heureka

How does this work ? 11/9-2/3+5/18

\(\begin{array}{|rcll|} \hline \frac{11}{9} - \frac{2}{3} + \frac{5}{18} &=& x \qquad &|\qquad \cdot 9 \\ 9\cdot \frac{11}{9} - 9\cdot \frac{2}{3} + 9\cdot \frac{5}{18} &=& 9x \\ 11 - 6 + \frac{5}{2} &=& 9x \\ 5 + \frac{5}{2} &=& 9x \\ \frac{10}{2} + \frac{5}{2} &=& 9x \\ \frac{15}{2} &=& 9x \qquad &|\qquad : 9\\ \frac{15}{2\cdot 9} &=& x \\ \frac{5}{2\cdot 3} &=& x \\ \frac{5}{6} &=& x \\ \frac{5}{6} &=& \frac{11}{9} - \frac{2}{3} + \frac{5}{18} \\ \hline \end{array}\)

4ab           2a             2b          2a

---------- - ---------- - ---------- = -----

a^2-b^2      a-b          a+b        a+b        is this correct?

\(\begin{array}{|rcll|} \hline && \frac{4ab}{a^2-b^2} -\frac{2a}{a-b} -\frac{2b}{a+b} \qquad &|\qquad a^2-b^2 = (a-b)(a+b)\\ &=& \frac{4ab}{(a-b)(a+b)} -\frac{2a}{a-b} -\frac{2b}{a+b} \\ &=& \frac{4ab}{(a-b)(a+b)} -\frac{2a}{a-b}\cdot \left( \frac{a+b}{a+b} \right) -\frac{2b}{a+b}\cdot \left( \frac{a-b}{a-b} \right)\\ &=& \frac{4ab-2a(a+b)-2b(a-b)}{(a-b)(a+b)} \qquad &|\qquad (a-b)(a+b)= a^2-b^2 \\ &=& \frac{4ab-2a(a+b)-2b(a-b)}{a^2-b^2} \\ &=& \frac{4ab-2a^2-2ab-2ab+2b^2}{a^2-b^2} \\ &=& \frac{4ab-4ab-2a^2+2b^2}{a^2-b^2} \\ &=& \frac{-2a^2+2b^2}{a^2-b^2} \\ &=& \frac{-2\cdot(a^2-b^2)}{a^2-b^2} \\ &=& -2 \\ \hline \end{array}\)

What  is  the determinant of  C?
\(C=\begin{bmatrix} 1 && 2 &&-2\\ 3 && 0 && -1\\ 5 && 3 && 2 \end{bmatrix}\)

\(\begin{array}{|rcll|} \hline C_{determinant} &=& \begin{vmatrix}1 && 2 &&-2\\ 3 && 0 && -1\\ 5 && 3 && 2 \end{vmatrix} \\ &=& 1\cdot 0 \cdot 2 + 2\cdot(-1)\cdot 5 + (-2)\cdot 3\cdot 3 \\ && - 5\cdot 0 \cdot (-2) -2\cdot 3 \cdot 2 -1\cdot 3 \cdot (-1) \\ &=& 2\cdot(-1)\cdot 5 + (-2)\cdot 3\cdot 3 -2\cdot 3 \cdot 2 -1\cdot 3 \cdot (-1) \\ &=& -10 -18 -12 +3 \\ &=& -37 \\ \hline \end{array} \)

How do i solve this equation?

5x/6 - 3/8 = 11/24 + 5x/12

\(\begin{array}{|rcll|} \hline x\cdot\frac56 - \frac38 &=& \frac{11}{24} + x\cdot\frac{5}{12} \qquad &| \qquad \cdot 12\\\\ x\cdot\frac56 \cdot 12 - \frac38 \cdot 12 &=& \frac{11}{24} \cdot 12 + x\cdot\frac{5}{12} \cdot 12\\\\ x\cdot 5 \cdot 2 - 3\cdot \frac{12}{8} &=& \frac{11}{2} + x\cdot 5\\\\ 10x- 3\cdot \frac{3}{2} &=& \frac{11}{2} + 5x \\\\ 10x - \frac{9}{2} &=& \frac{11}{2} + 5x \qquad &| \qquad -5x\\\\ 10x -5x -\frac{9}{2} &=& \frac{11}{2} \qquad &| \qquad + \frac{9}{2}\\\\ 10x -5x &=& \frac{11}{2} + \frac{9}{2}\\\\ 5x &=& \frac{11+9}{2} \\\\ 5x &=& \frac{20}{2} \\\\ 5x &=& 10 \qquad &| \qquad : 5 \\\\ x &=& \frac{10}{5} \\\\ \mathbf{ x } & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)

finding x and y, what is 4x+10y and 11x+5y. both are equal to 90.

determinant method:

\(\begin{array}{|rcll|} \hline x &=& \frac{ \begin{vmatrix} 90 & 10\\90 & 5\end{vmatrix} } { \begin{vmatrix} 4 & 10\\11 & 5\end{vmatrix} } \\\\ &=& \frac{ 90\cdot \begin{vmatrix} 1 & 10\\1 & 5\end{vmatrix} } { 4\cdot 5 - 11 \cdot 10 } \\\\ &=& \frac{ -90\cdot \begin{vmatrix} 10 & 1\\5 & 1\end{vmatrix} } { -90 } \\\\ &=& \begin{vmatrix} 10 & 1\\5 & 1\end{vmatrix} \\\\ &=& 10-5 \\\\ &=& \mathbf{5} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline y &=& \frac{ \begin{vmatrix} 4 & 90\\11 & 90\end{vmatrix} } { \begin{vmatrix} 4 & 10\\11 & 5\end{vmatrix} } \\\\ &=& \frac{ 90\cdot \begin{vmatrix} 4 & 1\\11 & 1\end{vmatrix} } { 4\cdot 5 - 11 \cdot 10 } \\\\ &=& \frac{ -90\cdot \begin{vmatrix} 1 & 4\\1 & 11\end{vmatrix} } { -90 } \\\\ &=& \begin{vmatrix} 1 & 4\\1 & 11\end{vmatrix} \\\\ &=& 11-4 \\\\ &=& \mathbf{7} \\ \hline \end{array}\)

Find all values of x for which (4x-5) to the power of 2x+5=1.

\(\begin{array}{|rclcrcll|} \hline && (4x-5)^{(2x+5)} &=& 1 && \qquad & | \qquad \ln() \\ && \ln( (4x-5)^{(2x+5)} ) &=& \ln(1) && \qquad & | \qquad \ln(1) = 0 \\ && \ln( (4x-5)^{(2x+5)} ) &=& 0 && \qquad & | \qquad \ln(a^b) = b\cdot \ln(a) \\ && (2x+5) \cdot \ln( 4x-5 ) &=& 0 \\ (2x+5) &=& 0 & or &\qquad \ln( 4x-5 ) &=& 0 \\ 2x &=& -5 & or &\qquad e^{\ln( 4x-5 )} &=& e^{0} \\ \mathbf{ x } & \mathbf{=} & \mathbf{-\frac52} & or &\qquad 4x-5 &=& 1 \\ && & &\qquad 4x &=& 6 \\ && & &\qquad x &=& \frac64 \\ && & &\qquad \mathbf{ x } & \mathbf{=} & \mathbf{\frac32} \\ \hline \end{array}\)

So the exponent must be zero => 2x+5 = 0 => x = -5/2

or the base must be 1 => 4x-5 = 1 => x = 3/2

Find all values of x for which (4x-5) to the power of 2x+5=1.

\(\begin{array}{|rclcrcll|} \hline && (4x-5)^{(2x+5)} &=& 1 && \qquad & | \qquad \ln() \\ && \ln( (4x-5)^{(2x+5)} ) &=& \ln(1) && \qquad & | \qquad \ln(1) = 0 \\ && \ln( (4x-5)^{(2x+5)} ) &=& 0 && \qquad & | \qquad \ln(a^b) = b\cdot \ln(a) \\ && (2x+5) \cdot \ln( 4x-5 ) &=& 0 \\ (2x+5) &=& 0 & or &\qquad \ln( 4x-5 ) &=& 0 \\ 2x &=& -5 & or &\qquad e^{\ln( 4x-5 )} &=& e^{0} \\ \mathbf{ x } & \mathbf{=} & \mathbf{-\frac52} & or &\qquad 4x-5 &=& 1 \\ && & &\qquad 4x &=& 6 \\ && & &\qquad x &=& \frac64 \\ && & &\qquad \mathbf{ x } & \mathbf{=} & \mathbf{\frac32} \\ \hline \end{array}\)

ln(1/904)/(ln(1)-ln(4))

\(\begin{array}{|rcll|} \hline && \frac{ \ln( \frac{1}{904} ) } { \ln(1)-\ln(4) } \qquad & | \qquad \ln(\frac{a}{b}) = \ln(a) - \ln(b) \\\\ &=& \frac{ \ln(1)-\ln(904) } { \ln(1)-\ln(4) } \qquad & | \qquad \ln(1) = 0\\\\ &=& \frac{ 0-\ln(904) } { 0-\ln(4) } \\\\ &=& \frac{ -\ln(904) } { -\ln(4) } \\\\ &=& \frac{ \ln(904) } { \ln(4) } \\\\ &=& \frac{ 6.80682936039 } { 1.38629436112 } \\\\ &=& 4.91008948121 \\ \hline \end{array}\)

Based on input of -6, 3 and 5 for a, b, and c respectively, what will be displayed?
 

It will be displayed: a + b + c = 4

integrating: 

\(\int dx ~ 2x \sin^{3}(x) \)

1. Let

\(\begin{array}{|rcll|} \hline \int dx ~ \sin^3(x) &=& \int dx ~ \sin^2(x)\cdot \sin(x) \\ &=& \int dx ~ [ 1-\cos^2(x) ]\cdot \sin(x) \\ && u = \cos(x) \\ && du = -\sin(x) dx \\ &=& \int du ~ \frac{1}{-\sin(x)} \cdot (1-u^2) \cdot \sin(x) \\ &=& -\int du ~ (1-u^2) \\ &=& -\int du +\int du ~ u^2 \\ &=& -u + \frac13 \cdot u^3 \\ &=& \mathbf{-\cos(x) + \frac13 \cdot \cos^3(x)} \\ \hline \end{array} \)

2. Let

\(\begin{array}{|rcll|} \hline \int dx ~ \cos^3(x) &=& \int dx ~ \cos^2(x)\cdot \cos(x) \\ &=& \int dx ~ [ 1-\sin^2(x) ]\cdot \cos(x) \\ && u = \sin(x) \\ && du = \cos(x) dx \\ &=& \int du ~ \frac{1}{\cos(x)} \cdot (1-u^2) \cdot \cos(x) \\ &=& \int du ~ (1-u^2) \\ &=& \int du - \int du ~ u^2 \\ &=& u - \frac13 \cdot u^3 \\ &=& \mathbf{\sin(x) - \frac13 \cdot \sin^3(x)} \\ \hline \end{array}\)

3. Let

\(\begin{array}{|rcll|} \hline \int dx ~ x \cdot \sin^3(x) \\ && u = x \\ && u' = 1 \\ && v = -\cos(x) + \frac13 \cdot \cos^3(x) \\ && v' = \sin^3(x) \\ &=& x \cdot [-\cos(x) + \frac13 \cdot \cos^3(x)] - \int dx ~ 1\cdot [-\cos(x) + \frac13 \cdot \cos^3(x)] \\ &=& -x \cdot \cos(x) + \frac13 \cdot x \cdot \cos^3(x) + \int dx ~ \cos(x) - \int dx ~ \frac13 \cdot \cos^3(x) \\ &=& -x \cdot \cos(x) + \frac13 \cdot x \cdot \cos^3(x) + \sin(x) - \frac13 \cdot \int dx ~ \cos^3(x) \\ &=& -x \cdot \cos(x) + \frac13 \cdot x \cdot \cos^3(x) + \sin(x) - \frac13 \cdot [ \sin(x) - \frac13 \cdot \sin^3(x)] \\ &=& -x \cdot \cos(x) + \frac13 \cdot x \cdot \cos^3(x) + \sin(x) - \frac13 \cdot \sin(x) + \frac19 \cdot \sin^3(x)] \\ &=& -x \cdot \cos(x) + \frac13 \cdot x \cdot \cos^3(x) + \frac23 \cdot \sin(x) + \frac19 \cdot \sin^3(x) \\\\ \int dx ~ 2\cdot x \cdot \sin^3(x) &=& -2x \cdot \cos(x) + \frac23 \cdot x \cdot \cos^3(x) + \frac43 \cdot \sin(x) + \frac29 \cdot \sin^3(x) \\ &=&\mathbf{\frac{1}{9} \cdot [ -18x \cdot \cos(x) + 6 x \cdot \cos^3(x) + 12 \cdot \sin(x) + 2 \cdot \sin^3(x) ] }\\ \hline \end{array} \)

4. Let

\(\begin{array}{|rcll|} \hline && \cos^3(x) = \frac14 \cdot [3\cdot \cos(x) + \cos(3x) ] \\ && \sin^3(x) = \frac14 \cdot [3\cdot \sin(x) - \sin(3x) ] \\ \hline \end{array}\)

5. Let

\(\begin{array}{|rcll|} \hline && \int dx ~ 2\cdot x \cdot \sin^3(x) \\ &=& \frac{1}{9} \cdot [ -18x \cdot \cos(x) + 6 x \cdot \cos^3(x) + 12 \cdot \sin(x) + 2 \cdot \sin^3(x) ] \\ &=& \frac{1}{9} \cdot \{ -18x \cdot \cos(x) + 6 x \cdot \frac14 \cdot [3\cdot \cos(x) + \cos(3x) ] + 12 \cdot \sin(x) + 2 \cdot \frac14 \cdot [3\cdot \sin(x) - \sin(3x) ] \} \\ &=& -2x \cdot \cos(x) + \frac16 \cdot x\cdot [3\cdot \cos(x) + \cos(3x) ] + \frac{4}{3} \cdot \sin(x) + \frac{1}{18} \cdot [3\cdot \sin(x) - \sin(3x) ] \\ &=& -2x \cdot \cos(x) + \frac12 \cdot x\cdot \cos(x) + \frac16 \cdot x\cdot \cos(3x) + \frac{4}{3} \cdot \sin(x) + \frac{1}{6} \cdot \sin(x) - \frac{1}{18} \cdot \sin(3x) \\ &=& \mathbf{ - \frac32 \cdot x\cdot \cos(x) + \frac16 \cdot x\cdot \cos(3x) + \frac32 \cdot \sin(x) - \frac{1}{18} \cdot \sin(3x) }\\ \hline \end{array} \)

or

\(\begin{array}{|rcll|} \hline && \int dx ~ 2\cdot x \cdot \sin^3(x) \\ &=& - \frac32 \cdot x\cdot \cos(x) + \frac16 \cdot x\cdot \cos(3x) + \frac32 \cdot \sin(x) - \frac{1}{18} \cdot \sin(3x) \\ &=& \mathbf{ \frac{1}{18} [ - 27 \cdot x\cdot \cos(x) + 3 \cdot x\cdot \cos(3x) + 27 \cdot \sin(x) - \sin(3x) ] } \\ \hline \end{array}\)