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kipiarov [429]
3 years ago
11

the importance of the agricultural revolution is A)soil erosion B)habitat destruction C)domestication of plants and animals D)al

l of the above
Advanced Placement (AP)
2 answers:
kap26 [50]3 years ago
7 0

Answer:

The answer is :

Explanation:

D)all of the above

viva [34]3 years ago
5 0
It is all of the above because the agriculture revolution impacted our current society through A, B and C. Soil erosion is wearing away the topsoil which was because of all the humans using it. Habitat destruction was a HUGE part (and still is) of society. The loss of animals that happened during this time was a lot and was mostly because of people taking there land away.
Domestication is the process of adapting wild plants and animals for human use. The animals where raised for food, clothes, medicine and other uses. All plants and animals in a domestic relationship must be raised by humans.

Anyway hope this helped
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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
3+ 2x - y=0<br> -3-7y= 10x
juin [17]
The answer for 3+2x-y=0 is
= 2x-y=-3


And for -3-7y=10x
=10x+7y=-3


Hope this helps
7 0
3 years ago
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