\left[\left(\sqrt{25}\times \left(-2\right)^3\right)\div \left(\sqrt{9}+\left(-2\right)^2+5^0\right)\right]^{-2}

asked by guest
on Mar 22, 2025 at 10:07 am



You asked:

Evaluate the expression: (25(−2)39+(−2)2+50)−2{\left( \frac{\sqrt{25} {\left( -2 \right)}^{3}}{\sqrt{9} + {\left( -2 \right)}^{2} + {5}^{0}} \right)}^{-2}

MathBot Answer:

(25(−2)39+(−2)2+50)−2=125=0.04{\left( \frac{\sqrt{25} {\left( -2 \right)}^{3}}{\sqrt{9} + {\left( -2 \right)}^{2} + {5}^{0}} \right)}^{-2} = \frac{1}{25} = 0.04

(25(−2)39+(−2)2+50)−2=1(25⋅(−2)3⋅19+(−2)2+50)2=1(5⋅(−2)3⋅19+(−2)2+50)2=1(5⋅−8⋅19+(−2)2+50)2=1(5⋅−8⋅13+(−2)2+50)2=1(5⋅−8⋅13+4+50)2=1(5⋅−8⋅13+4+1)2=1(5⋅−8⋅17+1)2=1(5⋅−88)2=1(5⋅−1)2=1(−5)2=125\begin{aligned}{\left( \frac{\sqrt{25} {\left( -2 \right)}^{3}}{\sqrt{9} + {\left( -2 \right)}^{2} + {5}^{0}} \right)}^{-2}&=\frac{1}{\left(\sqrt{25} \cdot \left(-2\right)^{3} \cdot \frac{1}{\sqrt{9} + \left(-2\right)^{2} + 5^{0}}\right)^{2}}\\&=\frac{1}{\left(5 \cdot \left(-2\right)^{3} \cdot \frac{1}{\sqrt{9} + \left(-2\right)^{2} + 5^{0}}\right)^{2}}\\&=\frac{1}{\left(5 \cdot -8 \cdot \frac{1}{\sqrt{9} + \left(-2\right)^{2} + 5^{0}}\right)^{2}}\\&=\frac{1}{\left(5 \cdot -8 \cdot \frac{1}{3 + \left(-2\right)^{2} + 5^{0}}\right)^{2}}\\&=\frac{1}{\left(5 \cdot -8 \cdot \frac{1}{3 + 4 + 5^{0}}\right)^{2}}\\&=\frac{1}{\left(5 \cdot -8 \cdot \frac{1}{3 + 4 + 1}\right)^{2}}\\&=\frac{1}{\left(5 \cdot -8 \cdot \frac{1}{7 + 1}\right)^{2}}\\&=\frac{1}{\left(5 \cdot \frac{-8}{8}\right)^{2}}\\&=\frac{1}{\left(5 \cdot -1\right)^{2}}\\&=\frac{1}{\left(-5\right)^{2}}\\&=\frac{1}{25}\end{aligned}

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