Vinculum

\(4x^2+5x+2-x^2+x+6 = 3x^2+6x+8\)

3(x^2+2x+(8/3))

3(x^2+2x+1-1+8/3)

3((x+1)^(2)-1+8/3)

3(x+1)^(2) + 5

-Vinculum

\(f(f(f(5)))=\)\(\frac{\left(\frac{\left(\frac{\left(x-1\right)}{\left(x+1\right)}-1\right)}{\left(\frac{\left(x-1\right)}{\left(x+1\right)}+1\right)}-1\right)}{\left(\frac{\left(\frac{\left(x-1\right)}{\left(x+1\right)}-1\right)}{\left(\frac{\left(x-1\right)}{\left(x+1\right)}+1\right)}+1\right)}\)

That simplifies to \(\frac{-x-1}{x-1}\)

Substituting 5 gives \(\frac{-\left(5\right)-1}{\left(5\right)-1}\)

=\(-3/2\)

So the guest in 3rd post was right

\(\displaystyle \prod_{n=0}^{99} 99^n\)

\(99^{4950}\)

As the guest before mentioned the last 2 digits are 01

-Vinculum

Hello Guest, this question has been answered in 2 other pages here:

Take a look(both have the same answers)!!

https://web2.0calc.com/questions/help-geometry_86109

https://web2.0calc.com/questions/diameter-of-a-right-cylindrical

-Vinculum

Hey this question has already been answered by CPhill

Take a look: https://web2.0calc.com/questions/diameter-of-a-right-cylindrical

-Vinculum

Thank you for checking Max!!!

-Vinculum

Hey Max, sorry for the confusion, I accidentally called the point (2, 1), P instead of A. I meant...

Let's say you got a vector, v = (1, 1)

We also have a point, A, at (2, 1). (This will be the tail of our vectors).

Then you got points X, Y, and Z (unknown points), which will be the head of our vectors.

Lets says that: 

Vectors AX = v, AY = -v, and AZ = 2v. What are points X, Y, and Z. 

I got: X: (3, 2) Y: (1, 0) and Z: (4, 3). Though I feel confident with my answer, I would just like to confirm with anyone else to make sure I got it right.

I did?

This question has already been answered

https://web2.0calc.com/questions/help-please_52172#r3

-Vinculum