What is the ratio of 25 and 11
1 answer:
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Check the picture below.
you can pretty much, without having to use the distance formula, see the diameter is 6 units long, thus the radius is half that or 3 then.
now, without having to use the midpoint formula, if we move from either endpoint inwards 3 units, we'll end up at the center, -5 -3, is just -8, so the center will be at -8, -3.
![\bf \textit{equation of a circle}\\\\ (x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2 \qquad center~~(\stackrel{-8}{{{ h}}},\stackrel{-3}{{{ k}}})\qquad \qquad radius=\stackrel{3}{{{ r}}} \\\\\\\ [x-(-8)]^2+[y-(-3)]^2=3^2\implies (x+8)^2+(y+3)^2=9](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%7B%7B%20h%7D%7D%29%5E2%2B%28y-%7B%7B%20k%7D%7D%29%5E2%3D%7B%7B%20r%7D%7D%5E2%0A%5Cqquad%20%0Acenter~~%28%5Cstackrel%7B-8%7D%7B%7B%7B%20h%7D%7D%7D%2C%5Cstackrel%7B-3%7D%7B%7B%7B%20k%7D%7D%7D%29%5Cqquad%20%5Cqquad%20%0Aradius%3D%5Cstackrel%7B3%7D%7B%7B%7B%20r%7D%7D%7D%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bx-%28-8%29%5D%5E2%2B%5By-%28-3%29%5D%5E2%3D3%5E2%5Cimplies%20%28x%2B8%29%5E2%2B%28y%2B3%29%5E2%3D9)
you can pretty much, without having to use the distance formula, see the diameter is 6 units long, thus the radius is half that or 3 then.
now, without having to use the midpoint formula, if we move from either endpoint inwards 3 units, we'll end up at the center, -5 -3, is just -8, so the center will be at -8, -3.
IMPORTANT:
If you're trying to factor a quadratic in Algebra I:
There are no two integers that can solve this problem!
Your quadratic is <em>prime</em>!
If you're trying to solve a quadratic (find x):
The factoring approach will not work for the same reasons listed above.
Try using splitting the middle or the quadratic formula instead.
Here's how you would solve it from a more advanced approach.
If you don't know what this stuff is, just ignore it.
ab = -18, a + b = -9
Find a in terms of b.
a = -9-b
Substitute this for a in the first equation.
(-9-b)b = -18
-9b-b² = -18
Multiply everything by -1 to get rid of all these negative signs.
b² + 9b = 18
Bring over that 18.
b² + 9b - 18 = 0
Apply the quadratic formula.
(a = 1, b = 9, c = -18)
If you need to write two distinct numbers, just write out one with a + and one with a - in place of the plus-minus sign.
If you're trying to factor a quadratic in Algebra I:
There are no two integers that can solve this problem!
Your quadratic is <em>prime</em>!
If you're trying to solve a quadratic (find x):
The factoring approach will not work for the same reasons listed above.
Try using splitting the middle or the quadratic formula instead.
Here's how you would solve it from a more advanced approach.
If you don't know what this stuff is, just ignore it.
ab = -18, a + b = -9
Find a in terms of b.
a = -9-b
Substitute this for a in the first equation.
(-9-b)b = -18
-9b-b² = -18
Multiply everything by -1 to get rid of all these negative signs.
b² + 9b = 18
Bring over that 18.
b² + 9b - 18 = 0
Apply the quadratic formula.
(a = 1, b = 9, c = -18)
If you need to write two distinct numbers, just write out one with a + and one with a - in place of the plus-minus sign.