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>
Domain of f = R .............
We use the Work formula to solve for the unknown in the problem which is W = F x d. First, we solve for the Net Force acting on the car. The Net Force is the summation of all forces acting on the object. For this case, we assume that Friction Force is negligible thus the Net Force is equal to:
F = mgsinα in terms of SI units and in terms of english units we have F = m(g/g₀)(sin α) where g₀ is the proportionality factor, 32.174 ft lb-m / lb-f s²
F = 2500 (32.174/32.174) (sin 12°) = 519.78 lb
W = Fd = 519.78 lb (400 ft) = 207912 ft - lb or 20800 ft-lb
33 * 3 + 3/3 = 100
I do not believe u have to use all of the operations because it says " at most once ".
The points L(10,9)L(10,9), M(10,-5)M(10,-5), N(-1,-5)N(-1,-5), and O(-1,9)O(-1,9) form rectangle LMNOLMNO. Which point is halfwa
Inessa [10]
You are trying to find the halfway point between OO and NN.
OO: (-1,9) NN: (-1,5)
The x-coordinate does not change, because in both instances it is -1. The y-coordinate is (9-5)/2 AWAY from each point. AKA the number that is equidistant from 5 and 9 (7).
