heureka

60liter = ... kg

$\boxed{~Liter \cdot Dichte = Gewicht~}$

Beispiel Wasser: 

$\begin{array}{rcl} Liter_{Wasser} \cdot Dichte_{Wasser} &=& Gewicht_{Wasser}\\\\ \end{array}\\ \begin{array}{rcl} & \boxed{~ \begin{array}{rcl} Dichte_{Wasser} = \frac{1\ kg}{1\ dm^3} \qquad 1\ Liter_{Wasser} = 1\ dm^3 \end{array}\\ ~}\\\\ \end{array}\\ \begin{array}{rcll} 60\ Liter_{Wasser} &=& 60\ dm^3 \\ 60\ dm^3 \cdot \frac{1\ kg}{1\ dm^3} &=& Gewicht_{Wasser}\\ Gewicht_{Wasser} & = & 60\ kg\\\\ 60\ Liter_{Wasser} &=& 60\ kg\\ \end{array}\\$

x+2y=6

-7x+3y=-8

$\begin{array}{rcll} (1) & x+2y &=& 6 \qquad & | \qquad \cdot 7 \\ (1) & 7x+14y &=& 42 \\ (2) & -7x+3y &=& -8 \\ \\ \hline \\ (1) +(2) & 7x - 7x + 14 y + 3 y &=& 42 - 8 \\ & 14 y + 3 y &=& 42 - 8 \\ & 14 y + 3 y &=& 34 \\ & 17 y &=& 34 \qquad & | \qquad : 17 \\ & y &=& \frac{ 34 }{17} \\ & \mathbf{y} & \mathbf{=} & \mathbf{2} \\ \\ \hline \\ (1) & x+2y &=& 6 \qquad & y = 2 \\ & x+2\cdot 2 &=& 6 \\ & x+4 &=& 6 \\ & x+4 &=& 6 \qquad & | \qquad -4 \\ & x &=& 6 -4 \\ & \mathbf{x} & \mathbf{=} & \mathbf{2} \\ \end{array}$

Herr Schmidt hat 8400 EUR bei seiner Bank zu einem Effektivzinssatz von 5.4 % bei monatlicher Zinszahlung angelegt. Wie viele Monate dauert es, bis sich das Kapital auf 24810 erhöht hat? Runden Sie das Ergebnis auf eine Nachkommastelle [z.B. 95,6 für 95,6 Monate].

$\small{ \begin{array}{rcll} 24810 &=& 8400 \cdot q^n \qquad & q = \left( 1+\frac{5,4}{100} \right) \qquad n = \text{ Anzahl Monate}\\ && \qquad & q = 1,054 \\\\ 24810 &=& 8400 \cdot 1,054^n \\ 8400 \cdot 1,054^n &=& 24810 \\ 1,054^n &=& \frac{ 24810 }{ 8400 } \qquad & | \qquad \log{} \\ \log{ ( 1,054^n ) } &=& \log{ ( \frac{ 24810 }{ 8400 } ) } \\ n\cdot \log{ ( 1,054 ) } &=& \log{ ( \frac{ 24810 }{ 8400 } ) } \\ n&=& \frac{ \log{ ( \frac{ 24810 }{ 8400 } ) } } { \log{ ( 1,054 ) } } \\ n&=& \frac{ \log{ ( 2,9535714286 ) } } { \log{ ( 1,054 ) } } \\ n&=& \frac{ 0,4703474782 } { 0,0228406109 } \\ n&=& 20,5925962643 \\ \end{array}\\ }$

Es dauert 20,6 Monate

A spherical balloon has a surface area of 88cm2. What is the diameter of the balloon to the nearest tenth of a centimetre?

$\begin{array}{rcl} &\text{Surface Area of a Sphere } \\ &\boxed{~ \begin{array}{rcl} A &=& 4 \pi r^2 \qquad \text{diameter } d = 2r \quad r = \frac{d}{2} \\ A &=& 4 \pi \left( \frac{d}{2} \right) ^2 \\ A &=& 4 \pi \frac{d^2}{4} \\ A &=& \pi d^2 \\ \end{array}\\ ~}\\\\ \end{array}\\ \begin{array}{rcll} A &=& \pi d^2 \\ \pi d^2 &=& A \\ d^2 &=& \frac{A}{\pi} \qquad & | \qquad \sqrt{}\\ d &=& \sqrt{\frac{A}{\pi}} \qquad & | \qquad A = 88\ cm^2\\ d &=& \sqrt{\frac{88\ cm^2}{\pi}} \\ d &=& \sqrt{\frac{88}{\pi}} \ cm\\ d &=& 5.2925674284 \ cm\\ \end{array}\\$

The diameter of the balloon to the nearest tenth of a centimetre is 5.3 cm

Find the line perpendicular to 3x+6y=12 that goes through (6,4)

$\begin{array}{rcl} 3x+6y &=& 12\\ 6y &=& 12-3x \\ y &=& \frac{12-3x}{6} \\ y &=& \frac{12}{6} -\frac{3x}{6} \\ y &=& 2 -\frac{x}{2} \\ y &=& -\frac{x}{2} + 2 \\ \end{array}\\$

$\begin{array}{rcl} \text{Formula } \boxed{~ \begin{array}{lrcl} y = mx+b \\\\ \dfrac{y-y_p}{x-x_p} = m_{\text{perpendicular}} \\ m_{\text{perpendicular}} = -\frac{1}{m} \end{array} ~}\\\\ \end{array}\\ \begin{array}{rcl} P(x_p,y_p) &=& (6,4) \\\\ y = -\frac{x}{2} + 2 \qquad m = -\frac12 \\ m_{\text{perpendicular}} = -\frac{1}{ -\frac12} &=& 2 \\ \dfrac{y-y_p}{x-x_p} = \dfrac{y-4}{x-6} &=& 2 \\ y-4 &=& 2\cdot (x-6) \\ y-4 &=& 2 x -12 \\ y &=& 2 x -12 +4 \\ y &=& 2 x - 8 \\ \end{array}\\$

Find the line perpendicular to y=-8/3 x+1 that goes through (8,9)

$\begin{array}{rcl} \text{Formula } \boxed{~ \begin{array}{lrcl} y = mx+b \\\\ \dfrac{y-y_p}{x-x_p} = m_{\text{perpendicular}} \\ m_{\text{perpendicular}} = -\frac{1}{m} \end{array} ~}\\\\ \end{array}\\ \begin{array}{rcl} P(x_p,y_p) &=& (8,9) \\\\ y = -\frac83 x +1 \qquad m = -\frac83 \\ m_{\text{perpendicular}} = -\frac{1}{-\frac83} &=& \frac38 \\ \dfrac{y-y_p}{x-x_p} = \dfrac{y-9}{x-8} &=& \frac38 \\ y-9 &=& \frac38\cdot (x-8) \\ y-9 &=& \frac38 x -3 \\ y &=& \frac38 x -3 +9 \\ y &=& \frac38 x + 6 \\ \end{array}\\ $

find the line perpendicular to y=2/5x-3 that goes through (4,4)

$\begin{array}{rcl} \text{Formula } \boxed{~ \begin{array}{lrcl} y = mx+b \\\\ \dfrac{y-y_p}{x-x_p} = m_{\text{perpendicular}} \\ m_{\text{perpendicular}} = -\frac{1}{m} \end{array} ~}\\\\ \end{array}\\ \begin{array}{rcl} P(x_p,y_p) &=& (4,4) \\\\ y = \frac25 x -3 \qquad m = \frac25 \\ m_{\text{perpendicular}} = -\frac{1}{\frac25} &=& -\frac52 \\ \dfrac{y-y_p}{x-x_p} = \dfrac{y-4}{x-4} &=& -\frac52 \\ y-4 &=& -\frac52\cdot (x-4) \\ y-4 &=& -\frac52 x +10 \\ y &=& -\frac52 x +10 +4 \\ y &=& -\frac52 x + 14 \\ \end{array}\\$

Find the line perpendicular to y=3x-7 that goes through (3,5)

$\begin{array}{rcl} \text{Formula } \boxed{~ \begin{array}{lrcl} y = mx+b \\\\ \dfrac{y-y_p}{x-x_p} = m_{\text{perpendicular}} \\ m_{\text{perpendicular}} = -\frac{1}{m} \end{array} ~}\\\\ \end{array}\\ \begin{array}{rcl} P(x_p,y_p) &=& (3,5) \\\\ y = 3x-7 \qquad m = 3 \\ m_{\text{perpendicular}} = -\frac{1}{m} &=& -\frac{1}{3} \\ \dfrac{y-y_p}{x-x_p} = \dfrac{y-5}{x-3} &=& -\frac{1}{3} \\ y-5 &=& -\frac{1}{3}\cdot (x-3) \\ y-5 &=& -\frac{x}{3}+1 \\ y &=& -\frac{x}{3}+1+5 \\ y &=& -\frac{x}{3}+6 \\ \end{array}\\$

six years ago Albert was 1/5 as old as his mother. his mother is now 3 times as old as alert. how old is his mother?

$\begin{array}{|l|r|r|} \hline & Albert(a) & Mother(x) \\ \hline \text{Now} & a & \quad (1) \quad x = 3a \\ -6\text{ years} & \quad (2) \quad a-6 = \frac15(x-6) & x -6 \\ \hline \end{array}\\ \begin{array}{lrcl} (1)& x &=& 3a \\ & a &=& \frac{x}{3} \\ \\ \hline \\ (2)& a-6 &=& \frac15 (x-6) \quad & | \quad a = \frac{x}{3}\\\\ & \frac{x}{3}-6 &=& \frac15 (x-6)\\\\ & \frac{x}{3}-6 &=& \frac{x}{5} - \frac{6}{5} \quad & | \quad -\frac{x}{5}\\\\ & \frac{x}{3}- \frac{x}{5}-6 &=& -\frac{6}{5}\\\\ & \frac{x}{3}- \frac{x}{5}-6 &=& -\frac{6}{5} \quad & | \quad +6 \\\\ & \frac{x}{3}- \frac{x}{5} &=& 6-\frac{6}{5} \\\\ & \frac{5x-3x}{15} &=& \frac{30-6}{5} \\\\ & \frac{2x}{15} &=& \frac{24}{5} \\\\ & \frac{2x}{15} &=& \frac{24}{5} \quad & | \quad :2 \\\\ & \frac{x}{15} &=& \frac{12}{5} \quad & | \quad \cdot 15 \\\\ & x &=& \frac{12\cdot 15}{5} \\\\ & x &=& 12\cdot 3 \\\\ & x &=& 36 \\\\ \end{array}\\ \text{His mother is } \mathbf{36} \text{ years old.}\\\\ \begin{array}{lrcl} \\ \hline \\ (1)& a &=& \frac{x}{3} \quad & | \quad x = 36\\\\ & a &=& \frac{36}{3} \\\\ & a &=& 12 \\\\ \end{array}\\ \text{Albert is } \mathbf{12} \text{ years old.}$

Right ABC has AB=3, BC=4, and AC=5. Square XYZW is inscribed in ABC with X and Y on AC, W on AB, and Z on BC. What is the side length of the square?

$\small{ \begin{array}{lcl} XW = WZ = x \qquad u = ZB \qquad v = WB\\ \end{array}\\ \begin{array}{lrcl} (1) & \frac{v}{x} &=& \frac{3}{5} \quad \rightarrow \quad v = \frac{3}{5} x\\ (2) & \frac{u}{x} &=& \frac{4}{5} \quad \rightarrow \quad u = \frac{4}{5} x\\\\ & \sin{(A)} &=& \frac{x}{3-v} \\ & \sin{(B)} = \sin{(90^{\circ}-A)} = \cos{(A)} &=& \frac{x}{4-u} \\ (3) & [\sin{(A)}]^2 + [\cos{(A)}]^2 =1 &=& \left(\frac{x}{3-v}\right)^2 + \left(\frac{x}{4-u}\right)^2 \\\\ \\ \hline \\ &\left( \dfrac{x}{3-v} \right)^2 + \left(\dfrac{x}{4-u} \right)^2 &=& 1 \\\\ & \dfrac{x^2}{ (3-v)^2 } + \dfrac{x^2}{ (4-u)^2 } &=& 1 \qquad v = \frac{3}{5} x \qquad u = \frac{4}{5} x\\\\ & \dfrac{x^2}{ \left(3-\dfrac{3}{5} x \right)^2 } + \dfrac{x^2}{ \left(4-\dfrac{4}{5} x \right)^2 } &=& 1 \\\\ & \dfrac{x^2}{ 3^2 \left(1-\dfrac{1}{5} x \right)^2 } + \dfrac{x^2}{ 4^2\left(1-\dfrac{1}{5} x \right)^2 } &=& 1 \\\\ & \dfrac{x^2}{ 3^2 } + \dfrac{x^2}{ 4^2 } &=& \left(1-\dfrac{1}{5} x \right)^2 \\\\ & \dfrac{x^2}{ 3^2 } + \dfrac{x^2}{ 4^2 } &=& \dfrac{ (5-x)^2 }{5^2}\\\\ & x^2 \left(\dfrac{1} { 3^2 }+ \dfrac{1}{ 4^2 } \right) &=& \dfrac{ (5-x)^2 }{5^2}\\\\ & x^2 \left(\dfrac{4^2+3^2} { 3^24^2 } \right) &=& \dfrac{ (5-x)^2 }{5^2} \qquad | \quad 3^2+4^2=5^2\\\\ & x^2 \left(\dfrac{5^2} { 3^24^2 } \right) &=& \dfrac{ (5-x)^2 }{5^2} \qquad | \quad 5^25^2=5^4\\\\ & x^2 \left( \dfrac{5^4} { 3^24^2 } \right) &=& (5-x)^2 \\\\ & x^2 \left( \dfrac{5^4} { 3^24^2 } \right) &=& 5^2-2\cdot 5 x + x^2 \\\\ & x^2 \left( \dfrac{5^4} { 3^24^2 } \right) -x^2 &=& 5^2-2\cdot 5 x \\\\ & x^2 \left( \dfrac{5^4} { 3^24^2 } -1 \right) &=& 5^2-2\cdot 5 x \\\\ & x^2 \left( \dfrac{5^4-3^24^2} { 3^24^2 } \right) &=& 5^2-2\cdot 5 x \\\\ & \mathbf{ x^2 \left( \dfrac{5^4-3^24^2} { 3^24^2 } \right) +2\cdot 5 x - 5^2 } & \mathbf{=}& \mathbf{ 0 }\\\\ \\ \hline \\ & x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \quad a= \frac{5^4-3^24^2} { 3^24^2 } \quad b=2\cdot 5 \quad c=-5^2 \\\\ & x^2 \left( \frac{5^4-3^24^2} { 3^24^2 } \right) +2\cdot 5 x - 5^2 &=& 0 \\\\ & x_{1,2} &=& {-2\cdot 5 \pm \sqrt{(-2\cdot 5)^2 - 4\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right) \cdot(-5^2) } \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-2\cdot 5 \pm \sqrt{2^25^2 + 2^25^2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right) } \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-2\cdot 5 \pm 2\cdot 5 \sqrt{1 + \left( \frac{5^4-3^24^2} { 3^24^2 }\right) } \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-2\cdot 5 \pm 2\cdot 5 \sqrt{\frac{3^24^2+ 5^4-3^24^2} { 3^24^2 } } \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-2\cdot 5 \pm 2\cdot 5 \sqrt{\frac{5^4} { 3^24^2 } } \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-2\cdot 5 \pm 2\cdot 5 \left( \frac{5^2} { 3\cdot 4 } \right) \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-10 \pm \frac{250} { 12 } \over 2\cdot \left( \frac{5^4-3^24^2} { 3^24^2 }\right)} \\\\ & x_{1,2} &=& {-10 \pm \frac{250} { 12 } \over \frac{962} { 144 } } \\\\ & x &=& {-10 + \frac{250} { 12 } \over \frac{962} { 144 } } \\\\ & x &=& {\frac{130} { 12 } \over \frac{962} { 144 } } \\\\ & x &=& \left( \frac{130} { 12 } \right) \cdot \left( \frac{ 144 }{962} \right) \\\\ & x &=& \left( \frac{130} { 12 } \right) \cdot \left( \frac{ 12^2 }{962} \right) \\\\ & x &=& \frac{130\cdot 12} { 962 } \\\\ & x &=& \frac{1560} { 962 } \\\\ & x &=& \frac{780} { 481 } \\\\ & x &=& \frac{60} { 37 } \\\\ & \mathbf{x} &\mathbf{=}& \mathbf{ 1.621\overline{621} } \end{array}\\ \text{The side length of the square is } \mathbf{1.621\overline{621}} }$