6.1 Areas Between Curves/10: Difference between revisions

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\begin{align}
\begin{align}
&1+\sqrt{x} = \frac{3+x}{3} \\
&1+\sqrt{x} = \frac{3+x}{3} \\
&= 1+\sqrt{x}-\frac{3+x}{3} = 0 \\
& 1+\sqrt{x}-\frac{3+x}{3} = 0 \\
&= \frac{3+3\sqrt{x}}{3}-\frac{3+x}{3} = 0 \\
& \frac{3+3\sqrt{x}}{3}-\frac{3+x}{3} = 0 \\
&= 3+3\sqrt{x}-3+x = 0 \\
& 3+3\sqrt{x}-3+x = 0 \\
&= 3\sqrt{x}+x = 0 \\
& 3\sqrt{x}+x = 0 \\
&= 3\sqrt{x} = -x \\
& 3\sqrt{x} = -x \\
&= 9x = x^2 \\
& 9x = x^2 \\
&= 9x-x^2 = 0 \\
& 9x-x^2 = 0 \\
&= x(9-x) = 0 \\
& x(9-x) = 0 \\
&= x = 0,9  
& x = 0,9  
\end{align}
\end{align}
</math>
</math>

Revision as of 04:44, 20 September 2022

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Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} &1+\sqrt{x} = \frac{3+x}{3} \\ & 1+\sqrt{x}-\frac{3+x}{3} = 0 \\ & \frac{3+3\sqrt{x}}{3}-\frac{3+x}{3} = 0 \\ & 3+3\sqrt{x}-3+x = 0 \\ & 3\sqrt{x}+x = 0 \\ & 3\sqrt{x} = -x \\ & 9x = x^2 \\ & 9x-x^2 = 0 \\ & x(9-x) = 0 \\ & x = 0,9 \end{align} }