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 believe the answer to your graphing piecewise function is the 2nd picture
Answer:
oh hey I gotchu-
first of all you are gonna make that into two rectangle by slicing the top off if that makes sense so you now have a large rectangle (11×12) and a small one (3×8) now you use the area formula for rectangles base×height so the small one becomes (3×8)=24 and the large one is (11×12)=132. now to make them into one shape for the total area you add those two values 132+24=156 so your total area is 156m²
F(x) = 500x-8000
500 times the number of years (x), subtracted by the original price (8000)
C
If you plug in the values in that equation, it works.
