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3 years ago

Antipsychotic medications may be used to treat

Advanced Placement (AP)

2 answers:

Anna [14]3 years ago

7 0

Schizophrenia is the answer.

statuscvo [17]3 years ago

4 0

Answer:

Antipsychotic medications may be used to treat schizophrenia.

Explanation:

Schizophrenia is the most common of psychotic diseases. At the population level, the incidence of schizophrenia in the world is about 0.4%.

Schizophrenia is diagnosed in people with very different symptoms. The main symptoms are a loss of a sense of reality, or psychoticism, due to hallucinations, delusions, impaired thinking and depression of emotions.

Schizophrenia does not mean that a person is in a psychotic state for the rest of his or her life, as at the onset of the illness, because schizophrenia has several medical conditions. Some have only one episode of psychosis that has led to the diagnosis of the disease, some have alternation of asymptomatic and psychotic episodes, and some have ongoing psychotic episodes.

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

Explanation:

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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 $(1 - 2\times t^2 })=0$

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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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