heureka

is it f(x)=3(5/3)^-x a exponential growth function?

$\begin{array}{rcll} f(x) &=& 3 \cdot \left( \frac53 \right)^{-x} \\ f(x) &=& 3 \cdot \frac{1} { \left( \frac53 \right)^x } \\ f(x) &=& 3 \cdot \frac{1} { \frac{5^x}{3^x} } \\ f(x) &=& 3 \cdot \frac{3^x} { 5^x } \\ f(x) &=& 3 \cdot \left( \frac35 \right)^x \\ \end{array}$

No it is a decaying function, because: $\frac35 < 1$

$y=a\cdot b^x\\ \text{Example: } y = 3 \cdot \left( \frac35 \right)^x$

when a > 0 and the b is between 0 and 1, the graph will be decreasing (decaying).

In the unit circle we find by using the periphery angle:

$\tan{( \frac{\theta}{2} ) } = \frac{ \sin{( \theta ) }}{ 1+\cos{(\theta ) }}$

49.

$\begin{array}{rcll} \tan{( \frac{\theta}{2} ) } &=& \frac{ \sin{( \theta ) }}{ 1+\cos{(\theta ) }} \qquad \sin{( \theta ) } = \frac35 \qquad \cos{( \theta ) } = \frac45 \\ \tan{( \frac{\theta}{2} ) } &=& \frac{ \frac35 }{ 1+\frac45 } \\ \tan{( \frac{\theta}{2} ) } &=& \frac35 \cdot \frac{ 1 }{ 1+\frac45 } \\ \tan{( \frac{\theta}{2} ) } &=& \frac35 \cdot \frac{ 1 }{ \frac{5+4}{5} } \\ \tan{( \frac{\theta}{2} ) } &=& \frac35 \cdot \frac{ 5 }{ 9 } \\ \tan{( \frac{\theta}{2} ) } &=& \frac39 \\ \tan{( \frac{\theta}{2} ) } &=& \frac13 \\ \end{array}$

Two candles are the same height but have different cross sections.......one candle takes 4 hours to burn down completely, while the other takes 7 hours to burn down completely......

The question is...assuming that they both burn at a steady rate...if we light both at the same time, how long will it take for one candle to be twice as tall as the other???

H = height to begin.

h = height during the burning

Candle 1:

$\begin{array}{rcl} h_1 = -\frac {H} {4\ \mathrm{hours}} \cdot t + H \end{array}$ 

Candle 2:

$\begin{array}{rcl} h_2 = -\frac {H} {7\ \mathrm{hours}} \cdot t + H \end{array}$

condition:

$\begin{array}{rcll} 2\cdot h_1 &=& h_2 \\ 2\cdot \left( -\frac {H} {4\ \mathrm{hours}} \cdot t + H \right) &=& -\frac {H} {7\ \mathrm{hours}} \cdot t + H \\\\ -\frac {2\cdot H} {4\ \mathrm{hours}} \cdot t + 2\cdot H &=& -\frac {H} {7\ \mathrm{hours}} \cdot t + H \qquad &| \qquad :H \\\\ -\frac {2} {4\ \mathrm{hours}} \cdot t + 2 &=& -\frac {1} {7\ \mathrm{hours}} \cdot t + 1 \\\\ -\frac {t} {2} + 2 &=& -\frac {t} {7} + 1 \\\\ -\frac {t} {2} + 1 &=& -\frac {t} {7} \\\\ \frac {t} {2}-\frac {t} {7} &=& 1 \\\\ t\left ( \frac {1} {2}-\frac {1} {7} \right) &=& 1 \\\\ t &=& \frac{1}{ \frac {1} {2}-\frac {1} {7} } \\\\ t &=& \frac{14}{ 7-2 } \\\\ t &=& \frac{14}{ 5 } \\\\ t &=& 2.8\ \mathrm{hours} \end{array}$

$\begin{array}{rcll} h_1 &=& -\frac {H} {4\ \mathrm{hours}} \cdot 2.8\ \mathrm{hours} + H \\ h_1 &=& -0.7 H + H \\ h_1 &=& 0.3H \\\\ h_2 &=& -\frac {H} {7\ \mathrm{hours}} \cdot 2.8\ \mathrm{hours} + H\\ h_2 &=& -0.4 H + H \\ h_2 &=& 0.6H \\\\ \dfrac{h_2}{h_1} &=& \dfrac{0.6H}{0.3H}= 2 \end{array}$

Welchen Beschleunigungsweg s legt ein Pkw ungefähr zurück, wenn er aus dem Stillstand in 10 s auf 90 km/h beschleunigt wird ?

$t=10\ s\qquad v = 90\ \mathrm{ \frac{km}{h} }$

I.  Zuerst die Geschwindigkeit v von $90\ \mathrm{ \frac{km}{h} }$ in  $\mathrm{ \frac{m}{s} }$  umrechnen.

$\begin{array}{rcl} v &=& 90\ \mathrm{ \frac{km}{h} } \cdot \frac{1\ h}{60\ \mathrm{Min.}} \cdot \frac{1\ Min.}{60\ s} \cdot \frac{1000\ m}{1\ \mathrm{km}} \\\\ v &=& \frac{ 90\cdot 1000 } {60\cdot 60}\ \mathrm{ \frac{m}{s} } \\\\ v &=& \frac{ 900 } { 36 }\ \mathrm{ \frac{m}{s} } \\\\ v &=& 25\ \mathrm{ \frac{m}{s} } \end{array}$

II.  Jetzt die Beschleunigung a ausrechnen

$\begin{array}{rcl} a &=& \frac{\Delta v}{\Delta t} \\\\ a &=& \frac{ 25\ \mathrm{ \frac{m}{s} } } {10\ s} \\\\ a &=& 2,5 \ \mathrm{ \frac{m}{s^2} } \end{array}$

III.  Den Beschleunigungsweg s ausrechnen

$\begin{array}{rcl} s &=& \frac12 \cdot a \cdot t^2 \\\\ s &=& \frac12 \cdot 2,5 \ \mathrm{ \frac{m}{s^2} } \cdot 10^2\ \mathrm{s^2} \\\\ s &=& \frac{250}{2} \ \mathrm{ m } \\\\ s &=& 125 \ \mathrm{ m } \\\\ \end{array}$

Der Beschleunigungsweg beträgt 125 m.

Find the nth term of this sequence:  8 15 22 29 36

$\begin{array}{lcrrcl} a_0 &=& 1 & & &=& 1+0\cdot 7 \\ && &+7& \\ a_1 &=& 8 & & &=& 1+1\cdot 7 \\ && &+7& \\ a_2 &=& 15 & & &=& 1+2\cdot 7 \\ && &+7& \\ a_3 &=& 22 & & &=& 1+3\cdot 7 \\ && &+7& \\ a_4 &=& 29 & & &=& 1+4\cdot 7 \\ && &+7& \\ a_5 &=& 36 & & &=& 1+5\cdot 7 \\ \dots \\ \mathbf{a_n} && & & &\mathbf{=}& \mathbf{1+n\cdot 7} \\ \end{array}$

What's 2.5! ?

$\begin{array}{rcll} 2.5! &=& \Gamma(3.5) \\ &=& \Gamma(2.5+1) \\ &=& 2.5\cdot\Gamma(2.5) \\ &=& 2.5\cdot\Gamma(1.5+1) \\ &=& 2.5\cdot 1.5\cdot\Gamma(1.5) \\ &=& 2.5\cdot 1.5\cdot\Gamma(0.5+1) \\ &=& 2.5\cdot 1.5\cdot 0.5\cdot\Gamma(0.5) \qquad \Gamma(0.5) = \sqrt{\pi}\\ &=& 2.5\cdot 1.5\cdot 0.5\cdot\sqrt{\pi}\\\\ &=& 1.875\cdot\sqrt{\pi}\\ &=& 1.875\cdot 1.77245385091\\ &=& 3.32335097045\\ \mathbf{2.5!} & \mathbf{=} & \mathbf{3.32335097045} \end{array}$

How do I find the geometric mean when there is an even number of terms (eg. 7 and 8 7 5, there are 4 terms)

$\begin{array}{rcll} \text{geometric mean } &=& \sqrt[n]{ x_1\cdot x_2\cdot x_3\cdot x_4 } \qquad x_1 = 7\qquad x_2 = 8\qquad x_3 = 7\qquad x_4 = 5\\ &=& \sqrt[4]{ 7\cdot 8\cdot 7\cdot 5 } \\ &=& \sqrt[4]{ 1960 } \\ \mathbf{ \text{geometric mean } } & \mathbf{=} & \mathbf{ 6.65371229032 } \end{array}\\$

if lnx=1/x, then what is ln1/x=?

$\begin{array}{rcll} \ln{(x)} =\frac{1}{x} \qquad \ln{( \frac{1}{x} )} = \ ? \end{array}$

$\begin{array}{rcll} \ln{( \frac{1}{x} )} &=& \ln{( 1 )} -\ln{( x )} \qquad & | \qquad \ln{( 1 )} = 0\\ &=& 0 -\ln{( x )} \\ &=& -\ln{( x )} \qquad & | \qquad \ln{(x)} =\frac{1}{x}\\ &=& -\frac{1}{x} \\ \mathbf{ \ln{( \frac{1}{x} )} } & \mathbf{=} & \mathbf{ -\frac{1}{x} } \end{array}$

excursus

$\begin{array}{rcll} \ln{(x)} = \frac{1}{x} \qquad x = \ ? \end{array}$

$\begin{array}{rcll} \ln{(x)} &=& \frac{1}{x} \\ \ln{(x^x)} &=& 1 \qquad & | \qquad e^{()} \\ \mathbf{x^x} &\mathbf{=}& \mathbf{e} \\\\ e^{\ln{(x^x)}}&=& e \qquad \ln{(x^x)} = 1\\ e^{ x\cdot \ln{(x)} } &=& e\\ x\cdot \ln{(x)} &=& 1 \qquad z = \ln{(x)} \\ x\cdot z &=& 1 \qquad e^z = x \\ e^z\cdot z &=& 1 \\\\ z &=& W(1) \\ e^{\ln{(x)}} &=& e^{W(1)}\\ x &=& e^{W(1)} \qquad \text{or}\\\\ x\cdot z &=& 1 \\ x &=& \frac{1}{z} \qquad z = W(1) \\ x &=& \frac{1}{W(1)}\\\\ \end{array}\\ \small{ W(1) = 0.5671432904097838729999686622103555497538157871865125081351310792230457930866\dots\\ x = \frac{1}{W(1)} = 1.7632228343518967102252017769517070804360179866674736\dots\\\\ W(1) = 0.5671432904\dots \text{ is called the omega constant }\\ \text{where } W(x) \text{ is the Lambert W-function } }$

Given a triangle with a=11, b=14, and $\mathbf{\alpha = 15^{\circ}}$, what is (are) the possible length(s) of c?

$\small{ \begin{array}{rcll} a^2 &=& b^2+c^2-2bc\cdot \cos{(\alpha)} \\ c^2-2b\cdot \cos{(\alpha)}\cdot c +b^2-a^2 &=& 0 \\\\ \boxed{~ \begin{array}{rcll} Ac^2+Bc+C = 0 \\ c = {-B \pm \sqrt{B^2-4AC} \over 2A} \end{array} ~}\\\\ c^2-2b\cdot \cos{(\alpha)}\cdot c +b^2-a^2 &=& 0 \qquad A=1 \qquad B = -2b\cdot \cos{(\alpha)} \qquad C = b^2-a^2 \\ c &=& {2b\cdot \cos{(\alpha)} \pm \sqrt{(2b\cdot \cos{(\alpha)})^2-4(b^2-a^2)} \over 2} \\ c &=& b\cdot \cos{(\alpha)} \pm \sqrt{ [ b\cdot \cos{(\alpha)} ]^2-(b^2-a^2)} \\ c &=& b\cdot \cos{(\alpha)} \pm \sqrt{b^2\cdot[ \cos^2{(\alpha)} -1] + a^2 } \\ c &=& b\cdot \cos{(\alpha)} \pm \sqrt{b^2\cdot[ 1-\sin^2{(\alpha)} -1] + a^2 } \\ c &=& b\cdot \cos{(\alpha)} \pm \sqrt{ a^2 - b^2\cdot \sin^2{(\alpha)} } \quad a=11 \quad b=14 \quad \alpha = 15^{\circ}\\ c &=& 14\cdot \cos{(15^{\circ})} \pm \sqrt{ 11^2 - 14^2\cdot \sin^2{(15^{\circ})} } \\ c &=& 14\cdot 0.96592582629 \pm \sqrt{ 107.870489571 } \\ c &=& 13.5229615680 \pm 10.3860719028\\\\ c_1 &=& 13.5229615680 + 10.3860719028\\ c_1 &=& 23.9090334709\\\\ c_2 &=& 13.5229615680 - 10.3860719028\\ c_2 &=& 3.13688966521 \end{array} }$

check:

$\begin{array}{rcll} c_1\cdot c_2 &=& (b-a)(b-a+2a)\\ c_1\cdot c_2 &=& (b-a)(b+a)\\ c_1\cdot c_2 &=& b^2-a^2 \quad a=11 \quad b=14 \\ c_1\cdot c_2 &=& 14^2-11^2 \\ c_1\cdot c_2 &=& 75 \\\\ c_1\cdot c_2 &=& 23.9090334709\cdot 3.13688966521 \\ c_1\cdot c_2 &=& 75\quad \text{ Okay}\\ \end{array}$