Answer: D
Step-by-step explanation:
all possible rational zeros are the factors of the last term divided by the coefficient of the first term
so it's (±1, ±3, ±9) / (±1, ±2)
(±1, ±3, ±9) / ±1 = ±1, ±3, ±9
(±1, ±3, ±9) / ±2 = ±1/2, ±3/2, ±9/2
--> ±1, ±3, ±9, ±1/2, ±3/2, ±9/2
The point of elimination is to have one variable so example
3x + 2y = 4
7X + 3y = 5
1) Try to find a way to have one variable
7(3x + 2y = 4) ----> 21x + 14y = 28
-3(7x+ 3y = 5) -----> -21x -9y = -15
2) add
0 + 5y = 13
3) solve for the last variable
5y = 13
y = 13/5 = 2 3/5 = 2.8
4) Subsitute the variable to get the other one
3x + 2(2.8) = 4
3x + 5.6 = 4
3x = -1.6
x = 0.533333333 (continue)
The intersection would be 0.53 with a line above the 3 and 2.8. (0.5333 , 2.8)
Hope you understand!!
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