Melody

Try watching this.

It matters to me Chris.

Thank you for presenting your question nicely :)

Please post your question one at a time.

and please type your worded question in so that they can be copied into the answer easily. 

Using the digits 1, 2, 3, 4, 5, how many even three-digit numbers less than 500 can be formed if each digit can be used more than once?

The hundreds digit can be 1,2,3 or 4   that is 4 choice       (less than 500)

the tens has 5 choices.               

The units digit has to be even so that is 2 choices

so there are  4*5*2=  40 possible numbers

without brackets your meaning is not clear

This is your question

2√52x^9/√13x =  2(√52)x^9/(√13)x =  2(sqrt52)x^9*x/(sqrt13) 

But I suspect that you actually mean

\(\frac{2*\sqrt{52}\;x^9}{\sqrt{13}\;x}\\ =\frac{2*2\;x^9}{x^1}\\ =4x^8 \)

I do not think that this is the whole question.

Yes i think that you are right.

I made a careless error in my last answer.

I have now edited that answer and hopefully it is now correct.

My answer does correspond with what you have said :)

|x-3|+|y-1|=5

\(If \;\;x\ge 3\;\;and\;\;y\ge1\;\;then\\ x-3+y-1=5\\ x+y=9\\ y=-x+9\\~\\ When\;x=3\quad y=-3+9=6 \quad (3,6)\\ When\;y=1\quad x+1=9\quad x=8 \quad (8,1)\\ \text{The segment of line joining (3,6) to (8,1) satisfies this} \)

------------

\(If \;\;x\ge 3\;\;and\;\;y\le1\;\;then\\ x-3-y+1=5\\ x-y=7 \\~\\ When\;x=3\quad 3-y=7 \quad y=-4\quad ( 3,-4) \\ When\;y=1\quad x-1=7 \quad x=8 \quad (8,1)\\ \text{The segment of line joining (3,-4) to (8,1) satisfies this}\)

-----

\(If \;\;x\le 3\;\;and\;\;y\ge1\;\;then\\ -x+3+y-1=5\\ -x+y=3\\ When\;x=3\quad y=6\quad \quad (3,6)\\ When\;y=1\quad x=-2 \quad (-2,1)\;\;\\ \text{So the points on the line segment joining (3,6) to (-2,1) are solution.} \)

---

\(If \;\;x\le 3\;\;and\;\;y\le1\;\;then\\ -x+3-y+1=5\\ -x-y=1\\ x+y=-1 \\~\\ When\;x=3\quad y=-4\quad (3,-4)\\ When\;y=1\quad x=-2 \quad (-2,1)\\ \text{The segment of line joining (3,-4) to (-2,1) satisfies this}\)

So the solution set is all the points on the perimeter of this square

It just takes practice :)