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Welche Längenabmessungen hat ein Blatt DIN A4

$$\small{\text{ (1) \quad x_0y_0=1\ m^2, \quad \boxed{y_0 = \dfrac{1}{x_0}} \qquad (2) \quad \dfrac{x_0}{y_0}=\dfrac{\dfrac{y_0}{2}}{x_0}=\dfrac{y_0}{2x_0}, \quad \left(\dfrac{x_0}{y_0} \right)^2=\dfrac{1}{2}, \quad \boxed{\dfrac{x_0}{y_0}=\dfrac{1}{\sqrt{2}} } $ }}$$

$$\small{\text{ Din A0: $ \boxed{y_0 = \dfrac{1}{x_0}}, \quad \boxed{\dfrac{x_0}{y_0}=\dfrac{1}{\sqrt{2}} }, \quad \dfrac{x_0}{\dfrac{1}{x_0}}=\dfrac{1}{\sqrt{2}}, \quad x_0^2=\dfrac{1}{\sqrt{2}}, $ }}\\ \small{\text{ $ x_0 = \dfrac{1}{\sqrt{\sqrt{2}}} = 2^{-\frac{1}{4}}=0,8409\ m $ }}\\ \small{\text{ $ y_0 = \dfrac{1}{x_0} = 2^{\frac{1}{4}}=1,1892\ m $ }}$$

$$\small{\text{ Din A1: }}\\ \small{\text{ $ x_1 = \dfrac{y_0}{2} = 2^{\frac{1}{4}-1}= 2^{-\frac{3}{4}} = 0,5946\ m $ }}\\ \small{\text{ $ y_1 = x_0 = 2^{-\frac{1}{4}}=0,8409\ m $ }}$$

$$\small{\text{ Din A2: }}\\ \small{\text{ $ x_2 = \dfrac{y_1}{2} = 2^{-\frac{1}{4}-1}= 2^{-\frac{5}{4}} = 0,4204\ m $ }}\\ \small{\text{ $ y_2 = x_1 = 2^{-\frac{3}{4}}=0,5946\ m $ }}$$

$$\small{\text{ Din A3: }}\\ \small{\text{ $ x_3 = \dfrac{y_2}{2} = 2^{-\frac{3}{4}-1}= 2^{-\frac{7}{4}} = 0,2973\ m $ }}\\ \small{\text{ $ y_3 = x_2 = 2^{-\frac{5}{4}}=0,4204\ m $ }}$$

$$\small{\text{ Din A4: }}\\ \small{\text{ $ x_4 = \dfrac{y_3}{2} = 2^{-\frac{5}{4}-1}= 2^{-\frac{9}{4}} = 0,2102\ m $ }}\\ \small{\text{ $ y_4 = x_3 = 2^{-\frac{7}{4}}=0,2973\ m $ }}$$

P.S.

$$\small{\text{ F$\ddot{u}$r x und y haben wir jeweils eine geometrische Reihe: }}\\ \small{\text{ F$\ddot{u}$r x gilt $ x_n = 2^{-\frac{1}{4}}* \left( 2^{-\frac{1}{2} } \right)^n $ das kann ausgearbeitet werden zu: $ x_n = 2^{ -(\frac{2n+1}{4}) } $ }}\\ \small{\text{ F$\ddot{u}$r y gilt $ y_n = 2^{\frac{1}{4}}* \left( 2^{-\frac{1}{2} } \right)^n $ das kann ausgearbeitet werden zu: $ y_n = 2^{ -(\frac{2n-1}{4}) } $ }}$$

Hi Alan,

Tiny Pic Upload Service says:

"This IP address has been banned for violating our Terms of Use" ????????????????????

Hi Melody,

i thank you very much.  Sorry i  can't  Insert pictures. I don't know why.

bye.

1,4,10,28,?,244

$$3^n + 1 - \left\{ \ \dfrac{ 1- \left[\ 1 \mod (n+1) \right \ ] } {n+1}\ \right\}$$

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                                 x/2    |          .                +

                                          |            r                      +

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                                          |           60      .                              +

            arc     (<----r/2---->+<----r/2----> C <------------r--------->) arc

                       <-      h   -> | <-   r - h    ->                            +

                                          |                                          +

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                              x/2       |                            +

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One can turn the equilateral triangle, thus one side is vertical, then the height (h) of the curve (chord) is half the radius.

What is the side length of an equilateral triangle that fits exactly inside a circle with a 9 inch diameter ?

I.

$$\cos{(60\ensurement{^{\circ}})}=\frac{r-h}{r}=\frac{1}{2} \quad | \quad \small{\text{h = the height of the chord}} \\ \Rightarrow \boxed{h=\dfrac{r}{2}}$$

II.

Intersecting chord theorem:

$$\small{\text{ $ h*(d-h)= (\frac{x}{2})*(\frac{x}{2}) = (\frac{x}{2})^2 $ }}\\ \Rightarrow \boxed{x = d* \frac{ \sqrt{3} } {2}= 4.5* \sqrt{3}}$$

21m , 20m , 29m.

(29-21)(29+21) = 20^2

8*50 = 20*20

8*50 = 400  equal!

Yes, it is a right triangle

What is biggest 4 digit number in base 23 ?

$$\small{\text{ The biggest 4 Digit number in base 23 is: }} \\ \small{\text{ $ 22*23^3+22*23^2+22*23^1+22*23^0 = 23^4-1 = 279840 $ }} \\\\ \small{\text{ Geometric sequence: $ s_4 = \dfrac{ a * (r^4-1) } { (r-1) } \quad a = 22, \ \quad r = 23 \qquad s_4 = \frac{ 22 * (23^4-1) } { ( 23-1) } = \frac{ 22 * (23^4-1) } { ( 22 ) } = (23^4-1) $ }}$$