Evaluate :
14+(0.01)−12−(27)23\sqrt{\dfrac{1}{4}} + (0.01)^{-\dfrac{1}{2}} - (27)^{\dfrac{2}{3}}41+(0.01)−21−(27)32
27 Likes
Simplifying the expression :
⇒14+(0.01)−12−(27)23=12+(1100)−12−(33)23=12+(1102)−12−32=12+(10−2)−12−33×23=12+10−2×−12−32=12+10−9=12+1=1+22=32=112.\Rightarrow \sqrt{\dfrac{1}{4}} + (0.01)^{-\dfrac{1}{2}} - (27)^{\dfrac{2}{3}} = \dfrac{1}{2} + \Big(\dfrac{1}{100}\Big)^{-\dfrac{1}{2}} - (3^3)^{\dfrac{2}{3}} \\[1em] = \dfrac{1}{2} + \Big(\dfrac{1}{10^2}\Big)^{-\dfrac{1}{2}} - 3^2 = \dfrac{1}{2} + (10^{-2})^{-\dfrac{1}{2}} - 3^{3 \times \dfrac{2}{3}} \\[1em] = \dfrac{1}{2} + 10^{-2 \times -\dfrac{1}{2}} - 3^2 \\[1em] = \dfrac{1}{2} + 10 - 9 \\[1em] = \dfrac{1}{2} + 1 \\[1em] = \dfrac{1 + 2}{2} \\[1em] = \dfrac{3}{2} \\[1em] = 1\dfrac{1}{2}.⇒41+(0.01)−21−(27)32=21+(1001)−21−(33)32=21+(1021)−21−32=21+(10−2)−21−33×32=21+10−2×−21−32=21+10−9=21+1=21+2=23=121.
Hence, 14+(0.01)−12−(27)23=112.\sqrt{\dfrac{1}{4}} + (0.01)^{-\dfrac{1}{2}} - (27)^{\dfrac{2}{3}} = 1\dfrac{1}{2}.41+(0.01)−21−(27)32=121.
Answered By
13 Likes
Simplify :
5n+3−6×5n+19×5n−5n×22\dfrac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^n - 5^n \times 2^2}9×5n−5n×225n+3−6×5n+1
(3x2)−3×(x9)23(3x^2)^{-3} \times (x^9)^{\dfrac{2}{3}}(3x2)−3×(x9)32
(278)23−(14)−2+50\Big(\dfrac{27}{8}\Big)^{\dfrac{2}{3}} - \Big(\dfrac{1}{4}\Big)^{-2} + 5^0(827)32−(41)−2+50
Simplify the following and express with positive index :
(3−42−8)14\Big(\dfrac{3^{-4}}{2^{-8}}\Big)^{\dfrac{1}{4}}(2−83−4)41
