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ddd [48]
3 years ago
11

2. You started hiking the Appalachian Trail at Springer Mountain in Georgia. You finally reach Maine, the last state on the Appa

lachian Trail. Wanting to know how much fur- ther you have to hike, you pull out a 1:500,000 scale map of the state. You measure 35.6 inches from the Maine border (where you are standing) to Baxter Peak at the top of Mount Katahdin (the end of the trail). How many miles do you have to go? Hint: How many inches are there in a mile?​
Advanced Placement (AP)
1 answer:
Citrus2011 [14]3 years ago
5 0

Answer:

280.934 miles

Explanation:

Given

Scale = 1:500,000

Required

Determine the equivalent of 35.6 inches using the scale map in miles

First, we need to determine the actual distance in inches

Since the scale is 1:500000;

The actual distance is as follows;

Actual Distance = 35.6 * 500,000 inches

Actual Distance = 17,800,000  inches

Next, we calculate the equivalent in miles

1 inch = 1.57828e-5 mile

Multiply both sides by 17,800,000  

1 inch * 17,800,000   = 17,800,000  * 1.57828e-5 mile

17,800,000  inches = 280.93434343 miles

17,800,000  inches = 280.934 miles (Approximated)

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The regions bounded by the graphs of y=x2 and y=sin2x are shaded in the figure above. What is the sum of the areas of the shaded
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The sum of the area of the shaded regions = 0.248685

Explanation:

The sum of the area of the shaded region is given as follows;

The point of intersection of the graphs are;

y = x/2

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∴ At the intersection, x/2 = sin²x

sinx = √(x/2)

Using Microsoft Excel, or Wolfram Alpha, we have that the possible solutions to the above equation are;

x = 0, x ≈ 0.55 or x ≈ 1.85

The area under the line y = x/2, between the points x = 0 and x ≈ 0.55, A₁, is given as follows

1/2 × (0.55)×0.55/2 ≈ 0.075625

The area under the line y = sin²x, between the points x = 0 and x ≈ 0.55, A₂, is given using as follows;

\int\limits {sin^n(x)} \, dx = -\dfrac{1}{n} sin^{n-1}(x) \cdot cos(x) + \dfrac{n-1}{n} \int\limits {sin^{n-2}(x)} \, dx

Therefore;

A_2 = \int\limits^{0.55}_0 {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_0 ^{0.55}

∴ A₂ =1/2 × ((0.55 - sin(0.55)×cos(0.55)) - (0 - sin(0)×cos(0)) ≈ 0.0522

The shaded area, A_{1 shaded} = A₁ - A₂ = 0.075625 - 0.0522 ≈ 0.023425

Similarly, we have, between points 0.55 and 1.85

A₃ = 1/2 × (1.85 - 0.55) × 1/2 × (1.85 - 0.55) + (1.85 - 0.55) × 0.55/2 = 0.78

For y = sin²x, we have;

A_4 = \int\limits^{1.85}_{0.55} {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_{0.55} ^{1.85} \approx 1.00526

The shaded area, A_{2 shaded} = A₄ - A₃ = 1.00526 - 0.78 ≈ 0.22526

The sum of the area of the shaded regions, ∑A = A_{1 shaded} + A_{2 shaded}

∴ A = 0.023425 + 0.22526 = 0.248685

The sum of the area of the shaded regions, ∑A = 0.248685

8 0
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