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Svetach [21]
3 years ago
10

Could I get the answer, all work and explanation for this problem?

Mathematics
1 answer:
ratelena [41]3 years ago
8 0
Okay so this is how i did it but idk if it’s right.
so i multiples the top and bottom by square root of 3 to get rid of the square root of 3
then i was left with 8-1/2(1) and got 7/2

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Please Help!
xz_007 [3.2K]

Answer:

See attached

Step-by-step explanation:

<em>Refer to attachment</em>

  • a. Law of Syllogism
  • b. No valid conclusion
  • c. Law of Detachment

8 0
3 years ago
Read 2 more answers
Simplify each each expression. If not possible, write simplified.
Harman [31]
1. 10x
2. 18g
3. 6m
4. 4p
5. -5y
6. 9w
7. c^2+3d
8. 3a^2+2b^2+6a
9. 3x^2+x
10. 16x-y+3
11. 68x^2-34x-1
12. 32
13. 32
14. -9
15. 3m+3n
16. 6x-6y
17. 15f+5
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
1 year ago
Graph the solution of the system of linear inequalities.<br> y≤−4x−4<br> y≥x+2
tamaranim1 [39]

Answer:

The graph in the attached figure

Step-by-step explanation:

we have

y\leq -4x-4 ----> inequality A

The solution of the inequality A is the shaded area below the solid line y=-4x-4

The y-intercept of the solid line is -4

The x-intercept of the solid line is -1

The slope of the solid line is negative

y\geq x+2 ----> inequality B

The solution of the inequality B is the shaded area above the solid line y=x+2

The y-intercept of the solid line is 2

The x-intercept of the solid line is -2

The slope of the solid line is positive

The solution of the system of inequalities is the shaded area between the two solids line

see the attached figure

3 0
2 years ago
Solve for x 5 – 57− 2 = 63 please help
umka2103 [35]
5x - 2x = 63 + 57
3x = 120
x = 40
4 0
3 years ago
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