Answer:
143cm
Step-by-step explanation:
Area=2(l+b)
Substitute the values for the variables.
66=2(l+22/5)
66=2xl+2x<u>22</u>
5
66=2l+<u>44</u>
5
Cross multiply
66x5=2l+44
330=2l+44
Collect like terms
-2l = 44-330
-2l = -286
<u>-2l </u>=<u> </u><u>-286</u>
-2 -2
l=143
The missing length is 143m.
Answer: Choice C

============================================================
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 
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
where 0 < x < 1
Let's compute the area of each general cross section.

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

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,

I used the rule that
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]](https://tex.z-dn.net/?f=%5Cdisplaystyle%200.5%5Cint_%7B-2%7D%5E%7B0%7De%5E%7Bu%7Ddu%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5Be%5Eu%2BC%5Cright%5D_%7B-2%7D%5E%7B0%7D%5C%5C%5C%5C%5C%5C%3D%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B%28e%5E0%2BC%29-%28e%5E%7B-2%7D%2BC%29%5Cright%5D%5C%5C%5C%5C%5C%5C%3D%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B1%20-%20%5Cfrac%7B1%7D%7Be%5E2%7D%5Cright%5D)
In short,
![\displaystyle \int_{0}^{1}e^{-2x}dx = \frac{1}{2}\left[1 - \frac{1}{e^2}\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B0%7D%5E%7B1%7De%5E%7B-2x%7Ddx%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B1%20-%20%5Cfrac%7B1%7D%7Be%5E2%7D%5Cright%5D)
This points us to choice C as the final answer.
Answer:
<h2>M(-6, 4)</h2>
Step-by-step explanation:
The formula of a midpoint of (x₁, y₁) and (x₂, y₂):

We have two points (-3, 7) and (-9, 1).
Substitute:

9514 1404 393
Answer:
WX = 33
(x, y) = (2, 10)
Step-by-step explanation:
The hash marks tell you WX is a midline, so has the measure of the average of the two bases.
WX = (PQ +SR)/2 = (27 +39)/2 = 66/2
WX = 33
__
The hash marks also tell you ...
PW = WS
y +4x = 18 . . . . . . substitute the given expressions
and also
QX = XR
2y +x = 22 . . . . . substitute the given expressions
__
If you solve the first equation for y, you get ...
y = 18 -4x
Substituting that into the second equation gives ...
2(18-4x) +x = 22
36 -7x = 22 . . . . . . . simplify
14 = 7x . . . . . . . . . . . add 7x-22 to both sides
2 = x . . . . . . . . . . . . divide by 7
y = 18 -4(2) = 10 . . . find y using the above relation
The values of x and y are 2 and 10, respectively.
__
My favorite "quick and dirty" way to solve a set of linear equations is using a graphing calculator. It works well for integer solutions.
Answer:
The smaller one is -6, the bigger one is -4.
Step-by-step explanation: