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2 years ago

2. What factors brought about the end of communism in Eastern Europe and the Soviet Union?

1 answer:

5 0

The Soviet Union experienced a crisis due to economic waste. The socio-cultural crisis in the USSR occurred because of class division, the low quality of people's lives, and the government was not allowed to develop people's creativity. In particular, the system named Glasnost and Perestroika which was proclaimed by President Gorbachev was the trigger for the explosion of social revolutions in Eastern European countries.

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Pull over to the closest edge of the road

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1. Which of the following statements about alcohol is correct?

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2 years ago

How to find out whether or not if the line is parallel,intersecting,perpendicular or coincident

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1 year ago

Hello, I need help with a calculus FRQ. My teacher has given a hint that this last part has to do with the intermediate value th

lesya [120]

Answer:

Yes, at a time t such that (√2)/2 ≤ t ≤ 2.

Explanation:

To answer the question

Therefore, where the domain of the function is the set of all real numbers x for which f(x) is a real number we have

For Chloe's velocity

$C(t) = t\times e^{4-t^2} \ for \ 0\leq t\leq 2$

Finding the boundaries of the function gives;

$0\times e^{4-0^2} = 0$ and $2\times e^{4-2^2} = 2$

At t = 1, we have $1\times e^{4-1^2} = e^{3} = 20.086$

We find the maximum point as follows;

$\frac{\mathrm{d} \left (t\times e^{4-t^2} \right )}{\mathrm{d} x}=0$

From which we have;

$\frac{\mathrm{} e^{4-t^2} - t\times e^{4-t^2} \times2\times t }{(e^{4-t^2} )^2}=0$

$e^{4-t^2} - t\times e^{4-t^2} \times2\times t }=0$

$e^{4-t^2}(1 - t\times2\times t })=0\\e^{4-t^2}(1 - 2\times t^2 })=0\\$

$e^{4-t^2}=0$ or $(1 - 2\times t^2 })=0$

∴ 1 = 2·t² and from which t = (√2)/2

Hence the function C(x) is decreasing from t = (√2)/2 to t = 2

For Brandon

For 0 ≤ t ≤ 1, 1 ≤ B(t) ≤8 and for 1 < t ≤ 2, 8 < B(t) ≤ 1.5

1 ≤ f(x) ≤ 1.5

Given that the function B(t) is differentiable, therefore, continuous, there exists a point at which the function C(t) and B(t) intersects given that;

For 0 ≤ t ≤ (√2)/2, 0 ≤ C(t) ≤ 23.416 for (√2)/2 < t ≤ 2, 23.416 > C(t) ≥ 2

and for 0 ≤ t ≤ 0 1 ≤ B(t) ≤ 8 and for 1 < t ≤ 2, 8 > B(t) ≥ 1.5

Therefore, the curves intersect at in between (√2)/2 ≤ t ≤ 2.

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3 years ago

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Answer:

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