gibsonj338

y=2x+3

First switch x and y around to get x=2y+3

Now solve for y

x-3=2y

(x-3)/2=y or y=(x-3)/2

2015 in binary is 11111011111

I do not know of an easy way to explain why 2015 in binary is 11111011111; however, I found a video that can explain it way better than I can.  Here is the web address:  https://www.khanacademy.org/math/pre-algebra/applying-math-reasoning-topic/alternate-number-bases/v/number-systems-introduction

$${\mathtt{1.98}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{4}}} = {\mathtt{0.000\: \!198}}$$

$${\frac{{\mathtt{1\,600}}}{{\mathtt{12}}}} = {\frac{{\mathtt{400}}}{{\mathtt{3}}}} = {\mathtt{133.333\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

$${{\mathtt{\pi}}}^{{\mathtt{x}}} = {\mathtt{41}}$$

Put in calculator and solve.

$${\sqrt{{\mathtt{8}}}}$$

$${\sqrt{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{2}}}}$$

$${\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}$$

$${\mathtt{\,-\,}}{\mathtt{30}}{\mathtt{\,\times\,}}{\mathtt{40}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}\left({\mathtt{5}}\right) = {\mathtt{Y}}$$

$${\mathtt{\,-\,}}{\mathtt{1\,200}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}\left({\mathtt{5}}\right) = {\mathtt{Y}}$$

$${\mathtt{\,-\,}}{\mathtt{1\,198}}{\mathtt{\,\small\textbf+\,}}\left({\mathtt{5}}\right) = {\mathtt{Y}}$$

$$-{\mathtt{1\,193}} = {\mathtt{Y}}$$

$0.80(1.5)+$1.90(3.25)

$1.20+1.90(3.25)

$1.20+$6.175

$7.375

$7.38

$${\mathtt{M}} = {\mathtt{100}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{M}}}{{\mathtt{2}}}}$$

$${\mathtt{M}}{\mathtt{\,-\,}}{\frac{{\mathtt{M}}}{{\mathtt{2}}}} = {\mathtt{100}}$$

$${\frac{{\mathtt{2}}{M}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{M}}}{{\mathtt{2}}}} = {\mathtt{100}}$$

$${\frac{{\mathtt{M}}}{{\mathtt{2}}}} = {\mathtt{100}}$$

$${\mathtt{M}} = {\mathtt{200}}$$

If you mean $${{\mathtt{9}}}^{\left({\frac{{\mathtt{5}}}{{\mathtt{2}}}}\right)}$$,you can write the expression as $${\sqrt{{{\mathtt{9}}}^{{\mathtt{5}}}}}$$which is $${\sqrt{{\mathtt{59\,049}}}}$$ which is $${\mathtt{243}}$$.

If you mean $${\frac{{{\mathtt{9}}}^{{\mathtt{5}}}}{{\mathtt{2}}}}$$,you can write the expression as $${\frac{{\mathtt{59\,049}}}{{\mathtt{2}}}}$$ which is $${\mathtt{29\,524.5}}$$.