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makkiz [27]
2 years ago
10

Identify and explain ONE way in which industrialization in the period circa 1750–1900 can be considered an economic turning poin

t in global history.
Advanced Placement (AP)
1 answer:
timama [110]2 years ago
4 0
Answer:
Industrialization turned the global history in terms of economic as given below:
Explanation:
With the arrival of the industrial revolution in Europe and America, it changed the social and economic conditions of the people. It increases the economy for various countries because it allowed more jobs and led to the formation of urbanization which led to the growth of population. Most of the industries were built based on making profits. Society was moving towards modernism. The change in the living style also led the people to seek an explanation.
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Your voltmeter and your teacher are both right, but the water analogy of electricity can only go so far. One big drawback is that, unlike with water, there is no absolute voltage. .... Wherever you put the black lead is defined as "0V". .... flowing (i.e. the circuit is broken) you will not get this change in voltage.

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Explain four reasons of preserving farm produce​
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Mrs young bought 4 packed of pencils to give to her students to start the year she has 25 students each pack of pencils has 8 in
Stolb23 [73]

Answer:

yes

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Sunog na nagsulong sa pagtatag ng pamahalaang sibil sa pilipinas
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Answer: Susog spooner
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2 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
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