All Answers 358098 Answers

I've changed it!

L'Hopital's rule:

limit_x→infinity(f(x)/g(x)) = limit_x→infinity(f`(x))/g`(x))

Here f`(x) = ln(2)*2^x + 1/x   and  g`(x) = 3*ln(2)*2^x - 1/x

as x gets large 1/x gets small so you are left with ln(2)*2^x/(3*ln(2)*2^x ) = 1/3

.

Yes thats what I'm seeing 

Thank you alan and melody

young lady, how dare you spread the secret formula of the products of Titanium Industries!!!!????

THAT IS DISGRACEFUL!

ROman, I believe she should spend the trip in Neu Alcatraz. It was a lot of fun last time I got to torture her! XD

What are mutually cotangent circles??

I have a number of 4 different sketches in my mind.  They would all give different answers:/

Oh.   Chris think it means that they all just touch one another externally.  

Is that what mutually cotangent circles are?

If x^3 = 219.6 then x is the cube root of 219.6.  x = 219.6^(1/3)

$${\mathtt{x}} = {{\mathtt{219.6}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} \Rightarrow {\mathtt{x}} = {\mathtt{6.033\: \!149\: \!844\: \!003\: \!693\: \!5}}$$

.

ah, what's this?

oh it s the god of mathematics! i'm so scared! run before he blasts numbers in your brains! (sarcasm)

ah, we have some interesting crime activity to catch

Thanks Melody for notice my wrong.

Alan, do you know how to do it with the change?

the answer is 1/3 if that helps.

Will check out L´Hopital´s rule, thanks for your help!

This means the exterior angle is 180° - 156° = 24° 

If there are n sides to the regular polygon then n*exterior angle = 360° so here we have n*24 = 360 or n = 360/24 = 15.

It is a 15-sided polygon

Great sketch CPhill,  you are getting much better at GeoGebra lately.  I'm impressed :))

$$\\\left(z-\frac{2}{\sqrt{z}}\right)\\\\The\:general\:term\:is\\\\
=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)\\
=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)\\
=(9Cr)\left(\frac{*(-2)^r}{z^(2r/2)*(z)^(r/2)}\right)\\
=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\18-3r=0\\r=6$$

This is the entire coding correct...

Hopefully it shows...My laptop might not be able to take in all this LaTeX.

\\\left(z-\frac{2}{\sqrt{z}}\right)\\\\The\:general\:term\:is\\\\
=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)\\
=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)\\
=(9Cr)\left(\frac{*(-2)^r}{z^(2r/2)*(z)^(r/2)}\right)\\
=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\18-3r=0\\r=6

XD

you sure you can get away easily, Heather Friedmann?

well think twice before leaving

we have protected you and have defended your darkest secrets and your own soul for almost 5 years, ever since the empire rose

and we will not stop rising now

we just held Nauseated captive, his reign has ended, and he is serving 2 weeks in Neu Alcatraz.

and now we are running after the Girl on Fire

if you leave now, then we will atack you on your birthday, where we will tell everyone all the secrets you've hidden for almost 5 years

and don't forget we are having our monthly party for all members of the empire, and this month will mark the 5 years of reign, on May 13th, 2010, is when the empire rose. and this one will be the best.

we went from a litttle gang to the Empire of Royality

and also, you dare release the company's secrets. you're mocking it, and that's unacceptable young lady

2^x increases much quicker than ln(x) as x gets large so the numerator tends to 2^x as x gets very large.

3*2^x also increases much quicker than ln(x) as x gets large, so the denominator tends to 3*2^x

The ratio therefore tends to 2^x/(3*2^x) = 1/3

oh sorry it will be 3*2^x

Is that a 6x on the bottom Headingnorth?

I probably can't do it anyway.  :(

The code didn't display...

$$\\=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\18-3r=0\\r=6$$

Here's the code Melody.

\\=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\18-3r=0\\r=6

The $$ at first i was trying to leave extra spaces between he words but I just stuck with \; .

I forgot to take them out also i was trying to leave bigger gaps between the equations I know you have to do \\ and with \\.

But how do you leave a slightly bigger gap.

LaTeX.

$$\\=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)\\
The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\
18-3r=0\\
r=6$$

It's alot easier seeing the actual coding underneath

Yes, Chris made a little mistake - it is no big deal.

Your answer is spot on Zac.   Thanks to both of you.   

inverse tan is the same as arc tan

just use  [2nd][atan] angle

Well yes, but this always happens to me when the code is incorrect,however when fixed the buffer goes...

But the thing is...I don't know whats wrong with the code..?

EDIT: Just refreshed page,you've already answered my question :)

Hi Heather, I look forward to reading your Channel 21 news :)

Yes in some countries a billion is  1 with 9 zeros and in others it is 1 with 12 zeros.

I think the 9 zeros is more common and it is the US one I believe.

MathsGod is British and I think that he is also correct. I think it England it is 1 with 12 zeros.

this is why scientific notation is so much better with very big and very tiny numbers :)

$${{\mathtt{4}}}^{{\mathtt{0.2}}} = {\mathtt{1.319\: \!507\: \!910\: \!772\: \!894\: \!3}}$$

this is already an approximation - if I round it off to 2 dec places I get   1.32

Now

$$4^{0.2}=4^{1/5}=\sqrt[5]{4}$$

Now id the answer really is  1.3195079 then    $$1.3195079^5$$    must equal  4

lets see if it does

$${{\mathtt{1.319\: \!507\: \!9}}}^{{\mathtt{5}}} = {\mathtt{3.999\: \!999\: \!836\: \!713\: \!459\: \!7}}$$    That is close enough to 4 to me    

$${\frac{{{\mathtt{301}}}^{{\mathtt{4}}}}{{\mathtt{10\,000}}}} = {\mathtt{820\,854.120\: \!1}}$$

the fraction equivalent of the decimal part is   1201/10000

The remainder is 1201