Melody

\(x = {1+\dfrac{\sqrt{2}}{1+\dfrac{\sqrt{2}}{1+...}}}\).

\(x=1+\frac{\sqrt2}{x}\\ x^2=x+\sqrt2\\ x^2-x=\sqrt2\\ \)

    \(\frac{1}{(x+1)(x-2)}\\ =\frac{1}{(x+1)(x-2)}\\ =\frac{1}{x^2-x-2}\\ =\frac{1}{\sqrt2-2}\\ =\frac{-1}{2-\sqrt2}\\ =\frac{-1}{2-\sqrt2}\times \frac{2+\sqrt2}{2+\sqrt2}\\ =\frac{-(2+\sqrt2)}{4-2}\\ =\frac{-2-\sqrt2}{2}\\ =\frac{2+\sqrt2}{-2}\\ \)  .

When your answer is in the form   \(\frac{A+\sqrt{B}}{C}\)   ,

A=+2,   B=+2,   C= -2

where A, B, and C are integers, what is |A|+|B|+|C|?

|A|+|B|+|C| = 2+2+2 = 6

Here is one, by GingerAle and Hectictar, that Hectictar just asked me to include.  Thanks Girls :)

This is the original thread but I will includ most of the salient coding underneath.

Hi Chris,

This question confused me but what you have done is certainly correct. :)

I have included the graph here.

For the benefit of others.

The inverse of a function is the reflection of that function in the line y=x.

That is why I included y=x on my graph.

(xi – x)2 where xi = 1, 2, 3, 4, 5

Mmm

Do you mean

\(\displaystyle\sum_{\xi=1}^5\;(\xi-5)^2\quad?\)

\(\displaystyle\sum_{\xi=1}^5\;(\xi-5)^2\\ =(1-5)^2+(2-5)^2+(3-5)^2+(4-5)^2+(5-5)^2\\ =16+9+4+1+0\\ =30\)

Thanks XSquared   

Chris has already answered and that is great but I have a suggestion.

How about people answer on a new question thread and then put in an answer link.

That way the answer is not right with the question. It is more like having to look up the back of the book for the answer.

This might not work for a lot of people and that is fine, it is just a suggestion   

Hi Jeff

You maths is fine but you have not made either x or y the subject.

Yours is fine, you only had to put x on the left hand side  

what is (y-5)/2<-x/3

Make x the subject:

\((y-5)/2<-x/3\\ \frac{3(y-5)}{2}<-x\\ \frac{-3(y-5)}{2}>x\\ x<\frac{-3(y-5)}{2}\\ or\\ x<\frac{3(5-y)}{2}\\\)

Make y the subject

\(\frac{y-5}{2}<\frac{-x}{3}\\ y-5<\frac{-2x}{3}\\ y<-\frac{2x}{3}+5\\ \)

I am just displaying the question better. 

\(\frac{2}{1 \cdot 2 \cdot 3} + \frac{2}{2 \cdot 3 \cdot 4} + \frac{2}{3 \cdot 4 \cdot 5} + \cdots\)

guest there really is not a lot of point posting if you cannot give any working.

28n^4+16n^3-80n^2

\(28n^4+16n^3-80n^2\\ =4n^2(7n^2+4n-20)\)

I could use the quadrative formula to solve it but I will see if I can figure it out another way.

I need 2 numbers that multiply to 7*-20  and add to give 4

One will need to be positive and the other negative and since 4 is positive, the bigger number will be positive.

7*-20 = -7*2*2*5 = 14* -10  that works, the numbers are  14 and -10

change 4n into  14n-10n

\(=4n^2(7n^2+4n-20)\\ =4n^2(7n^2+14n-10n-20)\\ \text{Now factor in pairs}\\ =4n^2[7n(n+2)-10(n+2)]\\ =4n^2(7n-10)(n+2)\\ \)

I think guest has done something similar to me only more simply but I have not looked at it closely.

My answer was mathematically precise.  There are no other possible answers.