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lorasvet [3.4K]
2 years ago
12

An initial deposit of $10000 is left in account for 5.5 years at 3.6% interest compounded continuously. What is the

Mathematics
2 answers:
aniked [119]2 years ago
4 0

Answer:

11980

Step-by-step explanation:

This is simple interest.

We use the formula i=prt. Interest = price x rate x time

You just plug in your numbers, 10000 is your price, 5.5 years is the time, and 3.6% interest is rate.

Now our equation looks like this: i = 10000 x 5.5 x 0.036

(i converted the percent to a decimal)

So the interest is 1980, but the question asks for the final value, so we add the interest to the initial deposit, 1980 + 10000, and we get 11980 for our final value.

I apologize if my answer is wrong, but if it isn't I hope I helped :)

nataly862011 [7]2 years ago
3 0

9514 1404 393

Answer:

  $12,190

Step-by-step explanation:

The formula for the balance in an account where continuous compounding occurs is ...

  A = Pe^(rt) . . . . where P is the principal invested at annual rate r for t years

  A = $10,000e^(0.036·5.5) = $10,000e^0.198

  A ≈ $12189.62 ≈ $12,190 . . . . rounded to the nearest dollar

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Answer: Choice C

\displaystyle \frac{1}{2}\left(1 - \frac{1}{e^2}\right)

============================================================

Explanation:

The graph is shown below. The base of the 3D solid is the blue region. It spans from x = 0 to x = 1. It's also above the x axis, and below the curve y = e^{-x}

Think of the blue region as the floor of this weirdly shaped 3D room.

We're told that the cross sections are perpendicular to the x axis and each cross section is a square. The side length of each square is e^{-x} where 0 < x < 1

Let's compute the area of each general cross section.

\text{area} = (\text{side})^2\\\\\text{area} = (e^{-x})^2\\\\\text{area} = e^{-2x}\\\\

We'll be integrating infinitely many of these infinitely thin square slabs to find the volume of the 3D shape. Think of it like stacking concrete blocks together, except the blocks are side by side (instead of on top of each other). Or you can think of it like a row of square books of varying sizes. The books are very very thin.

This is what we want to compute

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Apply a u-substitution

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du/dx = -2

du = -2dx

dx = du/(-2)

dx = -0.5du

Also, don't forget to change the limits of integration

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This means,

\displaystyle \int_{0}^{1}e^{-2x}dx = \int_{0}^{-2}e^{u}(-0.5du) = 0.5\int_{-2}^{0}e^{u}du\\\\\\

I used the rule that \displaystyle \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx which says swapping the limits of integration will have us swap the sign out front.

--------

Furthermore,

\displaystyle 0.5\int_{-2}^{0}e^{u}du = \frac{1}{2}\left[e^u+C\right]_{-2}^{0}\\\\\\= \frac{1}{2}\left[(e^0+C)-(e^{-2}+C)\right]\\\\\\= \frac{1}{2}\left[1 - \frac{1}{e^2}\right]

In short,

\displaystyle \int_{0}^{1}e^{-2x}dx = \frac{1}{2}\left[1 - \frac{1}{e^2}\right]

This points us to choice C as the final answer.

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Step-by-step explanation:

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