![\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} (y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\ \boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}})} \\ \end{array} \qquad \begin{array}{llll} vertex\ ({{ h}},{{ k}})\\\\ {{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------\\\\ y=\cfrac{1}{8}x^2\implies (y-0)=\cfrac{1}{8}(x-0)^2\implies 8(y-0)=(x-0)^2 \\\\\\ 4(2)(y-0)=(x-0)^2\impliedby \textit{that means, p = 2}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bparabola%20vertex%20form%20with%20focus%20point%20distance%7D%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%28y-%7B%7B%20k%7D%7D%29%5E2%3D4%7B%7B%20p%7D%7D%28x-%7B%7B%20h%7D%7D%29%20%5C%5C%5C%5C%0A%5Cboxed%7B%28x-%7B%7B%20h%7D%7D%29%5E2%3D4%7B%7B%20p%7D%7D%28y-%7B%7B%20k%7D%7D%29%7D%20%5C%5C%0A%5Cend%7Barray%7D%0A%5Cqquad%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0Avertex%5C%20%28%7B%7B%20h%7D%7D%2C%7B%7B%20k%7D%7D%29%5C%5C%5C%5C%0A%7B%7B%20p%7D%7D%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%0A%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Ay%3D%5Ccfrac%7B1%7D%7B8%7Dx%5E2%5Cimplies%20%28y-0%29%3D%5Ccfrac%7B1%7D%7B8%7D%28x-0%29%5E2%5Cimplies%208%28y-0%29%3D%28x-0%29%5E2%0A%5C%5C%5C%5C%5C%5C%0A4%282%29%28y-0%29%3D%28x-0%29%5E2%5Cimpliedby%20%5Ctextit%7Bthat%20means%2C%20p%20%3D%202%7D)
so... if you notice, the vertex is at h,k and that'd be the origin, 0,0
so...since the directrix is "p" units from the vertex, so it'd be 2 units from 0,0
now, the parabola has an equation with a positive leading term's coefficient, namely the 1/8 is positive, thus, the parabola is opening upwards, and the directrix is "outside" the parabola, so is below the vertex
that puts the directrix 2 units below 0,0
y = -2
Answer:
54
Step-by-step explanation:
6 x 9 = 54
she can there are
make 6 9 yards
ribbons
out of
a yard
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>
Yo vivo en Granada, una ciudad pequeña que tiene monumentos muy importantes como la Alhambra.
Answer:
Step-by-step explanation:
b) Parallelogram
c) Square
d) Parallelogram***
***Not too sure about this one.