heureka

28948022309329048855892746252171976963317496166410141009864396001978282409984

\(2^{254}=28948022309329048855892746252171976963317496166410141009864396001978282409984\)

whats the median

The median of a finite list of numbers can be found by arranging all the numbers from smallest to greatest.

If there is an odd number of numbers, the middle one is picked. For example, consider the set of numbers:

1, 3, 3, 6, 7, 8, 9

This set contains seven numbers. The median is the fourth of them, which is 6.

If there are an even number of observations, then there is no single middle value; the median is then usually defined to be the mean of the two middle values.] For example, in the data set:

1, 2, 3, 4, 5, 6, 8, 9

The median is the mean of the middle two numbers: this is (4 + 5) ÷ 2, which is 4.5

Was für eine Zahl ist das? 28948022309329048855892746252171976963317496166410141009864396001978282409984

n=1609:

28948022309329048855892746252171976963317496166410141009864396001978282409984

How to make numbers like 4,6 to 0,6 and 5,3 to 0,3 with a mathematical formula using only "classic" math?

4,6 mod 1 = 0,6

5,3 mod 1 = 0,3

confused

Matrix Rotation counterclockwise:

\(\begin{array}{|rcll|} \hline \begin{pmatrix} \cos(\varphi) & \sin (\varphi) \\ -\sin(\varphi) & \cos (\varphi) \\ \end{pmatrix} \stackrel{\varphi=180^{\circ}} \rightarrow \begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix} \\ \hline \end{array} \)

The point P becomes to P':

\(\begin{array}{|rcll|} \hline \binom{x}{y}\cdot \begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix} = \binom{-x}{-y} \\ \hline \end{array} \)

\(\text{Let}\ F =\binom{1}{5} \\ \text{Let}\ G =\binom{4}{-3} \)

\(\begin{array}{|rcll|} \hline F'=\binom{1}{5}\cdot \begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix} = \binom{-1}{-5} \\ G'=\binom{4}{-3}\cdot \begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix} = \binom{-4}{3} \\ \hline \end{array}\)

The answer is b.

What is 14 1/2 divided by three in fraction form

\(\begin{array}{|rcll|} \hline && \frac{14+\frac{1}{2}}{3} \\\\ &=& \frac{\frac{2\cdot14+1}{2}}{3} \\\\ &=& \frac{\frac{29}{2}}{3} \\\\ &=& \frac{29}{2\cdot 3} \\\\ &=& \frac{29}{6} \\ \hline \end{array} \)

write an equation to express the geometric sequence 10, 20, 40, 80...

\(\begin{array}{|rcll|} \hline r = \frac{20}{10} = \frac{40}{20} = \frac{80}{40} &=& 2 \\ a&=&10 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a_n &=& a\cdot r^{n-1} \\\\ a_n &=& 10\cdot 2^{n-1} \\ \hline \end{array}\)

check:

\(\begin{array}{|rcll|} \hline a_1 &=& 10\cdot 2^0 = 10 \\ a_2 &=& 10\cdot 2^1 = 20 \\ a_3 &=& 10\cdot 2^2 = 40 \\ a_4 &=& 10\cdot 2^3 = 80 \\ a_5 &=& 10\cdot 2^4 = 160 \\ \cdots \\ \hline \end{array}\)

Parallelschaltung von Widerständen:

Gleichung oder Formel für die Berechnung der Parallelschaltung zweier Widerstände R1 and R2:

\(\dfrac{1}{R_{ges}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}\)

oder

\(R_{ges} = R_1 || R_2 = \dfrac{R_1\cdot R_2}{R_1+R_2}\)

Mit der obigen Formel lassen sich die Umstellungen nach R₁ bzw. R₂ leichter vornehmen!

Wie stelle ich diese Formel nach R1 um R=(R1×R2)/(R1+R2)

\(\begin{array}{|rcll|} \hline R &=& \dfrac{R_1\cdot R_2}{R_1+R_2} \quad & |\quad \cdot (R_1+R_2) \\ R\cdot (R_1+R_2) &=& R_1\cdot R_2 \\ R\cdot R_1+ R\cdot R_2 &=& R_1\cdot R_2 \quad & |\quad - R\cdot R_1 \\ R\cdot R_2 &=& R_1\cdot R_2 - R\cdot R_1 \\ R\cdot R_2 &=& R_1\cdot (R_2 - R ) \quad & |\quad : (R_2 - R ) \\ \dfrac{R\cdot R_2}{R_2 - R} &=& R_1 \\ \hline \end{array} \)

Die Formelumstellung nach R₁ lautet: \(R_1 = \dfrac{R\cdot R_2}{R_2 - R} \)

666,26-30 Prozent =

\(\begin{array}{|rcll|} \hline && 666,26-30\ \% \\ &=& 666,26 - 666,26\cdot \frac{30}{100} \\ &=& 666,26 - 199,878 \\ &=& 466,382 \\ \hline \end{array}\)