rewrite each equation as a function of x.?
491024
y = f(x) = 3/4x + 11/4
665576
x^2<4x\n\ncase 1: x<=0\nno solution\n\ncase 2: x>0\ndivide by x\nx<4\n\ntherfore the solutions are 0<x<4
844394
If you have Microsoft Office Discs, then load the Equation Editor. It will allow you to write all of these.
990847
isolate the variable\n\n-7x+1+8x=x+1\n-7x+8x=x\nx=x\nx=1\n\nthe answer is c) 1\nnow give me 10 points please!!!
846527
You'll have to do this in more than one step.  Basically it will sort of have to be a program.  I would start with the restart; command and then  define the function (I would look this up in the help menu since it is dependent on 2 variables).  Then use a command to find the derivative.  Then a command to solve for the pairs (x,y) when the derivative is 0.  That will give you 1.  For 2 and 3 you will also have to do a series of commands.  Use your calc book to refresh your memory on the procedure of finding these, it should be a similar procedure in maple.
863644
take out 4x first\n4x(x2-6x+8)\n4x(x-4)(x-2)
906655
I assume that the last ex you wrote is e^x too, right?\ny' =  e^(e^x) + x [e^(e^x)] (e^x)\ny' = e^(e^x) [1+ x e^x]\n\nY''= e^(e^x)[ xe^x + e^x] + e^(e^x)  e^x [1 + xe^x]\ny''= e^(e^x)[  xe^x + e^x + 1 + xe^x]
88953
=\na(x-y)+x(x-y)\n=(a+x)(x-y)\ndone
518633
Call  Y of x like Y(x) \n\nY(x) = Y(x-1)+ (3*4^x)-1(1/9)x\nY(x) = Y(x-2)+ (3*4^(x-1))-1(1/9)(x-1) + (3*4^x)-1(1/9)x\nY(x) = Y(x-3)+ (3*4^(x-2))-1(1/9)(x-2) (3*4^(x-1))-1(1/9)(x-1) + (3*4^x)-1(1/9)x\n...\nY(x) = Y(1) + 3(4^x+...+4^2) - (1/9)(x+...+2)\nY(x) = Y(1) + 3*4[4^(x+1)-1]/[4-1] - [x(x+1)/2-1]/9\nY(x) = (sqrt3)/4 + 4^(x+2) - 4 - [x(x+1)/2-1]/9\nY(x) = 16 * 4^x - x(x+1)/18 - (37/9 - (sqrt3)/4)
408554
yes. take out an x \nyou answer should be x(x+3)
879176
= x(x+3) + 4(x+3)\n= x^2 + 3x +4x + 12\n= x^2 +7x + 12
913334
x^2 + 2x + 7 Remainder 10\n\ni can add the steps if you don't think i did it by hand
780298
first u distribute the - to the stuff in the parenthesis so you have\n7x^2-3+x^2\nthen combine like terms\n8x^2-3
518727
Multiply the first equation by 2.\n\nSubtract the second equation from the new first equation.\n\nSolve for y.\n\nTake your value for y and substitute it into the original first equation.\n\nSolve for x.
887664
let u = x^2\ndu = 2x dx\n\nThis turns the function in to:\n1 / (u^2 + 1) - which you can work from.
921998
is this for math?\nif so:\nTake the function c(x) and put it in for every x in the function d(x).\nSo, if \nc(x) = 2\nand \nd(x) = 3x^2 + 5x -1\nthe new equation would look like\nd(c(x)) = 3(2)^2 + 5(2) -1\nfor a final answer of:\n21
816542
Find derivatives with respect to each variable.