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Mariulka [41]
3 years ago
14

*TIMED*

Advanced Placement (AP)
1 answer:
MrMuchimi3 years ago
4 0

Answer:

attempt to treat a person with depression. Then, conclude the best course of treatment.

-Rational Emotive Behavior Therapy.

-Cognitive-Behavioral Therapy.

-Behavioral Therapy.

-Client-Centered

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Explanation:Its ball d

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Religion: Have humans upheld the covenants they made with God?
emmasim [6.3K]

Answer:

This question really requires me to think because I am not religious at all, but personally I dont think so because we as humans, or well the religious humans, they strive to god for practically everything, and they used ''God'' to justify all the wrong doings. A covenant from my knowledge is an agreement to something, so with religion its between someones God and the person who believes in them, so yeah religious humans they have kept it up.  Hopefully, me as a non religious person answering this question, will maybe help you to find the true answer if you are religious. Have a good day.

Explanation:

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2 years ago
Which of the following pairs of angles is not necessarily congruent
Alex

Answer:

E. \angle 3 and \angle 5.

Explanation:

According to the figure, we have the following conclusions:

1) \angle 1 and \angle 2 are vertical opposite angles. (m \angle 1 = m \angle 2)

2) \angle 2 and \angle 3 are internal alternate angles. (m \angle 2 = m\angle 3)

3) \angle 2 and \angle 4 are corresponding angles. (m\angle 2 = m\angle 4)

4) \angle 3 and \angle 4 are vertical opposite angles. (m \angle 3 = m \angle 4)

But, \angle 3 and \angle 5, since m \angle 3 + m\angle 6 + m\angle 5 = 180^{\circ}, where \angle 3 is an acute angle and \angle 6 is a right angle, and m \angle 4 + m \angle 5 = 90^{\circ}. Then,

m\angle 5 = 90^{\circ} -m\angle 4.

m\angle 5 = 90^{\circ}-m\angle 3

Hence, the right answer is E.

8 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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