Answer
Is x equal to negative 2?
Step-by-step explanation:
Answer:
<em>The second figure ( rectangle ) has a longer length of it's diagonal comparative to the first figure ( square )</em>
Step-by-step explanation:
We can't confirm the length of these diagonals based on the appearance of the figure, so let us apply Pythagorean Theorem;
This diagonal divides each figure ( square + rectangle ) into two congruent, right angle triangles ⇒ from which we may apply Pythagorean Theorem, where the diagonal acts as the hypotenuse;
5^2 + 5^2 = x^2 ⇒ x is the length of the diagonal,
25 + 25 = x^2,
x^2 = 50,
x = √50
Now the same procedure can be applied to this other quadrilateral;
3^2 + 7^2 = x^2 ⇒ x is the length of the diagonal,
9 + 49 = x^2,
x^2 = 58,
x = √58
<em>Therefore the second figure ( rectangle ) has a longer length of it's diagonal comparative to the first figure ( square )</em>
Answer:
Positive is 1
Negative is-1
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>
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,
<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:
Have a nice day!
