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3 years ago

9

The length of a rectangle is (x+5) and the width is (x+2). The area of the rectangle is 70 square feet. What is the length of th

e rectangle?

1 answer:

vaieri [72.5K]3 years ago

7 0

Answer:

The length of the rectangle is 30 feet.

Step-by-step explanation:

Because the equation for area is length times width, you would have to set the two equations to seventy and multiply them.

<em> (x+5) (x+2) = 70</em>

<em> </em><em>2x + 10 = 70</em>

Then, you would have to subtract ten from both sides, cancelling it from the left side.

<em> 2x = 60</em>

<em />

Then do the opposite and divide sixty by two, therefore getting a length of thirty feet.

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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.

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3 years ago

A track star runs two races on a certain day. The probability thathe wins the first race is 0.7, the probability that he wins th

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Answer:

a) 80% probability that he wins at least one race.

b) 30% probability that he wins exactly one race.

c) 20% probability that he wins neither race.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that he wins the first race.

B is the probability that he wins the second race.

C is the probability that he does not win any of these races.

We have that:

$A = a + (A \cap B)$

In which a is the probability that he wins the first race but not the second and $A \cap B$ is the probability that he wins both these races.

By the same logic, we have that:

$B = b + (A \cap B)$

The probability that he wins both races is 0.5.

This means that $A \cap B = 0.5$

The probability that he wins the second race is 0.6

This means that $B = 0.6$

$B = b + (A \cap B)$

$0.6 = b + 0.5$

$b = 0.1$

The probability that he wins the first race is 0.7.

This means that $A = 0.6$

$A = a + (A \cap B)$

$0.7 = a + 0.5$

$a = 0.2$

A) he wins at least one race.

This is

$P = a + b + (A \cap B) = 0.2 + 0.1 + 0.5 = 0.8$

There is an 80% probability that he wins at least one race.

B) he wins exactly one race.

This is

$P = a + b = 0.2 + 0.1 = 0.3$

There is a 30% probability that he wins exactly one race.

C) he wins neither race

Either he wins at least one race, or he wins neither. The sum of these probabilities is 100%.

From a), we have that there is an 80% probability that he wins at least one race.

So there is a 100-80 = 20% probability that he wins neither race.

6 0

3 years ago