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3 years ago

Mike travels 13 miles per 2 hours on his bike.

Mathematics

1 answer:

LiRa [457]3 years ago

4 0

Answer:

It took Markus half an hour to drive home from work. He averaged 34 miles per hour. How far does Markus live from his work?

Solution

We are given that it takes 1/2 an hour for the trip. This is a time:

t = 1/2

We are given that he averages 34 miles per hour. This is a rate:

r = 34

We are asked how few he has traveled. This is a distance. We use the d=rt equation:

d = rt

= (34)(1/2)

= 17

Markus lives 17 miles from work.

Now try one by yourself. If you want to see the answer, put your mouse on the yellow rectangle and the answer will appear.

Exercise 1

The current along the beach is moving towards the south at 1.5 miles per hour. If a piece of debris is placed into the water, how far will the current take it in 6 hours?

Step-by-step explanation:

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Need help with AP CAL

anzhelika [568]

Answer: Choice C

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

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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.

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Furthermore,

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