heureka

$$\dbinom {4} {2} * \dfrac{1}{2} * \dfrac{1}{2}* \dfrac{1}{2}* \dfrac{1}{2} = \dbinom {4} {2} * \left(\dfrac{1}{2}\right)^4=6*\dfrac{1}{16}=\dfrac{6}{16}=\dfrac{3}{8}=0.375$$

The probability is 37.5 %

... mod Button?

drop a long time the $$\div$$  Button. He will change to $$\mod$$  Button:

... asin Button?

drop a long time the $$\sin$$  Button. He will change to $$asin$$  Button:

 and so on...

I think "&" = "+"

$${\sqrt{{\mathtt{3.33}}{\mathtt{\,\small\textbf+\,}}{\mathtt{642.348}}}} = {\mathtt{25.410\: \!194\: \!804\: \!448\: \!076\: \!1}}$$

$${\sqrt{{{\mathtt{2.5}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{3.1}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{1.9}}}^{{\mathtt{2}}}}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}} = {\mathtt{0.5}}$$

$$\text{do } 2^{nd}\text{ - Button and use mod}$$

Hi CPhill,

Your correct answer is: $$2(3^5 -1 ) \textcolor[rgb]{1,0,0}{/} (3-1) = 242$$

It is equal to: $$s_n=\frac{a*(r^n-1)}{r-1}$$

here: $$s_5=\frac{2*(3^5-1)}{3-1}=242$$

Many Greetings

Heureka

hi Melody,

{rll...} is the alignment.

1. Field r (right-alignment) bevor first "&"

2. Field l (left-alignment) after first "&"

3. Field l (left-alignment) after second "&"

and so on...

r = right alignment

c = center alignment

l = left alignment

I hope, I could help you

heureka

5=11+2(x-1)    x?

$$\begin{array}{rll} 5&=&11+2(x-1)\quad | \quad -11\\ 5-11&=& 2(x-1)\\ -6&=& 2(x-1)\quad | \quad :2\\ -\frac{6}{2}&=& (x-1)\\ -3&=& x-1\quad | \quad +1\\ -3+1&=& x\\ -2&=& x\\ \end{array} \\ \boxed{x=-2}\\$$

Latex Code:

\begin{array}{rll}
5&=&11+2(x-1)\quad | \quad -11\\
5-11&=& 2(x-1)\\
-6&=& 2(x-1)\quad | \quad :2\\
-\frac{6}{2}&=& (x-1)\\
-3&=& x-1\quad | \quad +1\\
-3+1&=& x\\
-2&=& x\\
\end{array} \\
\boxed{x=-2}\\

Proof:

$$\begin{array}{rll} 5&=&11+2*(\textcolor[rgb]{1,0,0}{-2}-1)\\ 5&=&11+2*(-3)\\ 5&=&11-2*3\\ 5&=&11-6\\ 5&=&5\\ \end{array} \\$$

Latex code:

\begin{array}{rll}
5&=&11+2*(\textcolor[rgb]{1,0,0}{-2}-1)\\
5&=&11+2*(-3)\\
5&=&11-2*3\\
5&=&11-6\\
5&=&5\\
\end{array} \\

the forth root of 40000? $$\sqrt[4]{40000}=\;?$$

$$\sqrt[4]{40000}=$$

$$\\=\sqrt[2]{\sqrt[2]{40000}}\qquad 40000 = 4 * 100 * 100=2*2*100*100=2^2*100^2\\\\ =\sqrt[2]{\sqrt[2]{2^2*100^2}}\\\\ =\sqrt[2]{\sqrt[2]{2^2}\times\sqrt[2]{100^2}}\\\\ =\sqrt[2]{2*100}\qquad 2*100= 2 * 10 * 10=2*10^2 \\\\ =\sqrt[2]{2*10^2}\\\\ =\sqrt[2]{2}\times\sqrt[2]{10^2}}\\\\ =\sqrt2*10\\\\ =10\sqrt2$$

Latex code:

So we have:

$$\\\mbox{start} = a_1=4\\ \mbox{increase}=d=3\\ \mbox{term number}=n=50\\ \mbox{term 50}=a_n=a_{50}?\\ \boxed{a_n=a_1+(n-1)*d}\\ a_{50}=4+(50-1)*3\\ a_{50}=4+49*3\\ \boxed{a_{50}=151}$$

Latex code:

} = a_1=4\\
\mbox{increase}=d=3\\
\mbox{term number}=n=50\\
\mbox{term 50}=a_n=a_{50}?\\
\boxed{a_n=a_1+(n-1)*d}\\
a_{50}=4+(50-1)*3\\
a_{50}=4+49*3\\
\boxed{a_{50}=151}

Term 50 is 151