Verified answer

2nd : you don't shade any circle , rather everything above the line y = 3

2st ; either factor or use the quadratic formula...[ x - 5 ] [ x - 1 ] = 0 , solve

x² - 6x + 5 = 0 complete the square by adding 4 to both sides

x² - 6x + 9 = 4 factor the left side

(x - 3)^2 = 4 take the square root of both sides

x - 3 = +/- 2

x - 3 = 2 or x - 3 = -2

x = 5 or x = 1 answers

y - 2 > 1

y > 3 since it is greater then (not equal to), you do not shade in the circle

1. Write down two sets of parenthesis ( ) ( )

2. LOok at your first term, which is X2. X times X would equal X squared, so place X as the first term.

(X ) ( x )

3. next, you want to determine what terms will be multipled to give you a 5 on the outside of the equation, as well as a -6 in the middle.

What times what will give you a positive 5? These two terms added together should also give you a -6.

In this case, it will be a -5 and a -1.

You will now have: (X-5) (X-1)

For your second question, anything that could equal that number should be shaded. So if you put a circle around 3, and the point could equal 3... shade it.

Well to factorise I find a good method is to first get 2 brackets like this

(x )(x )

Then, find 2 numbers which add up to the middle x value (-6x) and multiply together to make the last number (5). In this case, -5 and -1. Then put them in the brackets.

(x-5)(x-1)=0

Solve separately for both brackets.

x-5=0

x=5

x-1=0

x=1

x is either 5 or 1

(x+5)(x+1)

X=-5 and -1

Y>3 is a line so I have no clue what the circle is you're talking about