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Standard XI
Applied Mathematics
Indices
Question
Simplify :
$$\left( \dfrac{1}{4}\right)^{-2}+\left( \dfrac 12\right)^{-2}+\left( \dfrac 13\right)^{-2}$$
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Solution
Verified by Toppr
Use, $$a^{m}=\left(\dfrac{1}{a}\right)^{-m}$$
$$\therefore \left( \dfrac {1}{4}\right)^{-2}+\left( \dfrac {1}{2}\right)^{-2}+\left( \dfrac {1}{3}\right)^{-2}=(4)^2+(2)^2+(3)^2=16+4+9=29$$
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Q5
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