y = -0.0002x^2 + 0.0412x + 0.1305

y=0.113

asked by guest
on Nov 18, 2024 at 11:13 am



You asked:

Solve the system of equations:\[\begin{aligned}y &= -\left( 0.0002 \cdot {x}^{2} \right) + 0.0412 x + 0.1305\\y &= 0.113\end{aligned}\]

MathBot Answer:

The system of equations has \(2\) solutions.

\[x = 103 - \frac{3 \sqrt{4754}}{2}, y = \frac{113}{1000}\]\[x = 103 + \frac{3 \sqrt{4754}}{2}, y = \frac{113}{1000}\]

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