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I got 32' 0.1"

Still got a bee in our bonnet about an incorrect answer I gave back in December, have we?

No, not at all. Your modified sodium chlorate keeps the bees out of my bonnet very well. That was one of the reasons I wanted more of it.

As flattering as it is that you've held on for so long, I'll have to ask you to get over it

I’ve held on because it is such an amazing discovery! Truely worthy of a revisit. It’s amazing how accidental chemical discoveries can change the world! I’m sure you will do something new, soon. Promoting the new product will be at the top of the list. The old ones fade, but never really end. At least not until the new and improved version is introduced.

I'm quite sure that you, yourself, are no angel.

My parole officer thinks I am an angel. Of course, the baseline for comparison isn’t exactly in the stratosphere. My GF thinks of me as an angel . . .though she prefaces it with “fallen” most of the time. Even so, it’s a title I carry with pride.

Narcissists are an endearing breed

I had to lookup “Narcissists” Yep, it fits. I must have a special version of that because it usually isn’t conducive for endearment. I saw several other psychological syndromes that often apply too. These mainly manifest when the CDD nausea flares.

Well, until your next great discovery, Sorasyn, take care.

5'2"+5"=5'7"

I suspect that you mean     (12^[3/5]*8^[3/5]) ^5 

$$\\(12^{3/5}*8^{3/5}) ^5\\\\
12^{(3/5)*5}*8^{(3/5)*5}}\\\\
12^3*8^3}\\\\
96^3\\\$$

You can finish it  :)

Mmm  a little confusion here I think :)

$$\\ln(x)=-3\\\\
e^{ln(x)}=e^{-3}\\\\
x=e^{-3}\\\\
x=\frac{1}{e^3}$$

$$\\\frac{-113}{24} + 3\frac{ 8}{9}\\\\
-\frac{113}{24}+3\frac{ 8}{9}\\\\
-4\frac{17}{24}+ 3\frac{ 8}{9}\\\\
-4\frac{17*3}{24*3}+ 3\frac{ 8*8}{9*8}\\\\
-4\frac{51}{72}+ 3\frac{ 64}{72}\\\\
-(4\frac{51}{72}- 3\frac{ 64}{72})\\\\
-(4\frac{51}{72}- 3-\frac{ 64}{72})\\\\
-(4- 3\frac{51}{72}-\frac{ 64}{72})\\\\
-(1\frac{51}{72}-\frac{ 64}{72})\\\\
-(\frac{72+51}{72}-\frac{ 64}{72})\\\\
-(\frac{72+51-64}{72})\\\\$$

$$\\-(\frac{59}{72})\\\\
-\frac{59}{72}\\\\$$

check

$${\mathtt{\,-\,}}{\frac{{\mathtt{113}}}{{\mathtt{24}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{8}}}{{\mathtt{9}}}} = {\mathtt{\,-\,}}{\frac{{\mathtt{59}}}{{\mathtt{72}}}} = -{\mathtt{0.819\: \!444\: \!444\: \!444\: \!444\: \!4}}$$

flag pole height 22'6"

What do you want us to do with this ?  

ans

does x(2) mean  x^2  which is     $${{\mathtt{x}}}^{{\mathtt{2}}}$$      ??

atan(3) =

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{3}}\right)} = {\mathtt{71.565\: \!051\: \!177\: \!078^{\circ}}}$$

$$\\ f(x)=1.2x^3-2^{1.5}x+\sqrt2\\\\
f'(x)=3*1.2x^2-2^{1.5}\\\\
f'(x)=3.6x^2-2\sqrt2\\\\$$

NOTE:

 $$\\y=ax^n\qquad then\qquad y'=n*a*x^{n-1}\\\\
so\;if\;
y=1.2x^3\qquad then\qquad y'=3*1.2*x^{2}=3.6x^2\\\\$$

hi ChaCha,

You are our youngest member  

P(0)=   $${{\mathtt{0.999}}}^{{\mathtt{3\,000}}} = {\mathtt{0.049\: \!712\: \!393\: \!998\: \!036\: \!3}}$$

P(1)=   nCr(3000,1)*0.001*0.999^2999

P(2)=    nCr(3000,2)*0.001^2*0.999^2998

P(3)=    nCr(3000,3)*0.001^3*0.999^2997

P(4)=    nCr(3000,4)*0.001^4*0.999^2996

P(5)=    nCr(3000,5)*0.001^5*0.999^2995

P(6)=    nCr(3000,6)*0.001^6*0.999^2994

work those out and add them up.

this calc was baulking - might do better in Wolfram|Alpha calcualtor

Hi GUH,

here is how it is done :)

See how I have factored into pairs first :)

$$\\8p^3-10p^2+12p-15\\\\
=2p^2(4p-5)+3(4p-5)\\\\
=(2p^2+3)(4p-5)\\\\$$

So we have

4500 = pi * r^2 * 4     divide both sides by 4 pi

4500/[4 pi] = r^2  = (1125/pi)   take the sq root of each side

√(1125/pi) = r = about 18.923 ft   (at most)

So the circular area is just pi * r^2  = pi *(18.923)^2 = about 1124.94 sq ft  (at most)

Here is a post I wrote a while back :)

This would be a really good answer anon except that 11-7=4     LOL

x-y=5      SO     x=5+y     (1)

4x+y=10     (2)

Sub (1) into (2)

4(5+y)+y=10

20+4y+y=10

5y=-10

y=-2

now sub y=-2 back into (1) to find x

8x - y= -3

-y=-8x-3

y=8x+3

f(x)=8x+3

$$\\\frac{(2+x-x^2)}{6+7x+x^2}\\\\
=\frac{-(x^2-x-2)}{x^2+7x+6}\\\\
=\frac{-(x-2)(x+1)}{(x+6)(x+1)}\\\\
=\frac{-(x-2)}{x+6}\\\\
=\frac{2-x}{x+6}\\\\$$

If you need help with that factorising then just ask how we did it and someone (maybe me or CPhill) will explain. 

85.44003745    Rustynut "just getting by"

I think having no visable end means that there is no beginning and no end to the shells

so the number of permutations  would be  (8-1)!

If there was a clasp then there would be 2 distinct ends and the numbero permutations would be 8! 

Thanks anon for that flash card site.  It looks like it could be really useful for students.

Do you use it yourself or did you just google this question?

Hi Ninja,

It is really great to see you back again. 

$$L=\frac{P-2W}{2}$$

$$\\V=\pi r^2*h\\
V=3.14 * 3.8^2*2.5\;\;\;cm^3$$