heureka

2 sinx cosx = sin x

\(\begin{array}{rcll} 2\cdot \sin{(x)}\cdot \cos{(x)} &=& \sin{(x)} \qquad & \boxed{~ \begin{array}{rcl} 2\cdot \sin{(x)}\cdot \cos{(x)} &=& \sin{(2x)} \end{array} ~}\\ \sin{(2x)}&=& \sin{(x)} \qquad & | \qquad \arcsin{()} \\ \arcsin{(\sin{(2x)})}&=& \arcsin{(\sin{(x)})} \\ 2x &=& x \qquad | \qquad - x\\ 2x-x &=& 0\\ \mathbf{x} &\mathbf{=}& \mathbf{0} \end{array}\)

aufgabe:subtrahiere von der zahl -17 die summe der zahlen 23&36.

\(-17- ( 23+36) \\ = -17 -59 \\ = -76\)

what is 1/8+7/12

\(\begin{array}{rcll} \frac{1}{8}+ \frac{7}{12} \\ &=& \frac{1}{8}\cdot \frac{3}{3}+ \frac{7}{12}\cdot \frac{2}{2} \\ &=& \frac{3}{24} + \frac{14}{24} \\ &=& \frac{3+14}{24} \\ &=& \frac{17}{24} \\ \mathbf{ \frac{1}{8}+ \frac{7}{12} } & \mathbf{=} & \mathbf{\frac{17}{24}} \\ \end{array}\)

Guten Abend Radix !

Den schönen Abend wünsche ich Dir auch!

Gruß heureka!

Given that sqrt40 = k sqrt10 Find the value of k and please explain

\(\begin{array}{rcll} \sqrt{40 } &=& k\cdot \sqrt{10} & \qquad |\qquad \text{square both sides} \\ (\sqrt{40 })^2 &=& (k\cdot \sqrt{10})^2 \\ 40 &=& k^2\cdot 10 & \qquad |\qquad :10 \\ \frac{40}{10} &=& k^2 & \\ 4 &=& k^2 & \\ k^2 &=& 4 &\qquad |\qquad \sqrt{()} \text{ both sides} \\ \sqrt{ k^2 } &=& \sqrt{4} & \\ \mathbf{k} & \mathbf{=} & \mathbf{2}& \\ \end{array}\)

(179)unten klein angeschrieben 10 = (?)2klein angeschrieben

Dezimal nach binär umwandeln.

\(\begin{array}{rclcrr} (179)_{10} &=& ( ? )_2\\ \hline & & & & & \text{Rest}\\ \hline 179 & : & 2 & = & 89 & 1\\ 89 & : & 2 & = & 44 & 1\\ 44 & : & 2 & = & 22 & 0\\ 22 & : & 2 & = & 11 & 0\\ 11 & : & 2 & = & 5 & 1\\ 5 & : & 2 & = & 2 & 1\\ 2 & : & 2 & = & 1 & 0\\ 1 & : & 2 & = & 0 & 1 \\ \hline (179)_{10} &=& ( 10110011 )_2\\ \end{array}\)

Solve for x: \(2x^2+3x-8=0\)

\(\begin{array}{rcl} \boxed{~ \begin{array}{rcl} ax^2+bx+c &=& 0\\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a}\\ \end{array} ~}\\ 2x^2+3x-8&=&0 \qquad a= 2 \quad b = 3 \quad c = -8\\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x &=& {-3 \pm \sqrt{3^2-4\cdot 2 \cdot(-8)} \over 2\cdot 2}\\ x &=& {-3 \pm \sqrt{9+ 64} \over 4}\\ x &=& {-3 \pm \sqrt{73} \over 4}\\ \hline x_1 &=& {-3 + \sqrt{73} \over 4}\\ x_1 &=&1.3860009363\\ \hline x_2 &=& {-3 - \sqrt{73} \over 4}\\ x_2 &=& -2.8860009363 \end{array}\)

A more important question in the same vein is:

On what day of the week will February 29, 3000 AD, fall??

Sorry no leap day there: \(\frac{3000}{400} = 7.5\)

The year is divisible by 4, but it is also divisible by 100, unfortunately, not also by 400.

No February 29, 3000 AD.

Leap day, 29th February, in 2004 fell on a Sunday. On what day of the week will the leap day in 2012 fall?

a. Monday

b. Friday

c. Wednesday

d. Thursday 

29th February, in 2004 ( Sunday )  \(\rightarrow\) 29th February, in 2012 ( ? )

Difference 8 Years in Days:

\(= 6\cdot 365\ \text{days} + 2 \cdot 366\ \text{days}\\ = 2922\ \text{days}\)

In weeks:

\(\frac{2922\ \text{days} } {7} = 417\ \text{weeks} \text{ and }\mathbf{ 3\ \text{days} }\)

Sunday + 3 days = Wednesday

29th February, in 2004 ( Sunday )  \(\rightarrow\) 29th February, in 2012 ( Wednesday )

c. Wednesday

What is the equation of the line that passes through the point (-2,10) and is parallel to the line y = -4x - 3?

\(\begin{array}{rcl} \boxed{~ \begin{array}{rcl} y &=& m\cdot x +b \\ \end{array} ~}\\ (1) & x_1 &=& -2 \\ & y_1 &=&10\\ \hline & y_1 &= &m \cdot x_1 + b\\ & 10 &= &m \cdot (-2) + b\\ \hline (2) & y &=& -4x-3\\ & \text{slope } m &=& -4 \qquad \text{parallel line}\\ &\mathbf{m} &\mathbf{=}& \mathbf{4}\\ \hline & 10 &= &(-4) \cdot (-2) + b\\ & 10 &= &8 + b\\ &2 &= & b\\ &\mathbf{b} &\mathbf{=}& \mathbf{2}\\ \hline \boxed{~ \begin{array}{rcl} y &=& m\cdot x +b \\ y &=& -4x + 2 \end{array} ~} \end{array}\)