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Cloud [144]
2 years ago
14

A power plant can generate 9000 megawatts of energy each year at full capacity. How much energy would be available each day to t

he citizens of Fremont?
Advanced Placement (AP)
1 answer:
Blababa [14]2 years ago
5 0

Answer:

24.65 megawatts.

Explanation:

Given that a power plant can generate 9000 megawatts of energy each year at full capacity, to determine how much energy would be available each day to the citizens of Fremont, the following calculation must be performed:

9,000 / 365 = X

24.65 = X

Thus, each day the citizens of Fremont have 24.65 megawatts of energy available.

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Explanation:

These are the reserve requirement, open market operations, the discount rate, and interest on excess reserves. These tools can either help expand or contract economic growth.

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Explain how the H-O theory can be used to explain the pattern of trade between two countries.​
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Answer: See explanation

Explanation:

The Heckscher-Ohlin model refers to an economic theory which states that countries will export the goods that they can produce efficiently and in large quantities while they'll import those that they are less efficient in producing.

According to the H-O theorem, the pattern of trade that exists between countries as a result of the characteristics that are possessed by the countries. In such case, a capital-abundant country can produce a capital intensive good efficiently and therefore should export the capital intensive good. Likewise, a labor-abundant country can produce labor intensive good efficiently and therefore should export the labor-intensive good.

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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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