The intersection between the curves are
3, 0
0, 3
The volume of the solids is obtained by
V = ∫ π [ (4 - (y-1)²)² - (3 - y)²] dy with limits from 0 to 3
The volume is
V = 108π/5 or 67.86
-
9² = 12² + 15² - 2 (12) (15) cos (B)
81 = 144 + 225 - 360 cos(B)
81 = 369 - 360 cos (B)
360 cos (B) = 369 - 81
360 cos (B) = 288
cos (B) = 0.8
-
Answer: Cosine B = 0.8
-
-
12² = 15² + 9² - 2 (15)(9) cos (A)
144 = 225 + 81 - 270 Cos A
144 = 306 - 270 Cos A
270 Cos A = 162
Cos A = 3/5 or 0.6
-
Answer: Cosine Angle A = 3/5
The initial dimenssions of the park lot are:
length: 140 ft
width: 90 ft
initial area: 140 * 90 = 12,600 ft^2
Area increased 29% = 12,600 * 1.29 = 16,254 ft^2
width of the strips: x
New length: 140 + x
New width: 90 + x
New area: (140+x)(90+x) = 16,254
Solution of the equation:
12600 + 230x + x^2 = 16254
=> x^2 + 230x - 3654 = 0
Use the quadratic formula.
x = {-230 +/- √[ 230^2 - 4*1*(-3654) ]} / 2 =
x = 14.92
The other solution is negative so it is discarded.
Answer: 15 ft
