If we assume the given segments are those from the vertices to the point of intersection of the diagonals, it seems one diagonal (SW) is 20 yards long and the other (TR) is 44 yards long. The area (A) of the kite is half the product of the diagonals:
... A = (1/2)·SW·TR = (1/2)·(20 yd)·(44 yd)
... A = 440 yd²
Answer:
4x+3y=68
9x+2y=77
plain= 5 shiny=16
Step-by-step explanation:
plain=x shiny=y
A^2 - 6a/a-6 • a+3/a
a(a - 6)/(a - 6) • a+3/a
a • (a + 3)/a
a(a+3)/a
a+3.
The correct solution to this question would be B. a+3.
3x + 1 + 5x = 7 + 15 + 7x
8x + 1 = 22 + 7x
x = 21
