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AlekseyPX
3 years ago
14

The lines graphed below are perpendicular. The slope of the red line is 3. What is the slope of the green line?

Mathematics
2 answers:
allochka39001 [22]3 years ago
7 0

→Answer:

-1/3

→Step-by-step explanation:

Well this is pretty simple.

2 lines that are perpendicular to each other are both reciprocals of each other.

So what’s the reciprocal of 3?

Ans: -1/3

Nastasia [14]3 years ago
6 0

Answer:

Hey there!

The slope of the green line is -1/3.

Perpendicular lines have opposite reciprocal slope.

Hope this helps :)

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120lbs Earth................................106lbs Venus
150lbs Earth...................................?Venus

Venus weight=150*106/120=132.50 lbs
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What is the equation of this line? <br>y=2×-3<br>y=-1/2×-3<br>y=-2×-3<br>y=1/2×-3​
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Answer:

First option, y = 2x - 3

Step-by-step explanation:

Remember the formula for a line is y=mx+b, where m is slope (slope = rise/run) and b is y-intercept (y-intercept is the point that crosses the y-axis)

Since the line on the graph crosses the y-axis at -3, that will be the y-intercept. Starting from that point, find another point. In this case you go upwards 2 units and to the right 1, so your slope will be 2/1 or just 2 .

Hope this helped :)

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Sweet Treats Bakery sells gourmet cookies. If three gourmet cookies cost $7.95, then
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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)

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

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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Move the shape down by one quadrant and flip it so where “U” was at, “W” is in its place and Vice versa
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