To finish the demonstration that the quadrilateral JKLM is a rhombus we need to prove that side JK is congruent with side LM.
The length of a segment with endpoints (x1, y1) and (x2, y2) is calculated as follows:
![\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
Substituting with points L(1,6) and M(4,2) we get:
![\begin{gathered} LM=\sqrt[]{(4-1)^2+(2-6)^2} \\ LM=\sqrt[]{3^2+(-4)^2} \\ LM=\sqrt[]{9+16^{}} \\ LM=5 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20LM%3D%5Csqrt%5B%5D%7B%284-1%29%5E2%2B%282-6%29%5E2%7D%20%5C%5C%20LM%3D%5Csqrt%5B%5D%7B3%5E2%2B%28-4%29%5E2%7D%20%5C%5C%20LM%3D%5Csqrt%5B%5D%7B9%2B16%5E%7B%7D%7D%20%5C%5C%20LM%3D5%20%5Cend%7Bgathered%7D)
Given that opposite sides are parallel, all sides have the same length, and, from the diagram, the quadrilateral is not a square, we conclude that it is a rhombus.
Answer:
5
Step-by-step explanation:
40/8 is 5 and that is my explanation
It should be 40 percent.
You can think of it as starting with 100 as in 100 dollars or 100 tic tacs
If you take away 25%, you take away 25 tic tacs and now you only have 75.
From your 75, you take away 20% which is equal to 15. 75- 15=60
If you look at the end result of 60, you can see that that is 100 - 40 so 40% of the total was taken off
Step-by-step explanation:
p/2 = 3/4 + p/3
p/2 - p/3 = 3/4
(3p - 2p )/6 = 3/4
p/6 = 3/4
multiply both sides by 6
p = 3×6/4
=18/4
=9/2
=4.5
8 + 5i. I just added the expression normally.
