<span>Sphere: (x - 4)^2 + (y + 12)^2 + (z - 8)^2 = 100
Intersection in xy-plane: (x - 4)^2 + (y + 12)^2 = 36
Intersection in xz-plane: DNE
Intersection in yz-plane: (y + 12)^2 + (z - 8)^2 = 84
The desired equation is quite simple. Let's first create an equation for the sphere centered at the origin:
x^2 + y^2 + z^2 = 10^2
Now let's translate that sphere to the desired center (4, -12, 8). To do that, just subtract the center coordinate from the x, y, and z variables. So
(x - 4)^2 + (y - -12)^2 + (z - 8)^2 = 10^2
(x - 4)^2 + (y - -12)^2 + (z - 8)^2 = 100
Might as well deal with that double negative for y, so
(x - 4)^2 + (y + 12)^2 + (z - 8)^2 = 100
And we have the desired equation.
Now for dealing with the coordinate planes. Basically, for each coordinate plane, simply set the coordinate value to 0 for the axis that's not in the desired plane. So for the xy-plane, set the z value to 0 and simplify. So let's do that for each plane:
xy-plane:
(x - 4)^2 + (y + 12)^2 + (z - 8)^2 = 100
(x - 4)^2 + (y + 12)^2 + (0 - 8)^2 = 100
(x - 4)^2 + (y + 12)^2 + (-8)^2 = 100
(x - 4)^2 + (y + 12)^2 + 64 = 100
(x - 4)^2 + (y + 12)^2 = 36
xz-plane:
(x - 4)^2 + (y + 12)^2 + (z - 8)^2 = 100
(x - 4)^2 + (0 + 12)^2 + (z - 8)^2 = 100
(x - 4)^2 + 12^2 + (z - 8)^2 = 100
(x - 4)^2 + 144 + (z - 8)^2 = 100
(x - 4)^2 + (z - 8)^2 = -44
And since there's no possible way to ever get a sum of 2 squares to be equal to a negative number, the answer to this intersection is DNE. This shouldn't be a surprise since the center point is 12 units from this plane and the sphere has a radius of only 10 units.
yz-plane:
(x - 4)^2 + (y + 12)^2 + (z - 8)^2 = 100
(0 - 4)^2 + (y + 12)^2 + (z - 8)^2 = 100
(-4)^2 + (y + 12)^2 + (z - 8)^2 = 100
16 + (y + 12)^2 + (z - 8)^2 = 100
(y + 12)^2 + (z - 8)^2 = 84</span>
5 x 3 = 15
Ty spent $15 on binders
Answer:
C. Ted will gain $7,532 more from the rent than from the appreciation.
Step-by-step explanation:
Yes
Here is your answer
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