The area of the shaded region is going to be the area of the rectangle minus the area of the square.
Area of a rectangle is L * W.
A = L * W
A = (x + 10)(2x + 5)
A = x(2x + 5) + 10(2x + 5)
A = 2x^2 + 5x + 20x + 50
A = 2x^2 + 25x + 50 .....this is the area of the rectangle
area of a square is : A = a^2...where a is one side
A = (x + 1)^2
A = (x + 1)(x + 1)
A = x(x + 1) + 1(x + 1)
A = x^2 + x + x + 1
A = x^2 + 2x + 1
now we subtract the area of the square from the area of the rectangle to get the area of the shaded region.
2x^2 + 25x + 50 - (x^2 + 2x + 1) =
2x^2 + 25x + 50 - x^2 - 2x - 1 =
x^2 + 23x + 49 <== the area of the shaded region
Minimum is (4,3) and Vertical intercept is (0,5) correct me if I’m wrong
Answer:
y = 46
Step-by-step explanation:
I'm assuming you wanted to solve for y.
x = 5
y = 8(5) + 6
y = 40 +6
y = 46
The two diagonals intersects at right-angle for a kite
the long diagonal also bisects the short diagonal so QP = PS = QS/2 = 6/2 = 3
QPR is a right-angle triangle with QR at 5 and QR at 3
RP = sqrt (5^2 - 3^) = 4m
