If \((8)x^2 = 128\) , then “x” is:
Explanation
A positive number can have a square root of either a positive or a negative number. Two negatives multiplied together = a positive number. In this case x = \(4^{2}\) or \(-4^{2}\)
\(\frac{{{6^4}}}{{{6^3} \times {6^2}}} = ?\)
If n is a positive integer divisible by 7, and if n < 70, what is the greatest possible value of n?
\( {\rm{2}}({\rm{5}} - \sqrt {{\rm{16}}} ) \div ({\rm{14}} - {\rm{12}}) \times {\rm{3}} = ?\)
Solve:\(3+6 x \leq 3 x-3\)