aysuccess99:
wow, great workings bro....wow!!! thank GOD o, i have gain something 2day.....and tanx 4 checking,correcting and answering d question...more wisdom, knowledge and understanding to ur brain, and more battery or charger 2 ur calculator, oh, i have 4goten dat u self gan u are a calculator bro, gud work sir...kudos to u.,

Ortarico:

hahahahahahahahahahahahahahahaha, IDONBILIVIT!

smurfy:

Believe me, you are.

smurfy: A man has 10 gallons of a 50% sulphuric acid mixture, and 20 gallons of a 20% solution, and 5 gallons of a 10% solution. He wants to use up all the 10% solution and make 15 gallons of 30% solution. How much of each solution should he use?

Alpha Maximus:

smurfy:

Believe me, you are.

smurfy: A man has 10 gallons of a 50% sulphuric acid mixture, and 20 gallons of a 20% solution, and 5 gallons of a 10% solution. He wants to use up all the 10% solution and make 15 gallons of 30% solution. How much of each solution should he use?

smurfy:

Stop smiling at me SIR and attempt my WORD PROBLEM, thou word problem lover.

Alpha Maximus: Alright, I think I've cracked it.....solution loading, Ortarico watch and learn! *cracks knuckles in preparation*

Alpha Maximus: Hmmmmm, dis is actually the right solution but this looks copied and pasted. I smell a Cotonou rat in your source of solution. You failed to give an in-depth explanation as to how you arrived at some points thus making the stench of this Cotonou rat even more repugnant!! I'll post mine anyways........

Ortarico:

Here is another trial;
Equation:
acid + acid + acid = acid
-----
0.10*5 + 0.50*x + 0.20(10-x) = 0.30*15
-------
Multiply thru by 100 to get:
10*5 + 50x + 20*10-20x = 30*15
--------
50 + 30x + 200 = 450
30x = 200
x = 20/3 = 6 2/3 gallons (amt. of 50% solution needed)
10-x = 3 1/3 gallons (amt. of 20% solution needed)

Since all the 10% solution is to be used up, the needed solution for the 50% solution is 6 2/3 gallons while for the 20% solution 3 1/3 gallons is needed.
-----------------------------------
Cheers

Alpha Maximus: The man requires a total amount of 15 gallons of a 30% mixture. Since he has already decided to use all of the 10% solution, of which there is 5 gallons, there remains a 10 gallons of a 'X'% solution of which is meant to be constituted by the untapped 50% and 20% solutions. The equation is Acid + Acid + Acid= Acid(obviously) The first acid amount is given by 0.1*5(the 10% mixture o 5 gallons)= 0.5.........the next acid amount is 0.5*X(the 50% mixture)=0.5X...........and the last amount is 0.20*(10-X)(the 20% mixture and the remnant of the 10 gallons)=2-0.2X . Adding the individual amounts and equating to the total of 0.30*15(15 gallons of 30%solution)=4.5, we have: 0.5 +0.5X+2-0.2X=4.5, 0.3X=4.5-2-0.5, 0.3x=2, X=2/0.3=6.6666667 gallons(for the 50% mixture) and 10-x=3.3333333gallons(for the 20%mixture).......Taadaaa!!

smurfy:

Correct. I won't say more than that.

**leaves thread with a big frown**

Ediss:
Please help me find dy/dx if f(x) = x^x^x meaning x to power x to power x. thanks

smurfy:

Here goes...

Y = x^x^x
y = x^(x^2)

smurfy:

Here goes...

Y = x^x^x
y = (x^x)^x
Let u = x^x
Take log of both sides:
log(u) = log(x^x)
log(u) = xlogx
d/dx[log(u)] = d/dx(xlogx)
1/u(du/dx) = logx + 1
du/dx = u(logx + 1)
du/dx = (x^x)(logx + 1)

Recall that y = u^x
y = u^x
logy = log(u^x)
logy = xlog(u)
d/du(logy) = d/du[xlog(u)]
dy/du(1/y) = x/u
dy/du = xy/u
dy/du = x(x^x^x)/(x^x)

Now, dy/dx = dy/du * du/dx

dy/dx = x(x^x^x)/(x^x) * (x^x)(logx + 1)

dy/dx = x(x^x^x) * (logx + 1)

benbuks: @master smurfy, ur boy dey hail..i'v bin. viewin d thread from page 0...omo i don taya o...bt i go continue..

benbuks: I sight my boss, @Dx..greetins 2u sir.

smurfy: Find dy/dx if y = (arcsinx)^cos^2(x).

benbuks: ...hmn, chairman , u sure say u no form dis question from your head.?.. Abi na txt buk u c yam frm?

smurfy:

I didn't form it.

Did you just say 'yam'? You mean you saw through what I ate this morning from that question? Genius!

doubleDx:

Hailings Chief Ben! It's been a while, how the Math solving nah? I greet you bruv....

smurfy: Find dy/dx if y = (arcsinx)^cos^2(x).

benbuks: ..wait o, is it (arcsinx)(cos^2 x)
or (arcsinx)^cos^2 x ?

smurfy:

arcsinx raised to power cosx squared

benbuks: ...take ln of both sides
we have, lny= cos^2 x ln(arcsinx)
y'/y =-2sin2xln(arcsinx) + cos^2 x/arcsinx.sqrt(1-x^2)...u cn finish up na' u b master, no nid 4 details abi?