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Ber [7]
3 years ago
15

The scale for a map is 20 miles = 1/2 inch. The distance between two towns on the map is 3 3/4 inches. What is the actual distan

ce between these towns?
Mathematics
1 answer:
zubka84 [21]3 years ago
4 0

Answer:

150 miles

Step-by-step explanation:

If 1/2 of an inch is 20 miles, we can assume that 1/4 of an inch is 10 miles, since half of 1/2 is 1/4 and 1/2 of 20 is 10.

Equipped with this information, we can divide 3 by 1/2 or multiply 3 by the reciprical, 2/1 to get how much we should multiply 20 by.

3*2=6

20*6=120

Then we can multiply 10 by 3 and add that on our answer, since there are still 3/4s left and 1/4=10 miles.

10*3=30

120+30=150

The actual distance is 150 miles.

Hope this helps!

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Convert the integral below to polar coordinates and evaluate the integral.
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Answer:

\int\limits_{0}^{4/\sqrt{2}}\int\limits_{y}^{\sqrt{16-y^2}}  xy \, dxdy =  \int\limits_{0}^{\pi/4} \, \int\limits_{0}^{4}    r^3 cos(\theta)\sin(\theta) \,\,  drd\theta = 16

Step-by-step explanation:

We are trying to evaluate this integral.

\int\limits_{0}^{4/\sqrt{2}}\,\,\int\limits_{y}^{\sqrt{16-y^2}}  xy \,\,dxdy

The first thing that we have to do is understand this region in the plane.

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Now we need to describe that same region in polar coordinates.

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Now remember that when we do the polar transformation we use the following formula

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Then our integral would be

\int\limits_{0}^{4/\sqrt{2}}\int\limits_{y}^{\sqrt{16-y^2}}  xy \, dxdy =  \int\limits_{0}^{\pi/4} \, \int\limits_{0}^{4}    r^3 cos(\theta)\sin(\theta) \,\,  drd\theta = 16

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