The gradient for a parallel line is the same.
perpendicular lines are negative reciprocals.
so all you have to do is re arrange it so y=Mx+b so the check out the gradients to what I listed above... remember y shouldn't have any number on it.
Answer:
41.7feet
Step-by-step explanation:
From the question we are given the following
angle of depression = 50°
Distance of the pole from the base of the feet = 35feet (Adjacent)
Required
height of the school (opposite)
Using the SOH CAH TOA identity
Tan theta = opp/adj
Tan 50 = H/35
H = 35tan 50
H = 35(1.1918)
H = 41.7feet
Hence the height of the school is 41.7feet
Answer:
15ft
Step-by-step explanation:
4/2=2
10/2=5
3/2=1.5
2x5x1.5=15ft
Hey There,
Your answer is:
C: Skewed
It changes direction and position.
Best Of Luck,
- kai -
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>
