Question
Simplify: $\Big(\frac{1}{4}\Big)^{-2}+\Big(\frac{1}{2}\Big)^{-2}+\Big(\frac{1}{3}\Big)^{-2}$
✓
Answer
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}$ [$\because$ a is non-zero integer]
$\therefore$ $\Big(\frac{1}{4}\Big)^{-2}+\Big(\frac{1}{2}\Big)^{-2}+\Big(\frac{1}{3}\Big)^{-2}$
$=(4)^2+(2)^2+(3)^2$
$=16+4+9$ $=29$
$\therefore$ $\Big(\frac{1}{4}\Big)^{-2}+\Big(\frac{1}{2}\Big)^{-2}+\Big(\frac{1}{3}\Big)^{-2}$
$=(4)^2+(2)^2+(3)^2$
$=16+4+9$ $=29$
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