so school is starting again .0.

Could I be walked through a few practice questions? I've forgotten everything over the summer and a refresher would be nice. I'm not looking for answers, just some help/pointers!

1. Solve for x. |2x - 4| >10
(for this one don't I have to do something with making the first half a -2x and positive 4?)

2. The variables x and y vary directly. Write an equation that relates x and y. Then, find y when x=-3.
x=4, y=-12
(not quite sure how to start/deal with this one)

3. determine the x-intercept and y-intercept of the line with the given equation. 10x+3y=20
(?)

4. What is the vertex of the parabola? y=-3(x-4)^2+2

I might ask a few more if anyone is willing to help! I've just forgotten how to do a majority of these over the summer and I'm scared to get called on in class and not know how to solve the problem haha
thanks!

4 answers

1. Solve for x. |2x - 4| >10
(for this one don't I have to do something with making the first half a -2x and positive 4?)

sort of
(2x-4) > 10 and -(2x-4)>10
2x >14 and 2x-4<-10
if you multiply both side of inequality by negative, reverse arrow

x > 7 and x < -3
2. The variables x and y vary directly. Write an equation that relates x and y. Then, find y when x=-3.
x=4, y=-12
(not quite sure how to start/deal with this one)
==================
y = k x
-12 = k (4)
k = -3
y = -3 x
y = -3*-3 = 9
3. determine the x-intercept and y-intercept of the line with the given equation. 10x+3y=20
(?)

x intercept when y = 0
so when x = 20/10 = 2
y intercept when x = 0
so when y = 20/3
4. What is the vertex of the parabola? y=-3(x-4)^2+2
=======================
LOL (x-4)^2 is the same as (-x+4)^2
so function symmetric about x = 4
(look at (x-4)^2 when x = 3 or x = 5)
so vertex at x = 4
then
y = 2
(4,2)