Power[\(40)1+16.8Power[z,-1]\(41)\(40)1+1.31Power[z,-1]\(41)z,-1]

asked by guest
on Nov 26, 2024 at 9:37 am



You asked:

Evaluate the expression: ((1+16.8z−1)(1+1.31z−1)⋅z)−1{\left( \left(1 + 16.8 {z}^{-1}\right) \left(1 + 1.31 {z}^{-1}\right) \cdot z \right)}^{-1}

MathBot Answer:

Evaluated



((1+16.8z−1)(1+1.31z−1)⋅z)−1=1z(1+131100z)(1+845z)\displaystyle {\left( \left(1 + 16.8 {z}^{-1}\right) \left(1 + 1.31 {z}^{-1}\right) \cdot z \right)}^{-1} = \frac{1}{z \left(1 + \frac{131}{100 z}\right) \left(1 + \frac{84}{5 z}\right)}

Expanded

((1+16.8z−1)(1+1.31z−1)⋅z)−1=1z+1811100+2751125z{\left( \left(1 + 16.8 {z}^{-1}\right) \left(1 + 1.31 {z}^{-1}\right) \cdot z \right)}^{-1} = \frac{1}{z + \frac{1811}{100} + \frac{2751}{125 z}}

Factored

((1+16.8z−1)(1+1.31z−1)⋅z)−1=500z(5z+84)(100z+131){\left( \left(1 + 16.8 {z}^{-1}\right) \left(1 + 1.31 {z}^{-1}\right) \cdot z \right)}^{-1} = \frac{500 z}{\left(5 z + 84\right) \left(100 z + 131\right)}

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