Evaluate:
(85×−32)+(−310×916)\Big(\dfrac{8}{5} \times \dfrac{-3}{2}\Big) + \Big(\dfrac{-3}{10} \times \dfrac{9}{16}\Big)(58×2−3)+(10−3×169)
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(8×−35×2)+(−3×910×16)=(−2410)+(−27160)=(−125)+(−27160)\Big(\dfrac{8 \times -3}{5 \times 2}\Big) + \Big(\dfrac{-3 \times 9}{10 \times 16}\Big)\\[1em] =\Big(\dfrac{-24}{10}\Big) + \Big(\dfrac{-27}{160}\Big)\\[1em] =\Big(\dfrac{-12}{5}\Big) + \Big(\dfrac{-27}{160}\Big)\\[1em](5×28×−3)+(10×16−3×9)=(10−24)+(160−27)=(5−12)+(160−27)
LCM of 5 and 160 is 2 x 2 x 2 x 2 x 2 x 5 = 160
=(−12×325×32)+(−27×1160×1)=(−384160)+(−27160)=(−384+(−27)160)=(−411160)=−291160=\Big(\dfrac{-12 \times 32}{5 \times 32}\Big) + \Big(\dfrac{-27 \times 1}{160 \times 1}\Big)\\[1em] =\Big(\dfrac{-384}{160}\Big) + \Big(\dfrac{-27}{160}\Big)\\[1em] =\Big(\dfrac{-384 + (-27)}{160}\Big)\\[1em] =\Big(\dfrac{-411}{160}\Big)\\[1em] =-2\dfrac{91}{160}=(5×32−12×32)+(160×1−27×1)=(160−384)+(160−27)=(160−384+(−27))=(160−411)=−216091
Hence, (85×−32)+(−310×916)=−291160\Big(\dfrac{8}{5} \times \dfrac{-3}{2}\Big) + \Big(\dfrac{-3}{10} \times \dfrac{9}{16}\Big) = -2\dfrac{91}{160}(58×2−3)+(10−3×169)=−216091
Answered By
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