All Answers 358338 Answers

C = pi * [ 9 in ]  ≈  28.27  in

Wow. This project needs to be funded so that children are aware of this resource and can see EXACTLY how the work was done to get the correct answer. I did not know this project was available and I think many people would agree that they are unaware as well. This forum is a novel and revolutionary idea which uses the application of communication technology and crowdsourcing to help increase overall knowledge and understanding of mathematics. Unbelievable. It's truly something amazing which has been put together. This idea needs to be spread 

Solution should contain a constant of integration.

That's nothing, Hawkeye.....I can type the Roman "0"...............

See???

Thank you very much CPhill.

Call the original amount    $30 + x....and x is the amount bet on Hot Feet

After the first race  he has  $2x  =  twice what he bet on Hot Feet

In the second race......he bets $30 on the Favorite   and [ 2x − 30]  on Lucky Horseshoe

And after the second race he has $[2x  − 30] *3   = $[6x  −  90]

In the 3rd race he bets $30  on  Wish Me Luck and $ [ 6x −  90 ] − 30   = $[ 6x − 120]  on  Bred for Speed

And after the last race he has

4 *  [ 6x − 120]   = 240

24x  −  480   = 240

24x  = 720

x  = $30

So.......the original amount  =  $30 + x  =  $30  + $30  =  $60

thanks Guest!

I actually have the solutions, but I expect to find better ones here because I noticed many times that some solving ways are better than some others :) 

Dear GA...... 

Originally....Melody's standards were much too high for her to be "bought off" with  cheap alcohol........however......things being what they are "Down Under,"  she's had to recently "revise" her moral codes........now???......money and/or cheap alcohol are perfectly acceptable forms of bribery.......

Sincerely,

Sir CPhill

There will always be the same number of black b***s in the first bag as red b***s in the second, regardless of how many black b***s you happen to transfer from the second to the first bag! Perhaps the best way to see this is by means of a simple example. If you removed 3 red and 9 black b***s from the second bag and placed them in the first bag, there will be 9 black b***s in the first bag. But, by taking only 3 red b***s across, you have left 9 red b***s behind in the second bag!.

I'll give this a shot

Afer taking 12 red b***s from the first bag and putting them into the second :

The first bag conains 18 b***s  [all red ]

And the second bag now contains :

 72 b***s..... 60 black,  and 12 red  =   5/6 black and 1/6 red

Assuming the second bag is well−mixed and taking 12 b***s from it and putting hem back into the  first bag.:

We should select 10 black b***s  (5/6)*12 and 2 red ones (1/6)*12 to be put back into the first bag

So the second bag [ theoreically] now contains   50 black and 10 red

And the first bag should now contain 10 black and 20 red

So.......the first bag should contain as many black b***s as the second bag contains red ones !!!

Get the answers by entering them into Wolfram/Alpha Computing Engine. If you need step by step solution, I will borrow my brother's computer and might be able to help you, since he is subscribed to it.

Evaluate the expression when x=3 

the anwser ?

Possible derivation:
d/dt(1 + 0.00003 t + 4.×10^-7 t^2)
Differentiate the sum term by term and factor out constants:
 = d/dt(1) + 0.00003 (d/dt(t)) + 4.×10^-7 (d/dt(t^2))
The derivative of 1 is zero:
 = 0.00003 (d/dt(t)) + 4.×10^-7 (d/dt(t^2)) + 0
Simplify the expression:
 = 0.00003 (d/dt(t)) + 4.×10^-7 (d/dt(t^2))
The derivative of t is 1:
 = 4.×10^-7 (d/dt(t^2)) + 1 0.00003
Use the power rule, d/dt(t^n) = n t^(n - 1), where n = 2: d/dt(t^2) = 2 t:
 = 0.00003 + 4.×10^-7 2 t
Simplify the expression:
Answer: |= 0.00003 + 8.×10^-7 t

Take the integral:
 integral1/(sin(x) + cos(x) + 1) dx
For the integrand 1/(sin(x) + cos(x) + 1), substitute u = tan(x/2) and du = 1/2 dx sec^2(x/2). Then transform the integrand using the substitutions sin(x) = (2 u)/(u^2 + 1), cos(x) = (1 - u^2)/(u^2 + 1) and dx = (2 du)/(u^2 + 1):
 = integral2/((u^2 + 1) ((2 u)/(u^2 + 1) + (1 - u^2)/(u^2 + 1) + 1)) du
Simplify the integrand 2/((u^2 + 1) ((2 u)/(u^2 + 1) + (1 - u^2)/(u^2 + 1) + 1)) to get 1/(u + 1):
 = integral1/(u + 1) du
For the integrand 1/(u + 1), substitute s = u + 1 and ds = du:
 = integral1/s ds
The integral of 1/s is log(s):
 = log(s) + constant
Substitute back for s = u + 1:
 = log(u + 1) + constant
Substitute back for u = tan(x/2):
 = log(tan(x/2) + 1) + constant
Which is equivalent for restricted x values to:
Answer: |= log(sin(x/2) + cos(x/2)) - log(cos(x/2)) + constant   P.S. log=ln

Happy New Year!!

Didn't even notice it is already 2017 until I see this post :P

Too focused on Maths recently :D

It is Okay you don't know how to use LaTeX, you still have 2 ways to give me the answer......

1) Do the LaTeX with assistance of my LaTeX tutorial posts in the 'Sticky Topics'

2) Write it on a paper, Take a photo, import in computer, and use the Image function of the forum to import the photo into the post.

Sorry Max: I can give you the answer but don't know how to use LaTex!!.

The answer is correct.

Btw Happy New Year :)

Thanks very much Max.

It is really great to know about all these wonderful internet tools.

I did not know about either of these sites before :)

Hi Max

separate variables to get

(1+x^2)dy = -2xydx 

1/y  dy      = (-2x dx)/(1+x^2)

intergrate lhs wrt y    and rhs wrt to x

ln y  = -  ln (1+x^2)    = ln(1/{1+x^2} )

anti log both sides

y= 1/(1+x^2)

and a happy new year

Also this awesome derivative calculator here!!

I did not pay for the W|A so I do not have access.

THANKS EP !!!

The integral calculator that EP found works and absolute treat!

Check it out !!!

Oh I just reread EP's suggestion.

Maybe they do it much better there.

I must check it out - thanks EP

Max, do you have access to Wolfram|Alpha,

the pay version?

I did it there.

I've downloaded the answer to my computer but Tiny Pics refuses to upload it - probably becasue it is very long .

So I can't post it easily.  

Do you want me to continue to try and post it here or have you got the problem sorted now.

(I have not personally worked through any of the mathematics for it.)