C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}
C = -\frac{\pi}{2}

\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}
6 0
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