1i) Let the numbers be n, n + 1, and n + 2. Then n + (n + 1) + (n + 2) = 126, 3n = 123, and n = 41. Therefore the integers are 41, 42, and 43.

1g) Let the amount in the tank before filling up be x (which could be 0). Then (12 - x)1.20 = 13.44, so x= 0.8. The tank was not empty; it held 0.8 gal of gas.

2i) Let the length of one diagonal be x. The two adjacent sides and the diagonal form a right triangle. Apply the Pythagorean theorem to write . The solution is x = 2, so the diagonal is 2 cm long.

2g)

3i) a.

3g) y - 1 = 3(x - 1), or y = 3x - 2 in slope-intercept form

4i) c.

4g) slope =; equation: y - 2 = ; in slope-intercept form

5i) a.

5g)

6i) c.

6g)

7i) b.

7g) a.

8i)

x = 3 or x = -1.

8g) ; t = 0 and t = 2
The projectile will return to the ground after 2 seconds.

9i) c.

9g) Let the numbers be n and n + 1. So the numbers are 15 and 16.

10i) b.

10g)

11i) a.

11g) b.

12i) c.

12g) Let the shorter side be x. Then the other side is x + 7. By the Pythagorean theorem, . The solution is x = 8, so the side lengths are 8 cm and 15 cm.

13i) a.

13g)

14i) c.

14g) 

y = 2x and x = -3, so y = -6.

15i) a.

15g) By adding the first and second equations we get (*) 4a + 3c = 14; by adding the second and third equations we get (**) 4a - 3c = 4. Now add (*) and (**) to get 8a = 18; a = ¹⁸/₈ = ⁹/₄.

16i) c.

16g) , so the parabola opens upward with the vertex at (-2, 1).

17i) 1. b.  2. a.

17g)

18i)

18g)

-1<x<6; The integers that satisfy this inequality are 0, 1, 2, 3, 4, and 5. There are six integers that satisfy the condition.

19i) c.

19g) There are 9 desired outcomes: (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), and (6, 6). The total number of possibilities is 36, so the probability is ⁹/₃₆ = ¹/₄.

20i) a.

20g)

There are 3 students without any siblings.