heureka

2045617563423098x9.6 = 19637928608861740.8

10,248,191,152,060,862 x2.4 = 24595658764946068.8

A bag of sweets contains mints and toffees only.
There are 21 mints in the bag.
One quarter of the sweets are toffees.
Calculate the total number of sweets in the bag.

Let x = mints

Let y = toffees

Let s = the total number of sweets in the bag.

$\begin{array}{|rcll|} \hline s &=& x + y \quad & | \quad x = 21 \\ s &=& 21 + y \quad & | \quad y = \frac14 s \\ s &=& 21 + \frac14 s \quad & | \quad \cdot 4 \\ 4s &=& 4\cdot 21 + s \quad & | \quad -s \\ 4s - s &=& 4\cdot 21 \\ 3s &=& 4\cdot 21 \quad & | \quad :3 \\ s &=& \frac{4\cdot 21}{3} \\ s &=& 4\cdot 7 \\ s &=& 28 \\ \hline \end{array}$

There are 28  sweets in the bag.

21 mints and (28-21) = 7 toffees.

Area Between Curves.

Formula:

$\begin{array}{|rcll|} \hline A = \displaystyle |~ \int \limits_{a}^{b} ~ (~f(x)-g(x)~)\ dx ~| \\ \hline \end{array}$

Ich verwende als Formel für die logistische Gleichung f(t)= S/(1+a*e^-xt)

Für S habe ich 200 gegeben, für a = 19,
für f(t) = 150 und für t=8
wie stelle ich das nun schrittweise nach x um?

$\begin{array}{|rcll|} \hline f(t)&=& \frac{S} {1+a\cdot e^{-xt}} \quad &| \quad \cdot (1+a\cdot e^{-xt}) \\ f(t)\cdot (1+a\cdot e^{-xt})&=& S \quad &| \quad : f(t) \\ 1+a\cdot e^{-xt} &=& \frac{S}{f(t)} \quad &| \quad -1 \\ a\cdot e^{-xt} &=& \frac{S}{f(t)} -1 \quad &| \quad : a \\ e^{-xt} &=& \frac{1}{a}\cdot \left(\frac{S}{f(t)} -1 \right) \quad & | \quad \text{logarithmieren auf beiden Seiten }\\ \ln( e^{-xt}) &=& \ln( \frac{1}{a}\cdot \left(\frac{S}{f(t)} -1 \right) ) \\ -xt\cdot \ln( e ) &=& \ln( \frac{1}{a}\cdot \left(\frac{S}{f(t)} -1 \right) ) \quad & | \quad \ln(e) = 1\\ -xt &=& \ln( \frac{1}{a}\cdot \left(\frac{S}{f(t)} -1 \right) ) \quad &| \quad : (-t) \\\\ \mathbf{ x }&\mathbf{ = }& \mathbf{ \frac{ \ln( \frac{1}{a}\cdot \left(\frac{S}{f(t)} -1 \right) ) } {-t} } \quad & | \quad S=200, ~ f(t)=150,~a=19,~t=8 \\\\ x &=& \frac{ \ln( \frac{1}{19}\cdot \left(\frac{200}{150} -1 \right) ) } {-8} \\ x &=& \frac{ \ln( \frac{1}{19}\cdot \left(\frac{4}{3} -1 \right) ) } {-8} \\ x &=& \frac{ \ln( \frac{1}{19}\cdot \left(\frac{4-3}{3} \right) ) } {-8} \\ x &=& \frac{ \ln( \frac{1}{19}\cdot \frac{1}{3} ) } {-8} \\ x &=& \frac{ \ln( \frac{1}{57} ) } {-8} \\ x &=& \frac{ \ln(1)-\ln(57) } {-8} \quad & | \quad \ln(1) = 0\\ x &=& \frac{ 0-\ln(57) } {-8} \\ x &=& \frac{ -\ln(57) } {-8} \\ x &=& \frac{ \ln(57) } { 8} \\ x &=& \frac{ 4.04305126783 } { 8} \\ x &=& 0.50538140848 \\ \hline \end{array}$

sqrt(5x-4)-sqrt(x)=2

$\begin{array}{|rcll|} \hline \sqrt{5x-4}-\sqrt{x} &=& 2 \quad &| \quad +\sqrt{x} \\ \sqrt{5x-4} &=& 2 +\sqrt{x} \quad &| \quad \text{square both sides} \\ 5x-4 &=& (2 +\sqrt{x})^2 \\ 5x-4 &=& 4+4\cdot \sqrt{x} + x \quad &| \quad -x \\ 4x-4 &=& 4+4\cdot \sqrt{x} \quad &| \quad :4 \\ x-1 &=& 1+ \sqrt{x} \quad &| \quad -1\\ x-2 &=& \sqrt{x} \quad &| \quad \text{square both sides} \\ (x-2)^2 &=& x\\ x^2-4x+4 &=& x \quad &| \quad -x \\ x^2-5x+4 &=& 0\\\\ x &=& \frac{5\pm \sqrt{25-4\cdot 4} } {2} \\ x &=& \frac{5\pm \sqrt{25-16} } {2} \\ x &=& \frac{5\pm \sqrt{9} } {2} \\ x &=& \frac{5\pm 3 } {2} \\\\ x_1 &=& \frac{5 + 3 } {2} \\ x_1 &=& \frac82 \\ \mathbf{x_1} & \mathbf{=} & \mathbf{4} \\\\ x_2 &=& \frac{5 - 3 } {2} \\ x_2 &=& \frac22 \\ \mathbf{x_2} & \mathbf{=} & \mathbf{1} \quad \text{ no solution}\\ \hline \end{array}$

log₄x+log₅(x-6)=2

$x\approx 8.179270752839169853762493233675433392045\dots$

The sum of the angle measures of a polygon with n sides is 1260.

Find n.

$\begin{array}{|rcll|} \hline (n-2)\cdot 180 &=& 1260 \quad &| \quad : 180 \\ n-2 &=& \frac{1260}{180} \\ n-2 &=& 7 \quad &| \quad +2 \\ \mathbf{n} & \mathbf{=} & \mathbf{9} \\ \hline \end{array}$

log(3x-9)=log(2x+6)

$\begin{array}{|rcll|} \hline \log(3x-9)&=& \log(2x+6) \\ 3x-9 &=& 2x+6 \quad &| \quad -2x \\ x -9 &=& 6 \quad &| \quad +9 \\ x &=& 15 \\ \hline \end{array}$

Am 3.12.16 23:47stand eine Frage zum Thema Pysik-Dynamik im Forum. Wo kann man sie noch mal lesen?