## A1

1. (a) 216 (b) 4/9 (c) -1 (d) 0.09 (e) (2.0)4 = 16 (f) 26 = 64 (g) (2 ■ 3 • 4)2 = (24)2 = 576 (h) 66 = 46656

2. (a) 153 (b) (-1/3)3 (c) 10"' (d) 10"7 (e) t6 (f) (a - b)3 (.g)a2b4 (h)(-a)3

4. (a) 26 = 64 (b) 64/27 (c) 8/3 (d) x9 (e) y12 (f) 8x3y3

5. (a) 0 (b) Undefined, (c) 0 (d) 0 (e) 1 (f) Undefined, (g) 1 (h) 1

6. (a) x = 5 (b) x = 0 (c)x = 3 (d)x = 4 (e) x = S (f)x = 0

7. (a) False. 35 = 243, 53 = 125 (b) False. (52)3 = 56, whereas 52' = 58. (c) True. 0a")« = (a?Y (d) True. O3 - 4° = 0 • 1 = 0 (e) False. 0~2 is not defined, (f) False. (5 +1)2 = (12)2 = 144, 52 + 72 = 25 + 49 = 74

(g) False. The correct value of the ratio is (2x + 4)/2 = x + 2. (h) True. Both equal Ix - 2y. (i) True.

8. (a) False, a0 = 1. (b) True. c~n = 1/c" for all c ^ 0. (c) True. am ■ am = am+m = a2™. (d) False, unless m = 0. ambm = (ab)m. (e) False (unless m = 1). For example, (a + b)2 is equal to a2 + lab + b2. (f) False (unless ambn = 1). For example, a2b3 is not equal to (ab)2+3 = (ab)6 = a6b6.

9. (a) x3y3 = (.xy)3 = 33 = 27 (b) (ab)4 = (-2)4 = 16 (c) (a20)0 = 1, for all a ^ 0. (d) 2n is 0, ±2. ±4 so (-1)2" = [(-l)2]" = 1" = 1

(e) x3y3 = (x-ly~1)-3 = 3"3 = 1/27 (f) (x~3)6U2)2 = = x"14 =

(x7)-2 = 2"2 = 1/4 (g) (z/xy)6 = (xy/zr6 = [(.xy/zT2]3 = 33 = 27

(h) {abcf = (fl-'fc-'c-')-4 = (1/4)"4 = 44 = 256

10. (a) 16x4 (b) 4 (c) 6xyz (d) a21b9 (e) a3' (f) x"15 (g) a4 (h) 5"«

13. (a) Given a constant interest rate of 11% per year, then in 8 years, an initial investment of 50 francs will be worth 50 • (1.11)8 % 115.23 francs.

(b) Given a constant interest rate of 12% per year, then in 20 years, an initial investment of 10,000 rand will be worth 10,000- (1.12)20 ^ 96,462.93 rand.

(c) 5000 • (1.07)-10 ^ 2541.75 crowns is what you should have deposited 10 years ago in order to have 50Q0 crowns today, given the constant interest rate of 7%.

14. 2]0 = 1024 and 103 = 1000. So 230 = (210)3 is bigger than (103)3 = 109. A calculator should say that 230 = (1024)3 = 1,073,741,824.

1. (a) 3 (b) 40 (c) 10 (d) 5 (e) 1/6 (f) 0.7 (g) 1/10 (h) 1/5

3. (a) |V7 (b) 4 (c) 1^6 (d) 1 (e) ±V6 (f) 2V2^/y (g) V2I/2 (h) a: + jx _

4. (a) =, by [A.6]. Also, both expressions = 20. (b) In fact, V25 + 16 =

= V25 + VI6 (c) In fact, (a1/2+fc1/2)2 = a+2a1^1/2 +£ = a + only when ab = 0. (d) =. In fact, (JaTb)-1 = [(a + ¿)1/2]"] = (a + b)'l/2

1. (a) 1 (b) 6 (c) —18 (d) —18 (e)3x + 12 (f)45*-27>- (g) 3 (h) 0 (i)-l

2. (a) 3a2 - 5Z> (b) -2x2 + 3x + Ay (c) t (d) 2r3 - 6r2j + 3rs2 + 2j3

3. (a) —3n2 + 6n — 9 Q>)x5 + x2 (c)4n2-lln + 6 (d) -18a32>3 + 30a3fc2

4. (a)a2-a Qo)x2+Ax-2\ (c)-3+3-^2 (d)3-2v/5 (e) x3-3x2+3a:-1

5. (a) 2x (b) a2 - Aab + Ab2 (c) \x2 - ±y2 (d) -;t2y - 3* - 2 (e) x2 + (a +*b)x + ab (f) x3 - 6*2y + 12*y2 - 8y3

6. (a)2r3-5r2+4r-l (b)4 (c) ;t2+2xy+2xz+y2+2yz+:r (d) 4jc>-+4jcz

7. (a)9;c2 + 12*y+4y2 (b) 5 + 2^6 (c) 9h2-48ku + 64u2 (d)w2-25u2

8. 500 (Note that (252)2- (248)2 = (252 + 248)(252 - 248) = 500-4 = 2000.)

9. (a);t4-2x2y2+y4 (b) 1/2 (c) a2-2ab+b2+2a-2b+\ (d) a-lVab+b (e) 1 (f) n4 — An3 + 6n2 —4/7+1

10. (a) acx2 + (ad + bc)x + bd (b) 4 - f2 (c) a2 + b2 + c2 + lab + 2ac + 2be (d)a10-b10 (e) 2715 + 1 (f) u4 - lu2v2 + v4

11. In the first figure, the (big) square has sides of length a + b, so its area is (a+b)2. The four rectangular parts have a combined area a2+ab+ab+b2 = a~ + lab + b2. The two ways of computing the area must give the same result, so (a + b)2 = a2 + lab + b1. The interpretation of the second figure is similar.

12. (a10 — bl0)/(a —b) =a9 + a*b + cPb2 + a6b3 + a5b4jra4b5 + a3b6 + a2b7 + ab8 + b9

1.(a)2-2-7 a a-b-b-b (b)2-2(x + 2y-6z) (c)2x(x-3y) (d) 2aab-b(3a + 2b) (e) lx(x — ly) (f) 5x-y-y(l-3jc)(1+3jc) (g) (A+b)(A-b) (h) 3(x + l)(x — 2)

2. (a) (x - 2)(x - 2) (b) 2 ■ lts(t - Is) (c) 2-2(2a + b)(2a + b) (d) 5xx(x + V2y)(jc - V2y)

2. (a)* = 3 (b) x = —7 (c) x = -28/11 (d)x=5/ll (e) x = 1 (f)x = 121

4. (a) 2x + 5 = x — 3. Solution: X = -8. (b) x + (x + l) + (x + 2) = 10 + 2*. Solution: x = 7, so that the numbers are 7, 8, and 9. (c) If x is Ann's regular hourly wage, then 38* + (48 - 38)2* = 812. Solution: x = \$14. (d) 15,000 • 10% + a: • 12% = 2100. Solution: x = 5000. (e) \x + \x + 1000 = x. Solution: x = 12,000.

5. (a) y = 17/23 (b) * = -4 (c) z = 4 (d) p = 15/16

6. 10 minutes. (If x is the number of liters per minute from the first hosepipe, the two others give 2x¡3 and x/3 liters per minute. The number of minutes needed to fill the pool with all three hosepipes in use is given by the expression 20x/(x + 2x/3 +x/3), which is 10.)

1. (a), (b), (d), (f), and (h) are true; (c), (e), and (g) are false.

2. (a) x <-9 (b) Satisfied for all at. (c) x <25/2 (d) x < 19/7 (e) t > -1/4 " (f) * < -5 or x > -4

3. (a) -2 < x < 1 (b) x < -4 or x > 3 (c) -5 < a < 5 (d)-1 < x < -2 (e) n > 160 or n < 0 (f) 0 < g < 2 (g) p > -1 and p # 2

(h) -4 < n < -10/3 (i) -1 < x < 0 or 0 < x < 1. (Hint: x4 - x2 = x2(x + l)(x- 1).)

4. (a) a: > 1 or x < —4 (b) x > -4 and x ^ 1 (c) x < 1 or 2 < x < 3

(d) x < 1 and x ^ 1/5 (e)l/5<x<l (f) x < 0 (g) -3 < x < -2 or * > 0 (h) -5 < * < 1 (Hint: x2 + 4x - 5 = (x + 5)(x - 1).)

(i) -6 < x < 0 or x > 3. (Hint: -\x2 -x2 + 6x = -\x(x + 6)(x - 3).)

5. (a) -41/6 <x <2/3 (b) x < -1/5 (c) -1 < x < 0

6. (a) Yes. (b) No, put * = 1 /2, for example, (c) No, not for x <0.

(d) Yes. because the inequality is equivalent to x2 — 2xy + y2 > 0, or (x - y)- > o, and this inequality is satisfied for all x and y.

7. (a) \$120 + 0.161 x (b) Smallest number of calls: 300. Largest number of calls: 400.

8. (a) Between 39.2°F and 42.8°F (b) Between 2.2°C and 4.4°C, approximately.