6.2 Volumes/25: Difference between revisions

From Mr. V Wiki Math
Jump to navigation Jump to search
No edit summary
No edit summary
Line 7: Line 7:


\begin{align}
\begin{align}
2x^2 + 3(x-1)(x-2) & = 2x^2 + 3(x^2-3x+2)\\ \nonumber &= 2x^2 + 3x^2 - 9x + 6\\ &= 5x^2 - 9x + 6
2x^2 + 3(x-1)(x-2) & = 2x^2 + 3(x^2-3x+2)\\ &= 2x^2 + 3x^2 - 9x + 6\\ &= 5x^2 - 9x + 6
\end{align}  
\end{align}  
</math>
</math>

Revision as of 03:02, 12 September 2022

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 R + f(y) = 1 R = 1-f(y) r = 1 \begin{align} 2x^2 + 3(x-1)(x-2) & = 2x^2 + 3(x^2-3x+2)\\ &= 2x^2 + 3x^2 - 9x + 6\\ &= 5x^2 - 9x + 6 \end{align} }

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} \pi\int_0^1\left[(1)^2-(1-y^2)^2\right]dy & = \pi\int_0^1\left[(1-(1-2y^2+y^4)\right]dy = \pi\int_0^1\left[(2y^2-y^4)\right]dy \\[2ex] &= \pi\left[\frac{2y^3}{3}-\frac{y^5}{5}\right]\Bigg|_0^1 \\[2ex] &= \pi\left[\frac{2}{3}-\frac{1}{5}\right]= \pi\left[\frac{10}{15}-\frac{3}{15}\right] \\[2ex] &= \frac{7\pi}{15} \end{align} }