re write the equation as a function of x?
924395
-3x + 4y = 11  ~add 3x to both sides~\n4y = 3x + 11  ~ divide by 4 on both sides~\ny = 3/4x + 11/4
961595
umm..are you missing an equals sign somewhere or am i really dumb? how do you figure out what x is unless you have both sides of the equation?
720847
Use logarithms. If you haven't learned them yet, it just helps to solve equations with powers.\na = b^x\nlna = ln(b^x) = xlnb\nx = lna/lnb\nx = ln(10)/ln(32.6087)\nx = 0.6607928852\nYou can use your calculator to check. I guarentee that it works.
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.
60538
if [ -7x + x^2 =0 ]\nthe you will get y = 7/x
991696
To define an equation, you can use the syntax\nf := x -> (k*x*(1-x));\nWhich is creating a function defined on a variable x.\nThen, you can set the value of k by saying \nk := 2;\nIf you then say\nf(.1);\nit will give you the value of f(.1).  You can repeat this with the value you get from this.
618071
simplify to f(x)=(sqrt(x-2)(x-3))/(x-4)(x+5)\nthen sub in 1/x\nto get (sqrt(1/x -2)(1/x -3))/(1/x -4)(1/x +5)
909269
Take the derivative of the original function. For your given example, if I remember my calculus correctly:\nf(x)=y\nf'(x)=1\nGraph the new equation.
1041686
That's easy:   f'(x)=\n\n2*(x)^(3/2) - 3 - (x)^(1/2)\n--------------------------------\n6*(x)^(5/2)
363348
Are we solving for x?
852324
We have f(x) = (x)^(7x) = e^(7x*ln(x)). So, f'(x) = e^(7x*ln(x)) * d/dx(7x*ln(x) = . (x)^(7x) *7 *(x *1/x + ln(x) = 7* (1+lnx) * (x)^(7x)
998890
If you mean to solve it, it would be this:\nF:xtimesx=x^2\nO:xtime6.5\nI:xtimes-4.5\nL:-4.5times6.5\nEquals:x^2+2x-29.25
26362
Part 1:\n\nWe can rewrite y^x as (e^ln(y))^x which is e^(xln(y))\n\n∂f/∂x f(x,y)= ∂f/∂x (x^y) + ∂f/∂x(y^x)\n\n∂f/∂x f(x,y)= yx^(y-1) + ln(y)e^(xln(y))\n\n∂f/∂x f(x,y)= yx^(y-1) + ln(y)y^x\n\n\nPart 2: There is no simple function which is the integral of e^(x^2).
689918
use  power rule and product rule\nf'(x)=[1/5]*5*x^4 + 2x(x^2-x-9)+ \n(x^2+9)(2x-1) + 0=\n\nx^4 + 2x^3 - 2x^2 -18x + 2x^3 - x^2 +18x -9 =\n\nx^4 + 4x^3 - 3x^2 -9
887664
let u = x^2\ndu = 2x dx\n\nThis turns the function in to:\n1 / (u^2 + 1) - which you can work from.
80570
f(x) = x^2 - 1\nIf you replace x with x + h, then\nf(x + h) = (x + h)^2 - 1\nYou will NOT do any operations when changing the function. Instead, just SUBSTITUTE. (A technique is putting parentheses in the variable to change, except if it already has its own parentheses. Then you may exchange the variables inside the parentheses with what you want to substitute)\nf(x) = x^2 - 1\nPutting the parentheses, we get\nf(x) = (x)^2 - 1\nThen replace x with x + h\nf(x + h) = (x + h)^2 - 1\nHope this helps ;-)
853022
It should be a function because one x input value gives only one y output value. Try the vertical-line test. But the function does not have an inverse though.