Answer:
I hope this helps
Ex: vertical translation |x+5|-2 the graph will shift down 2 units. Or the same as (x+5)+2 in this case the graph will shift up 2 units.
Horizontal translation |x+5|-2 the graph will shift to the left 5 units. Or same as (x-5)-2 in this case the graph will shift over to the right 5 units.
Step-by-step explanation:
Vertical translation is when a graph is being shifted up or down.
Horizontal translation is when a graph is being shifted over left or right.
There are two ways to do this but the way I prefer is to make one of the equations in terms of one variable and then 'plug this in' to the second equation. I will demonstrate
Look at equation 1,

this can quite easily be manipulated to show

.
Then because there is a y in the second equation (and both equations are simultaneous) we can 'plug in' our new equation where y is in the second one

which can then be solved for x since there is only one variable

and then with our x solution we can work out our y solution by using the equation we manipulated

.
So the solution to these equations is x=-2 when y=6
The answer is the 3rd option
30-60-90 triangle for reference
Remember soh cah toa
O:opposite, a:adjacent, h:hypotenuse
S:sin c:cosine t:tangent
Hope this helped :)