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Sladkaya [172]
3 years ago
7

What was John C. Calhoun’s term for the 1828 Tariff?

Advanced Placement (AP)
1 answer:
galben [10]3 years ago
4 0

Answer:

Tariff of Abominations

Explanation:

John C. Calhoun’s term  for the 1828 Tariff was the Tariff of Abominations

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During Excavation, grading, and other site work, the ___ is the permanent reference point from which all elevation information i
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I can’t figure this problem out
Natalka [10]

Answer:

Explanation:

Alright so the way to do this is to use properties of integrals to make our life easier.

So we have:

\int\limits^4_1 {(3f(x)+2)} \, dx

So lets break this up into two different integrals that represent the same area.

\int\limits^4_0 {f(x)} \, dx - \int\limits^1_0 {f(x)} \, dx = \int\limits^4_1 {f(x)} \, dx

Lets think about what is going on up there. The integral from four to zero gives us the area under the curve of f(x) from four to zero. If we subtract this from the integral from one to zero (the area under f from one to zero) we are left with the area under f from four to one! Hence:

\int\limits^4_1 {f(x)} \, dx

But since we have these values we can say that:

-3 - 2 = -5

Which means that \int\limits^4_1 {f(x)} \, dx = -5

So now we can evaluate \int\limits^4_1 {(3f(x)+2)} \, dx

Lets first break up our integrand into two integrals

\int\limits^4_1 {(3f(x)+2)} \, dx = 3\int\limits^4_1{f(x)} \, dx + 2\int\limits^4_1 {} \, dx

Now we can evaluate this:

We know that \int\limits^4_1 {f(x)} \, dx = -5

So:

3(-5)+2[x] where x is evaluated at 4 to 1 so

-15 + 2(3)

So we are left with -15 + 6 = -9

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3 years ago
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