No, mx+ cx= (m+c)x. To solve (m+ c)x= y- c, you get x= (y- c)/(m+ c)ii) y=mx+cx+c
y-c = mx+cx
I'm not sure where to go after this, I would come out with a final answer of y-c/mc = 2x?
I don't know what you mean by this. Is that supposed go be y= ((x+ c)/a)/b or y= x+ (c/a)/b?iii) y=x+c/a/b - once again i'm not even sure where to go with this to start with
The equation x^2= y- c has two solutions, x= sqrt(y-c), which you give and x= -sqrt(y- c).iv) y=x^2+c
[quote]v) y=e^Lx -1
y+1 = e^Lx
not sure where to go with this, I imagine you use logs somehowAssuming you mean e^(Lx), yes. The definition of the (natural) logarithm is that it is the inverse function to e^x: log(e^x)= x. So if you take the logarithm of both sides, log(y+ 1)= log(e^(Lx))= Lx. Now divide both sides by L.
So am I! I have no idea what "cos" and "sin", could possibly mean without an argument! Is it possible you meant cos^2(x)+ sin^2(x) or xsin^2(x)+ xcos^2(x)? If it is either of those, use the fact that sin^2(x)+ cos^2(x)= 1.vi) y=xcos^2 + xsin^2
Not sure where to begin
All help is apprecaited! many thanks