Melody

I am not really sure about the notation that is needed.

but

P(N rooms empty) = $$^5C_N * 0.4^N * 0.6^{5-N}$$           where     $$0\le N\le5\;\;\;and\;\;\;N\in Z$$ 

Thanks Radix,      

I'll just take a look   

$$\\\frac{x^2-1}{x^2-3x+2}>0\\\\
\frac{(x-1)(x+1)}{(x-1)(x-2)}>0\\\\
$NOTE: x cannot be 1 or 2$\\\\
\frac{(x+1)}{(x-2)}>0\\\\
\frac{(x-2)(x+1)}{1}>0\\\
(x-2)(x+1)>0\\\\
consider\;\;y=(x-2)(x+1)\\\\
$This is a concave up parabola with roots at x=2 and x=-1$\\\\
$It will be above the x axis (positive) at the two ends$\\\\
$roughly sketch it to convince yourself$\\\\
$therefore true for$\\\\
x<-1\;\;\;and \;\;\;x>2$$

$$\\\frac{(e^x+7)}{(3e^{-x}+5)}=3\\\\
(e^x+7)\div (3e^{-x}+5)=3\\\\
(e^x+7)\div (\frac{3}{e^{x}}+5)=3\\\\
(e^x+7)\div (\frac{3+5e^{x}}{e^{x}})=3\\\\
(e^x+7)\times (\frac{e^{x}}{3+5e^{x}})=3\\\\
(e^x+7)\times e^{x}=3(3+5e^{x})\\\\
(e^x)^2+7e^{x}=9+15e^{x}\\\\
(e^x)^2-8e^{x}-9=0\\\\$$

$$\\let\;y=e^x\\\\
y^2-8y-9=0\\\\
(y-9)(y+1)=0\\\\
y=9\;\;or\;\;y=-1\\\\
e^x=9\;\;or\;\;e^x=-1\\\\
e^x>0\;\;$for all real x so $ \\\\
e^x=9\\\\
x=ln9\\\\
x=ln3^2\\\\
x=2ln3$$

There are 6 numbers after the point so the 1with 6 zeros will be on the bottom.

0.000023 = 23/1000000

HAPPY BIRTHDAY JILLYBEAN  

  HAVE A REALLY GREAT DAY !!   

There are a lot of us so I thought I would make some cupcakes for everyone to share :))

Hi Sabi,

Yes i guess I missed a couple of steps, :)

$$\frac{(1-x)}{(x-2)}>0$$

FIRST:   I must state that x-2 cannot be zero so  x cannot equal 2

Now I am allowed to multiply both sides by whatever I want and in so doing I will cancel the denpominator out.

If this was an equation (not an inequality) then I would multiply both sides by (x-2)  BUT that causes some problems here because x-2 could be negative or positive.

It would make life easier if we multiply by something that HAS to be positive.

So I multiplied by   $$(x-2)^2$$

$$\\\frac{(1-x)}{(x-2)}\times \frac{(x-2)^2}{1}>0\times \frac{(x-2)^2}{1}\\\\
\frac{(1-x)}{1}\times \frac{(x-2)}{1}>0\\\\
\frac{(1-x)(x-2)}{1}>0\\\\
(1-x)(x-2)>0$$

Now  the rest is the same as before :)

If you need more explanation - just ask :)

$$\\(1-x)(x-2)>0\\
$If you let $ y=(1-x)(x-2)\\
$then the statement will be true when $y>0\\
y=(1-x)(x-2)\\
y=-1x^2+3x-2\\
$this is a parabola- the coefficient of $ x^2$ is -1$\\
$So it is concave down.$\\
$The middle bit will be above the x axis where y is positive.$\\
$The roots are x=1 and x=2$\\
SO\\
(1-x)(x-2)>0\quad when\;\;1

Go TR  

Thanks Chris :))

When working in base ten, the first place after the decimal point is  tenths, the second is tenths squared,  the third is tenths ^3  etc

When working in base eight, the first place after the decimal point is  eigths, the second is eigths squared,  the third is eigths ^3  etc

So I needed to change 59/10^2   to something over eight to the power of some positive integer.

I just chose to do it to three decimal places so I used 8^3.

Does that make sense?

You are very welcome    

with this calc put the first one enter it likie this

asin(25/30)

and this will be the output :)

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{25}}}{{\mathtt{30}}}}\right)} = {\mathtt{56.442\: \!690\: \!238\: \!079^{\circ}}}$$