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.
I would say 8 oz is the answer
The answer is 4 days & 14hrs, because a day has 24 hours, so 110 divided 24 equals 4 and an remainder of 14, so you get 4 days and 14 hours. If you can, plz mark mine the brainliest, and i would really appreciate it.
Hope it works.
Sin=95/100 you should Find an other edge if we say y is an other edge. so,100^2=95^2+y^2
Y=31.22
Cosx=31.22/100=
O.3122
Answer:
Step-by-step explanation:
Thr answer is the slope.
