heureka

$$log_4(4) = 1$$

$$\\\log_4(x)=\sqrt{\log_4(x)} \\
\\
\quad x=4 \Rightarrow \log_4(4)=\sqrt{\log_4(4)}\Rightarrow 1 = \sqrt1 = 1$$

$$\boxed{x=4}$$

...or

$$\\\log_4(x)=\sqrt{\log_4(x)} \quad | \quad 1^2\\ \\
\log_4(x)\times\log_4(x)=\log_4(x) \quad | \quad :\log_4(x)\\ \\
\log_4(x) = 1\quad | \quad 4^x\\ \\
4^{\log_4(x)} =x= 4^1 \\
\boxed{x=4}$$

1. $$2^{nd} \; Button$$
2. $$\% \; Button$$

 $$\text{oder halte lange den }\boxed{1/x \; Button} \text{ , er wechselt dann zum }\boxed{\% \; Button}$$

$$\\\$4.50+ \$3.50+ \$5.50 +\$2.90 = \$16.40\\
\\
\dfrac{\$16.40}{4}=\$4.10\\
\\
\text{The average of }\; \$4.50, \$3.50, \$5.50 \text{ and } \$2.90 \text{ is } \$4.10$$

Mr. Tanaka = T

Son = S

years = y

$$\\T+y=(S+y)*3\\
T+y= 3S+3y \quad | \quad -3S\\
T+y-3S=3y \quad | \quad -y\\
T-3S=2y \quad | \quad :2 \\
y=\dfrac{T-3S}{2}\\
y=\dfrac{35-3*5}{2}\\
y=\dfrac{35-15}{2}\\
y=\dfrac{20}{2}\\
\boxed{y=10}$$

In 10 years: Mr Tanaka is 45 and his son is 15.

15 * 3 = 45

$$\\4x^2-4x-27 \\
=4(x^2-x)-27\\
=4\left(x^2-x+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\right) -27\\
=4\left( \left( x-\frac{1}{2}\right)^2-\frac{1}{4}\right) -27\\
=4\left(x-\frac{1}{2}\right)^2 - \frac{4}{4}-27\\
=4\left(x-\frac{1}{2}\right)^2 - 1-27\\
=4\left(x-\frac{1}{2}\right)^2 -28\\$$

• 6 notebooks + 6 highlighters = $23.06
• 1 notebook = $3.48
• 6 notebooks = 6 * $3.48 = $20.88
• $20.88 + 6 highlighters = $23.06   | - $20.88
• 6 highlighters = $23.06 - $20.88 = $2.18
• 1 highlighter = $2.18/6 = $0.36333333333

1 highlighter costs $0.3633333333333

Hi Melody,

here is the code:

\documentclass[7pt,a4paper,landscape]{article}
\usepackage[utf8]{inputenc}
\usepackage{pgfplots}
\usepackage{qtree}

\begin{document}
\Tree [
[
[
[[ F\\F\\F\\F ].($\frac{1}{2}$)\\F [ F\\F\\F\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\F [[ F\\F\\B\\F ].($\frac{1}{2}$)\\F [ \fbox{  \textcolor[rgb]{1,0,0}{FFBB}} ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\F
[
[[ F\\B\\F\\F ].($\frac{1}{2}$)\\F [ \fbox{ \textcolor[rgb]{1,0,0}{FBFB}} ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\F [[ \fbox{ \textcolor[rgb]{1,0,0}{FBBF}} ].($\frac{1}{2}$)\\F [ F\\B\\B\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\F
[
[
[[ B\\F\\F\\F ].($\frac{1}{2}$)\\F [  \fbox{ \textcolor[rgb]{1,0,0}{BFFB}} ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\F [[ \fbox{ \textcolor[rgb]{1,0,0}{BFBF}} ].($\frac{1}{2}$)\\F [ B\\F\\B\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\F
[
[[\fbox{ \textcolor[rgb]{1,0,0}{BBFF}} ].($\frac{1}{2}$)\\F [ B\\B\\F\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\F [[ B\\B\\B\\F ].($\frac{1}{2}$)\\F [ B\\B\\B\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\B
].Start\\(F=Front\\B=Back)
\end{document}

heureka

Hi Melody,

I use Latex \usepackage{qtree}

Sorry, i can't answer your second question.

6 times you have exactly two tails: FFBB, FBFB, FBBF, BFFB, BFBF, BBFF

And the Probability is  $$6*\left(\dfrac{1}{2}\times\dfrac{1}{2}\times\dfrac{1}{2}\times\dfrac{1}{2} \right)$$

Quadrant:   I    II   III   IV

sin(x)         +    +    -     -

cos(x)         +    -    -     +

3. Quadrant  sin X< 0 and cos X <0