The squares are shown in the attached picture.
As you can see, JC is half the diagonal of ABCD and JH is half the diagonal of EFGH.
In order to find the area of the shaded figure, we need to subtract the area of the white square (EFGH) from the area of the big square (ABCD).
The area of a square know the diagonal is given by the formula:
A = d² ÷ 2
A(ABCD) = (2×JC)² ÷ 2
= (2×9)² ÷ 2
= 162 cm²
A(EFGH) = (2×JH)² ÷ 2
= (2×4)² ÷ 2
= 32 cm²
Therefore:
A = A(ABCD) - <span>A(EFGH)
= 162 - 32
= 130 cm</span>²
The area of the shaded region is
130 cm².
Answer:
m=-2
Step-by-step explanation:
As the product of the roots of a quadratic equation is c/a in ax^2+bx+c=0
here a=2, b=+8, c=-m^3
Given c/a=4
-m^3/2=4
-m^3=8
m^3= -8
m=-2.
The answer is 1.27777778 (:
-3x + y = 4
Solve for y
y = 3x + 4
Plug this y value into the other equation.
-9x + 5y = -1
-9x + 5(3x + 4) = -1
Distribute 5(3x + 4)
-9x + 15x + 20 = -1
Subtract 20 from both sides
-9x + 15x + 20 - 20 = -1 - 20
-9x + 15x = -21
Combine like terms
6x = -21
Divide both sides by 6
6/6 x = -21/6
1x = -21/6
x = -7/2
x = -3.5 or -7/2
Plug this value into an equation to solve for y.
-3(-3.5) + y = 4
10.5 + y = 4
Subtract both sides by 10.5
10.5 - 10.5 + y = 4 - 10.5
y = -6.5
Check this answer by plugging the value of x into the other equation.
-9(-3.5) + 5y = -1
31.5 + 5y = -1
Subtract both sides by 31.5
31.5 - 31.5 + 5y = -1 - 31.5
5y = -32.5
Divide both sides by 5
y = -6.5 or -13/2
Therefore the solution is ( -3.5 , -6.5 )