Try a few terms!

a1=1/(1-x), a2=1/(1-a1)→1/(1-1/(1-x))→1-1/x, a3=1/(1-a2)→1/(1-(1-1/x))→x,

continuing in this fashion we find

a4=1/(1-x), a5=1-1/x, a6=x ...etc.

So a107=...

(I’m sure you can take it from here)

I think this should probably read as:

\(\text{if } a_1=\frac{1}{1-x},a_2=\frac{1}{1-a_1}, \text{ and }a_n=\frac{1}{1-a_{n-1}}\text{, for }n\ge2, x\ne0,\text{ find }a_{107}\text{ in simplest form}\)

My attempt at a solution:

.

The difference is given by p(x^{2} - (x-2)^{2}) + q(x - (x-2)) or (4x-4)p + 2q

For this to equal (2x - 1)2 we must have 4p = 4 or p= 1 and 2q - 4p = -2, so 2q - 4 = -2 or q = 1

p+q = 2

"If A its nxn Matrix and A^4 - A^3 + A^2 - A + Ιnxn = 0nxn, prove Α^-1 = -Α^4"

"What is the value of k if point P(k, k sqrt(3)) is 4 units from the line 5x+3y=12?"

Presumably, you want to know the values of each of the angles. If so:

Drawing the triangle might help:

(and, no, it's not "adjacent").

In general lg(u^{a}) = a*lg(u). When a is -1 then lg(u^{-1}) = -1*lg(u) or lg(1/u) = -lg(u)

Here's my take:

As follows:

You can do matrix calculations on the home page calculator here. See the following example:

You might be making this more complicated than necessary!

As \(x\rightarrow -\infty\) the 2 in the numerator is negligible compared with 9x^{2}, and the -9 in the denominator is negligible compared with the 2x. So:

\(\lim_{x\rightarrow -\infty}\frac{\sqrt{9x^2+2}}{2x-9}\rightarrow\lim_{x\rightarrow -\infty}\frac{\sqrt{9x^2}}{2x}\rightarrow \lim_{x\rightarrow -\infty}\frac{3|x|}{2x}\)

Now, when x is negative \(\frac{|x|}{x}=-1\) so the limit is -3/2.