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Serggg [28]
3 years ago
7

Calculus 2

Mathematics
1 answer:
krek1111 [17]3 years ago
5 0

A simple way to see what was done is to add and subtract -1 from the numerator:

\dfrac{u^2-1+1}{u^2-1}=\dfrac{u^2-1}{u^2-1}+\dfrac1{u^2-1}=1+\dfrac1{u^2-1}

(provided that u^2-1\neq0, or u\neq\pm1)

###

Suppose you had a slightly more complex integrand, like

\dfrac{u^3}{u^2-1}=u+\dfrac u{u^2-1}

How do we know that? (Assume we don't already know the previous result, so that it's not just a matter of multiplying both sides by u.) Simple polynomial division:

u^3=\boxed{u}_{\,q}\cdot u^2, and u(u^2-1)=u^3-u. Subtracting this from u^3 gives a remainder of u^3-(u^3-u)=\boxed{u}_{\,r}, so

\dfrac{u^3}{u^2-1}=\boxed{u}_{\,q}+\dfrac{\boxed{u}_{\,r}}{u^2-1}

(where q and r denote "quotient" and "remainder")

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2 years ago
Select the scenario that correctly represents the given graph.
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Answer:

C. The vertical distance of a roller coaster from the ground changes with respect to its horizontal distance from the beginning of the ride.

Step-by-step explanation:

See how the graph is in a "U" shape? That tells us that it first decreases and then goes back up at some point.

The way you would find this answer is to go through each scenario (A, B, C, and D) and think about which thing would make sense to go down and then up.

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Solve the following linear programming problem by applying the simplex method to the dual problem.
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Answer:

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Step-by-step explanation:

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3 years ago
Three more than twice the sum of a number and half the number equals 105. Which equation models this relationship?
AlekseyPX

Answer:2(x+1/2x)+3=105

Step-by-step explanation: that is how i got my answer. If not a similar answer i would go with 2x+3=3x+105.

You can automatically eliminate the first answer. The second one has a random 5, so eliminated. and the last one just isn't correct.

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3 years ago
Airen’s grandparents deposited $1300 in a mutual fund earning 6% interest compounded annually. Write an equation to represent ho
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~~~~~~ \textit{Compound Interest Earned Amount \underline{in 18 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$1300\\ r=rate\to 6\%\to \frac{6}{100}\dotfill &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &18 \end{cases}

A=1300\left(1+\frac{0.06}{1}\right)^{1\cdot 18}\implies A=1300(1.06)^{18}\implies A\approx 3710.64 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y = 1300(1.06)^x~\hfill

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2 years ago
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