Answer:
Area of the shaded region = 23.33 in²
Step-by-step explanation:
Area of a sector = 
Where θ = Central angle subtended by an arc
r = radius of the circle
Area of the sector BCD = 
= 52.36 in²
Area of equilateral triangle BCD = 
= 
=
in²
= 43.30 in²
Area of the shaded portion in ΔBCD = 52.36 - 43.3
= 9.06 in²
Area of sector CAD = 
= 39.27 in²
Area of right triangle CAD = 
= 
=
= 25 in²
Area of the shaded part in the ΔACD = 39.27 - 25
= 14.27 in²
Area of the shaded part of the figure = 9.06 + 14.27
= 23.33 in²
Standard Equation for trig function: y = A cos (B (x - H))+ K
*Here, we are just dealing with A (amplitude) and B (period)
Period = 2π/ b
8 = 2π/ b
Solve for b: 2π/8 or π/4
Equation: y = 2 cos (π/4 (x))
H would equal to 4 (8+4=12)
Using the quotient rule
dy/dx = (x^-1) * (3x^2 -5 ) - (x^3 - 5x) * 2x
----------------------------------------------
(x^2 - 1)^2
You are correct
Lets x = width
length = x + 4 (4 meters longer than wide)
A = L * W
192 = x ( x +4)
192 = x^2 + 4x
x^2 + 4x - 192 = 0
(x +16)(x-12) = 0
x - 12 = 0, x = 12
x + 16 = 0, x = -16
so width x = 12
length = 12 + 4 = 16 (4 meters longer than wide)
answer. J
16