<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>
<span>280
I'm assuming that this question is badly formatted and that the actual number of appetizers is 7, the number of entres is 10, and that there's 4 choices of desserts. So let's take each course by itself.
You can choose 1 of 7 appetizers. So we have
n = 7
After that, you chose an entre, so the number of possible meals to this point is
n = 7 * 10 = 70
Finally, you finish off with a dessert, so the number of meals is:
n = 70 * 4 = 280
Therefore the number of possible meals you can have is 280.
Note: If the values of 77, 1010 and 44 aren't errors, but are actually correct, then the number of meals is
n = 77 * 1010 * 44 = 3421880
But I believe that it's highly unlikely that the numbers in this problem are correct. Just imagine the amount of time it would take for someone to read a menu with over a thousand entres in it. And working in that kitchen would be an absolute nightmare.</span>
Step-by-step explanation:commutative property
hope it helps
Answer:
x= -4/9
Step-by-step explanation:
HOPE THIS HELPS!!!! :)
<3333333333
No of out comes =216
favorable outcome=3
<span>probability that 5 will not appear on the die=3/216</span>
