heureka

Given g(x)= 18^x

First Question: Find the derivative of  $g(x) = 18^x$

$\begin{array}{|rcll|} \hline g(x) &=& 18^x \quad & | \quad \ln() \text{ both sides } \\ \ln(~g(x)~) &=& \ln(18^x) \quad & | \quad \text{Formula: } ln(a^b) = b\cdot ln(a)\\ \ln(~g(x)~) &=& x\cdot \ln(18) \quad & | \quad \text{derivate both sides } \text{Formula: } (~ln(~g(x)~)~)' = \frac{g'(x)}{g(x)} \\ \frac{g'(x)}{g(x)} &=& ln(18) \quad & | \quad \cdot g(x) \\ g'(x) &=& g(x) \cdot ln(18) \quad & | \quad g(x) = 18^x \\\\ \mathbf{g'(x)} & \mathbf{=} & \mathbf{18^x\cdot ln(18) } \\ \hline \end{array}$

The same answer.

Using the same formula: $\sin(\varphi) = 2\cdot \sin{ \frac{\varphi}{2} } \cdot \cos{ \frac{\varphi}{2} }$

Given $\sin ( \frac{\pi } {10}) = \frac{( -1 + \sqrt{ 5}) }{4}$

Find $\sin( \frac{\pi } {5})$

Formula:
$\begin{array}{|rcll|} \hline \sin(\varphi) &=& 2\cdot \sin{ \frac{\varphi}{2} } \cdot \cos{ \frac{\varphi}{2} } \\ \cos{\frac{\varphi}{2}} &=& \sqrt{1-\sin^2{\frac{\varphi}{2}}} \\\\ \sin(\varphi) &=& 2\cdot \sin{ \frac{\varphi}{2} } \cdot \sqrt{1-\sin^2{\frac{\varphi}{2}}} \\ \hline \end{array}$

We set:

$\begin{array}{|rcll|} \hline \sin{\varphi} &=& \sin( \frac{\pi } {5})\\\\ \sin{\frac{ \varphi } {2} } &=& \sin( \frac{\pi } {10})\\ &=& \frac{ \sqrt{ 5}-1 }{4} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \sin(\varphi) &=& 2\cdot \sin{ \frac{\varphi}{2} } \cdot \sqrt{1-\sin^2{\frac{\varphi}{2}}} \\ \sin( \frac{\pi } {5}) &=& 2\cdot \frac{ \sqrt{ 5}-1 }{4} \cdot \sqrt{1- \left(\frac{ \sqrt{ 5}-1 }{4}\right)^2 } \\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{1- \left(\frac{ \sqrt{ 5}-1 }{4}\right)^2 } \\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{16- (\sqrt{5}-1)^2 }{16} }\\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{16- (5-2\sqrt{5}+1) }{16} }\\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{16- 5+2\sqrt{5}-1) }{16} }\\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{10+2\sqrt{5} }{16} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ (\sqrt{5}-1)^2 }{4} \cdot \frac{(10+2\sqrt{5}) }{16} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ (\sqrt{5}-1)^2\cdot (10+2\sqrt{5}) } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ (5-2\sqrt{5}+1)\cdot (10+2\sqrt{5}) } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ 60+12\sqrt{5}-20\sqrt{5}-4\cdot 5 } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ 40-8\sqrt{5} } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ 8\cdot 5-8\sqrt{5} } {8\cdot 8} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{\frac{8\cdot 5}{8\cdot 8} -\frac{8\sqrt{5}}{8\cdot 8} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{\frac{5}{8} -\frac{\sqrt{5}}{8} }\\ \hline \end{array}$

Thank you all very much for your help guys! But I must solve it using the half- angle formulas. Sorry I forgot to mention that, I was in a rush. Also with correct brackets it is sin (pi/5)

Also the problem says that given $\sin ( \frac{\pi } {10}) = \frac{( -1 + \sqrt{ 5}) }{4}$, find $\sin ( \frac{\pi } {5})$

Formula:

$\begin{array}{|rcll|} \hline \sin(2\varphi) &=& 2\cdot \sin{\varphi} \cdot \cos{\varphi} \\ \cos{\varphi} &=& \sqrt{1-\sin^2{\varphi}} \\\\ \sin(2\varphi) &=& 2\cdot \sin{\varphi} \cdot \sqrt{1-\sin^2{\varphi}} \\ \hline \end{array}$

We set:

$\begin{array}{|rcll|} \hline \sin{2\varphi} &=& \sin( \frac{\pi } {5})\\\\ \sin{\varphi} &=& \sin( \frac{\pi } {10})\\ &=& \frac{ \sqrt{ 5}-1 }{4} \\ \hline \end{array}$

We have:

$\begin{array}{|rcll|} \hline \sin(2\varphi) &=& 2\cdot \sin{\varphi} \cdot \sqrt{1-\sin^2{\varphi}} \\ \sin( \frac{\pi } {5}) &=& 2\cdot \frac{ \sqrt{ 5}-1 }{4} \cdot \sqrt{1- \left(\frac{ \sqrt{ 5}-1 }{4}\right)^2 } \\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{1- \left(\frac{ \sqrt{ 5}-1 }{4}\right)^2 } \\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{16- (\sqrt{5}-1)^2 }{16} }\\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{16- (5-2\sqrt{5}+1) }{16} }\\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{16- 5+2\sqrt{5}-1) }{16} }\\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{10+2\sqrt{5} }{16} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ (\sqrt{5}-1)^2 }{4} \cdot \frac{(10+2\sqrt{5}) }{16} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ (\sqrt{5}-1)^2\cdot (10+2\sqrt{5}) } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ (5-2\sqrt{5}+1)\cdot (10+2\sqrt{5}) } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ 60+12\sqrt{5}-20\sqrt{5}-4\cdot 5 } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ 40-8\sqrt{5} } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ 8\cdot 5-8\sqrt{5} } {8\cdot 8} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{\frac{8\cdot 5}{8\cdot 8} -\frac{8\sqrt{5}}{8\cdot 8} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{\frac{5}{8} -\frac{\sqrt{5}}{8} }\\ \hline \end{array}$

2^x * 5^y = 100

2^y * 5^x = 1000

x+y = ?

$\begin{array}{rcll} 2^x \cdot 5^y &=& 100 \\ 2^x \cdot 5^y &=& 10^2 \\\\ \hline \\ 2^y \cdot 5^x &=& 1000 \\ 2^y \cdot 5^x &=& 10^3 \\\\ \hline \\ (2^x \cdot 5^y)\cdot (2^y \cdot 5^x) &=& 10^2 \cdot 10^3 \\ 2^x \cdot 2^y \cdot 5^y \cdot 5^x &=& 10^{2+3} \\ 2^x \cdot 2^y \cdot 5^y \cdot 5^x &=& 10^{5} \\ 2^{x+y} \cdot 5^{y+x} &=& 10^{5} \\ (2\cdot 5)^{x+y} &=& 10^{5} \\ 10^{x+y} &=& 10^{5} \\\\ \mathbf{x+y} & \mathbf{=} & \mathbf{ 5 } \\ \end{array}$

what is the exact value of sin (pi/5) ?

Formula:

$\begin{array}{|rcll|} \hline \sin(5\varphi) = 16 \sin^5(\varphi) - 20 \sin^3(\varphi) + 5\sin(\varphi) \\ \hline \end{array}$

Ansatz:

$\small{ \begin{array}{|lrcl|} \hline (1): & \sin(360^{\circ}) = 0 = \sin(5\cdot 72^{\circ}) = 16 \sin^5(72^{\circ}) - 20 \sin^3(72^{\circ}) + 5\sin(72^{\circ})\\ & 16 \sin^5(72^{\circ}) - 20 \sin^3(72^{\circ}) + 5\sin(72^{\circ}) &=& 0 \quad & | \quad : \sin(72^{\circ}) \\ & 16 \sin^4(72^{\circ}) - 20 \sin^2(72^{\circ}) + 5 &=& 0 \quad &| \quad x = \sin^2(72^{\circ}) \\ & 16 x^2 - 20 x + 5 &=& 0 \\\\ (2): & \sin(180^{\circ}) = 0 = \sin(5\cdot 36^{\circ}) = 16 \sin^5(36^{\circ}) - 20 \sin^3(36^{\circ}) + 5\sin(36^{\circ}) \\ & 16 \sin^5(36^{\circ}) - 20 \sin^3(36^{\circ}) + 5\sin(36^{\circ}) &=& 0 \quad & | \quad : \sin(36^{\circ}) \\ & 16 \sin^4(36^{\circ}) - 20 \sin^2(36^{\circ}) + 5 &=& 0 \quad &| \quad x = \sin^2(36^{\circ}) \\ & 16 x^2 - 20 x + 5 &=& 0 \\ \hline \end{array} }$

The quadratic polynomial $16 x^2 - 20 x + 5$ has following roots: $<~ \sin^2(72^{\circ}),\ \sin^2(36^{\circ}) ~>$

The general quadratic equation is: ${\displaystyle ax^{2}+bx+c=0.}$

The quadratic formula is: ${\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.}$

$\begin{array}{|rcll|} \hline x &=& \displaystyle { \frac {20\pm {\sqrt {20^{2}-4\cdot 16 \cdot 5}}}{2\cdot 16}} \\\\ &=& \frac {20\pm {\sqrt { 400 - 320 }}} {32} \\\\ &=& \frac {20\pm {\sqrt { 80 }}} {32} \\\\ &=& \frac {20\pm {\sqrt { 16\cdot 5 }}} {32} \\\\ &=& \frac {20\pm 4 {\sqrt { 5 } }} {32} \\\\ &=& \frac {5\pm {\sqrt { 5 } }} {8} \\\\ \sqrt{x} &=& \sqrt{ \frac {5\pm {\sqrt { 5 } }} {8} } \\ \hline \end{array}$

Because $\sin(72^{\circ}) > \sin(36^{\circ})$
we have:

$\begin{array}{|rcll|} \hline \sin(72^{\circ}) = \sqrt{ \frac {5 + {\sqrt { 5 } }} {8} } \\ \sin(36^{\circ}) = \sqrt{ \frac {5 - {\sqrt { 5 } }} {8} } \\ \hline \end{array}$

x      y
5      8
9    10

11  13
17  19

Use the Paired Data Set.  
Find the equation for the least-squares regression line

$\begin{array}{|c|r|r|r|r|} \hline & x & y & x\cdot y & x\cdot x \\ \hline 1 & 5 & 8 & 40 & 25 \\ \hline 2 & 9 & 10 & 90 & 81 \\ \hline 3 & 11 & 13 & 143 & 121 \\ \hline 4 & 17 & 19 & 323 & 289 \\ \hline \text{sum} & 42 & 50 & 596 & 516 \\ \hline \end{array}$

We find for the values:

$\begin{array}{lcr} \hline N &=& 4\\ \sum{X} &=& 42 \\ \sum{Y} &=& 50 \\ \sum{XY} &=& 596 \\ \sum{X^2} &=& 516 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \text{Slope(b)} &=& \displaystyle \frac{N\cdot \sum{XY} - (\sum{X})\cdot (\sum{Y}) } { N\cdot \sum{X^2} - (\sum{X})^2 } \\\\ &=& \displaystyle \frac{4\cdot 596 - (42)\cdot (50) } { 4\cdot 516 - (42)^2 } \\\\ &=& \frac{2384 - 2100 } { 2064 - 1764 } \\\\ &=& \frac{ 284 } { 300 } \\\\ &=& 0.94666666667 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \text{Intercept(a)} &=& \displaystyle \frac{ \sum{Y} - b\cdot (\sum{X}) } { N } \\\\ &=& \displaystyle \frac{ 50 - 0.94666666667\cdot (42) } { 4 } \\\\ &=& \frac{ 50 - 0.94666666667\cdot (42) } { 4 } \\\\ &=& \frac{ 50 - 39.76 } { 4 } \\\\ &=& \frac{ 10.24 } { 4 } \\\\ &=& 2.56 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \text{Regression Equation(y)} &=& \displaystyle a + bx \\\\ y &=& 2.56 + 0.94\overline{6}\cdot x \\ \hline \end{array}$

5x-y+4z=14

-5x-y-2z=-10

-5x+3y-6z=-18

Can I have help solving this by linear combination?

5x-y+4z=14

-5x-y-2z=-10

-5x+3y-6z=-18

Solving this by linear combination?

$\begin{array}{|lrcll|} \hline (1): & 5x - y + 4z & = & 14 \\ (2): & -5x - y - 2z & = & -10 \\ (3): & -5x +3y - 6z & = & -18 \\\\ \hline \\ (1) + (2): & 5x - y + 4z -5x - y - 2z &=& 14 - 10 \\ & - 2y + 2z &=& 4 \quad & | \quad :2 \\ & - y + z &=& 2 \\ (I): & z &=& 2 + y \\\\ (1) + (3): & 5x - y + 4z -5x +3y - 6z &=& 14 - 18 \\ & 2y - 2z &=& -4 \quad & | \quad : (-2) \\ & -y + z &=& 2 \\ (II): & z &=& 2 + y \quad & | \quad \mathbf{(II)= (I)\ !!!} \\\\ \hline \\ (1): & 5x - y + 4z & = & 14 \quad & | \quad z=2+y \\ & 5x - y + 4\cdot(2+y) & = & 14 \\ (III): & 5x +3y & = & 6 \\\\ (2): & -5x - y - 2z & = & -10 \quad & | \quad z=2+y \\ & -5x - y - 2\cdot(2+y) & = & -10 \\ & -5x -3y & = & -6 \quad & | \quad \cdot(-1) \\ (IV): & 5x +3y & = & 6 \quad & | \quad \mathbf{(IV)= (III)\ !!!} \\\\ \hline \end{array}$

Solution:

$\text{Let } t \in R \text{ and we set } z = t$

$\begin{array}{|lrcll|} \hline (I): & z &=& 2 + y \quad & | \quad z=t \\ & t &=& 2 + y \\ & y &=& t-2 \\\\ (III): & 5x +3y & = & 6 \quad & | \quad y = t-2 \\ & 5x +3\cdot (t-2) & = & 6 \\ & 5x +3t - 6 & = & 6 \\ & 5x +3t & = & 12 \\ & 5x & = & 12 - 3t \\ & 5x & = & 12 - 3t \quad & | \quad : 5 \\ & x & = & \frac{12}{5} - \frac{3}{5}t \\ \hline \end{array}$

summary:
$\begin{array}{|rcll|} \hline t \in R \\ x &=& \frac{12}{5} - \frac{3}{5}t \\ y &=& t-2\\ z &=& t \\ \hline \end{array}$

The moon takes 328 days to orbit earth 12 times.

How many times will it orbit earth in 246 days

$\begin{array}{|rcll|} \hline \dfrac{12\ \text{Orbits} }{328\ \text{days}} \cdot 246\ \text{days} = 9 \ \text{Orbits} \\ \hline \end{array}$

the segment from (1,3) to (-2,1) is extended twice its own length

find the terminal point

$\begin{array}{|rcll|} \hline &&\displaystyle 3 \cdot \left[ \binom{-2}{1} - \binom{1}{3} \right] + \binom{1}{3} \\\\ &=& \displaystyle 3 \cdot \binom{-2-1}{1-3} + \binom{1}{3} \\\\ &=& \displaystyle 3 \cdot \binom{-3}{-2} + \binom{1}{3} \\\\ &=&\displaystyle \binom{-9}{-6} + \binom{1}{3} \\\\ &=&\displaystyle \binom{-9+1}{-6+3} \\\\ &=& \displaystyle \binom{-8}{-3} \\\\ \hline \end{array}$