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:
8
Step-by-step explanation:
That is the quetion that sal khan explaind in the vid BOI
<h3>
- - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - </h3>
➷ You need to know these rules to do this:
Two different signs next to each other, make a negative sign
Two signs that are the same, make a plus
The answer = 7
➶ Hope This Helps You!
➶ Good Luck (:
➶ Have A Great Day ^-^
↬ ʜᴀɴɴᴀʜ ♡
Answer:
C.
Step-by-step explanation:
1 X 26 = 26
2 X 26 = 52
3 X 26 = 78
how is tat algebra???
D u got to add a lil bit and that you do division
