MaxWong

\(\dfrac{1}{2} = \dfrac{1}{1}r+\dfrac{1}{2}r+\dfrac{1}{3}r\\ 3 = 6r + 3r + 2r\\ 11r = 3\\ r = \dfrac{11}{3}\)

Or you mean

\(\dfrac{1}{2}=\dfrac{1}{1r}+\dfrac{1}{2r}+\dfrac{1}{3r}\\ \dfrac{r}{2} = 1+\dfrac{1}{2}+\dfrac{1}{3} =\dfrac{11}{6}\\ 3r = 11\\ r=\dfrac{11}{3}\)

The answers are the same, never mind

\(\begin{array}{rll}4^x&=&32^{-x-2}\times8^{x -3}\\4^x&=&\dfrac{8^x\times 8^{-3}}{32^x\times32^ 2}\\4^x&=&(\dfrac{1}{4})^x(\dfrac{1}{8^3\times 32^2}) \\\left(\dfrac{4}{\frac{1}{4}}\right)^x&=&\dfrac{1}{2^{19}}\\ 16^x&=&2^{-19}\\ 2^{4x}&=&2^{-19}\\ 4x &=& -19\\ x&=&-\dfrac{19}{4}\end{array}\)

A method without logarithms. Similar to Omi67's answer.

If you think this is simple you won't ask this here.

Nvm I will answer you.

\(\dfrac{20}{4-(10-8)}\\ =\dfrac{20}{4-2}\\ =\dfrac{20}{2}\\ =10\)

\(\sqrt{2x+4}=4\\ 2x+4=4^2\\ 2x = 4^2 - 4 = 12\\ x = 12/2 = 6\)

Q:tangent a = 1.49, find a

A: a = arctangent 1.49 = 56.13 degrees or 0.98 radians.

40 M x 1M

= 40Trillion.

= 40,000,000,000,000

That was a typing mistake. Look at the original question.

4x + 3 = 72

4x + 3 - 3 = 72 - 3

4x = 69

x = 17.25

Everything is correct that way 

Sorry for ignoring radians.

I automatically read it as x degrees when x > 2*pi without a question.

x + .04x = 1.04x

mean of 25 29 35 = (25+29+35)/3 = 29 2/3