Watch your language...... Respect other languages.
~The smartest cookie in the world
1. Press the '\(\sum\)LaTeX' button.
2. Type \dfrac{a}{b} and press OK
3. \(\dfrac{a}{b} \) will appear.
\(1\dfrac{1}{3}\div \dfrac{2}{3}\\ = \dfrac{4}{3}\div \dfrac{2}{3}\\ = 4 \div 2\\ = 2\)
Round 0.52173913043 to a whole number.
Round it up at the tenths.
= 1.
Because Maths is really so fun!!!!!
~ The smartest cookie in the world
Example:
\(2\log_2(x+1) - \log_2x = 2\\ \log_2(x+1)^2 -\log_2x=2\\ \log_2\left(\dfrac{(x+1)^2}{x}\right) = 2\\ \dfrac{(x+1)^2}{x} = 4\\ x^2 + 2x + 1 = 4x\\ x^2 - 3x + 1 = 0\\ x = \dfrac{-(-3)\pm \sqrt{(-3)^2-(4)(1)(1)}}{2(1)}=\dfrac{3\pm \sqrt5}{2}\\ x = \dfrac{3+\sqrt5}{2} \text{ or }\dfrac{3-\sqrt5}{2}\text{(rejected)}\\ x = \dfrac{3+\sqrt5}{2}\)
Try to do your own question now!!
Ok.......
I shouldn't have answered this question.......
\(\sqrt[3]{\dfrac{\left(2n^0\right)^4\cdot 2m^2n^4\cdot m^{-1}n^3}{m^{-4}n^4}}\\ =\left(\dfrac{\left(2n^0\right)^4\cdot 2m^2n^4\cdot m^{-1}n^3}{m^{-4}n^4}\right)^{1/3}\\ =\left(2^5\cdot m^5n^3\right)^{1/3}\\ =\sqrt[3]{32}\cdot m^{5/3}n\\ =2\sqrt[3]{4}\cdot m^{5/3}n\)
Ok....... I am not very familiar with probability...... Calculus is my thing!!
OK....... What is that!?!?
binom of (5+5, 5) thing!?!?
I just treated it as a simple probability problem!!
