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geometric sequence 

$a_n=2^{5-2n}$

$\begin{array}{lcl} n=1: \qquad a_1 = 2^{5-2\cdot 1} =2^{3}=8\\ n=2: \qquad a_2 =2^{5-2\cdot 2} =2^{5-4}=2^1=2\\ n=3: \qquad a_3 = 2^{5-2\cdot 3}=2^{5-6}=2^{-1}=\frac12\\ \hline \\ r = \frac{a_3}{a_2}=\frac{\frac12}{2}=\frac14\\ r = \frac{a_2}{a_1}=\frac{2}{8}=\frac{1}{4}\\ \hline \\ n=10: \qquad a_{10}=2^{5-2\cdot 10}=2^{5-20}=2^{-15}=\frac{1}{2^1} \end{array}$

A worker is paid 0.05 on the first day, 0.10 on the second day, 0.20 on the third day and 0.40 on the fourth day, and so on. How much money in total does the work earn after working 21 days?

without mistake:

all days together:

geometric sequence

$a_1 = 0.05$

$r = 2$

$\begin{array}{rcll} s &=& a_1 \left( \frac{1-r^{21}}{1-r} \right) \\ s &=& a_1 \left( \frac{r^{21}-1}{r-1} \right) \\ s &=& 0.05 \left( \frac{2^{21}-1}{2-1} \right) \\ s &=& 0.05 \left( \frac{2^{21}-1}{1} \right) \\ s &=& 0.05 \cdot ( 2^{21} - 1) \\ s &=& 0.05 \cdot 2097151 \\ s &=& 104857.55 \end{array}$

A worker is paid 0.05 on the first day, 0.10 on the second day, 0.20 on the third day and 0.40 on the fourth day, and so on. How much money in total does the work earn after working 21 days?

$\begin{array}{llll} \text{after } 1 \text{ day }:& 0.05 \\ \text{after } 2 \text{ days }: & 0.10 &=& 0.05 \cdot 2 \\ \text{after } 3 \text{ days }: & 0.20 = 0.10\cdot 2 = 0.05 \cdot 2 \cdot 2 &=& 0.05 \cdot 2^2\\ \text{after } 4 \text{ days }: & 0.40 = 0.20\cdot 2 =0.05 \cdot 2^2 \cdot 2 \cdot 2 &=& 0.05 \cdot 2^3\\ \dots \\ \text{after } 21 \text{ days }:&&& 0.05 \cdot 2^{21-1} = 0.05 \cdot 2^{20} = 0.05\cdot 1048576 = 52428.80\\ \end{array}$

all days together:

geometric sequence

$a_1 = 0.05$

$r = 2$

$\begin{array}{rcll} s &=& a_1 \left( \frac{1-r^{21}}{1-r} \right) \\ s &=& a_1 \left( \frac{r^{21}-1}{r-1} \right) \\ s &=& 0.05 \left( \frac{2^{21}-1}{2-1} \right) \\ s &=& 0.05 \left( \frac{2^{21}-1}{1} \right) \\ s &=& 0.05 \cdot 2^{21} \\ s &=& 0.05 \cdot 2097152 \\ s &=& 104857.60 \end{array}$

Suppose you drop a ball from a window 40 meter above the ground.The ball bounces 70% of the previous height each bounce.What is the total number of meters the ball travels between the time it dropped and the tenth bounce?

geometric sequence
$a_1 = 40\ m \\ r = 70\ \% = 0.7$

$\begin{array}{rcll} s &=& a_1 \left( \frac{1-r^{11}}{1-r} \right) \\ s &=& 40 \left( \frac{1-0.7^{11}}{1-0.7} \right) \\ s &=& 40 \left( \frac{1-0.01977326743}{1-0.7} \right) \\ s &=& 40 \left( \frac{0.98022673257}{0.3} \right) \\ s &=& 40 \cdot 3.26742244190 \\ s &=& 130.696897676\ m \end{array}$

h(x)= (e^x)^(1/2)

h'(x)=?

$\begin{array}{rcll} h(x) &=& (e^x)^{\frac12} \\ h'(x) &=& \frac12 \cdot (e^x)^{(\frac12-1)} \cdot e^x \\ h'(x) &=& \frac12 \cdot (e^x)^{(\frac12-1)} \cdot (e^x)^1 \\ h'(x) &=& \frac12 \cdot (e^x)^{(\frac12-1+1)} \\ h'(x) &=& \frac12 \cdot (e^x)^{\frac12} \\ h'(x) &=& \frac12 \cdot e^{ (\frac{x}{2}) } \\ h'(x) &=& \frac { e^{ (\frac{x}{2})} }{2} \end{array}$

How do u put this in slope form

(10,33) (5,15)

ex: y=mx+b

$\text{Point 1: } ( x_1 = 10,\ y_1 = 33) \\ \text{Point 2: }(x_2 = 5,\ y_2 = 15 )$

$\begin{array}{rcll} m &=& \frac{y_2-y_1}{x_2-x_1} \\\\ m &=& \frac{15-33}{5-10} \\\\ m &=& \frac{-18}{-5} \\\\ m &=& \frac{18}{5} \\\\ \mathbf{m} &\mathbf{=}& \mathbf{3.6} \\\\ \hline \\ \frac{y-y_1}{x-x_1} &=& m\\\\ \frac{y-33}{x-10} &=& 3.6 \qquad & | \qquad \cdot (x-10) \\\\ y-33 &=& 3.6\cdot( x-10 ) \qquad & | \qquad +33 \\\\ y &=& 3.6\cdot( x-10 ) +33 \\\\ y &=& 3.6\cdot x -3.6\cdot 10 +33 \\\\ y &=& 3.6\cdot x -36 +33 \\\\ \mathbf{y} &\mathbf{=}& \mathbf{ 3.6\cdot x -3 } \end{array}$

A worker is paid 0.05 on the first day, 0.10 on the second day, 0.20 on the third day and 0.40 on the fourth day, and so on. How much money in total does the work earn after working 21 days?

$\small{ \begin{array}{llll} \text{after } 1 \text{ day }:& 0.05 \\ \text{after } 2 \text{ days }: & 0.10 &=& 0.05 \cdot 2 \\ \text{after } 3 \text{ days }: & 0.20 = 0.10\cdot 2 = 0.05 \cdot 2 \cdot 2 &=& 0.05 \cdot 2^2\\ \text{after } 4 \text{ days }: & 0.40 = 0.20\cdot 2 =0.05 \cdot 2^2 \cdot 2 \cdot 2 &=& 0.05 \cdot 2^3\\ \dots \\ \text{after } 21 \text{ days }:&&& 0.05 \cdot 2^{21-1}\\ &&& = 0.05 \cdot 2^{20} = 0.05\cdot 1048576 = 52428.80\\ \end{array} }$

He earn after 21 days 52428.80

                   x

Lim           _____   = ?

                    x2-1

x→(infinity sign)

$\begin{array}{rcll} \lim \limits_{x\to \infty} { \left( \dfrac{x } { x^2-1 } \right) } &=& \lim \limits_{x\to \infty} { \left( \dfrac{ x } { x^2-1 } \frac{:x^2}{:x^2} \right) } \\ &=& \lim \limits_{x\to \infty} { \left( \dfrac{ \frac{x}{x^2} } { \frac{x^2-1}{x^2} } \right) } \\ &=& \lim \limits_{x\to \infty} { \left( \dfrac{ \frac{x}{x^2} } { \frac{x^2}{x^2} - \frac{1}{x^2} } \right) } \\ &=& \lim \limits_{x\to \infty} { \left( \dfrac{ \frac{1}{x} } { 1 - \frac{1}{x^2} } \right) } \\ &=& \dfrac{ 0 } { 1 - 0 } \\ &=& \dfrac{ 0 } { 1 } \\ \lim \limits_{x\to \infty} { \left( \dfrac{x } { x^2-1 } \right) } &=& 0 \\ \end{array}$

You are jumping on a trampoline. Your height y (in feet) above the trampoline after t seconds can be represented by y=-16t^2+24t. How many seconds are you in the air?

$\begin{array}{rcll} y&=&-16t^2+24t \qquad y(t) = 0\\ -16t^2+24t &=& 0 \\ t(\underbrace{-16\cdot t + 24}_{=0} ) &=& 0 \\\\ -16\cdot t + 24 &=& 0 \\ -16\cdot t + 24 &=& \qquad & | \qquad -24\\ -16\cdot t &=& - 24 \qquad & | \qquad : (-16)\\ t &=& \frac{-24}{-16} \\ t &=& \frac{24}{16} \\ t &=& \frac{3}{2} \\ t &=& 1.5\ \text{seconds} \\ \end{array}$

How to find percent of discount

regular price: $42

Sale price: $36

$\begin{array}{rcll} $36 &=& $42 \cdot ( 100\ \% - x ) \qquad & | \qquad : $42\\ \frac{$36} {$42} &=& 100\ \% - x \\ \frac{36} {42} &=& 100\ \% - x \qquad & | \qquad +x\\ x + \frac{36} {42} &=& 100\ \% \qquad & | \qquad - \frac{36} {42}\\ x &=& 100\ \% - \frac{36} {42} \qquad & | \qquad 100\ \% = 1\\ x &=& 1 - \frac{36} {42} \\ x &=& 1 - \frac{18} {21} \\ x &=& \frac{21 -18} {21} \\ x &=& \frac{3} {21} \\ x &=& \frac{1} {7} \\ x &=& 0.14285714286\\ \mathbf{ x } & \mathbf{=} & \mathbf{ 14.285714286 \ \% } \end{array}$

The discount is 14.29 %