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sveticcg [70]
3 years ago
12

Help! Show work pleasewill mark brainliest ​

Mathematics
1 answer:
Arada [10]3 years ago
6 0

Answer:

x = 62.23 ft

y = 47.67 ft    

Step-by-step explanation:

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Jack ran 4 kilometers each day for 3 days. How many meters did jack run in 3 days?​
Novosadov [1.4K]

Answer:

12,000

Step-by-step explanation:

4 x 1000 x 3

7 0
3 years ago
Read 2 more answers
Please help! I have this question and one more I'm about to post, if you can answer them that would be most appreciated. These a
dimaraw [331]
D is the answer to this problem
4 0
3 years ago
What is the depth and volume of this rectangular prism in cubes and cubic inches?
Inessa05 [86]
Height = 3 cubes
Width = 5 cubes
Depth = 4 cubes

Number of Cubes = 3*4*5 = 60

Each cube is .2 inches on each side

So, it's volume = 1 * .6 * .8 = .48 cubic inches

Double - Check
Each cube is .2*.2*.2 = <span> <span> <span> 0.008 </span> </span> </span> cubic inches
There are 60 cubes so the volume =
60 * .008 = .48 cubic inches CORRECT !!




6 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
URGENT EXPLANATION AND SOLVING
Alex787 [66]
\cfrac{a}{\sin(a)} = \cfrac{b}{\sin(b)} &#10;

\cfrac{31}{\sin(42)} = \cfrac{37}{\sin(x)} &#10;

31\sin(x) = 37 \sin(42)

\sin(x) = \cfrac{37 \sin(42) }{31}

x = \sin^{-1}(\cfrac{37 \sin(42) }{31} )

x = 53 \textdegree

Answer: 53°
7 0
3 years ago
Read 2 more answers
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