True
because 2 is a prime number, and it is divisible by 2
What was your question that was never answered?
Answer:
a = 0
Step-by-step explanation:
The cube root (radical) is equivalent to x^(1/3). When that is divided by x^(1/3), the result is ...
(x^(1/3))/(x^(1/3)) = x^(1/3 -1/3) = x^0
Comparing that to x^a, we find a=0.
_____
The applicable rules of exponents are ...
![\sqrt[n]{x^m}=x^{\frac{m}{n}}\\\\\dfrac{x^a}{x^b}=x^{a-b}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5Em%7D%3Dx%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%5C%5C%5C%5C%5Cdfrac%7Bx%5Ea%7D%7Bx%5Eb%7D%3Dx%5E%7Ba-b%7D)
Answer:
B. x < -8 or x > 8
Step-by-step explanation:
You can use process of elimination to solve this problem by going through every solution and testing them out, but let's jump right to B.
Process:
You know that since the inequality states that x^2 has to be greater than 64, x has to be more than 8, or less than -8.
This is because 8^2 = 64, and -8^2 = 64, and the inequality requires the answer to be more than 64.
Looking at B., you can see that if x is < -8, the square of, for example, -9, would be 81. This is greater than 64, so this works!
Now, B. also has an alternative. The 'or' is a major clue to which is the correct answer, since the square root of any number can be positive or negative. (-8^2 = 8^2)
The 'or' states that x must be greater than 8. So, for example, if we take the square of 10, we get 100, and that is also greater than 64.
We've proven that this solution is accurate for both parts, so it is definitely the one we want!
Hope this helps!
I strongly believe that the answer is d. But I really believe that is the answer
