can someone help my with calculus asap?
546678
Hopefully you'll find this site helpful.\n\nhttp://www.sosmath.com/calculus/calculus.html\n\nThere is a lot of information and example problems on limits and derivatives.  Good luck with your calculus class!
741999
whats the equation or formula u need help with?
160821
1. Dy = 1/2 A*t^2 + Vy*t, Vy = 0 (speed is horizontal),\nso t = sqrt(66m/(1/2*9.8m/s^2)) = 3.67 s.\n   Dx = Vx*t = 6 m/s * 3.67 s = 22 m.\n\n2.(a)  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso t = sqrt(1.150m/(1/2*g)) = 0.484 s\n  (b)  Dx = Vx*t, so Vx = Dx/t = 0.385m/0.484s = 0.795 m/s\n\n3.  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso Dy = 1/2*9.8m/s*(2.9s)^2 = 41.2 m\n\n   Dx = Vx*t = 2.4 m/s * 2.9 s = 6.96 m\n\n4. Dy = 0, Dx = 97 m, a = 50 degrees\n   Dx = Vx*t = V*cos(a)*t, \nso t = Dx/(V*cos(a))\n   -1/2*g*t^2 + Vyt = Dy = 0 ( Vy = V*sin(a))\ndivide by t, so -1/2*g*t + V*sin(a) = 0, or\nt = V*sin(a)/(1/2*g)).  Equate the two expressions for t:\nV*sin(a)/(1/2*g)=Dx/(V*cos(a)), multiply both sides by V\nV^2*sin(a)/(1/2*g) = Dx/cos(a)\nV= sqrt(Dx/cos(a)/sin(a)*(1/2*g))\nV = 31.1 m/s
896136
try http://www.webmath.com/\nit gives you help on math on almost any time of math.
1136463
It's been a while since I had calculus, but I think it's something like this:\n\nYou'd have to take the integral from 0 to k (treat k as a constant). When you have that, take that value (which is equivalent to the area) and set it equal to 3. Then, algebraically, solve for k.
129427
Get "Calculus for Dummies."  It is an amazing prep book.
21911
Try "Calculus" - By Gilbert Strang\nThis is available along with Student study guide &\nInstructor's manual!\n\nhttp://ocw.mit.edu/ans7870/textbooks/Strang/strangtext.htm
30751
The Edifier has it right, but darned if I can follow the notation. You really need to see this written out like a textbook to understand what's going on.\n\nClick the link below in "sources" to see the answer hosted at my website www.calculus-help.com. Right now, Yahoo! answers, as far as I can tell, doesn't allow graphics, which makes it tricky to answer ANY math questions.\n\nPlease don't post math answers, guys, if the poster knows the answer but can't get there, and you get something completely different.
392166
Try this guy:\n\nhttp://dli.taft.cc.ca.us/mathvideo/Calculus/Calculas.htm
460400
You would need to know how to integrate (that is, find anti-derivatives). The proceedure forthis type of problem can usually be understood after first semester calculus. The specific integrations may need some second semester calculus.
986880
Do you want how to derive them or what the derivatives are?\n\nF(x)=a^x  ---> f'(x) = a^x (ln a)\n\nf(x) = sin x ---> f'(x) = cos x\n         cos x      --->           -sin x\n         tan x       ---->           sec^2 (x)\n         sec x      ---->           sec x tan x\n         csc x      ---->           - csc x cot x\n         cot x       ---->            - csc^2 (x)
635193
Drop me your email. I will do one and send it to you. Easier than explaining...just look at my formulas...will only take a minute...really...\n\nCheck your email\n\nIs what I gave you what you want?
428665
This one is fairly clear.  Hope it helps.\n\nhttp://www.math.hmc.edu/calculus/tutorials/substitution/
154380
I posted these sources earlier to someone else who was looking for math help.\n\nGood Luck!!
805503
need to take the derivative vector:\n( -3 sin t, 4 cost)\nwhen t= pi/4, the vector is: \n(- 3 sin(pi/4), 4 cos(pi/4)) = ( -3/sqrt(2), 4/sqrt(2) )\nnow, the slope of the tangent line is:\n (4/sqrt(2) ) /  ( -3/sqrt(2) ) =-4/3\nso the equation of the tangent line is:\ny- 4sin(pi/4) = -4/3 (x-3cos(pi/4)) \ny-4/sqrt(2) = -4/3( x-3/sqrt(2) )\n'
183373
Try the inside cover of your Calculus book - it usually summarizes the rules for derivatives and some key integrals.
515516
goto purple math.com it will hopefully help you