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pishuonlain [190]
3 years ago
15

¿A qué disciplina artística pertenecen las obras que serán exhibidas en el museo?

Advanced Placement (AP)
1 answer:
dexar [7]3 years ago
8 0
Sorry no entiendo? Puedes explicar mas?
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Open book tests are about problem solving and applying information. true, false?
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Read 2 more answers
(a) Use the climatograph above to identify the following.
Alex73 [517]

Answer:

(i) Identify the month with the least amount of rainfall in this temperate grassland.

The month with the least amount of rainfall is January, when only around 20 mm of rain fall in the month. In the climatograph, rainfall is represented by the bars, while temperature is represented by the line.

(ii) Identify the maximum mean monthly temperature in this temperate grassland.

The maximum mean monthly temperature occurs at the end of June and the start of July. The mean monthly temperature reaches over 25 degrees centigrades over this time period.

(iii) A climatograph for a tropical grassland or savanna would look different from the climatograph shown for a temperate grassland. Describe one difference in the patterns or trends in the temperature data in a tropical grassland compared to those of a temperate grassland.

The climatograph for the tropical grassland would show similar patterns for rainfall, but higher mean monthly temperatures. Tropical grasslands are generally almost as dry as temperate grasslands, but they are much warmer over the year (lacking a proper winter for example).

3 0
3 years 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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