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2 years ago

What are the similarities and differences between neurotransmitters and hormones

1 answer:

8 0

Neurotransmitters work locally and their actions are very fast. Both hormones and neurotransmitters act as chemical messengers in the body. However, neurotransmitters' actions are short-lived while hormones act for longer periods of time. Neurotransmitters are delivered through the bloodstream, whereas hormones are found primarily in the synaptic cleft. Hormones are mostly released by glands, whereas neurotransmitters are made in the cell body of the neuron.

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Which of the following is most likely to produce a destabilizing effect that results in uncontrolled growth

DanielleElmas [232]

Hello there :)

The answer to your question is oncogene.

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3 years ago

If ABD measures 48 degrees .what is the measure of DBC?

pashok25 [27]

The measure of DBC is 78
Have a great day! :)

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3 years ago

Read 2 more answers

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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3 years ago

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Arte-miy333 [17]

Answer:

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2 years ago

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Twelve year old Jamie has decided that he wants to be a doctor when he grows up. From the viewpoint of each of the following per

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Humanistic Bc he is told to be something high paying in society

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3 years ago