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4 years ago

What is the name of the body's largest artery?

Advanced Placement (AP)

1 answer:

aliina [53]4 years ago

3 0

The largest artery is the abdominal aorta.

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A key drawback with the use of tests to screen individuals for desirable characteristics is that

DochEvi [55]

Answer:

A key drawback with the use of tests to screen individuals for desirable characteristics is that It is a time-consuming procedure.

It is very important for the employees to know exactly what they must look into and assess the characteristics accurately. This process is very time consuming and takes up a lot of patience and resources to carry the tests.

A study that would need to occur quickly cannot indulge into this testing process because of it being very tedious and time consuming.

3 0

4 years ago

Saan nag mumula telang pinya jusi at sinamay

nordsb [41]

What? Is this a serious question?

6 0

3 years ago

Do anyone else need help with anything

Kisachek [45]

Answer

The classroom had 35 glue sticks. If the ratio of glue sticks to glue bottles was 5:2, how many glue bottles did the classroom have? _

Explanation:

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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 $2\times e^{4-2^2} = 2$

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.

8 0

3 years ago

What does a perceived lack of ability to influence government and politics reflect?

Alenkinab [10]

Political efficacy is the ability to influence government and politics, a perceived lack of political efficacy reflects a lack of faith in government, where citizens feel that their actions do not affect government and their political leaders. Efficacy normally increases with age and when people feel low political efficacy they tend to feel less engaged and have less strong views.

5 0

3 years ago