heureka

what is c in the formula c^2 = b^2 +a^2 if a=6 and b=10 ?

$$\small{\text{$
\begin{array}{rcl}
c^2 &=& b^2 + a^2 \quad | \quad \mathbf{b = 10} \rm{~~and~~} \mathbf{a = 6} \\
c^2 &=& 10^2 + 6^2 \\
c^2 &=& 100 + 36 \\
c^2 &=& 136 \quad | \quad \pm\sqrt{} \\
c &=& \pm\sqrt{136} \\
c_1 &=& +\sqrt{136} = 11.6619037897\\
c_2 &=& -\sqrt{136} = -11.6619037897\\
\end{array}
$}}$$

Wie berechne ich den Winkel Alpha bei Tangens? Sprich ich habe tan(alpha) = 0,06 (da in dieser Aufgabe die Neigung maximal 60% betragen darf). Dazu verwende ich doch den arctan?

Neigung maximal 60% :

$$\small{\text{$
\mathbf{60~\% = \dfrac{60}{100}}
$}}\\\\
\small{\text{$
\tan{ ( \alpha \ensurement{^{\circ}} ) } = \dfrac{60}{100}= \dfrac{6}{10}=0,6
$}}\\\\
\small{\text{$
\mathbf{
\alpha\ensurement{^{\circ}} = \arctan{ (0,6 ) } = 30.9637565321 \ensurement{^{\circ}}
}
$}}\\\\
\small{\text{$
\mathbf{
\alpha\ensurement{^{\circ}}\approx 30.9638 \ensurement{^{\circ}}
}
$}}\\\\$$

One circle has a circumference of 12pi cm. Another circle has a circumference of 32pi cm. What is the ratio of the radius of the smaller circle to the radius of the larger circle ?

Diameter D = 32 cm  ->  R ( larger circle ) = 16 cm

diameter  d = 12 cm  ->  r ( smaller circle ) = 6 cm

$$\dfrac{r}{R}=\dfrac{6}{16}=\dfrac{3}{8}=0.375$$

The ratio of the radius of the smaller circle to the radius of the larger circle = 0.375

There are 5 girls and 5 boys in a chess club. The club holds a round-robin tournament in which every player plays against every other player exactly once. In how many games does a girl play against a boy ?

$$\small{\text{$
\mathbf{ \rm{all ~games}=\binom{10}2 = 45 }
$}} \\
\small{\text{$
\mathbf{ \rm{only~ boys~games}=\binom52 = 10 }
$}} \\
\small{\text{$
\mathbf{ \rm{only~ girls~games}=\binom52 = 10 }
$}} \\
\small{\text{$
\mathbf{ \rm{one ~girl~one~boy~ games}=45-10-10 = 25 }
$}}$$

$$\small{\text{$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
& \textcolor[rgb]{150,0,0}{Girl~1}
& \textcolor[rgb]{150,0,0}{Girl~2}
& \textcolor[rgb]{150,0,0}{Girl~3}
& \textcolor[rgb]{150,0,0}{Girl~4}
& \textcolor[rgb]{150,0,0}{Girl~5}
& \textcolor[rgb]{0,0,150}{Boy~1}
& \textcolor[rgb]{0,0,150}{Boy~2}
& \textcolor[rgb]{0,0,150}{Boy~3}
& \textcolor[rgb]{0,0,150}{Boy~4}
& \textcolor[rgb]{0,0,150}{Boy~5} \\
\hline
\textcolor[rgb]{150,0,0}{Girl~1}& - & \textcolor[rgb]{150,0,0}{GG} & \textcolor[rgb]{150,0,0}{GG} & \textcolor[rgb]{150,0,0}{GG} & \textcolor[rgb]{150,0,0}{GG} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} \\
\textcolor[rgb]{150,0,0}{Girl~2}& & - & \textcolor[rgb]{150,0,0}{GG} & \textcolor[rgb]{150,0,0}{GG} & \textcolor[rgb]{150,0,0}{GG} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} \\
\textcolor[rgb]{150,0,0}{Girl~3}& & & - & \textcolor[rgb]{150,0,0}{GG} & \textcolor[rgb]{150,0,0}{GG} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} \\
\textcolor[rgb]{150,0,0}{Girl~4}& & & & - & \textcolor[rgb]{150,0,0}{GG} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} \\
\textcolor[rgb]{150,0,0}{Girl~5}& & & & & - & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} & \textcolor[rgb]{150,0,0}{G}\textcolor[rgb]{0,0,150}{B} \\
\textcolor[rgb]{0,0,150}{Boy~1} & & & & & & - & \textcolor[rgb]{0,0,150}{BB} & \textcolor[rgb]{0,0,150}{BB} & \textcolor[rgb]{0,0,150}{BB} & \textcolor[rgb]{0,0,150}{BB} \\
\textcolor[rgb]{0,0,150}{Boy~2} & & & & & & & - & \textcolor[rgb]{0,0,150}{BB} & \textcolor[rgb]{0,0,150}{BB} & \textcolor[rgb]{0,0,150}{BB} \\
\textcolor[rgb]{0,0,150}{Boy~3} & & & & & & & & - & \textcolor[rgb]{0,0,150}{BB} & \textcolor[rgb]{0,0,150}{BB} \\
\textcolor[rgb]{0,0,150}{Boy~4} & & & & & & & & & - & \textcolor[rgb]{0,0,150}{BB} \\
\textcolor[rgb]{0,0,150}{Boy~5} & & & & & & & & & & - \\
\hline
\end{array}$}}$$

A circular flower bed has a diameter of 2.3 m. What is the length of the edge of the bed ?

The length of the edge of the flower bed $$\small{\text{$
\mathbf{= d\cdot \pi = 2.3\rm{~m~}\cdot \pi = 7.23 \rm{~m} }
$}}$$

$$\small{\text{$
\mathbf{\pi \approx 3.14159265359}
$}}$$

(6x^-2y^4)/2x  ?

$$\small{\text{$
\begin{array}{rcl}
\dfrac{ 6 x^{-2} y^4 }{ 2 x } \\\\
&=& \dfrac{ 6 }{ 2 }\cdot \dfrac{ x^{-2} }{ x } \cdot y^4\\\\
&=& 3 \cdot \dfrac{ 1 }{ x^{2}x } \cdot y^4\\\\
&=& 3 \cdot \dfrac{ 1 }{ x^{3} } \cdot y^4\\\\
&=& 3 \cdot \dfrac{ y^4 }{ x^{3} } \\\\
&=& \dfrac{ 3y^4 }{ x^{3} }
\end{array}
$}}$$

Find x if f(x) = 2x + 7 and f(x) = -1.

$$\\ f(x) = 2x+7 \qquad | \qquad f(x) = \textcolor[rgb]{1,0,0}{-1} \\\\
\textcolor[rgb]{1,0,0}{-1} = 2x+7\\\\
2x = -8\\\\
x=-4$$

Find f(-7) if f(x) = 2x + 8.

$$\\ f(x) = 2x+8 \qquad | \qquad x = \textcolor[rgb]{1,0,0}{-7} \\\\
f(\textcolor[rgb]{1,0,0}{-7}) = 2\cdot (\textcolor[rgb]{1,0,0}{-7}) + 8\\\\
f(\textcolor[rgb]{1,0,0}{-7}) = -14+ 8\\\\
f(\textcolor[rgb]{1,0,0}{-7}) = -6$$

If f(x) = x - 5, then match each of the following.

f(-1) = -1 - 5 = -6

f(0) = 0 - 5 = -5

f(1) = 1 - 5 = -4

f(2) = 2 - 5 = -3

f(5) = 5 - 5 = 0

f(8) = 8 - 5 = 3

Find g(1) if g(x) = x ² + 1.

$$\\ g(x) = x^2+1 \qquad | \qquad x = \textcolor[rgb]{1,0,0}{1} \\\\
g(\textcolor[rgb]{1,0,0}{1}) = \textcolor[rgb]{1,0,0}{1}^2 + 1\\\\
g(\textcolor[rgb]{1,0,0}{1}) = 2$$

The problem is "Find f(2) if f(x) = (x + 1)²

$$\\ f(x) = (x + 1)^2 \qquad | \qquad x = \textcolor[rgb]{1,0,0}{2} \\\\
f(\textcolor[rgb]{1,0,0}{2}) = (\textcolor[rgb]{1,0,0}{2} + 1)^2\\\\
f(\textcolor[rgb]{1,0,0}{2}) = 3^2\\\\
f(\textcolor[rgb]{1,0,0}{2}) = 9$$

arrange the following 1/4 2/3 7/12 5/6 4/9

$$\\ \frac14 \cdot \frac99 = \frac9{36}\\\\
\frac23 \cdot \frac{12}{12} = \frac{24}{36}\\\\
\frac7{12} \cdot \frac33 = \frac{21}{36}\\\\
\frac56 \cdot \frac66 = \frac{30}{36}\\\\
\frac49 \cdot \frac44 = \frac{16}{36}\\\\
\rm{~~so~we~have~~}
\frac14 ~~
\frac49 ~~
\frac7{12} ~~
\frac23 ~~
\frac56$$