If you can separate the figures and the segment, or angle in question, is the same, the reflexive property of congruence can be used in the proof. ... Separating the two triangles, you can see Angle Z is the same angle for each triangle. Since they are the same angle, the angle is congruent to itself.
Answer:
332=y
Step-by-step explanation:
<h2>
Hello!</h2>
The answer is:
The second option,
![(\sqrt[m]{x^{a} } )^{b}=\sqrt[m]{x^{ab} }](https://tex.z-dn.net/?f=%28%5Csqrt%5Bm%5D%7Bx%5E%7Ba%7D%20%7D%20%29%5E%7Bb%7D%3D%5Csqrt%5Bm%5D%7Bx%5E%7Bab%7D%20%7D)
<h2>
Why?</h2>
Discarding each given option in order to find the correct one, we have:
<h2>
First option,</h2>
![\sqrt[m]{x}\sqrt[m]{y}=\sqrt[2m]{xy}](https://tex.z-dn.net/?f=%5Csqrt%5Bm%5D%7Bx%7D%5Csqrt%5Bm%5D%7By%7D%3D%5Csqrt%5B2m%5D%7Bxy%7D)
The statement is false, the correct form of the statement (according to the property of roots) is:
![\sqrt[m]{x}\sqrt[m]{y}=\sqrt[m]{xy}](https://tex.z-dn.net/?f=%5Csqrt%5Bm%5D%7Bx%7D%5Csqrt%5Bm%5D%7By%7D%3D%5Csqrt%5Bm%5D%7Bxy%7D)
<h2>
Second option,</h2>
![(\sqrt[m]{x^{a} } )^{b}=\sqrt[m]{x^{ab} }](https://tex.z-dn.net/?f=%28%5Csqrt%5Bm%5D%7Bx%5E%7Ba%7D%20%7D%20%29%5E%7Bb%7D%3D%5Csqrt%5Bm%5D%7Bx%5E%7Bab%7D%20%7D)
The statement is true, we can prove it by using the following properties of exponents:

![\sqrt[n]{x^{m} }=x^{\frac{m}{n} }](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5E%7Bm%7D%20%7D%3Dx%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%20%7D)
We are given the expression:
![(\sqrt[m]{x^{a} } )^{b}](https://tex.z-dn.net/?f=%28%5Csqrt%5Bm%5D%7Bx%5E%7Ba%7D%20%7D%20%29%5E%7Bb%7D)
So, applying the properties, we have:
![(\sqrt[m]{x^{a} } )^{b}=(x^{\frac{a}{m}})^{b}=x^{\frac{ab}{m}}\\\\x^{\frac{ab}{m}}=\sqrt[m]{x^{ab} }](https://tex.z-dn.net/?f=%28%5Csqrt%5Bm%5D%7Bx%5E%7Ba%7D%20%7D%20%29%5E%7Bb%7D%3D%28x%5E%7B%5Cfrac%7Ba%7D%7Bm%7D%7D%29%5E%7Bb%7D%3Dx%5E%7B%5Cfrac%7Bab%7D%7Bm%7D%7D%5C%5C%5C%5Cx%5E%7B%5Cfrac%7Bab%7D%7Bm%7D%7D%3D%5Csqrt%5Bm%5D%7Bx%5E%7Bab%7D%20%7D)
Hence,
![(\sqrt[m]{x^{a} } )^{b}=\sqrt[m]{x^{ab} }](https://tex.z-dn.net/?f=%28%5Csqrt%5Bm%5D%7Bx%5E%7Ba%7D%20%7D%20%29%5E%7Bb%7D%3D%5Csqrt%5Bm%5D%7Bx%5E%7Bab%7D%20%7D)
<h2>
Third option,</h2>
![a\sqrt[n]{x}+b\sqrt[n]{x}=ab\sqrt[n]{x}](https://tex.z-dn.net/?f=a%5Csqrt%5Bn%5D%7Bx%7D%2Bb%5Csqrt%5Bn%5D%7Bx%7D%3Dab%5Csqrt%5Bn%5D%7Bx%7D)
The statement is false, the correct form of the statement (according to the property of roots) is:
![a\sqrt[n]{x}+b\sqrt[n]{x}=(a+b)\sqrt[n]{x}](https://tex.z-dn.net/?f=a%5Csqrt%5Bn%5D%7Bx%7D%2Bb%5Csqrt%5Bn%5D%7Bx%7D%3D%28a%2Bb%29%5Csqrt%5Bn%5D%7Bx%7D)
<h2>
Fourth option,</h2>
![\frac{\sqrt[m]{x} }{\sqrt[m]{y}}=m\sqrt{xy}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5Bm%5D%7Bx%7D%20%7D%7B%5Csqrt%5Bm%5D%7By%7D%7D%3Dm%5Csqrt%7Bxy%7D)
The statement is false, the correct form of the statement (according to the property of roots) is:
![\frac{\sqrt[m]{x} }{\sqrt[m]{y}}=\sqrt[m]{\frac{x}{y} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5Bm%5D%7Bx%7D%20%7D%7B%5Csqrt%5Bm%5D%7By%7D%7D%3D%5Csqrt%5Bm%5D%7B%5Cfrac%7Bx%7D%7By%7D%20%7D)
Hence, the answer is, the statement that is true is the second statement:
![(\sqrt[m]{x^{a} } )^{b}=\sqrt[m]{x^{ab} }](https://tex.z-dn.net/?f=%28%5Csqrt%5Bm%5D%7Bx%5E%7Ba%7D%20%7D%20%29%5E%7Bb%7D%3D%5Csqrt%5Bm%5D%7Bx%5E%7Bab%7D%20%7D)
Have a nice day!
The first one is negative six and the second one is positive six
Answer: The result is one half (
)
Step-by-step explanation:
We have the following expression:

Since both fractions have the same denominator, we can just add both numerators and keep the denominator:

Dividing numerator and denominator by
:
This is the result