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JulsSmile [24]
3 years ago
11

If someone has a current gpa of 4.33, what does that mean? Do they have a good change to go to a competitive school? (Just based

on the gpa)
Advanced Placement (AP)
2 answers:
elena-s [515]3 years ago
8 0
Yes it actually does because in most schools you have to have a gpa of 4.0
kakasveta [241]3 years ago
5 0
A GPA is a grade point average, so it is supposed to be out of 4, but smarter people who take honors or AP can have a max of out of 5. A GPA as 4.33 is very good and you can. I can tell you this from an experience because I had a 4.73 GPA and I got in Stanford so go for it!
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Evaluate the extent to which United States foreign policy in the period from 1970 to 1991 contributed to the end of the Cole War
mr_godi [17]

Answer:

The Kennedy and Johnson administrations advocated a "flexible response" to containing communism, supporting a failed attempt by Cuban exiles to overthrow Fidel Castro, issuing a naval blockade with the threat of nuclear weapons during the Cuban Missile Crisis and deploying troops to prevent the spread of communism in South Vietnam, a decade-long struggle that caused domestic turmoil in the U.S. Containment also took place in more subtle ways. In the 1970s, President Nixon attempted to ease tensions with the Soviet Union. Nixon visited communist China and engaged in several diplomatic meetings with the Soviet leader. By the end of the decade, tensions once again escalated as the Soviet Union invaded Afghanistan. When President Reagan took office he denounced the Soviet Union as the "Evil Empire" and dramatically increased military budgets in an attempt to "win" the Cold War. Despite Reagan's contentious rhetoric, tensions between the two superpowers calmed in the late 1980s. Soviet leader adopted friendly relations with the west and instituted liberal domestic reforms through glasnost and perestroika. Reagan, Gorbachev, and British Prime Minister Margaret Thatcher met repeatedly to find common ground as the decade came to a close. In the end, the struggling Soviet economy led to the end of the Cold War. Weakened, the Soviets lost control of much of Eastern Europe by 1990. The fall of the Berlin Wall in November 1989 and the collapse of the Soviet Union in 1991 amounted to the end of the Cold War.

Explanation:

5 0
1 year ago
Which of the following is true?
Stells [14]

hey buddy here is your answer

when object produces light energy,they sometimes produces heat energy too

8 0
2 years ago
Read 2 more answers
Ito ang bunga ng pagsasama ng salik ng produksyon
ioda

Answer:

output ang tawag sa bunga ng pagsasama ng salik ng edukasyon

3 0
2 years ago
AP Calculus AB - Differentiating using the Product Rule
faltersainse [42]

Answer:

  h'(t) = (√t)(1 -5t²)/(2t)

Explanation:

First of all, your product rule needs to be written correctly.

  (f(x)g(x))' = f'(x)g(x) +f(x)g'(x)

You have ...

h(t)=\sqrt{t}(1-t^2)\\\\f(t)=\sqrt{t} \qquad f'(t)=\dfrac{1}{2\sqrt{t}}=\dfrac{\sqrt{t}}{2t}\\\\g(t)=1-t^2 \qquad g'(t)=-2t\\\\h'(t)=f'(t)g(t)+f(t)g'(t)=\dfrac{\sqrt{t}}{2t}(1-t^2)+\sqrt{t}(-2t)\\\\h'(t)=\dfrac{\sqrt{t}(1-t^2-4t^2)}{2t}=\dfrac{\sqrt{t}(1-5t^2)}{2t}

4 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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