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Find the Distance between points (-2, -9) and (4, -3) to the nearest tenth.
Show work. (help please?)

We have the points P₁ and P₂ :

$\begin{array}{|rcll|} \hline P_1 &=& (x_1, y_1) \\ &=& (4, -3) \\\\ P_2 &=& (x_2, y_2) \\ &=& (-2, -9) \\ \hline \end{array}$

Formula to find the Distance between these Points is:

$\begin{array}{|rcll|} \hline && \sqrt{(x_1-x_2)^2+ (y_1-y_2)^2} \\\\ &=& \sqrt{[4-(-2)]^2+ [-3-(-9)]^2} \\\\ &=& \sqrt{(4+2)^2+ (-3+9)^2} \\\\ &=& \sqrt{6^2+ 6^2} \\\\ &=& \sqrt{2\cdot 6^2} \\\\ &=& \sqrt{2}\cdot \sqrt{6^2} \\\\ &=& \sqrt{2}\cdot 6 \\\\ &=& 1.41421356237\cdot 6 \\\\ &=& 8.48528137424 \\ \hline \end{array}$

The distance to the nearest tenth is 8.5

How do I find out what

$\sqrt {i}$

is?

$\begin{array}{rcll} \sqrt{z} = \sqrt{i} = \ ? \qquad z = i \end{array}$

$\text{Formula}: \begin{array}{|rcll|} \hline z &=& a +i\cdot b & |z| &=& \sqrt{a^2+b^2} & |z| = \text{ magnitude of z } \\ z &=& 0 +i\cdot 1 & |z| &=& \sqrt{0^2+1^2} & \quad a = 0 \quad b= 1\\ && &&=& \sqrt{1} \\ && &&=& 1 \\ \varphi &=& \arctan(\frac{b}{a}) \\ \varphi &=& \arctan(\frac{1}{0}) \quad \Rightarrow \varphi = \frac{\pi}{2} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline r= \sqrt{z} &=&\sqrt{i} \\ &=& \sqrt{|z|} \cdot \left( \cos( \frac{\varphi}{2} ) + i\cdot \sin(\frac{\varphi}{2}) \right) \\ &=& \sqrt{1} \cdot \left( \cos( \frac{\varphi}{2} ) + i\cdot \sin(\frac{\varphi}{2}) \right) \\ &=& 1 \cdot \left( \cos( \frac{\varphi}{2} ) + i\cdot \sin(\frac{\varphi}{2}) \right) \\ &=& \cos( \frac{\frac{\pi}{2}}{2} ) + i\cdot \sin(\frac{\frac{\pi}{2}}{2}) \\ &=& \cos( \frac{\pi}{4} ) + i\cdot \sin(\frac{\pi}{4}) \\ &=& \frac{\sqrt{2}} {2} + i\cdot \frac{\sqrt{2}} {2} \\\\ \mathbf{r_1=\sqrt{i}} &\mathbf{=}& \mathbf{ \frac{\sqrt{2}} {2} + i\cdot \frac{\sqrt{2}} {2} } \\\\ r_2 &=& -r_1 \\ \mathbf{r_2=\sqrt{i}} &\mathbf{=}& \mathbf{ - \frac{\sqrt{2}} {2} - i\cdot \frac{\sqrt{2}} {2} } \\\\ \hline \end{array}$

The square root has two solutions!

Here is my formula without compute  angle $\varphi$ :

It doesn't work if $a \ge 0 \land b = 0$ .

Then you can use the normal $\sqrt{a}.$

$\begin{array}{|rcll|} \hline r_1 = \sqrt{z} = \frac{1}{ \sqrt{2} } \cdot \left( \dfrac{b}{\sqrt{|z|-a}} + i \cdot \sqrt{|z|-a} \right); \qquad r_2 = -r_1 \quad & | \quad z =a+i\cdot b \quad |z| = \sqrt{a^2+b^2}\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline z &=& i \\ z &=& 0 +i\cdot 1 & |z| = \sqrt{0^2+1^2}= 1 \qquad a = 0 \quad b= 1 \\\\ r_1 &=& \sqrt{i} = \frac{1}{ \sqrt{2} } \cdot \left( \dfrac{1}{\sqrt{1-0}} + i \cdot \sqrt{1-0} \right) \\ &=& \sqrt{i} = \frac{1}{ \sqrt{2} } \cdot \left( \dfrac{1}{1} + i \cdot 1 \right) \\ \mathbf{r_1=\sqrt{i}} &\mathbf{=}& \mathbf{ \frac{1}{ \sqrt{2} } \cdot \left( 1 + i \right) } \\\\ r_2 &=& -r_1 \\ \mathbf{r_2=\sqrt{i}} &\mathbf{=}& \mathbf{ -\frac{1}{ \sqrt{2} } \cdot \left( 1 + i \right) } \\ \hline \end{array}$

Because $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ we have the same solutions, see above.

simplify 4√32+5√12-5√50+7√75

$\begin{array}{|rcll|} \hline && 4\sqrt{32}+5\sqrt{12}-5\sqrt{50}+7\sqrt{75} \\ &=& 4\sqrt{16\cdot 2}+5\sqrt{4\cdot 3}-5\sqrt{25\cdot 2}+7\sqrt{25\cdot 3} \\ &=& 4\sqrt{16}\sqrt{2}+5\sqrt{4}\sqrt{3}-5\sqrt{25}\sqrt{2}+7\sqrt{25}\sqrt{3} \quad & | \quad \sqrt{16}= 4 \quad \sqrt{4}=2 \quad \sqrt{25} = 5 \\ &=& 4\cdot 4\cdot \sqrt{2}+5\cdot 2 \cdot \sqrt{3}-5\cdot 5 \cdot \sqrt{2}+7\cdot 5 \cdot \sqrt{3} \\ &=& 16 \sqrt{2}+10 \sqrt{3}-25\sqrt{2}+35\sqrt{3} \\ &=& 45\sqrt{3} - 9\sqrt{2} \\ \hline \end{array}$

What is 

$i^i$ ?

Formula:

$\begin{array}{|rcll|} \hline z &=& a+i\cdot b \\ \ln(z) &=& \ln(|z|) + i \cdot Arg(z) \qquad\qquad Arg(z) = \varphi \\ \ln(a+i\cdot b ) &=& \ln(\sqrt{a^2+b^2}) + i \cdot \arctan(\frac{b}{a}) \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline i^i &=& e^{i\cdot \ln(i) } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline z &=& i \\ &=& 0+i\cdot 1 \qquad & | \qquad a = 0 \qquad b = 1\\ \ln(i) = \ln( 0+i\cdot 1 ) &=& \ln(\sqrt{0^2+1^2}) + i \cdot \arctan(\frac{1}{0}) \\ &=& \ln(1) + i \cdot \arctan(\frac{1}{0}) \qquad & | \qquad \ln(1) = 0\\ &=& 0 + i \cdot \arctan(\frac{1}{0}) \qquad & | \qquad \arctan(\frac{1}{0}) = \frac{\pi}{2} \\ &=& 0 + i \cdot \frac{\pi}{2} \\ &=& i \cdot \frac{\pi}{2} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline i^i &=& e^{i\cdot \ln(i) } \\ &=& e^{i\cdot i \cdot \frac{\pi}{2} } \\ &=& e^{i^2 \cdot \frac{\pi}{2} } \quad & | \quad i^2 = -1 \\ &=& e^{-\frac{\pi}{2} } \\ &=& 0.20787957635\dots \\ \hline \end{array}$

ich hätte eine kleine Formel zum umstellen bei der ich nicht ganz durchblicke.

Ich würde mich freuen, wenn mir jemand die umgestelle Formel + Rechenweg zeigen könnte.

V = (Ce+Cp) / (Ca+Cp)

Cp=?

$\begin{array}{|rcll|} \hline V &=& \dfrac{C_e+C_p } { C_a+C_p} \quad & | \quad \cdot (C_a+C_p) \\\\ V\cdot (C_a+C_p) &=& C_e+C_p \\\\ V\cdot C_a+ V\cdot C_p &=& C_e+C_p \quad & | \quad -C_p \\\\ V\cdot C_a+ V\cdot C_p -C_p &=& C_e \quad & | \quad - V\cdot C_a \\\\ V\cdot C_p -C_p &=& C_e - V\cdot C_a \\\\ C_p\cdot ( V - 1) &=& C_e - V\cdot C_a \quad & | \quad : ( V - 1) \\\\ \mathbf{C_p} & \mathbf{=} & \mathbf{ \dfrac{ C_e - V\cdot C_a } { V - 1} } \\ \hline \end{array}$

I have having problem with this pi notation.

I saw this on Internet: $a^{2n} - b^{2n} = (a-b)[\displaystyle\prod^{n}_{c=0}(a+b)]$

I know the Pi notation means product, but the expression after the Pi notation doesn't have the variable 'c' in it,

so how does that Pi notation work??

Algebraic Identity:

$\begin{array}{|rcll|} \hline a^{n}-b^{n} &=& \displaystyle (a-b) \left[ \sum\limits_{k=0}^{n-1} a^{n-1-k}\cdot b^{k} \right] \\ \text{or} \\ a^{2n}-b^{2n} &=& \displaystyle (a-b) \left[ \sum\limits_{k=0}^{2n-1} a^{2n-1-k}\cdot b^{k} \right] \\ \hline \end{array}$

For Example:

$\begin{array}{|rcll|} \hline a^2-b^2 &=& (a-b)\cdot (a+b) \\ a^3-b^3 &=& (a-b)\cdot (a^2+ab+b^2) \\ a^4-b^4 &=& (a-b)\cdot (a^3+a^2b+ab^2+b^3) \\ \hline \end{array}$

what is a multiple of 4 between 20 and 30

$\begin{array}{|rrr|l|} \hline & & & \text{multiple of } 4 \text{ between }\color{red}{20} \color{black}\text{ and } \color{red}{30}\\ \hline 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12 \\ 13 & 14 & 15 & 16 \\ 17 & 18 & 19 & \color{red}20 \\ 21 & 22 & 23 & \color{red}24 \\ 25 & 26 & 27 & \color{red}28 \\ 29 & 30 & 31 & 32 \\ \hline \end{array}$

If i have 1$ increasing 1% every day for one year, how do i calculate it in %? 100(1+0.01)^365??

$\small{ \begin{array}{|lrcll|} \hline 1. \text{ day} \\ \quad $1 + $1 * 1\ \% \\ \quad = $1 + $1\cdot 0.01 = $1\cdot (1+ 0.01) = \mathbf{$1\cdot (1.01)} \\ \hline 2. \text{ day}\\ \quad $1\cdot (1.01) + $1\cdot (1.01) * 1\ \% \\ \quad = $1\cdot (1.01) + $1\cdot (1.01)\cdot 0.01 = $1\cdot (1.01)\cdot (1+ 0.01) = $1\cdot (1.01)\cdot (1.01) = \mathbf{$1\cdot (1.01)^2 }\\ \hline 3. \text{ day}\\ \quad $1\cdot (1.01)^2 + $1\cdot (1.01)^2 * 1\ \% \\ \quad = $1\cdot (1.01)^2 + $1\cdot (1.01)^2\cdot 0.01 = $1\cdot (1.01)^2\cdot (1+ 0.01) = $1\cdot (1.01)^2\cdot (1.01) = \mathbf{$1\cdot (1.01)^3} \\ \hline \cdots \\ \hline 365. \text{ day}\\ \quad $1\cdot (1.01)^{364} + $1\cdot (1.01)^{364} * 1\ \% \\ \quad = $1\cdot (1.01)^{364} + $1\cdot (1.01)^{364}\cdot 0.01 = $1\cdot (1.01)^{364}\cdot (1+ 0.01) = $1\cdot (1.01)^{364}\cdot (1.01) = \mathbf{$1\cdot (1.01)^{365}} \\ \hline \end{array} }$

So we have:

$\begin{array}{|rcll|} \hline $1\cdot (1.01)^{365} = $1\cdot 37.7834343329 = $37.78 \\ \hline \end{array}$

Mein Hauptwert C ist 1.810,
den ich beibehalten musst.
Dieser Wert ergibt sich durch die Division von Wert A 1.810 und Wert B 1.
Wert A wächst mit Wert B. Bedeutet, wenn ich Wert B auf 1.1 setze dann wächst Wert A auf 1.991.
Wenn ich nun Wert A 1.991 durch Wert B dividieren ergibt sich wieder Wert C von 1.810.
Wenn ich allerdings Wert B auf 0.2 setze sinkt Wert A ebenfalls auf 0.941 mit.
Dividiere ich nun Wert A 0.941 und Wert B 0.2 bekomme ich den Wert 4.705.
Meine Frage: Wie komme ich wieder auf meinen Hauptwert C von 1.810 wenn der Wert A unter 1 sinkt?

C = 1,810

$\begin{array}{|rcll|} \hline C &=& C \cdot \frac{B}{B} \qquad & \text{ oder } \qquad 1.810 = 1.810 \cdot \frac{B}{B}\\ C &=& \frac{C \cdot B}{B} \qquad & \text{ oder } \qquad 1.810 = \frac{1.810 \cdot B}{B}\\ C &=& \frac{A}{B} \qquad & \text{ oder } \qquad 1.810 = \frac{(1.810 \cdot B)}{B}\\ A &=& C \cdot B \qquad & \text{ oder } \qquad A = 1.810 \cdot B \\ \hline \end{array}$

Beispiel 1:

$\begin{array}{|rcll|} \hline B &=& 1.1 \\ A &=& 1.810 \cdot 1.1 \\\\ \frac{A}{B} &=& \frac{1.810 \cdot 1.1}{1.1} \\ &=& \frac{1.991}{1.1} \\ &=& 1.810 \\ \hline \end{array}$

Beispiel 2:

$\begin{array}{|rcll|} \hline B &=& 0.2 \\ A &=& 1.810 \cdot 0.2 \\\\ \frac{A}{B} &=& \frac{1.810 \cdot 0.2}{0.2} \\ &=& \frac{0.362}{0.2} \\ &=& 1.810 \\ \hline \end{array}$

The supply function and demand function for the sale of a certain type of DVD player are given by
S(p)=140140e^(0.005p) and D(p)=490490e^(-0.002p),
where S(p) is the number of DVD players that the company is willing to sell at price p and
D(p) is the quantity that the public is willing to buy at price p.
Find p such that D(p)=S(p).
This is called the equilibrium price.

$\begin{array}{|rcll|} \hline S(p) &=& D(p) \\ 140140\cdot e^{0.005p} &=& 490490\cdot e^{-0.002p} \quad & | \quad : 140140 \\ e^{0.005p} &=& \frac{490490}{140140} \cdot e^{-0.002p} \\ e^{0.005p} &=& 3.5 \cdot e^{-0.002p} \quad & | \quad \cdot e^{0.002p} \\ e^{0.005p}\cdot e^{0.002p} &=& 3.5 \cdot e^{-0.002p+0.002p} \\ e^{0.005p+0.002p} &=& 3.5\cdot e^0 \quad & | \quad e^0 = 1 \\ e^{0.007p} &=& 3.5 \quad & | \quad \ln() \text{ both sides } \\ \ln(~e^{0.007p}~) &=& \ln(3.5) \\ 0.007 p\cdot \ln(e) &=& \ln(3.5) \quad & | \quad \ln(e) = 1\\ 0.007\cdot p &=& \ln(3.5) \quad & | \quad : 0.007 \\ p &=& \frac{ \ln(3.5) }{ 0.007 } \\ p &=& \frac{ 1.25276296850 }{ 0.007 } \\ \mathbf{p} &\mathbf{=} & \mathbf{178.966138356} \\ \hline \end{array}$

The equilibrium price is 178.97