18q+18=180
-18. -18
18q=162
/18. /18
q=9
3q=27
Answer:
640 units squared
Step-by-step explanation:
The shape below is a trapezoid.
The formula for the area of a trapezoid is average of the bases times the height.
The bases are the parallel lines that make up 2 of the sides of the trapezoid.
The height is the line dotted line that creates right angles and is the distance between the two bases.
The bases are 24 and 40.
To find the average, we add them together and divide by 2
(24+40)/2=64/2=32
Now, we multiply 32 by the height, which is given as the length of 20.
32*20=640
the area of the trapezoid is 640 units squared
Hope this helps!
Answer:
2.2
Step-by-step explanation:
A kilogram is 2.2lbs so if you weigh yourself and it says you weigh 60k then you multiply 60 times 2.2 you get 132lbs.
To calculate the distance between two points on the coordinate system you have to use the following formula:
![d=\sqrt[]{(x_1-x_2)^2+(y_1-y_2)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%5B%5D%7B%28x_1-x_2%29%5E2%2B%28y_1-y_2%29%5E2%7D)
Where
d represents the distance between both points.
(x₁,y₁) are the coordinates of one of the points.
(x₂,y₂) are the coordinates of the second point.
To determine the length of CD, the first step is to determine the coordinates of both endpoints from the graph
C(2,-1)
D(-1,-2)
Replace the coordinates on the formula using C(2,-1) as (x₁,y₁) and D(-1,-2) as (x₂,y₂)
![\begin{gathered} d_{CD}=\sqrt[]{(2-(-1))^2+((-1)-(-2))}^2 \\ d_{CD}=\sqrt[]{(2+1)^2+(-1+2)^2} \\ d_{CD}=\sqrt[]{3^2+1^2} \\ d_{CD}=\sqrt[]{9+1} \\ d_{CD}=\sqrt[]{10} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20d_%7BCD%7D%3D%5Csqrt%5B%5D%7B%282-%28-1%29%29%5E2%2B%28%28-1%29-%28-2%29%29%7D%5E2%20%5C%5C%20d_%7BCD%7D%3D%5Csqrt%5B%5D%7B%282%2B1%29%5E2%2B%28-1%2B2%29%5E2%7D%20%5C%5C%20d_%7BCD%7D%3D%5Csqrt%5B%5D%7B3%5E2%2B1%5E2%7D%20%5C%5C%20d_%7BCD%7D%3D%5Csqrt%5B%5D%7B9%2B1%7D%20%5C%5C%20d_%7BCD%7D%3D%5Csqrt%5B%5D%7B10%7D%20%5Cend%7Bgathered%7D)
The length of CD is √10 units ≈ 3.16 units
We want to find one-half of the reciprocal of 7/sqrt(98). Let's write down an expression for this:

We can rewrite 98 into 


The square root of 49 is 7



This should be your answer. Let me know if you need any clarifications, thanks!