Y = x^2 - 6y + 8
y = (x - 3)^2 - 9 + 8
y = (x -3 )^2 - 1
vertex is at ( 3, -1)
2x^2 - y when x + 3 and y = -2
2(3)^2 - (-2)
2(9) + 4
18 + 4
22
Hope this helps,
Brian
Answer:
all reals except 3 & 4
Step-by-step explanation:
recall you cannot have a 0 in the denominator.
any values of x that would produce a 0 in the denominator are excluded from the domain
factor 
Factors -3, -4
(x-3)(x-4)
x cannot be 3 or 4
5 is A, because it's the only number that is whole when square rooted.
6 is confusing because it only gives you one number per answer, unless you are to add the two distances together? Then it would be 7 because X's distance is +4 and Y's distance is +3
I do not have an answer for 7
8 is 3.7 (or C)
9 is an easy one. Find out what the L and W is, and since it's a square we only need to find one of them. We square root the area and we are given 14.96662955, if rounded we get 15. So the answer is A
10 I don't understand the "estimate" for a square root number, I was not taught that.
