A is x times larger than B
4*10^-3=x times larger than 4*10^-4
4*10^-3=x*4*10^-4
divide both sides by 4
10^-3=x*10^-4
divide both sides by 10^-4
10^1=x
10^1=10
answer is 10 times or C
<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>
Answer:
n =24
Step-by-step explanation:
pretend n is x its just a normal fraction lol
Given :
C, D, and E are col-linear, CE = 15.8 centimetres, and DE= 3.5 centimetres.
To Find :
Two possible lengths for CD.
Solution :
Their are two cases :
1)
When D is in between C and E .
. . .
C D E
Here, CD = CE - DE
CD = 15.8 - 3.5 cm
CD = 12.3 cm
2)
When E is in between D and C.
. . .
D E C
Here, CD = CE + DE
CD = 15.8 + 3.5 cm
CD = 19.3 cm
Hence, this is the required solution.
The number is = 6
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