heureka

1/2+ 1 1/3+1 5/12  What does that equal?

$$\\{1\over 2}+ 1 {1\over 3}+1 {5\over 12}\\\\
={1\over 2}+ 1 +{1\over 3}+1 +{5\over 12}\\\\
=2+{1\over 2}+{1\over 3}+{5\over 12}\\\\
=2+
{1\over 2}
\times(
{3\over3}\times
{12\over12}
)
+
{1\over 3}
\times(
{2\over2}\times
{12\over12}
)
+
{5\over 12}
\times(
{2\over2}\times
{3\over3}
)
\\\\
=2+{5*12+5*6\over 2*3*12}\\\\
=2+{6\over6}\times{5*2+5\over 2*3*2}=2+{5*2+5\over 2*3*2}\\\\
=2+{15\over 12}=2+{3*5\over3*4}\\\\
=2+({3\over3})\times{5\over 4}=2+{5\over4}\\\\
=2+1+{1\over 4}=3+{1\over4}\\\\
=3{1\over4}$$

What are the factors of 45

$$\begin{array}{rcrcl}
1&\times&45&=&45\\
3&\times&15&=&45\\
5&\times&9&=&45\\
\left{(}
9&\times&5&=&45\\
15&\times&3&=&45\\
45&\times&1&=&45
\right{)}
\end{array}$$

The factors of 45 are: 1, 3, 5, 9, 15, 45

$$\\
\tan{(40\ensuremath{^\circ})}={20\;cm \over l-7.3\;cm}\\\\
(l-7.3\;cm)={20\;cm \over \tan{(40\ensuremath{^\circ})} }\\\\
l=7.3\;cm+{20\;cm \over \tan{(40\ensuremath{^\circ})} }\\\\
l=7.3\;cm + 23.8\;cm=31.1\;cm$$

The pendulum is 31.1 cm long

Vielen Dank.

Gruß Heureka  :-D

$$\\\boxed{FWHM=
a\times\left[
1-e^{-(B \times D)}
\right]}\\ \\
\dfrac{FWHM}{a}=1-e^{-(B\times D)} \qquad | \quad +e^{-(B \times D)}\\\\
e^{-(B \times D)}+\dfrac{FWHM}{a} =1 \qquad | \quad -\dfrac{FWHM}{a}\\\\
e^{-(B \times D)}=1-\dfrac{FWHM}{a} \qquad | \quad \ln()\\\\
\ln{(
e^{-(B \times D)}
)}
=
\ln{(
1-\dfrac{FWHM}{a})}$$

$$\\
-(B\times D) \ln{e}=\ln{(1-\dfrac{FWHM}{a})} \qquad | \quad \ln{e}=1 !\\\\
-(B \times D)=\ln{(1-\dfrac{FWHM}{a})} \qquad | \quad : -B$$

$$\\D = -\dfrac{1}{B}* \ln{\left(1-\dfrac{FWHM}{a}\right)}\\\\
D = -\dfrac{1}{B}* \ln{\left(\dfrac{a-FWHM}{a}\right)}\\\\
D = -\dfrac{1}{B}* \left(\ln{\left({a-FWHM}\right)-\ln{a}} \right)\\\\
D = -\dfrac{\ln{\left({a-FWHM}\right)-\ln{a}}}{B}\\\\
\boxed{D = \dfrac{ \ln{a}-\ln{\left({a-FWHM}\right) }}{B}}$$

(1/2)^3 * 512 - (-216)   ?

$$\\({1\over 2})^3 * 512 - (-216)\\\\
={1\over 2^3} * 2^9 +216\\\\
={2^9\over 2^3} +216\\\\
=2^{9-3} +216\\\\
=2^6 +216\\\\
=64 +216\\\\
=280$$

1/a-1/c=1/f   c=?

$$\\{1\over a} -{1\over c}= {1\over f} \quad |\quad +{1\over c}-{1\over f} \\\\
{1\over a} -{1\over f}={1\over c} \quad | \quad \leftrightarrow\\\\
{1\over c}={1\over a} -{1\over f}\\\\$$

$$\\{1\over c}={1\over a} \times{f\over f} -{1\over f}\times{a\over a }\\\\
{1\over c}={f-a\over af} \quad | \quad \updownarrow \\\\
\boxed{c ={af \over f-a } }$$

Zeile2:  11/350 10^6  AAsss/gmm ?

Hallo Roman:

$$\frac{AAsss}{gmm }=\frac{A^2s^3}{g\;m^2 }=\frac{10^3\textcolor[rgb]{1,0,0}{\not}{g}}{Kg}\frac{A^2s^3}{\textcolor[rgb]{1,0,0}{\not}{g}\;m^2 }=10^3\frac{A^2s^3}{Kg\;m^2 } = 10^3\;S$$    

  $$=\dfrac{1}{R}$$

$$\begin{array}{rcr}
\textcolor[rgb]{1,0,0}{1}\times5\times7=1*35&=&35\\
\textcolor[rgb]{1,0,0}{2}\times5\times7=2*35&=&70\\
\textcolor[rgb]{1,0,0}{3}\times5\times7=3*35&=&105\\
\textcolor[rgb]{1,0,0}{4}\times5\times7=4*35&=&140\\
\textcolor[rgb]{1,0,0}{5}\times5\times7=5*35&=&175\\
...&=&...\\
\textcolor[rgb]{1,0,0}{n}\times5\times7=n*35&\approx&2009\\
\end{array}$$

$$n\approx\dfrac{2009}{5\times7}=\dfrac{2009}{35}=57.4$$

 Between 1 and 2009, how many numbers are multiples of 5 and 7:

$$\boxed{n=57}$$

$$V_1=\dfrac{4}{3}\pi r_1^3$$

$$V_2=\dfrac{4}{3}\pi r_2^3 \quad | \quad r_2=2r_1$$

$$V_2=\dfrac{4}{3}\pi (2r_1)^3= \underbrace{\dfrac{4}{3}\pi r_1^3}_{V_1}*\underbrace{2^3}_8$$

$$V_2=8V_1=8*71\;cm^3=568\;cm^3$$