heureka

x/y=2 remainder 8

x=?

y=?

if x and y are  integer and y > 8 :

$\begin{array}{|rcll|} \hline n \in N \\ x &=& 8+2\cdot n \\ y &=& n \\ \hline \end{array}$

$\begin{array}{|r|l|l|r|c|} \hline n & x=8+2\cdot n & y = n & \frac{x}{y} & \text{quotient} & \text{remainder} \\ \hline 9 & x=8+2\cdot 9=26 & y = 9 & \frac{26}{9}& 2 & 8 \\ 10 & x=8+2\cdot 10=28 & y = 10 & \frac{28}{10}& 2 & 8 \\ 11 & x=8+2\cdot 11=30 & y = 11 & \frac{30}{11}& 2 & 8 \\ 12 & x=8+2\cdot 12=32 & y = 12 & \frac{32}{12}& 2 & 8 \\ 13 & x=8+2\cdot 13=34 & y = 13 & \frac{34}{13}& 2 & 8 \\ 14 & x=8+2\cdot 14=36 & y = 14 & \frac{36}{14}& 2 & 8 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ \hline \end{array}$

constant

(-5x - 8a) - (8a - 2x)

$\begin{array}{|rcll|} \hline && (-5x - 8a) - (8a - 2x) \\ &=& -5x - 8a - (8a - 2x) \\ &=& -5x - 8a + (-1)\cdot (8a - 2x) \\ &=& -5x - 8a + (-1)\cdot 8a + (-1)\cdot (-2x) \\ &=& -5x - 8a -8a + 2x \\ &=& -5x + 2x - 8a -8a \\ &=& -3x - 16a \\ \hline \end{array}$

Was ergibt 141 : ( - 94/7)?

94/7 ist ein Bruch.

$\begin{array}{|rcll|} \hline && \displaystyle \frac {141} { \frac{-94}{7} } \\\\ &=& \displaystyle \frac {141} { -\frac{94}{7} } \\\\ &=& \displaystyle -\frac {141} { \frac{94}{7} } \\\\ &=& \displaystyle -141\cdot \frac{7}{94} \\\\ &=& \displaystyle - \frac{141\cdot 7}{94} \\\\ &=& \displaystyle - \frac{987}{94} \\\\ &=& -10,5 \\\\ \mathbf{\displaystyle \frac {141} { \frac{-94}{7} }} &\mathbf{=} & \mathbf{ \displaystyle -10,5 }\\ \hline \end{array}$

Solve the following differential equation:
y'' + y = 0,

y(0)=2,

y'(0)=1.

$\begin{array}{|rcll|} \hline y(x)'' + y(x) &=& 0 \\ y(x)'' &=& -y(x) \\ \hline \end{array}$

We search a function which arises twice derived again, but with a negative algebraic sign.

There occurs very fast sin (x) or cos (x).

Example 1:

$\begin{array}{|rcll|} \hline y(x) &=& c_1 \cdot \sin(x) \\ y'(x) &=& c_1 \cdot \cos(x) \\ y''(x) &=& -c_1 \cdot \sin(x) \\\\ \Rightarrow y''(x) &=& -y(x) \\ \hline \end{array}$

Example 2:

$\begin{array}{|rcll|} \hline y(x) &=& c_2 \cdot \cos(x) \\ y'(x) &=& -c_2 \cdot \sin(x) \\ y''(x) &=& -c_2 \cdot \cos(x) \\\\ \Rightarrow y''(x) &=& -y(x) \\ \hline \end{array}$

The solution exists of a functional family and becomes unequivocal by initial conditions.

$\begin{array}{|rcll|} \hline y_{family}(x) &=& c_1 \cdot \sin(x) + c_2 \cdot \cos(x) \\ y'(x) &=& c_1 \cdot \cos(x) - c_2 \cdot \sin(x) \\ y''(x) &=& -c_1 \cdot \sin(x) - c_2 \cdot \cos(x) \\ &=& - (c_1 \cdot \sin(x) + c_2 \cdot \cos(x)) \\ &=& -y(x) \ \checkmark \\ \hline \end{array}$

computation of c₁ and c₂

$\begin{array}{|rcll|} \hline y(x) &=& c_1 \cdot \sin(x) + c_2 \cdot \cos(x) \quad & | \quad y(0) = 2 \\ 2 &=& c_1 \cdot \sin(0) + c_2 \cdot \cos(0) \\ 2 &=& 0 + c_2 \cdot 1 \\ c_2 &=& 2 \\\\ y'(x) &=& c_1 \cdot \cos(x) - c_2 \cdot \sin(x) \quad & | \quad y'(0) = 1 \\ 1 &=& c_1 \cdot \cos(0) - c_2 \cdot \sin(0) \\ 1 &=& c_1 \cdot 1 - 0 \\ c_1 &=& 1 \\ \hline \end{array}$

The function y(x):

$\begin{array}{|rcll|} \hline y(x) &=& c_1 \cdot \sin(x) + c_2 \cdot \cos(x) \quad & | \quad c_1 = 1 \quad c_2 = 2 \\\\ y(x) &=& 1 \cdot \sin(x) + 2 \cdot \cos(x) \\ \mathbf{y(x)} & \mathbf{=} & \mathbf{ \sin(x) + 2 \cdot \cos(x) } \\ \hline \end{array}$

how do you do a negative power or minus a power

$\begin{array}{|rcll|} \hline a^{-n} &=& \dfrac{1}{a^n} \\\\ \dfrac{1}{a^{-n}} &=& a^n \\\\ \left( \dfrac{a}{b} \right)^{-n} &=& \left( \dfrac{b}{a} \right)^n \\ \hline \end{array}$

a vendor sells a tool chest at $65.
She makes a 30% profit on each of the tool chests.
How much does she pay the manufacturer for each tool set?

$\begin{array}{|rcll|} \hline $ 65 &=& $65 \cdot 100\ \% \\ $ 65 &=& $65 \cdot (30\ \% + 70\ \%) \\ $ 65 &=& \underbrace{$65 \cdot 30\ \%}_{\text{profit}} + \underbrace{ $65 \cdot 70\ \% }_{\text{manufacturer}} \\ \hline \end{array}$

$\begin{array}{rcll} \text{manufacturer} &=& $65 \cdot 70\ \%\\ &=& $65 \cdot 0.7 \\ &=& $45.5 \\ \end{array}$

$\begin{array}{rcll} \text{profit} &=& $65 \cdot 30\ \%\\ &=& $65 \cdot 0.3 \\ &=& $19.5 \\ \end{array}$

$\begin{array}{|rcll|} \hline $ 65 &=& \underbrace{$19.5}_{\text{profit}} + \underbrace{ $45.5 }_{\text{manufacturer}} \\ \hline \end{array}$

A man crosses the 250-meter wide road in 75 seconds | The man speed km / h is?

(एक आदमी 250 मीटर चौडी सडक 75 सेकण्ड में पार कर लेता है | उस आदमी की गति km/h है ?)

$\begin{array}{|rcll|} \hline && \dfrac{250\ m}{75\ s} \\\\ &=& \dfrac{250\ m}{75\ s} \cdot \dfrac{1\ km}{1000\ m } \cdot \dfrac{60\ s}{ 1\ min } \cdot \dfrac{60\ min}{ 1\ h } \\\\ &=& \dfrac{250 }{75 } \cdot \dfrac{60\cdot 60}{1000} \cdot \dfrac{km}{h} \\\\ &=& \dfrac{250 }{75 } \cdot 3.6 \cdot \dfrac{km}{h} \\\\ &=& 12\ \dfrac{km}{h} \\ \hline \end{array}$

The speed of the man is $12\ \dfrac{km}{h}$

आदमी की गति है  $12\ \dfrac{km}{h}$

Find the mean of these five daily temperatures: −4°, −6°, 2°, −5°, 3°.

$\begin{array}{|rcll|} \hline &&\text{Mean } \{ -4^{\circ}, -6^{\circ}, 2^{\circ}, -5^{\circ}, 3^{\circ} \} \\\\ &=& \dfrac{-4^{\circ}+ (-6^{\circ})+ 2^{\circ}+ (-5^{\circ})+ 3^{\circ}}{5} \\ \\ &=& \dfrac{-10^{\circ}}{5} \\ \\ &=& -2^{\circ} \\ \hline \end{array}$

Hello, i have this problem
$\frac{(1-\sqrt{2})+(1-\sqrt{2}i)}{1-\sqrt{2}i}$
, how would i go at simplifying this.

$\begin{array}{|rcll|} \hline && \frac{(1-\sqrt{2})+(1-\sqrt{2}i)}{1-\sqrt{2}i} \\ &=& \frac{1-\sqrt{2}} {1-\sqrt{2}i} + \frac{1-\sqrt{2}i} {1-\sqrt{2}i} \\ &=& \frac{1-\sqrt{2}} {1-\sqrt{2}i} + 1 \\ &=& \frac{(1-\sqrt{2})} {(1-\sqrt{2}i)}\cdot \frac{(1+\sqrt{2}i)}{(1+\sqrt{2}i)} + 1 \\ &=& \frac{1\cdot (1+\sqrt{2}i)-\sqrt{2}(1+\sqrt{2}i)} {(1-\sqrt{2}i)(1+\sqrt{2}i)} +1 \\ &=& \frac{ 1+\sqrt{2}i-\sqrt{2} -\sqrt{2}\sqrt{2}i} {(1-\sqrt{2}i)(1+\sqrt{2}i)} +1 \\ &=& \frac{ 1-\sqrt{2}+\sqrt{2}i -2i} {(1-\sqrt{2}i)(1+\sqrt{2}i)} +1 \\ &=& \frac{ 1-\sqrt{2}+(\sqrt{2}-2)i } {(1-\sqrt{2}i)(1+\sqrt{2}i)} +1 \\ &=& \frac{ 1-\sqrt{2}+(\sqrt{2}-2)i } {[1^2-(\sqrt{2}i)^2]} +1 \\ &=& \frac{ 1-\sqrt{2}+(\sqrt{2}-2)i } {1 - 2i^2} +1 \quad & | \quad i^2 = -1\\ &=& \frac{ 1-\sqrt{2}+(\sqrt{2}-2)i } {1 -2\cdot (-1)} +1 \\ &=& \frac{ 1-\sqrt{2}+(\sqrt{2}-2)i } {1+2} +1 \\ &=& \frac{ 1-\sqrt{2}+(\sqrt{2}-2)i } {3} +1 \\ &=& \frac{ 1-\sqrt{2}+(\sqrt{2}-2)i } {3} +\frac33 \\ &=& \frac{ 4-\sqrt{2}+(\sqrt{2}-2)i } {3} \\ &=& \frac{ 4-\sqrt{2}+(\sqrt{2}-2)i } {3} \\ &=& \frac{ 4-\sqrt{2}}{3}+\frac{(\sqrt{2}-2) } {3}i \\ &=& 0.861928812542301650399437... - 0.195262145875634983732770... i \\ \hline \end{array}$

The graph of a transformed exponential function has the following characteristics:

horizontal asymptote at y = -9 passes through the points (-4, -8) and (-1, 18)

What are the coordinates of the x-intercept?

Formula:

Transformed Exponential Function in the Form $y = b^{x–h} + k$
h = Horizontal Shift
k = Vertical Shift

Horizontal asymptote at $y = -9, \text{ so } k = -9$

$\begin{array}{|lrcll|} \hline \text{Point 1 }~ (-4,-8): & y &=& b^{x–h} - 9 \\ & -8 &=& b^{-4–h} - 9 \\ & 1 &=& b^{-4–h} \quad & | \quad \ln() \text{ both sides}\\ & \ln(1) &=& \ln(b^{-4–h}) \\ & \ln(1) &=& (-4–h)\cdot \ln(b) \quad & | \quad \ln(1) = 0 \\ & 0 &=& (-4–h)\cdot \ln(b) \quad & | \quad : \ln(b) \\ & 0 &=& -4–h \quad & | \quad +h \\ & \mathbf{h} & \mathbf{=} & \mathbf{ -4 } \\\\ \text{Point 2 }~ (-1,18): & y &=& b^{x–h} - 9 \\ & 18 &=& b^{-1–h} - 9 \\ & 27 &=& b^{-1–h} \quad & | \quad h = -4 \\ & 27 &=& b^{-1+4} \\ & 27 &=& b^{3} \quad & | \quad 27 = 3^3 \\ & 3^3 &=& b^{3} \quad & | \quad \sqrt[3]{} \text{ both sides}\\ & 3 &=& b \\ & \mathbf{b} & \mathbf{=} & \mathbf{3} \\ \hline \end{array}$

The Transformed Exponential Function is  $y = 3^{x+4} - 9$

The coordinates of the x-intercept:

$\begin{array}{|rcll|} \hline y &=& 3^{x+4} - 9 \quad & | \quad \mathbf{y = 0} \\ 0 &=& 3^{x+4} - 9 \quad & | \quad +9 \\ 9 &=& 3^{x+4} \quad & | \quad 9 = 3^2 \\ 3^2 &=& 3^{x+4} \\ 2 &=& x+4 \quad & | \quad -4 \\ 2-4 &=& x \\ -2 &=& x \\ \mathbf{x} & \mathbf{=} & \mathbf{-2} \\ \hline \end{array}$

The coordinates of the x-intercept is (-2,0)

The graph is: