heureka

the following system of equations is shown below. 2x+3y=4 x - y = -3 which of the below is the solution to this system? A) (-13,10) B) (-1,2) C) (1,-2) D) (2,3)

\(\small{ \begin{array}{lrcll} (1) & 2x+3y &=& 4 \qquad & | \qquad -2x\\ & 3y &=& 4-2x \qquad & | \qquad :3\\ & y &=& \frac43-\frac23x \\\\ (2) & x - y &=& -3 \qquad & | \qquad \cdot (-1)\\ & -x + y &=& 3 \qquad & | \qquad +x\\ & y &=& 3+x \\ \hline \\ (2) = (1) \\ & y = 3+x &=& \frac43-\frac23x \\ & 3+x &=& \frac43-\frac23x \qquad & | \qquad +\frac23x\\ & 3+x+\frac23x &=& \frac43 \qquad & | \qquad -3\\ & x+\frac23x &=& \frac43 -3\\ & \frac33x+\frac23x &=& \frac43 -\frac33\cdot 3\\ & \frac33x+\frac23x &=& \frac43 -\frac93\\ & \frac53x &=& -\frac53 \qquad & | \qquad \cdot 3\\ & 5x &=& -5 \qquad & | \qquad :5\\ & x &=& -\frac55 \\ & \mathbf{ x } & \mathbf{=} & \mathbf{-1} \\\\ & y &=& 3+x \\ & y &=& 3-1 \\ & \mathbf{ y } & \mathbf{=} & \mathbf{2} \\\\ \end{array} }\)

B) (-1,2) is the answer

what does e mean in math?

The number e is an important mathematical constant that is the base of the natural logarithm.

It is approximately equal to  2,71828 18284 59045 23536 02874 71352 66249 77572 47093 69995 …

Sometimes called Euler's number after the Swiss mathematician Leonhard Euler.

Also like \(\pi\), e is transcendental

Determine if the two equations are parallel, perpendicular, or the same line.  If, perpendicular, at what point does the two lines intersect?

\(\small{ \begin{array}{lrclll} (1) & 2x+3y = -2 \qquad \Rightarrow \qquad y &=& -\frac23 x - \frac23 & m_1 = -\frac23 & b_1 = -\frac23 \\ (2) & x+y= 0 \qquad \Rightarrow \qquad y &=& -x & m_2 = -1 & b_2 = 0 \\ \end{array} }\)

1. Parallel ?

Parallel, if  \(\small{m_1 = m_2 \text{ and } b_1 \ne b_2}\)

We have: \(\small{m_1 \ne m_2 \qquad -\frac23 \ne -1 } \qquad \Rightarrow {\color{red}parallel~ no} \)

2. The same line ?

The same line, if  \(\small{m_1 = m_2 \text{ and } b_1 = b_2}\) 

We have:

\(\small{m_1 \ne m_2 \qquad  -\frac23 \ne -1 } \qquad \Rightarrow {\color{red}the~ same~ line~ no}\)\(\small{b_1 \ne b_2 \qquad  -\frac23 \ne 0 } \qquad \Rightarrow {\color{red}the~ same~ line~ no}\) 

3. Perpendicular ?  

Perpendicular, if  \(\small{m_1 = -\frac{1}{m_2}}\)  

We have: \(\small{m_1 \ne -\frac{1}{m_2} \qquad  -\frac23 \ne -\frac{1}{-1} } \qquad \Rightarrow {\color{red}perpendicular~ no}\)

x^2-2y^2-3=0

\(\begin{array}{rcll} x^2-2y^2-3 &=& 0 \qquad | \qquad + 3 \\ x^2-2y^2 &=& 3 \qquad | \qquad : 3 \\ \frac{x^2}{3}-\frac{2y^2}{3} &=& 1 \\ \frac{x^2}{3}-\frac{ y^2}{\frac32} &=& 1 \\ \end{array}\)

parameter notation hyperbola \(\boxed{~ \begin{array}{rcll} \frac{x^2}{3}-\frac{ y^2}{1.5} &=& 1 \\ \end{array} ~}\)

I saw in a magazine that there was a contest or something about determining the 3 last digits of 3^123456, is there anyone that could help with this? 

\(\small{ \begin{array}{rcll} 3^{123456} \pmod {1000} & \equiv & ? \\ \end{array} }\)

\(\boxed{~ \small{ \begin{array}{rcll} a^{\varphi{(m)}} &= 1 \pmod {m}, \qquad \text{ if } gcd(a,m) = 1 \end{array} } ~}\)

\(\small{a = 3}\)

\(\small{m = 1000 = 2^3 \cdot 5^3}\)

1.)  \(\small{ gcd(3,1000) = 1 }\)

2.)

 \(\small{ \begin{array}{rcll} \varphi{(1000)} &=& 1000 \cdot \left( 1-\frac{1}{2} \right)\cdot \left( 1-\frac{1}{5} \right) \\ \varphi{(1000)} &=& 1000 \cdot \left( \frac12 \right)\cdot \left( \frac{4}{5} \right) \\ \varphi{(1000)} &=& 400 \\ \end{array} }\)

3.)  \(\small{ \begin{array}{rcll} 3^{400} &= 1 \pmod {1000} \\ \end{array} }\)

4. Split 123456 in units of 400

\(\small{ \begin{array}{rcll} 123456 &=& 400 \cdot 308 + 256 \\ 3^{123456} \pmod {1000} &=& 3^{400 \cdot 308 + 256} \pmod {1000} \\ &=& 3^{400 \cdot 308} \cdot 3^{256} \pmod {1000} \\ &=& ( \underbrace{3^{400}}_{=1\pmod {1000} } )^{308} \cdot 3^{256} \pmod {1000} \\ &=& 1^{308} \cdot 3^{256} \pmod {1000} \\ &=& 1 \cdot 3^{256} \pmod {1000} \\ &=& 3^{256} \pmod {1000} \qquad | \qquad 3^{10} = 49 \pmod {1000}\\ &=& 3^{10\cdot 25+6} \pmod {1000}\\ &=& 3^{10\cdot 25} \cdot 3^6 \pmod {1000}\\ &=& (3^{10})^{25} \cdot 3^6 \pmod {1000}\\ &=& (49)^{25} \cdot 3^6 \pmod {1000}\\ \end{array} } \)

\(\small{ = 1311081016089963454886184368275274375604160521\pmod {1000}\\ = 521 }\)

whats is 2817291cm*1212km

\(\small{ \begin{array}{rcll} 2817291\ cm \cdot 1212\ km &=& 2817291\ cm \cdot \frac{1\ m}{100\ cm} \cdot \frac{1\ km}{1000\ m}\cdot 1212\ km \\ &=& 28.17291\ km \cdot 1212\ km \\ &=& 34145.56692\ km^2 \\ \end{array} }\)

what is 2*4-76+1/23

The numbers 1, 2, 3, 4, 5, 6, 7, 8, and 9 are arranged in a list so that each number is either greater than all the numbers that come before it or is less than all the numbers that come before it. For example, 4, 5, 6, 3, 2, 7, 1, 8, and 9 is one such list: notice that (for instance) the 6 is greater than all the numbers that come before it, and the 2 is less than all the numbers that come before it.

How many such lists of the numbers  are possible?

Be sure to include complete explanations with your answer, using complete sentences. Imagine you were going to show your solution to a classmate, and try to write your solution so that he or she could understand it without doing any extra 

Ich suche den Lösungsweg von arctan(1/1+x²).

Mein Ansatz war erstmal Arctan abzuleiten auf 1/1+x² und dann den Ausdruck in der Klammer mit einzusetzen so das 1/1+(1/1+x²)² raus kommt. Nur weiss ich jetzt nicht weiter, die Lösung habe ich hier doch der Rechenweg wird mir nicht begreiflich.

Dein Ansatz ist richtig!

Gesucht ist vermutlich die Ableitung von \(\arctan \left(\frac{1}{1+x^2} \right)\)
Wir substituieren \(z =  \frac{1}{1+x^2} \) und erhalten  \(y = \arctan(z)\)
Nun berechnen wir die 1. Ableitung mit der Kettenregel. 
Äußere Ableitung mal der inneren Ableitung \(y' = [\arctan(z)]' \cdot z'\)

\(\begin{array}{rcll} [\arctan(z)]' &=& \frac{1}{1+z^2} \\ &=& \frac{1}{1+ \left( \frac{1}{1+x^2} \right) ^2} \\ &=& \frac{1}{1+ \frac{1}{\left( 1+x^2\right) ^2} } \\ &=& \frac{1}{ \frac{\left( 1+x^2\right) ^2+1}{\left( 1+x^2\right) ^2} } \\ &=& \frac{ \left( 1+x^2\right) ^2} { \left( 1+x^2\right) ^2+1} \end{array}\)

\(\begin{array}{rcll} z' &=& \left[ \frac{1}{1+x^2} \right]' \\ &=& \left[ (1+x^2)^{-1} \right]' \\ &=& (-1)\cdot \left[ (1+x^2)^{-2} \right] \cdot 2x\\ &=& -\frac{ 2x }{ (1+x^2)^2 } \\ \end{array}\)

\(\begin{array}{rcll} y' &=& [\arctan(z)]' \cdot z' \\ y' &=& \frac{ \left( 1+x^2\right) ^2} { \left( 1+x^2\right) ^2+1} \cdot [ -\frac{ 2x }{ (1+x^2)^2 } ] \\ y' &=& -\frac{ 2x} { \left( 1+x^2\right) ^2+1} \\ \mathbf{y'} & \mathbf{=} & \mathbf{-\frac{ 2x} { x^4+2x^2+2} } \\ \end{array}\)

960^2=720^2+c^2−2(720)(c)*cos(20) solve for c with steps

This is a triangle ( cos - rule )

We have \(a = 960,\ b = 720,\ A = 20^{\circ}\)

1. sin - rule B?

\(\begin{array}{rcll} \frac{ \sin{(B)} } {720} &=& \frac{ \sin{ ( 20^{\circ} )} } {960} \\ \sin{(B)} &=& \frac{720}{960} \cdot \sin{(20^{\circ} ) } \\ \sin{(B)} &=& 0.75 \cdot 0.34202014333\\ \sin{(B)} &=& 0.25651510749\\ B &=& \arcsin{ (0.25651510749) } \\ B &=& 14.8633809491^{\circ} \\ \end{array}\)

2.  C?

\(\begin{array}{rcll} C &=& 180^{\circ} - (A+B) \\ C &=& 180^{\circ} - (20^{\circ}+14.8633809491^{\circ}) \\ C &=& 180^{\circ} - 34.8633809491^{\circ} \\ C &=& 145.136619051^{\circ} \\ \end{array}\)

3. sin - rule c?

\(\begin{array}{rcll} \frac{c}{ \sin{(C)} } &=& \frac{960}{ \sin{ ( 20^{\circ} )} } \\ c &=& 960 \cdot \frac{\sin{(C)}}{\sin{ ( 20^{\circ} )}} \\ c &=& 960 \cdot \frac{\sin{(145.136619051^{\circ} )}}{\sin{ ( 20^{\circ} )}} \\ c &=& 960 \cdot \frac{0.57162157869}{0.34202014333} \\ c &=& 960 \cdot 1.67130968701 \\ c &=& 1604.45729953 \\ \end{array}\)