Answer:
√23
Step-by-step explanation:
When you are given two side lengths of a right triangle, you use the Pythagorean theorem to find the third side: a² + b² = c², where c is the hypotenuse (the longest side).
All you have to do is plug the given information in:
Remember, 13 is the hypotenuse for this triangle.
12² + b² = 13²
Simplify:
144 + b² = 169
Subtract 144 from both sides:
b² = 169-144
b² = 23
Square root both sides:
b = √23
And that's your answer!
To simplify
![\sqrt[4]{\dfrac{24x^6y}{128x^4y^5}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B24x%5E6y%7D%7B128x%5E4y%5E5%7D%7D)
we need to use the fact that
![\sqrt[4]{x^4}=|x|](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E4%7D%3D%7Cx%7C)
Why the absolute value? It's because 
.
We start by rewriting as
![\sqrt[4]{\dfrac{2^23x^6y}{2^6x^4y^5}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B2%5E23x%5E6y%7D%7B2%5E6x%5E4y%5E5%7D%7D)
![\sqrt[4]{\dfrac{2^23x^4x^2y}{2^42^2x^4y^4y}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B2%5E23x%5E4x%5E2y%7D%7B2%5E42%5E2x%5E4y%5E4y%7D%7D)
Since 
, we have 
, and the above reduces to
![\sqrt[4]{\dfrac{3x^2y}{2^4y^4y}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B3x%5E2y%7D%7B2%5E4y%5E4y%7D%7D)
Then we pull out any 4th powers under the radical, and simplify everything we can:
![\dfrac1{\sqrt[4]{2^4y^4}}\sqrt[4]{\dfrac{3x^2y}{y}}](https://tex.z-dn.net/?f=%5Cdfrac1%7B%5Csqrt%5B4%5D%7B2%5E4y%5E4%7D%7D%5Csqrt%5B4%5D%7B%5Cdfrac%7B3x%5E2y%7D%7By%7D%7D)
![\dfrac1{|2y|}\sqrt[4]{3x^2}](https://tex.z-dn.net/?f=%5Cdfrac1%7B%7C2y%7C%7D%5Csqrt%5B4%5D%7B3x%5E2%7D)
where 
allows us to write 
, and this also means that 
. So we end up with
![\dfrac{\sqrt[4]{3x^2}}{2y}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B4%5D%7B3x%5E2%7D%7D%7B2y%7D)
making the last option the correct answer.
Answer:
Congruent sides or segments have the exact same length. Congruent angles have the exact same measure. For any set of congruent geometric figures, corresponding sides, angles, faces, etc. are congruent.
Step-by-step explanation:
comment how it helps
(x-h)^2 + (y-k)^2 = r^2
(x-2)^2 + (y-0)^2 = (rt(5))^2
-h = -2
h = 2
k=0
Center(2, 0)
