There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).
1
If: u = 1+lnx Then: x = (u-1)/(ln)
ln(x+1)=0 eln(x+1) = x+1 = e0 = 1 tama ba ako Sharon Perez?
The equation is: ln(1+tx)=tx-(h/g)x^2 BTW
There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).
1
If: u = 1+lnx Then: x = (u-1)/(ln)
you need to know natural logarithms3e to the 2x-1 power = 8(2x-1) ln e = ln (8/3)ln e = 1(2x-1) = ln(8/3) = 0.982x = 1.98x = 0.99
-1 plus 1 = 0 plus 1 =1
ln(x+1)=0 eln(x+1) = x+1 = e0 = 1 tama ba ako Sharon Perez?
a+a=1 2a=1 a=1/2
The equation is: ln(1+tx)=tx-(h/g)x^2 BTW
i have no clue what the answer is
x = 5, y = -1
That factors to (a + 1)(a + b) a = -1, -b b = -a
e2x + ex = 0e2x = -exe2x / ex = -1ex = -1x = ln(-1)Actually, that's just a restatement of the original problem, but that'sas far as you can go with it, because there's no such thing as the log(or the ln) of a negative number. Your equation has no solution.