1i) a.

1g) (-4) + (+6) + (-|-3|) + [- (-5)] = (-4) + (+6) + (-3) + (+5) = +4

2i) c.

2g) 12 - (8 ÷ 4) x 2 -(-2 )(-1) = 12 - 4 - 2 = 6

3i) b.

3g) Leftover area = area of the cardboard (50 x 60) - total area of the box: [2 x  (12 x 15 + 12 x 20 + 15 x 20)] = 3000 - 1440 = 1560; 1560 cm²

4i) c.

4g) (-1)³ (-2)² (-1) = (-1)(+4) (-1) = +4

5i) a.

5g) b.

6i) b.

6g) a.

7i) a.

7g) Let the number of teachers be t. Then , so t = 27.

8i) b.

8g) d.

9i) The ordered pairs of (0, -2), (1, 1), and (-1, -5) satisfy the equation and are the coordinates of points on the line. (There are many other possibilities.)

9g) They intersect at (-1, 4).

10i) c.

10g) If the enrollment was p, then So p = 55,000 students.

11i)

11g)

12i) a.

12g) c.

13i) d.

13g) P (yellow, then red or blue) =

14i) b.

14g) f(x) = -3x + 10
f(x² + 2) = -3(x² + 2) + 10
f(x² + 2) = -3x² - 6 + 10
f(x² + 2) = -3x² + 4

15i) c.

15g) The equations are r = 2w + 3 and r + w = 78.

To solve for r:	w = 78 - r
	r = 2(78 - r) + 3
	r = 156 - 2r + 3
	r = 159 - 2r
	3r = 159
	r = 53

To solve for w:	53 + w = 78
	w = 78 - 53
	w = 25
	25 wrens, 53 robins

16i)

16g)

17i) Let the length of one side be x. Then two consecutive sides and the diagonal form a right triangle. Apply the Pythagorean theorem to write x² + x² = 64. The solution is The length of a side is

17g) Note that a right triangle forms. Use the Pythagorean theorem to find the required length: 12,000 ft.

18i) slope = -⁴/₃; equation: y - 4 = -⁴/₃(x + 3) or y = -⁴/₃x

18g)

19i) a.

19g) Let the needed time be t. So 12 5 = 6 t, and t = 10; 10 hours.

20i) (no real number solutions)

20g) x² + (x - 14)² = 26². The positive solution of the equation is x = 24. The lengths of the sides are 24 cm and 10 cm.