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Question

Find the value of: $$(3^0 + 4^{-1}) \times 2^2$$


Solution

$$ (3^0 + 4^{-1}) \times 2^2 = \left ( 1 + \cfrac{1}{4} \right ) \times 4 = \cfrac{4 + 1}{4} \times 4$$$$ = \cfrac{5}{4} \times 4 = 5$$

Mathematics

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