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>
Answer:
k= -8
Explaining:
Move variable to the left-hand side and change its sign.
2.25k - 6.5k - 20= 6.5k - 6.5k + 14
Since two opposites add up to zero, remove them from the expression.
2.25k - 6.5k - 20= 14
Collect like terms!
-4.25k = 14 + 20
Add the numbers
-4.25k = 35
Divide both sides of the equation by -4.25
So you will get the answer: k= -8
Answer:
Basically ur wrong
Step-by-step explanation:
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
There are 11 numbers on cards.
6 cards contain even numbers.
Answer: p(even) = 6/11
There is only one card with a 7. After the first card is picked and returned, there are 11 cards in the bag again.
p(even then 7) = p(even) * p(7) = 6/11 * 1/11 = 6/121
Answer: p(even then 7) = 6/121
