heureka

Given that x is 25% less than s and y is 20% great than x. Find y in terms of s.

\(\begin{array}{rcl} y\cdot(100 \%+20\%) &=& x \\ x\cdot(100 \%-25\%) &=& s \\ \hline y\cdot(100 \%+20\%)\cdot(100 \%-25\%) &=& s \\ y\cdot( \frac{120}{100} )\cdot( \frac{75}{100} ) &=& s \\ y\cdot 1.2 \cdot 0.75 &=& s \\ y\cdot 0.9 &=& s \qquad & | \quad : 0.9\\ y &=& \frac{ s } {0.9} \\ y &=& \frac{ 10 } { 9 } s \\ y &=& 1.1\overline{1} s \\ \end{array}\)

z+iz= 3+7i How do you get z?

\(\begin{array}{lrcl} & z+iz &=& 3+7i \\ & z(1+i) &=& 3+7i \\\\ \text{We set } & z = a+bi \\ & (a+bi)\cdot (1+i) &=& 3+7i \\ & a + ai+bi+bi^2 &=& 3+7i \qquad i^2 = -1\\ & a + ai+bi -b &=& 3+7i \\ & (a -b)+ (a+b) i &=& 3+7i \\\\ \text{We compare } &(1)\quad (a -b) &=& 3 \\ &(2)\quad (a+b) &=& 7 \\\\ & (a-b)+(a+b) &=& 3+7 \\ & a-b+a+b &=& 10 \\ & 2a &=& 10 \\ & \mathbf{a} & \mathbf{=} & \mathbf{5} \\\\ & (a+b)-(a-b) &=& 7-3 \\ & a+b-a+b &=& 4 \\ & 2b &=& 4 \\ & \mathbf{b} & \mathbf{=} & \mathbf{2} \\\\ & \mathbf{ z = a+bi }&\mathbf{=}& \mathbf{5+2i} \end{array}\)

The starships Enterprise and Intrepid, after refits, are undergoing flight tests.

One such test involves them flying, in opposite directions, around a large circular course.

Enterprise and Intrepid are at opposite ends of a diameter, when, after James T KIrk has said ready steady go,  Enterprise sets off in a clockwise direction and Intrepid in an anticlockwise direction. Flying at constant speeds, they first pass each other when the Enterprise has travelled 20 light years. They pass each other again when the Intrepid has travelled on a further 10 light years.

What is the circumference of the circular course ?

I. First pass each other:

\(\small{ \begin{array}{llcl} & \text{Way Enterprise } = w_1 &=& 20\ \text{Ly} \\ & \text{Way Intrepid} = w_2 \\ & \text{Speed Enterprise } = v_1 \\ & \text{Speed Intrepid} = v_2 \\ & \text{Circumference } = 2\pi r \\ \end{array}\\ \begin{array}{llcl} \text{velocity time rule }& \\ & w_1 &=& v_1\cdot t_1 \\ & w_2 &=& v_2\cdot t_1 \\ & t_1 &=& \frac{w_1}{v_1} = \frac{w_2}{v_2} \\ \hline (1)& w_1+w_2 &=& \frac{ 2\pi r }{2} = \pi r \\ & w_2 &=& \pi r - w_1 \\ (2)& w_1 &=& \frac{v_1}{v_2} \cdot w_2 \\ & w_1 &=& \frac{v_1}{v_2} \cdot (\pi r - w_1) \\ & \dfrac{v_1}{v_2} &=& \dfrac{w_1}{\pi r - w_1} \\ \end{array} }\)

II. Second pass each other:

\(\small{ \begin{array}{llcl} & \text{Way Enterprise } = W_1 \\ & \text{Way Intrepid} = W_2 &=& 10\ \text{Ly} \\ \end{array}\\ \begin{array}{llcl} \text{velocity time rule }& \\ & W_1 &=& v_1\cdot t_2 \\ & W_2 &=& v_2\cdot t_2 \\ & t_2 &=& \frac{W_1}{v_1} = \frac{W_2}{v_2} \\ \hline (3)& W_1+W_2 &=& 2\pi r \\ & W_1 &=& 2\pi r - W_2 \\ (4)& W_2 &=& \frac{v_2}{v_1} \cdot W_1 \\ & W_2 &=& \frac{v_2}{v_1} \cdot (2\pi r - W_2) \\ & \dfrac{v_1}{v_2} &=& \dfrac{2\pi r - W_2}{W_2} \\ \end{array} }\)

III. conclusion

\(\small{ \begin{array}{rcl} \dfrac{v_1}{v_2} = \dfrac{w_1}{\pi r - w_1} &=& \dfrac{2\pi r - W_2}{W_1} \\\\ \dfrac{w_1}{\pi r - w_1} &=& \dfrac{2\pi r - W_2}{W_2} \\\\ w_1\cdot W_2 &=& (2\pi r - W_2)(\pi r - w_1) \\ w_1\cdot W_2 &=& 2\pi^2 r^2 - 2\pi r\cdot w_1+ W_2\pi r+ w_1\cdot W_2\\ w_1\cdot W_2 &=& 2\pi^2 r^2 -r(2\pi w_1+W_2\pi)+ w_1\cdot W_2\\ 2\pi^2 r^2 -r(2\pi w_1+W_2\pi) &=& 0\\ 2\pi^2 r^2 &=& r(2\pi w_1+W_2\pi) \qquad | \qquad : r\\ 2\pi^2 r &=& 2\pi w_1+W_2\pi \qquad | \qquad : \pi\\ 2\pi r &=& 2 \cdot w_1+W_2 \qquad | \qquad w_1 = 20\ \text{Ly} \qquad W_2 = 10\ \text{Ly}\\ 2\pi r &=& 2 \cdot (20\ \text{Ly} )+10\ \text{Ly} \\ \mathbf{ 2\pi r } & \mathbf{=} & \mathbf{ 50\ \text{Ly} } \\ \end{array} }\)

The circumference of the circular course is 50 light years.

ax^2 +bx +c has 2 solutions, x1 and x2  x1=(-b+√Δ)/2a and x2=(−b−√Δ)2a. Is it possible that x1 + x2 = x1/x2? YES

New edit, without mistake:

\(\small{ \begin{array}{lrcll} (1) & x_1+x_2 &=& k & \qquad \rightarrow \qquad x_2 = k-x_1\\ (2) & \frac{x_1}{x_2} &=& k & \qquad \rightarrow \qquad x_1 = k \cdot x_2 \qquad \rightarrow \qquad x_1 = k \cdot (k-x_1)\\ \\ \hline \\ (2) & x_1 &=& k \cdot (k-x_1) \\ & x_1 &=& k^2 -k \cdot x_1 \\ & x_1 +k \cdot x_1 &=& k^2 \\ & x_1(1 +k ) &=& k^2 \\ & \mathbf{x_1} & \mathbf{=} & \mathbf{ \frac{k^2}{1 +k} }\\ \\ \hline \\ (1) &\frac{x_1}{x_2} &=& k \\ & x_2 &=& \frac{x_1}{k} \\ & x_2 &=& \frac{\frac{k^2}{1 +k}}{k} \\ & \mathbf{x_2} & \mathbf{=} & \mathbf{ \frac{k}{1 +k} } \\ \\ \hline \end{array} }\\ \small{ \begin{array}{rcll} (x-x_1)(x-x_2) &=& 0 \\ (x - \frac{k^2}{ 1 + k } )(x - \frac{k}{ 1 + k } ) &=& 0 \\ x^2-x\cdot( \frac{k}{ 1+k } + \frac{k^2}{ 1+k } ) + \frac{k^3}{ (1+k)^2 } &=& 0 \\ x^2-x\cdot[ \frac{k}{ 1+k }( 1+k ) ] + \frac{k^3}{ (1+k)^2 } &=& 0 \\ x^2-x\cdot k + \frac{k^3}{ (1+k)^2 } &=& 0 \qquad | \qquad \cdot(1+k)^2 \\ \underbrace{(1+k)^2 }_{=a}\cdot x^2 \underbrace{- k \cdot (1+k)^2 }_{=b}\cdot x + \underbrace{k^3}_{=c} &=& 0 \\ \\ \hline \\ \end{array} }\)

Example 1:

\(\begin{array}{rcll} a=(1+k)^2 \qquad b = -k \cdot (1+k)^2 \qquad c = k^3 \\ \end{array}\\ \begin{array}{rcll} \\ k & = & 1 \\ a & = & (1+1)^2 = 4 \\ b & = & -1\cdot (1+1)^2 = -4\\ c & = & 1^3 = 1\\\\ 4x^2-4x+1 &=& 0 \\ ax^2+bx+c &=& 0 \\ x_{1,2} &=& {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x_{1,2} &=& {4 \pm \sqrt{(-4)^2-4\cdot 4\cdot 1} \over 2\cdot 4} \\ x_{1,2} &=& {4 \pm \sqrt{0} \over 8 } \\ x_{1,2} &=& \frac{4}{8} \\ x_{1,2} &=& \frac{1}{2} \\ x_1 = \frac{1}{2} &\text{ or }& x_2 = \frac{1}{2} \\\\ x_1+x_2 &=& \frac{1}{2}+\frac{1}{2} = 1 \\ \frac{x_1}{x_2} &=& \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \end{array}\)

Example 2:

\(\begin{array}{rcll} a=(1+k)^2 \qquad b = -k \cdot (1+k)^2 \qquad c = k^3 \\ \end{array}\\ \begin{array}{rcll} \\ k & = & 2 \\ a & = & (1+2)^2 = 9 \\ b & = & -2\cdot (1+2)^2 = -18\\ c & = & 2^3 = 8\\\\ 9x^2-18x+8 &=& 0 \\ ax^2+bx+c &=& 0 \\ x_{1,2} &=& {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x_{1,2} &=& {18 \pm \sqrt{(-18)^2-4\cdot 9\cdot 8} \over 2\cdot 9} \\ x_{1,2} &=& {18 \pm \sqrt{324-288} \over 18 } \\ x_{1,2} &=& {18 \pm \sqrt{36} \over 18 } \\ x_{1,2} &=& {18 \pm 6 \over 18 } \\ x_1 = \frac{18+6}{18} &\text{ or }& x_2 = \frac{18-6}{18} \\ x_1 = \frac{24}{18} &\text{ or }& x_2 = \frac{12}{18} \\\\ x_1 = \frac{4}{3} &\text{ or }& x_2 = \frac{2}{3} \\\\ x_1+x_2 &=& \frac{4}{3}+\frac{2}{3} = \frac{6}{3} = 2 \\ \frac{x_1}{x_2} &=& \frac{\frac{4}{3}}{\frac{2}{3}} = \frac{4}{2} = 2 \end{array}\)

Suppose you use this formula to model the sunrise, where t is the time after midnight and m is the number of months after January 1st. What happens on September 1st? (Hint: September is the ninth month. Substitute the appropriate value for m and solve for t(m)). 

t(m) = 1.665 sin π/6 (m+3) + 5.485

I assume:
\(\begin{array}{lrcll} & t(m) &=& 1.665 \cdot \sin{ \left( \frac{\pi\cdot (m+3)}{6} \right) }+ 5.485 \\\\ m = 9 & t(9) &=& 1.665 \cdot \sin{ \left( \frac{\pi\cdot (9+3)}{6} \right) }+ 5.485 \\ & t(9) &=& 1.665 \cdot \sin{ \left( \frac{\pi\cdot (12)}{6} \right) }+ 5.485 \\ & t(9) &=& 1.665 \cdot \sin{ ( 2 \pi ) }+ 5.485 \qquad \sin{ ( 2 \pi )} = 0\\ & t(9) &=& 1.665 \cdot 0 + 5.485 \\ & t(9) &=& 5.485 \\\\ & t(9) &=& 5:[(5.485-5)\cdot 60]\ \mathrm{am} \\ & t(9) &=& 5:[29.1]\ \mathrm{am} \end{array}\)

B. The sun rises at about 5:29 am

ax^2 +bx +c has 2 solutions, x1 and x2  x1=(-b+√Δ)/2a and x2=(−b−√Δ)2a. Is it possible that x1 + x2 = x1/x2? YES

\(\small{ \begin{array}{lrcll} (1) & x_1+x_2 &=& k & \qquad \rightarrow \qquad x_2 = k-x_1\\ (2) & \frac{x_1}{x_2} &=& k & \qquad \rightarrow \qquad x_1 = k \cdot x_2 \qquad \rightarrow \qquad x_1 = k \cdot (k-x_1)\\ \\ \hline \\ (2) & x_1 &=& k \cdot (k-x_1) \\ & x_1 &=& k^2 -k \cdot x_1 \\ & x_1 +k \cdot x_1 &=& k^2 \\ & x_1(1 +k ) &=& k^2 \\ & \mathbf{x_1} & \mathbf{=} & \mathbf{ \frac{k^2}{1 +k} }\\ \\ \hline \\ (1) &\frac{x_1}{x_2} &=& k \\ & x_2 &=& \frac{x_1}{k} \\ & x_2 &=& \frac{\frac{k^2}{1 +k}}{k} \\ & \mathbf{x_2} & \mathbf{=} & \mathbf{ \frac{k}{1 +k} } \\ \\ \hline \end{array} }\)

\(\small{ \begin{array}{rcll} (x-x_1)(x-x_2) &=& 0 \\ (x - \frac{k^2}{ 1 + k } )(x - \frac{k}{ 1 + k } ) &=& 0 \\ x^2-x\cdot( \frac{k}{ 1+k } + \frac{k^2}{ 1+k } ) + \frac{k^3}{ (1+k)^2 } &=& 0 \\ x^2-x\cdot[ \frac{k}{ 1+k }( 1+k ) ] + \frac{k^3}{ (1+k)^2 } &=& 0 \\ x^2-x\cdot k + \frac{k^3}{ (1+k)^2 } &=& 0 \qquad | \qquad \cdot(1+k)^2 \\ \underbrace{(1+k)^2 }_{=a}\cdot x^2 \underbrace{- k \cdot (1+k)^2 }_{=b}\cdot x + \underbrace{k^3}_{=c} &=& 0 \\ \\ \hline \\ \end{array} }\)

Example 1:

\(\begin{array}{rcll} a=(1+k)^2 \qquad b = -k \cdot (1+k)^2 \qquad c = k^3 \\ \end{array}\\ \begin{array}{rcll} \\ k & = & 1 \\ a & = & (1+1)^2 = 4 \\ b & = & -1\cdot (1+1)^2 = -4\\ c & = & 1^3 = 1\\\\ 4x^2-4x+1 &=& 0 \\ ax^2+bx+c &=& 0 \\ x_{1,2} &=& {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x_{1,2} &=& {4 \pm \sqrt{(-4)^2-4\cdot 4\cdot 1} \over 2\cdot 4} \\ x_{1,2} &=& {4 \pm \sqrt{0} \over 8 } \\ x_{1,2} &=& \frac{4}{8} \\ x_{1,2} &=& \frac{1}{2} \\ x_1 &=& \frac{1}{2} \text{ or } x_1 &=& \frac{1}{2} \\\\ x_1+x_2 &=& \frac{1}{2}+\frac{1}{2} = 1 \\ \frac{x_1}{x_2} &=& \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \end{array}\)

Example 2:

\(\begin{array}{rcll} a=(1+k)^2 \qquad b = -k \cdot (1+k)^2 \qquad c = k^3 \\ \end{array}\\ \begin{array}{rcll} \\ k & = & 2 \\ a & = & (1+2)^2 = 9 \\ b & = & -2\cdot (1+2)^2 = -18\\ c & = & 2^3 = 8\\\\ 9x^2-18x+8 &=& 0 \\ ax^2+bx+c &=& 0 \\ x_{1,2} &=& {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x_{1,2} &=& {18 \pm \sqrt{(-18)^2-4\cdot 9\cdot 8} \over 2\cdot 9} \\ x_{1,2} &=& {18 \pm \sqrt{324-288} \over 18 } \\ x_{1,2} &=& {18 \pm \sqrt{36} \over 18 } \\ x_{1,2} &=& {18 \pm 6 \over 18 } \\ x_1 = \frac{18+6}{18} &\text{ or }& x_1 = \frac{18-6}{18} \\ x_1 = \frac{24}{18} &\text{ or }& x_1 = \frac{12}{18} \\\\ x_1 = \frac{4}{3} &\text{ or }& x_1 = \frac{2}{3} \\\\ x_1+x_2 &=& \frac{4}{3}+\frac{2}{3} = \frac{6}{3} = 2 \\ \frac{x_1}{x_2} &=& \frac{\frac{4}{3}}{\frac{2}{3}} = \frac{4}{2} = 2 \end{array}\)

Kannst du da nicht vielleicht Zwischenschritte beim Umformen der V(x) Formel mit angeben?

Ich versteh gar nicht wie du darauf kommst.. :(

\(\begin{array}{rcll} V = \pi\cdot r^2 \cdot h \qquad r = \frac{x}{2} \qquad h = y\\\\ V&=& \pi\cdot r^2 \cdot h \qquad & | \qquad r = \frac{x}{2} \\\\ V&=& \pi\cdot (\frac{x}{2})^2 \cdot h \\\\ V&=& \pi\cdot (\frac{x}{2})^2 \cdot h \qquad & | \qquad h = y \\\\ V&=& \pi\cdot (\frac{x}{2})^2 \cdot y \\\\ V&=& \pi\cdot (\frac{x}{2})^2 \cdot y \qquad & | \qquad y=\frac{U-2x}{2} \\\\ V(x)&=& \pi\cdot (\frac{x}{2})^2 \cdot (\frac{U-2x}{2}) \\\\ V(x)&=& \pi\cdot \frac{x^2}{2^2} \cdot (\frac{U-2x}{2}) \\\\ V(x)&=& \pi\cdot \frac{x^2}{4} \cdot (\frac{U-2x}{2}) \\\\ V(x)&=& \pi\cdot \frac{x^2}{4\cdot 2} \cdot (U-2x) \\\\ V(x)&=& \pi\cdot \frac{x^2}{8} \cdot (U-2x) \\\\ V(x)&=& \frac{\pi}{8} \cdot x^2\cdot (U-2x) \\\\ V(x)&=& \frac{\pi}{8} \cdot x^2 \cdot U - \frac{\pi}{8}\cdot x^2 \cdot(2x) \\\\ V(x)&=& \frac{\pi}{8} \cdot x^2 \cdot U - \frac{\pi}{8}\cdot(2x) \cdot x^2 \\\\ V(x)&=& \frac{\pi}{8} \cdot x^2 \cdot U - \frac{\pi}{8}\cdot 2 \cdot x^3 \\\\ \end{array} \)

26C5/67C5 =

\(\begin{array}{rcll} 26C5 &=& \frac{26!}{5!\cdot (26-5)! } \\ &=& \frac{26!}{5!\cdot 21! } \\ &=& \frac{21!\cdot 22\cdot 23\cdot 24\cdot 25\cdot 26}{5!\cdot 21! } \\ &=& \frac{22\cdot 23\cdot 24\cdot 25\cdot 26}{5! } \\\\ 67C5 &=& \frac{67!}{5!\cdot (67-5)! } \\ &=& \frac{67!}{5!\cdot 62! } \\ &=& \frac{62!\cdot 63\cdot 64\cdot 65\cdot 66\cdot 67}{5!\cdot 62! } \\ &=& \frac{63\cdot 64\cdot 65\cdot 66\cdot 67}{5! } \\\\ \frac{ 26C5 } {67C5} &=& \dfrac{ \frac{22\cdot 23\cdot 24\cdot 25\cdot 26}{5!} } { \frac{63\cdot 64\cdot 65\cdot 66\cdot 67}{5!} }\\ &=& \frac{22\cdot 23\cdot 24\cdot 25\cdot 26}{5!} \cdot \frac{5! }{63\cdot 64\cdot 65\cdot 66\cdot 67}\\ &=& \frac{22\cdot 23\cdot 24\cdot 25\cdot 26}{63\cdot 64\cdot 65\cdot 66\cdot 67}\\ &=& \frac{7893600 } { 1158917760 } \\ &=& \frac{ 115 } { 16884 } \\ &=& 0.00681118218 \end{array}\)

sqrt(50)-2sqrt(72)+3(sqrt(18)) = ?

\(\begin{array}{rcll} \sqrt{50}-2\sqrt{72}+3\sqrt{18} &=& \sqrt{50}-2\sqrt{4\cdot18}+3\sqrt{18}\\ &=& \sqrt{50}-2\cdot2\sqrt{18}+3\sqrt{18}\\ &=& \sqrt{50}-4\sqrt{18}+3\sqrt{18}\\ &=& \sqrt{50}-\sqrt{18}\\ &=& \sqrt{2\cdot5^2}-\sqrt{2\cdot3^2}\\ &=& 5\sqrt{2}-3\sqrt{2}\\ &=& 2\sqrt{2}\\ \end{array}\)

26C5/67C5 =

\(\begin{array}{rcll} \frac{ 26C5 } {67C5} &=& \frac{ \binom{26}{5} }{ \binom{67}{5} } \\\\ &=& \frac{ \frac{26}{5} \cdot \frac{25}{4}\cdot \frac{24}{3}\cdot \frac{23}{2}\cdot \frac{22}{1} }{ \frac{67}{5} \cdot \frac{66}{4}\cdot \frac{65}{3}\cdot \frac{64}{2}\cdot \frac{63}{1} } \\\\ &=& \dfrac{ 26 \cdot 25 \cdot 24 \cdot 23 \cdot 22 } { 67 \cdot 66 \cdot 65 \cdot 64 \cdot 63 } \\\\ &=& \dfrac{7893600 } { 1158917760 } \\\\ &=& \dfrac{ 115 } { 16884 } \\\\ &=& 0.00681118218 \end{array}\)