55.6*10^3=0.2729*v^2+(11.6*10^7/v^2)

asked by guest
on Nov 19, 2024 at 8:30 am



You asked:

Solve the equation 55.6⋅103=0.2729v2+11.6⋅107v255.6 \cdot {10}^{3} = 0.2729 {v}^{2} + \frac{11.6 \cdot {10}^{7}}{{v}^{2}} for the variable vv.

MathBot Answer:

The 4 solutions to the equation are: v=−20069502729−518529592729≈−45.914536v=20069502729−518529592729≈45.914536v=−20069502729+518529592729≈−449.03172v=20069502729+518529592729≈449.03172\begin{aligned}v &= - 200 \sqrt{\frac{6950}{2729} - \frac{5 \sqrt{1852959}}{2729}} \approx -45.914536\\v &= 200 \sqrt{\frac{6950}{2729} - \frac{5 \sqrt{1852959}}{2729}} \approx 45.914536\\v &= - 200 \sqrt{\frac{6950}{2729} + \frac{5 \sqrt{1852959}}{2729}} \approx -449.03172\\v &= 200 \sqrt{\frac{6950}{2729} + \frac{5 \sqrt{1852959}}{2729}} \approx 449.03172\end{aligned}

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