MaxWong

One billion, seven hundred and fifty eight million, five hundred thousand.

Same as Guest #1's answer.

8 - w + 4

= 12 - w 

;D

\((r+2i)(r-2i)\)

coefficient in front of x³ = 3 - 4 = -1.

coefficient in front of x = 7 - 12 = -5.

How did I get that?

First, ask, how many ways you get the form of ax³ if you pick 2 terms and multiply?

First way,3 times x³

Second way, -4x times x²

No other ways.

So the coefficient of x³ = 3 + (-4) = 3 - 4 = -1

You do the coefficient of x one as above too.

\(\sqrt{\dfrac{1}{96}}\times \sqrt3\\ =\sqrt{\dfrac{3}{96}}\\ =\sqrt{\dfrac{1}{32}}\\ =\dfrac{\sqrt{1}}{\sqrt{32}}\\ =\dfrac{1}{4\sqrt{2}}\\ =\dfrac{1\times\sqrt2}{4\sqrt2 \times \sqrt2}\\ =\dfrac{\sqrt2}{8}\)

\(f(x)= 3x-2x^2\)

\(f'(x)=\displaystyle\lim_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}\\ \;\;\;\;\;\;\;\;= \displaystyle\lim_{h\rightarrow 0}\dfrac{(3(x+h)-2(x+h)^2)-(3x-2x^2)}{h}\\ \;\;\;\;\;\;\;\;=\displaystyle\lim_{h\rightarrow 0} \dfrac{3x+3h-2x ^2-4xh-2h^2-3x+2x^2}{h}\\ \;\;\;\;\;\;\;\;=\displaystyle\lim_{h\rightarrow 0}\dfrac{3h-4xh-2h^2}h\\ \;\;\;\;\;\;\;\;=\displaystyle\lim_{h\rightarrow 0} (3-4x-2h)\\ \;\;\;\;\;\;\;\;= 3-4x\)

-(-6)¹= -(-6)=+6

2/3 approximately.

6666666666666667/10000000000000000 exactly.

\((\frac{1}{5}+\sqrt{\frac{19}{10i}})^2\\ = \frac{1}{25}+2\times \frac{1}{5}\times \sqrt{\frac{19}{10i}}+\frac{19}{10i}\\= \frac{1}{25}+\frac{2}{5}\times\frac{\sqrt{19}}{\sqrt{10}}\times \frac{\sqrt{2}}{1+i}+\frac{19}{10i}\)

Because sqrt i = (1+i)/ sqrt2

http://web2.0calc.com/questions/what-is-square-root-of-i

\(=\frac{1}{25}+\frac{2\sqrt{38}}{(5\sqrt{10})(1+i)}+\frac{19}{10i}\)

\(=\frac{1}{25}+\frac{2\sqrt{-380}}{50i(1+i)}+\frac{95+95i}{50i(1+i)}\)

\(=\frac{1}{25}+\frac{(2\sqrt{380}+95)i+95}{50i-50}\)

You can do the last half yourself.

\(2C^2=2(-2)^2=2\times 4 = 8\)