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Levart [38]
3 years ago
6

___ is a federal aid program that provides undergraduate and graduate students with opportunities for part-time jobs

Advanced Placement (AP)
2 answers:
klemol [59]3 years ago
6 0
Your answer should be “C”
Feliz [49]3 years ago
3 0

Answer:

<h2>The answer is <u>C</u></h2>

Explanation:

<h3>Federal work-study is jobs that help students earn money to pay for college or career school. Undergraduate and graduate students with work-study jobs will work part-time on or off campus while enrolled.</h3>
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2 years ago
Solve the equation 2(3x+1) = 6x + 14
vova2212 [387]
2(3x + 1) = 6x + 2

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4 0
3 years ago
Allie is shopping when she finds a pair of running shoes priced at $90. When Allie uses her debit card to pay, it is declined be
Nadusha1986 [10]

Answer:

The economic concept of scarcity.

Explanation:

In economics, <em>scarcity</em><em> </em>represents the phenomenon of <em>limitless</em> <em>wants</em> suppressed by <em>limited</em><em> </em><em>resources</em>.

In this case, Allie feels she needs $90 shoes while she has not got the resources required to buy them.

This typical economic problem can be solved by moderating one's wants and clearly identifying what is priority from what is not, then intelligently making decisions on what available resources should be spent.

7 0
3 years ago
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
2 years ago
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