Equation 1.
20÷4=5
0×3=0
5-5=0
Equation 2.
44÷4=11
2×3=6
11-5=6
Equation 3.
56÷4=14
3×3=9
14-5=9
Answer:
She owe $37736.96 after 9 years .
Step-by-step explanation:
Debra borrowed $8000 at a rate of 18% compounded semiannually
We are supposed to find how much will she owe after 9 years
Principal = 8000
Rate of interest = 18% =0.18
No. of compounds per year = 2
Time = 9 years
Formula : 
Substitute the values in the formula :

A= 37736.96
Hence She owe $37736.96 after 9 years .
It doesn't matter. It's the same distance from zero
9514 1404 393
Answer:
A, C
Step-by-step explanation:
The attached graph shows which lines go through the given point. They are ...
y = 1/2x -1 . . . . 1st selection
y = -1/6x +3 . . . 3rd selection
__
The equations can be found algebraically by substituting the given point in the equation and seeing if the result is a true statement.
a) 2 = (1/2)(6) -1 = 3 -1 . . . true
b) 2 = -3(6) . . . . false
c) 2 = -1/6(6) +3 = -1 +3 . . . true
d) 2 = 2/3(6) -1 = 4 -1 . . . . false
e) 2 = 4(6) -2 = 24 -2 . . . . false
f) 2 = -3/2(6) +6 = -9 +6 . . . . false