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ser-zykov [4K]
3 years ago
13

Two cars leave the same parking lot, with one heading north and the other heading'east. After several minutes, the northbound ca

r has
traveled 12 kilometers and the eastbound car has traveled 9 kilometers, measured in a straight line. How far apart are the two cars? Round
to the nearest tenth, if a decimal.
Mathematics
1 answer:
Alika [10]3 years ago
6 0

Answer:

I DONT KNOW DO YOU?

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Wewaii [24]

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x=23°

Step-by-step explanation:

93°+(3x+18)°=180°

93°+3x+18°=180°

111°+3x=180°

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NEED THE EQUATION
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Step-by-step explanation:

7 0
3 years ago
Need help with AP CAL
anzhelika [568]

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

\displaystyle \int_{0}^{1}e^{-2x}dx\\\\

Apply a u-substitution

u = -2x

du/dx = -2

du = -2dx

dx = du/(-2)

dx = -0.5du

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

  • If x = 0, then u = -2x = -2(0) = 0
  • If x = 1, then u = -2x = -2(1) = -2

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.

5 0
2 years ago
3 divided by 11/12 <br> I need the answer ASAP
s344n2d4d5 [400]

36/11

Calculator solves everything

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