Answer:
Step-by-step explanation:
-2x + 3 = 0
-2x = -3
x = 3/2
-x^2 + 2x + 3 = (-3/2)^2 + 2 × 3/2 + 3
= 9/4 + 3 + 3
= 9/4 + 6
= 33/4
Hope this helps
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Its not far apart if it 15 or more yes it far
<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>
9514 1404 393
Answer:
BC = 5
Step-by-step explanation:
Of course, this geometry program can tell you the length of BC.
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If you follow directions, you get a right triangle BCF that has leg lengths 3 and 4. The Pythagorean theorem then tells you the length of hypotenuse BC is ...
BF = 4 -1 = 3
FC = 4 -0 = 4
BC² = BF² +FC²
BC² = 3² +4² = 9 +16 = 25
BC = √25
BC = 5
