Answer:
the answer is b
Step-by-step explanation:
trust me im looking at the answer rn
The paraboloid meets the x-y plane when x²+y²=9. A circle of radius 3, centre origin.
<span>Use cylindrical coordinates (r,θ,z) so paraboloid becomes z = 9−r² and f = 5r²z. </span>
<span>If F is the mean of f over the region R then F ∫ (R)dV = ∫ (R)fdV </span>
<span>∫ (R)dV = ∫∫∫ [θ=0,2π, r=0,3, z=0,9−r²] rdrdθdz </span>
<span>= ∫∫ [θ=0,2π, r=0,3] r(9−r²)drdθ = ∫ [θ=0,2π] { (9/2)3² − (1/4)3⁴} dθ = 81π/2 </span>
<span>∫ (R)fdV = ∫∫∫ [θ=0,2π, r=0,3, z=0,9−r²] 5r²z.rdrdθdz </span>
<span>= 5∫∫ [θ=0,2π, r=0,3] ½r³{ (9−r²)² − 0 } drdθ </span>
<span>= (5/2)∫∫ [θ=0,2π, r=0,3] { 81r³ − 18r⁵ + r⁷} drdθ </span>
<span>= (5/2)∫ [θ=0,2π] { (81/4)3⁴− (3)3⁶+ (1/8)3⁸} dθ = 10935π/8 </span>
<span>∴ F = 10935π/8 ÷ 81π/2 = 135/4</span>
Given:
2 parallelograms with an area of 9 1/3 yd²
height of each parallelogram is 1 1/3 yd
Area of parallelogram = base * height
We need to divide the combined area into two to get each parallelogram's base.
9 1/3 = ((9*3)+1)/3 = 28/3
28/3 ÷ 2 = 28/3 * 1/2 = 28/6 yd² or 4 4/6 yd² ⇒ 4 2/3 yd²
Area of each parallelogram is 4 2/3 yd²
4 2/3 yd² = base * 1 1/3 yd
14/3 yd² ÷ 4/3 yd = base
14/3 yd² x 3/4 yd = base
14*3 / 3*4 = base
42 / 12 = base
3 6/12 yd = base
or 3 1/2 yd = base
a) the base of each parallelogram is 3 1/2 yards
b) we can assume that the two parallelograms form a rectangle.
area of a rectangle is length times width.
length is 3 1/2 yds * 2 = 7 yds
width is 3 1/2 yds
Area of rectangle = 7 yds * 3 1/2 yds
Area = 7 yd * 7/2 yd
Area = 7*7 / 2 yd²
Area = 49 / 2 yd²
Area = 24 1/2 yd²
(32x⁴ - 12x² + 7x) / 2x
32x⁴/2x - 12x²/2x + 7x/2x
16x³ - 6x + 7/2