Calc question HELP please?
928672
let the base of the isosceles triangle be x\nheight of the triangle=h\ngiven the radius of the circle is 4\n(4-h)^2+(x/2)^2=4^2\n16-8h+h^2+x^2/4=16\n4h^2-32h+x^2=0\nx=rt(32-4h^2)\narea=1/2*x*h\n=1/2*h[2rt(8-h^2)]\ndA/dh=0 for max\nafter finding h sub to get x
492690
S(I)+3f (g)--> SF (g)\n..........2...........6
79903
Here is an excellent site:\nhttp://www.sosmath.com/calculus/calculus.html\n\nVery thorough explanations and examples provided.
782468
1.ac\n\n2.-3r\n\n3.3m/-2r\n\n4.(x-4)\n\n5.(a-b)(b-c)
566883
First, since 1/x is not defined for x = 0, before you can show  anything about your functions, you have to define values for g(0) and f(0). I'm going to assume that either the original statement of the problem had that or that you are supposed to find the value that makes the problem solvable.\n\nTo be true you have to define the problems with f(x) = 0 and g(x) = 0\n\n\nFor the continutiy of f(x) given epsilon > 0 find delta > 0 such that |f(x) - f(0)| < epsilon for all |x-0| < delta.\n\n|f(x) - f(0)| = |xsin(1/x) - 0| < |x| = |x - 0|\n\nGiven epsilon, pick delta = epsilon.\nIf |x-0| < delta, then |f(x) - f(0)| < epsilon.\n\n\nAs for the differentiability of g(x),\n\nf(0+h) - f(0)) = (0+h)^2sin(1/(0+h)) - 0\n\nlim(x->0)(f(0+h) - f(0))/h = \nlim(x->0)((0+h)^2sin(1/(0+h)) - 0)/h = \nlim(x->0) h^2(sin(1/h)/h = lim hsin(1/h) = 0\n\nSo the limit exists and is 0.
892157
wait 1 sec.\nOk this is reasonably difficult, you have to use the chain rule and the quotient rule.\nUsing the chain rule:\nderivative = [ (1/2) * ( (2x + 3) / (7x+4) ) ^ (-1/2) ]  *  [d/dx (2x+3) / (7x+4) ]\n\nSo now you to use the quotient rule to find:\nd/dx (2x+3) / (7x+4) = (14x+8 - 14x + 21) / (7x+4)^2\n= 29 / (7x+4)^2\n\nSo the derivative is [1/2 * ((2x+3)/(7x+4))^(-1/2)]*[29 / (7x+4)^2]\n\n\nBy the way i got the thumbs down before I wrote all this up, so it doesn't mean it's wrong.
1032260
1. if x=f(t) and y=f(t) then\n\nx^4+y^4=6y\n\nd/dt { x^4+y^4=6y}\n\n4x^3 (dx/dt) + 4y^3(dy/dt) =6(dy/dt)\n\nand you have to have the x=f(t) & y=f(t) in order to substitute for the derivatives.\n\n2. if y=f(x) and x=f(t) then\n\nx^4+y^4=6y\n\nd/dt { x^4+y^4=6y}\n\n4x^3 (dx/dt) + 4y^3(dy/dx)(dx/dt) =6(dy/dt)(dx/dt)\n\nthen substitute for the derivatives of the functions y=f(x) & x=f(t)
714589
x/(x+3) - 2(x-1)/(x+3)(x+2)\n\nx(x+2) - 2 (x-1)\n--------------------\n   (x+3) (x+2)\n\n\nx^2 +2\n----------\n(x+3)(x+2)
879979
f(x)=(x^2-2x)(x^2+2.44x+1.4884)\n =(x^4+2.44x^3+1.4884x^2-2x^3-4.88x^2- 2.9768x)\n =(x^4 + 0.44x^3 - 3.3916x^2 - 2.9768x)\n\nF(x)=(x^5)/5+(0.44x^4)/4-(3.3916x^3)/3-             (2.9768x^2)/2+c\n\nam i really asked a Q?
923039
email me, calcu_lust@yahoo.com, i'll write out a nice solution.\n\nEdit: Go to my website, and aforementioned nice solution is typed out as a .pdf for you to download, with all the correct symbols and such:\nhttp://geocities.com/calcu_lust/yahooans.pdf
858889
QUESTION 1:\n\nAsk yourself: WHICH TRIG FUNCTION IS OPPOSITE OVER ADJACENT?  IT IS THE TANGENT FUNCTION, RIGHT?\n\nThe bottom of the tree is 25 feet away from a point that measures 30 degrees from ground level to the top of the tree.\n\nAnswer: tan30degrees = x/25feet...choice 1.\n\nNotice that they are NOT asking you to solve for x.  Get it?  \n\nQUESTION 2:\n\nYou need to write the given equation 3x - 2y = 12 in the y =mx + b\nform.\n\nLet's do that right now.\n\nWe have:\n\n3x - 2y = 12 \n\nI will subtract -3x from both sides of the equation.\n\nWe get:\n\n-2y = -3x + 12...See what I did?  \n\nNext:\n\nDivide BOTH sides of the equation by -2 to find the value of y.\n\nWe get this:\n\ny = (-3x + 12)/-2\n\nWe can reduce the equation to this:\n\ny = 3/2 (x) - 6\n\nSlope is 3/2 and the y-intercept is -6.\n\nFinal answer: choice 1\n\nGuido
490047
I think a calculator would be great for this one! It sounds perfect for a math wiz to try...Good luck! MayMay
745882
C is correct but not complete... only the 3 is factored out... the real answer would be:\n\n3(x-1)(x+4)
953139
f'(x)=(4/3)x^(1/3) +(4/3)x^(-2/3)
501445
Try this website. http://www.webmath.com/index3.html
121355
well if your taking the derivative, then there should be no h's left over at the end, since they go to zero with the limit.  The derivative of that function is [5(5x+2)+25x]/(5x+2)^2.  I'll leave it to you to simplify.
969104
okay,\n\ncos(x-y)      cos(x)*cos(y) + sin(x)*sin(y)\n----------  =   ----------------------------------- \ncos(x+y)     cos(x)*cos(y) - sin(x)*sin(y)\n\n(multiply and divide the fraction by cos(x)*cos(y))\n\n     1 + tan(x)*tan(y)\n=   ---------------------\n     1 -  tan(x)*tan(y)