Carcass wrote:
\((x^y) \times (x^2)=x^6\) and \((2z)^{-2} = \frac{1}{36}\)
Quantity A |
Quantity B |
\(\frac{1}{y}\) |
\(\frac{1}{z}\) |
There are
lots of problems with this question
Given: \((2z)^{-2} = \frac{1}{36}\)
Rewrite as: \(\frac{1}{(2z)^2} = \frac{1}{36}\)
Simplify: \(\frac{1}{4z^2} = \frac{1}{36}\)
From this we can conclude that: \(4z^2 = 36\)
Divide both sides by 4 to get: \(z^2 = 9\)
So, z = 3 OR z = -3
Given: \((x^y) \times (x^2)=x^6\)
One possible solution here is: x = 5 and
y = 4, since (5^4)(5^2) = 5^6
Another possible solution is: x = 0 and
y = 99, since (0^99)(0^2) = 0^6
Another possible solution is: x = 1 and
y = -7, since (1^77)(1^2) = 1^6
.
.
.etc
Since
y can have infinitely many values, the correct answer here is D
Cheers,
Brent
_________________
Brent Hanneson - founder of Greenlight Test Prep