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patriot [66]
3 years ago
9

According to jefferson, who had the right to crest a government

Advanced Placement (AP)
1 answer:
Misha Larkins [42]3 years ago
3 0

the people had a right to create a government. If the government does not serve the people, then they can create a new government

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At a minimum, you should always have these two versions of a résumé saved:
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Answer: D. Electronic and Scannable

Explanation: because I took the test and got the answer right

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Adolescent egocentrism is characterized by:
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A a heightened self consciousness
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hey guys I have a hard question for my college pre help I have done this question but I want to see if you can get it
myrzilka [38]

Replacing <em>F</em> with <em>F</em> + 1 gives

5/9 (<em>F</em> + 1 - 32) = 5/9 (<em>F</em> - 32) + 5/9 = <em>C</em> + 5/9

so a temperature increase of 1º F translates to a 5/9º increase in Celsius, so I is true.

Replacing <em>F</em> with <em>F</em> + 5/9 gives

5/9 (<em>F</em> + 5/9 - 32) = 5/9 (<em>F</em> - 32) + (5/9)² = <em>C</em> + 25/81

so increasing the temperature by 5/9º F amounts to a 25/81º increase in Celsius, so III is false.

Solve for <em>F</em> in terms of <em>C</em> :

<em>F</em> = 9/5 <em>C</em> + 32

Replacing <em>C</em> with <em>C</em> + 1 gives

9/5 (<em>C</em> + 1) + 32 = 9/5 <em>C</em> + 32 + 9/5 = <em>F</em> + 9/5

and 9/5 = 1.8, so a 1º C increase translates to a 1.8º F increase, so II is also true and the answer is D.

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3 years ago
Explain how the environment can impact and influence how people and societies live?
Goshia [24]

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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

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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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