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2x=1.5(x+6)   I know it equals 18, but how? Where did I go wrong ?

2x = 1.5  * ( x + 6 )  |   * 2

4x =  3    * ( x + 6 )

4x =  3 * x   + 3 * 6

4x = 3x + 18      |    -3x

4x - 3x = 18

1x = 18

x = 18

Vater und Sohn spazieren durch Thalagu. Vater hat eine Schrittlänge von 80 cm, der Junior von 60cm.

I.   Wann treten die beiden als erstes Mal wieder gemeinsam auf ?

$$\small{\text{ Den kleinsten gemeinsamen Vielfachen (kgV) der beiden }}\\ \small{\text{ Zahlen 80 und 60 kann man mit der Primfaktorzerlegung }}\\ \small{\text{ der beiden Zahlen bestimmen. }}$$

Die Primfaktorzerlegung von 80 und 60:

$$\small{\begin{array}{l|rclclclcl}\text{Vater Schrittl}\ddot{a}\text{nge:} & 80 &=& 2^{\textcolor[rgb]{1,0,0}{4}} &*& 3^{0} &*&5^{\textcolor[rgb]{1,0,0}{1}} && \\\text{Sohn Schrittl}\ddot{a}\text{nge:}& 60 &=& 2^{2}&*& 3^{\textcolor[rgb]{1,0,0}{1}} &*& 5^{\textcolor[rgb]{1,0,0}{1}} && \\\hline \text{Den jeweils gr} \ddot{o}\ss \text{ten Exponenten }\\ \text{ von jeder Primzahl nehmen } &\text{kgV}&=& 2^{\textcolor[rgb]{1,0,0}{4}}&*&3^{\textcolor[rgb]{1,0,0}{1}} &*&5^{\textcolor[rgb]{1,0,0}{1}} && = 16*3*5=240 \end{array}}}$$

Nach 240 cm treten die beiden als erstes Mal wieder gemeinsam auf.

II.   Beim wie vielten Schritt ist das beim Vater bzw Sohn ?

  $$\\\small{\text{ Schritte Vater $=\dfrac{240}{80}=3$ }} \\\\ \small{\text{ Schritte Sohn $=\dfrac{240}{60}=4$ }} \\$$

How did you get from [b₁(a₁ - d)]*r^n-1    to     [b₁(a₁ - d)] / (1 - r)    ???

Hi CPhill,

first i tell you about der geometric sequence of beta:

$$\small{\text{ The geometric sequence is $ b_1 = \underbrace{9*(\frac{2}{3})^0}_{=9}, \ b_2 = \underbrace{9*(\frac{2}{3})^1}_{=6}, \ b_3 = \underbrace{9*(\frac{2}{3})^2}_{=4}, \ b_4 = \underbrace{9*(\frac{2}{3})^3}_{=\frac{8}{3}}, \ \dots \ ,\ b_n = 9*(\frac{2}{3})^{n-1} $ }} \\ \small{\text{ The sum $s=\frac{9}{1-\frac{2}{3}}$\ , now we set $\ b_1 = 9 $ and $\ r = \frac{2}{3} $ and see why }}\\\\ \small{\text{ b_1 = b_1, \ b_2 = b_1 r, \ b_3 = b_1r^2, \ b_4 = b_1r^3, \ \dots \ ,\ b_n = b_1r^{n-1} $ }} \\ \small{\text{ The sum is $ s = b_1 + b_1 r + b_1r^2 + b_1*r^3 + b_1r^4 + \ \dots $ }} \\ \small{\text{ $ \begin{array}{rcll} \hline s &=& b_1 \ + & b_1 r + b_1r^2 + b_1*r^3 + b_1r^4 + \ \dots \\ rs &=& & b_1 r + b_1r^2 + b_1*r^3 + b_1r^4 + \ \dots \\ \hline s-rs &=& b_1 & + 0 + 0 + 0 + 0 +\dots \\ s(1-r) &=& b_1 & \\ s &=& \dfrac{b_1}{1-r} & \end{array} $ }} \\ \boxed{ \small{\text{ So, if we have a Geometric Sequence $ b_n = b_1r^{n-1} $ the infinite sum $ s = \dfrac{b_1}{1-r} $ }} }$$

Now we look to Gamma Sequence.  There is a constant:     $$b_1(a_1-d)=g_1$$ : 

  $$\boxed{ \small{\text{ The Geometric Sequence Gamma first part is $ g_{n_{First Part}} = g_1r^{n-1} $ the infinite sum $ s = \dfrac{g_1}{1-r} $ }} }\\\\ \small{\text{ $ s = \dfrac{g_1}{1-r} =\dfrac{b_1(a_1-d)}{1-r}$ }}\\\\ \small{\text{ set $ b_1 = 9, \ a_1 = 10,\ d = -2, \ r =\frac{2}{3} $ }}\\ \small{\text{ we have the sum of Gamma Sequence first part: }}\\ \small{\text{ $ 108*(\frac{2}{3})^0 + 108*(\frac{2}{3})^1 + 108*(\frac{2}{3})^2 +108*(\frac{2}{3})^3 + \dots $ }} \\ \small{\text{ and the infinite sum $s = \frac{108}{1-\frac{2}{3}}$ and $g_{1_{first part}}=108,\ g_{2_{first part}}=108*\frac{2}{3}, \ \dots \ ,\ g_{n_{first part}}=108*(\frac{2}{3})^{n-1} $ }}\\$$

P.S.

$$\\\small{\text{ The finite sum: $ s_n=a_1 + a_1\cdot r + a_1\cdot r^2 + a_1\cdot r^3 + \dots +a_1\cdot r^{n-1} $ is $ s_n = \dfrac{a_1(1-r^n)}{1-r} $ }}\\\\ \small{\text{ The infinite sum: $ s_n=a_1 + a_1\cdot r + a_1\cdot r^2 + a_1\cdot r^3 +a_1\cdot r^4 \dots $ }}\\\\ \small{\text{ with $ s_n = \dfrac{a_1(1-r^n)}{1-r} $ is $s=\lim\limits_{n\to \infty} s_n = \dfrac{a_1}{1-r}$,}}\\\\ \small{\text{ if $\quad -1 }}$$

A cylindrical can has a radius of 3.5cm and a volume of 416cm cubed. work out the height of the can give your answer correct 2  decimal place

$$V = \pi r^2 h \\\\ h = \dfrac{V}{\pi r^2} \quad | \quad r = 3.5\ cm \quad V=416\ cm^3 \\\\ h = \dfrac{416}{\pi (3.5)^2} = 10.81 \ cm \\$$

$1200[{[1 + (.07/12) ] 180 – 1} / (.07/12)] = ?

Do you mean:

$$\small{\text{ $r = \frac{7}{12}\ \% $ and after $n = 180 $ months, the sum $ s = \$ \ 1200 * \left( \dfrac{ (1+r)^n - 1 } { r } \right) $ }}\\ \small{\text{ $ s = \$ \ 1200 * \left( \dfrac{ \left( 1+ \dfrac{0.07}{12} \right)^{180} - 1 } { \dfrac{0.07}{12} } \right) = \$ \ 1200 * \left( \dfrac{ 2.84894673087 - 1 }{ 0.00583333333 }\right) $ }}\\\\ \small{\text{ $ = \$ \ 1200 * 316.962296721 = \$ \ 380354.76 $ }}$$

(-5,-5) parallel to 3x+5y=8

Parallel line through Point (-5,-5) :  3x+5y= 3*(-5) + 5*(-5)    3x+5y= -15-25     $$\boxed{ 3x+5y= -40 }$$

limit x to infinity 2x+3/5x+7

$$\lim\limits_{x \to \infty} \dfrac{ 2x+3 }{ 5x+7 } \\\\ =\lim\limits_{x \to \infty} \dfrac{ x \left( 2+ \dfrac{3}{x} \right) }{ x \left( 5+ \dfrac{7}{x} \right) } \\ \\ =\lim\limits_{x \to \infty} \dfrac{ \left( 2+ \dfrac{3}{x} \right) }{ \left( 5+ \dfrac{7}{x} \right) } \quad | \quad \lim\limits_{x \to \infty} \left(\dfrac{3}{x} \right) = 0 \qquad \lim\limits_{x \to \infty} \left(\dfrac{7}{x} \right) = 0\\ \\\\ = \dfrac{2+0}{5+0} = \dfrac{2}{5}$$

$$\small{\text{Geometric sequence : $ a_1=a,\ a_2=a*r,\ a_3=a*r^2,\ a_4=a*r^3,\ a_5=a*r^4,\ \dots $ }}\\ \small{\text{ $ a_1=a=3 \qquad a = 3 $ }}\\ \small{\text{ $ a_4=a*r^3 = 3*r^3 =192 \qquad 3*r^3 =192 \qquad r^3 =64 \qquad r =4 $ }}\\ \small{\text{ $ a_3=a*r^2 = 3*4^2 = 3*16 = 48$ }}$$

∫ 2cos(3+π/2) dx

I.

If you mean :

$$\small{\text{ $ \begin{array}{rcl} \hline \int 2\cos{ (3+ \frac{\pi}{2} ) } \ dx \\ = 2\cos{ (3+ \frac{\pi}{2} ) } \int{dx} \\ = 2\cos{ (3+ \frac{\pi}{2} ) } x + c \\ \hline \end{array} $ }}$$

II.

If you mean :

$$\small{\text{ $ \begin{array}{rcl} \hline \int 2 \cos{ (x ) } \ dx \\ = 2 \int \cos{(x)} \ dx \\ = 2 \sin{(x)} +c\\ \hline \end{array} $ }}$$

III.

If you mean:

$$\small{\text{ $ \begin{array}{rcl} \hline \int 2 \cos{ (x + \frac{\pi}{2} ) } \ dx \\ = 2 \int \cos{ (x + \frac{\pi}{2} ) } \ dx \quad | \quad u=x+\frac{\pi}{2} \quad du=dx \\ = 2 \int \cos{(u)} \ du \\ = 2 \sin{(u)} +c\\ = 2 \sin{(x+\frac{\pi}{2} )} +c\\ \hline \end{array} $ }}$$

What are the coordinates of a point if the equation is   -3x-3y=3  and  y=-5x-17  ?

$$\begin{array}{lrcrr|} \hline (1) & y &=&\textcolor[rgb]{1,0,0}{ -5x-17 }&\\ \hline (2) & -3x-3y &=& 3 &\qquad | \quad :(-3)\quad \\ (2) & x+y &=& -1 &\\ (1) \text{ in } (2) & x+\textcolor[rgb]{1,0,0}{(-5x-17)} &=& -1 & \\ & x-5x-17 &=& - 1 & \\ & -4x-17 &=& -1 & \qquad | \quad *(-1)\quad \\ & 4x+17 &=& 1 & \\ & 4x+17 &=& 1 & \qquad | \quad -17 \quad \\ & 4x &=& 1-17 & \\ & 4x &=& -16 & \\ & 4x &=& -16 & \qquad | \quad :4 \quad \\ & x &=& -4 &\\ \hline \text{ in } (2) & -4+y &=& -1 & \qquad | \quad +4 \quad \\ & y &=& -1+4 & \\ & y &=& 3 &\\ \hline \end{array}$$

x = -4  and y = 3