heureka

Triangle RST has vertices R (-3,4), S (-1,0), T (3,2). Find the length of the altitude from S to RT.

$\boxed{~ \begin{array}{rcl} \vec{RT} &=& \vec{R} - \vec{T} = \binom{-3}{4} - \binom{3}{2} = \binom{-6}{2}\\\\ \overline{RT}=|\vec{RT}| &=&\sqrt{ \vec{RT}\cdot \vec{RT} }\\ \overline{RT} &=& \sqrt{ \binom{-6}{2} \binom{-6}{2} }\\ \overline{RT} &=& \sqrt{36+4 }\\ \overline{RT} &=& \sqrt{40}\\ \overline{RT} &=& \sqrt{4\cdot 10}\\ \mathbf{\overline{RT}}& \mathbf{=} & \mathbf{2\sqrt{10} }\\ \end{array} ~}$

$\boxed{~ \begin{array}{rcl} \vec{ST} &=& \vec{S} - \vec{T} = \binom{-1}{0} - \binom{3}{2} = \binom{-4}{-2}\\\\ \overline{RT} \cdot h &=& |\vec{RT} \times \vec{ST}|\\ h &=& \frac{|\vec{RT} \times \vec{ST}|}{\overline{RT}}\\ h &=& \frac{ \left| \binom{-6}{2} \times\binom{-4}{-2} \right| }{2\sqrt{10}}\\ h &=& \frac{(-6)\cdot (-2) - (2) \cdot (-4) }{2\sqrt{10}}\\ h &=& \frac{ 12 + 8 }{2\sqrt{10}}\\ h &=& \frac{ 20 }{2\sqrt{10}}\\ h &=& \frac{ 10 }{ \sqrt{10} } \cdot \frac{ \sqrt{10} }{ \sqrt{10} }\\ h &=& \frac{ 10 }{ 10 } \cdot \sqrt{10} \\ \mathbf{h} &\mathbf{=}& \mathbf{\sqrt{10}} \\ \end{array} ~}$

( 4 1/3 )  /  ( 5 1/6 ) = ?

$\begin{array}{rcl} \frac{ ( 4 \frac13 ) }{ ( 5 \frac16 ) }\\ &=& \frac{ \frac{13}{3} }{ \frac{31}{6} }\\ &=& \frac{13}{3} \cdot \frac{6}{31} \\ &=& \frac{13}{3} \cdot \frac{2\cdot 3}{31} \\ &=& 13 \cdot \frac{2}{31} \\ &=& \frac{26}{31} \\ &=& 0.83870967742\\ \mathbf{\frac{ ( 4 \frac13 ) }{ ( 5 \frac16 ) } }& \mathbf{=} & \mathbf{0.83870967742}\\ \end{array}$

x-y=1

y-2z=1

3x-4z=7

x=?

y=?

z=?

$\begin{array}{lrcll} (1) & x-y&=&1 \\ & x &=& 1+y \\ \hline (2) & y-2z&=&1 \\ & y &=& 1+2z \\ \hline (3) & 3x-4z&=&7\\ & 3x &=& 7 + 4z &\qquad x = 1+y\\ & 3(1+y) &=& 7 + 4z &\qquad y = 1+2z \\ & 3[1+(1+2z)] &=& 7 + 4z \\ & 3(1+1+2z) &=& 7 + 4z \\ & 3(2+2z) &=& 7 + 4z \\ & 6+6z &=& 7 + 4z \qquad &| \qquad -4z \\ & 6+6z-4z &=& 7 \\ & 6 + 2z &=& 7 \qquad &| \qquad -6 \\ & 2z &=& 7 -6\\ & 2z &=& 1 \qquad &| \qquad :2 \\ & \mathbf{z} &\mathbf{=}& \mathbf{0.5}\\ \hline (2) & y &=& 1+2z \\ & y &=& 1+2(0.5)\\ & y &=& 1+1\\ & \mathbf{y} &\mathbf{=}& \mathbf{2}\\ \hline (1) & x &=& 1+y \\ & x &=& 1+2\\ & \mathbf{x} &\mathbf{=}& \mathbf{3} \end{array}$

-1 1/2 + 5 7/8 = ?

$\begin{array}{rcl} -1 \frac12 + 5 \frac78 \\ &=& - 1 - \frac12 + 5+\frac78 \qquad | \qquad \frac78 = \frac{4+3}{8} = \frac48+\frac38 = \frac12+\frac38\\ &=& - 1 - \frac12 + 5+\frac12+\frac38\\ &=& - 1 + 5+\frac38\\ &=& 4+\frac38\\ &=& 4\frac38\\ \mathbf{-1 \frac12 + 5 \frac78 } & \mathbf{=} & \mathbf{4\frac38}\\ \end{array}$

the time in hours and minutes between 12:20 and 23:00

$\begin{array}{rcl} 23:00 &=& 22:60 \\\\ \hline 22&:&60 \\ -12&:&20 \\ \hline =10&:&40 \end{array}$

Hallo! hoffe es kennt sich jemand aus: $\begin{array}{rcl} \int \sqrt[5]{ x(\frac{1}{x}-5 )}\ dx \end{array}$

$\begin{array}{rcl} \int \sqrt[5]{ x(\frac{1}{x}-5 )}\ dx \\ &=& \int \sqrt[5]{ (1-5x )}\ dx \\ &=& \int (1-5x )^\frac15\ dx \\\\ \qquad \text{ Substitution } z&=& 1-5x\\ dz &=& -5\ dx\\ dx &=& -\frac15 \ dz\\\\ \int \sqrt[5]{ x(\frac{1}{x}-5 )}\ dx &=& \int z^{\frac15}\cdot ( -\frac15 \ dz ) \\ &=& -\frac15 \int z^{\frac15}\ dz \quad \boxed{~ \text{Formel } \int x^n \ dx = \frac{x^{n+1}}{n+1} ~}\\ &=& -\frac15\left( \frac{ z^{\frac15 + 1} } { \frac15 + 1 } \right)+c\\ &=& -\frac15\left( \frac{ z^{\frac65} } { \frac65 } \right)+c\\ &=& -\frac15\cdot \frac56 z^{\frac65}+c\\ &=& -\frac16 z^{\frac65}+c \qquad z= 1-5x\\ &=& -\frac16 (1-5x)^{\frac65}+c\\ \mathbf{ \int \sqrt[5]{ x(\frac{1}{x}-5 )}\ dx } & \mathbf{=} & \mathbf{-\frac16 (1-5x)^{\frac65} +c} \end{array}$

-3x-2y=-12

9x+6y=-9

$\text{Determinante }~D=\begin{vmatrix} -3 && -2 \\ 9 && 6 \end{vmatrix} =(-3)\cdot 6 - 9\cdot (-2)= -18+18 = 0\\ D = 0 \qquad \Rightarrow \qquad \text{ no solution }$

$d=\sqrt{x^2+\frac{b^2}{a^2}}$

Can you, and how do you simplify this further?

$d=\sqrt{x^2+\frac{b^2}{a^2}} \\ d=\sqrt{x^2+\left(\frac{b}{a} \right)^2 } \\$

if n= -2 then what is the answer to the sum 2n2 + 3n

$\begin{array}{lcl} 2n^2 + 3n \qquad &| &\qquad n=-2 \\ 2\cdot (-2)^2 + 3\cdot (-2) \qquad &|& \qquad (-2)^2=(-2)\cdot(-2)=+4\\ &=& 2\cdot 4 -6 \\ &=& 8-6 \\ &=& 2 \end{array}$

x*(x+1) - 2*(x+1) = 4

Solve x.

$\begin{array}{rcl} x\cdot (x+1) - 2\cdot (x+1) &=& 4 \\ x^2+x-2x-2 &=& 4\\ x^2-x-6 &=& 0\\ \boxed{~ x = {-b \pm \sqrt{b^2-4ac} \over 2a}~ }\\ x^2-x-6 &=& 0\\ x &=& {1 \pm \sqrt{1-4(-6)} \over 2} \\ x &=& {1 \pm \sqrt{1+24} \over 2} \\ x &=& {1 \pm \sqrt{25} \over 2} \\ x &=& {1 \pm 5 \over 2} \\ \hline x_1 &=& {1 + 5 \over 2}\\ x_1 &=& {6 \over 2}\\ \mathbf{x_1} &\mathbf{=} & \mathbf{3}\\ \hline x_2 &=& {1 - 5 \over 2}\\ x_2 &=& {-4 \over 2}\\ \mathbf{x_2} &\mathbf{=} & \mathbf{-2}\\ \end{array}$