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:
(a) x > 4 (b) y < -2
Step-by-step explanation:
Domain is referring to the x-values while the range is referring to the y-values.
Since the function (the line) has a circle at the point (4, -2), the function will be exclusive at that coordinate.
The line goes to infinity for the x-values from 4, so you write x > 4 or ∞ > x > 4.
Similarly, the line goes to infinity for the y-values from -2, so you write y < -2 or -∞ < y < -2.
Answer:
A (1,1) Aef(1,1)
B(2,3) Bef(3,2)
C(3,2) Cef(2,3)
D(2,1) Def(1,2)
Step-by-step explanation:
S(x) = x - 7; t(x) = 4x² - x + 3
(t · s)(x) = (4x² - x + 3)(x - 7) = (4x²)(x) + (4x²)(-7) - (x)(x) - (x)(-7) + (3)(x) + (3)(-7)
= 4x³ - 28x² - x² + 7x + 3x - 21 = 4x³ - 29x² + 10x - 21