We have [ ( -3 - (-1) , (-2 - 6) ] = [ -2 , - 8 ]
The length of this veector = ll a ll = √ [ ( -2)^2 + (-8)^2 ] = √ [ 4 + 64 ] = √68 =
√4 * √17 =
2√17
r = √ [ (-3)^2 + (3√3)^2 ] = √ [ 9 + 27] = √36 = 6
θ = arctan (3√ 3 / -3) = arctan (-√ 3) = -pi/ 3 + pi = 2pi/3
z = 6 cis (2pi/3)
(x -3)^2 +y^2 = 9
( rcos θ - 3)^2 + (rsinθ)^2 = 9
r^2cos^2θ - 6rcosθ + 9 + r^2 sin^2θ = 9 subtract 9 from both sides
r^2 ( sin^2 θ + cos ^2 θ) - 6r cosθ = 0
r^2 (1) = 6r cosθ divide out r
r = 6cosθ
We have all the possible arrangements of this set
{right, right, right, right, right, up, up, up up }
We can either choose any 5 of the 9 places to put the "rights" or any 4 of the 9 places to place the "ups"
So
C(9,5) = C(9,4) = 126 paths
Oops....I forgot we needed a sq ft answer
I gave my answer in sq in
1 sq ft =144 in^2
18in ^2 /144 in^2 = 1/8 sq ft
The other leg = sqrt [ (√97)^2 - 4^2] = sqrt [ 97 - 16] = sqrt 81 = 9
The area is =
(1/2) (product of the leg lengths) =
(1/2) (4 * 9) =
(1/2) (36) =
18
Look at the top (or bottom) base of the figure....what do they look like????
That is your answer
I'll let you compute this, MCP
Vcone = (1/3) * pi * Radius^2 * Height
I'll verify your answer, if you want me to....
C is correct
(This is true by Cavilieri's Theorem which says that two solids have the same volume if their altitudes are equal and all plane sections parallel to their bases and at equal distances from their bases are equal. )
We can use the tangent-secant theorem that says that
PT^2 = PA ( PA + BA)
45^2 = 25 (25 + BA)
2025 = 25^2 + 25BA
2025 = 625 + 25BA
2025 - 625 = 25BA
1600 = 25BA divide both sides by 25
64 = BA
PB = (PA + BA) = (25 + 64) = 89
