Answer:
The answer is that Regina is absolutely right
Step-by-step explanation:
Hello!
The answer is that Regina is absolutely right, this is because the fundamental theorem of algebra states that every polynomial equation has the same number of solutions as its highest exponent, for this case it is 2. Therefore, there are two values of X that cause the equation to result in zero
On a basic parallelogram, opposite sides are parallel, diagonals bisect each other and, consecutive angles are supplementary. However on a parallelogram diagonals aren't congruent.
So the answer is D
Answer:
Step-by-step explanation:
When you take the n-th root of a number, you can rewrite the expression by taking it to the 1/n-th power. For example:
![\sqrt[n]{x} =x^\frac{1}{n}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%7D%20%3Dx%5E%5Cfrac%7B1%7D%7Bn%7D)
For the first expression, we can use this proprtery to get:
![\sqrt[5]{a^x} =(a^x)^\frac{1}{5}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Ba%5Ex%7D%20%3D%28a%5Ex%29%5E%5Cfrac%7B1%7D%7B5%7D)
Using exponent rules, you can combine the exponents by simply multiplying them to get:
![a^\frac{x}{5}](https://tex.z-dn.net/?f=a%5E%5Cfrac%7Bx%7D%7B5%7D)
Moving on to the second expression. It is now the square root, or equivalently a 1/2 power. If we break up the terms under the radical into powers of 2, we can cancel a lot of the terms:
![\frac{\sqrt{81a^3b^1^0} }{\sqrt{3}a } =\frac{\sqrt{81a^2*a*b^1^0} }{\sqrt{3}a }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B81a%5E3b%5E1%5E0%7D%20%7D%7B%5Csqrt%7B3%7Da%20%7D%20%3D%5Cfrac%7B%5Csqrt%7B81a%5E2%2Aa%2Ab%5E1%5E0%7D%20%7D%7B%5Csqrt%7B3%7Da%20%7D)
The a^2 and b^10 can be taken out of the radical because they have perfect roots:
![\frac{\sqrt{81a^2*a*b^1^0} }{\sqrt{3}a }=ab^5\frac{\sqrt{81a} }{\sqrt{3}a }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B81a%5E2%2Aa%2Ab%5E1%5E0%7D%20%7D%7B%5Csqrt%7B3%7Da%20%7D%3Dab%5E5%5Cfrac%7B%5Csqrt%7B81a%7D%20%7D%7B%5Csqrt%7B3%7Da%20%7D)
The square root of 81 has a perfect root of 9. We have:
![ab^5\frac{\sqrt{81a} }{\sqrt{3}a }=9ab^5\frac{\sqrt{a} }{\sqrt{3}a }](https://tex.z-dn.net/?f=ab%5E5%5Cfrac%7B%5Csqrt%7B81a%7D%20%7D%7B%5Csqrt%7B3%7Da%20%7D%3D9ab%5E5%5Cfrac%7B%5Csqrt%7Ba%7D%20%7D%7B%5Csqrt%7B3%7Da%20%7D)
You can divide 9 and the square root of 3 by breaking up 9 into a product:
![\frac{(3*3)ab^5}{\sqrt{3} } \frac{\sqrt{a} }{a }=3\sqrt{3} ab^5\frac{\sqrt{a} }{a }](https://tex.z-dn.net/?f=%5Cfrac%7B%283%2A3%29ab%5E5%7D%7B%5Csqrt%7B3%7D%20%7D%20%5Cfrac%7B%5Csqrt%7Ba%7D%20%7D%7Ba%20%7D%3D3%5Csqrt%7B3%7D%20ab%5E5%5Cfrac%7B%5Csqrt%7Ba%7D%20%7D%7Ba%20%7D)
Simply by cancelling the 'a' terms to get:
![3\sqrt{3} ab^5\frac{\sqrt{a} }{a }={3\sqrt{3}}\sqrt{a}b^5=3b^5\sqrt{3a}](https://tex.z-dn.net/?f=3%5Csqrt%7B3%7D%20ab%5E5%5Cfrac%7B%5Csqrt%7Ba%7D%20%7D%7Ba%20%7D%3D%7B3%5Csqrt%7B3%7D%7D%5Csqrt%7Ba%7Db%5E5%3D3b%5E5%5Csqrt%7B3a%7D)