Answer:
A(2,2)
Step-by-step explanation:
Let the vertex A has coordinates 
Vectors AB and AB' are perpendicular, then

Vectors AC and AC' are perpendicular, then

Now, solve the system of two equations:

Subtract these two equations:

Substitute it into the first equation:

Then

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)
Answer:
You subtract because you're taking away money when you're withdrawing.
Step-by-step explanation:
- The numbers to the <u>right</u> of 0 are <u>positive</u>. The numbers to the <u>left</u> of 0 are <u>negative</u><u>.</u> 0 is neither positive nor negative.

The multiple zero is x = -4.
The multiplicity is 2 because there are two values of x.
18 ÷ 3 is 6, which means that Daniel is 6, and there is a 12 year gap between Steve and Daniel.
12 x 2 is 24, so therefore when Steve is 24, Daniel will be 12, so Steve is twice as old as Daniel:)