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butalik [34]
3 years ago
15

Raise the Driving Age to Eighteen! After reading Mr. J. Brown's letter, dated August 15, I feel compelled to write and let your

readers know how I feel about raising the driving age from 16 to 18. Teenagers should not be allowed to drive until they are 18. Only high school graduates should be given a driver's license, and only after successfully completing a driver's education class during high school. Right now, 15 year-olds can get a practice permit and try for their license at 16. Eighteen is a better age. At 18, people are more mature and better able to make mature decisions. Not all 18 year-olds are mature, but the majority of them are. At least we are more mature than 15 year-olds. What is the main purpose of this passage? A) to warn older drivers to be careful when they drive B) to convince readers to support raising the driving age C) to entertain teens with stories of safe drivers D) to support insurance claims against teens who get into accidents
Advanced Placement (AP)
2 answers:
Neko [114]3 years ago
5 0

Answer:

B) to convince readers to support raising the driving age

Explanation:

The author's purpose in this passage is to convince readers to support raising the driving age. The author explicitly says several times that teenagers below the age of eighteen are not mature enough to be allowed to drive; therefore, the age at which teenagers are allowed to drive needs to be raised from sixteen to eighteen.

<h2>Hope this helps!</h2>
olya-2409 [2.1K]3 years ago
3 0
The answer is B) to convince readers to support raising the driving age.
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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
Alex Ar [27]

Answer:

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

y = sin²x

∴ 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
3 years ago
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