Answer:
what is the question
Step-by-step explanation:
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:
cosx= 35. Use Trignometrical identity cosx = √1−sin2x . cos x = √1−1625 = √925 = 35 to be the ...
Missing: =0.82 | Must include: =0.82
Step-by-step explanation:
The Answer is "X=2" awodkawopdawdawd
Answer:
<h2>See below</h2>
Step-by-step explanation:
I can't drag and drop the graphs, but I can graph the equations shown. Then, all you will have to do is match the graphs shown the the graph that I will provide.
<h3>EQUATION 1: y = x² - 2</h3>
Graph Properties:
Opens up
Vertex is 0, -2
Axis of symmetry is x = 0
Graph photo shown in file called equation 1 graph
<h3>EQUATION 2: y = 2x²</h3>
Graph Properties:
Opens up
Vertex is 0, 0
Axis of symmetry is x = 0
Graph photo shown in file called equation 2 graph
<h3>EQUATION 3: y = (x - 2)²</h3>
Graph Properties:
Opens up
Vertex is 2, 0
Axis of symmetry is x = 2
Graph photo shown in file called equation 3 graph
I'm always happy to help :)
