Solution Manual For Finite Mathematics and Applied Calculus, 8th Edition by Stefan Waner,q Steven Co

Solutions Section 0.1 Section 0.1

1. 2(4 + (−1))(2 ⋅ −4) = 2(3)(−8) = (6)(−8) = −48

2. 3 + ([4 − 2] ⋅ 9) = 3 + (2 × 9) = 3 + 18 = 21

3. 20∕(3 * 4) − 1 = 20 − 1 = 53 − 1 = 23 12

4. 2 − (3 * 4)∕10 = 2 − 12 = 2 − 65 = 45 10

5.

3 + ([3 + (−5)]) 3−2×2 3 + (−2) 1 = = = −1 3−4 −1

7. (2 − 5 * (−1))∕1 − 2 * (−1) 2 − 5 ⋅ (−1) = − 2 ⋅ (−1) 1 2+5 = +2=7+2=9 1

6.

12 − (1 − 4) 2(5 − 1) ⋅ 2 − 1 12 − (−3) 15 = = =1 16 − 1 15

8. 2 − 5 * (−1)∕(1 − 2 * (−1)) 5 ⋅ (−1) =2− 1 − 2 ⋅ (−1) 5 11 −5 =2− +2=2+ = 1+2 3 5

2 × (−1) 2 2 2×1 2 = = =1 2 2

10. 2 + 4 ⋅ 3 2 = 2 + 4 × 9 = 2 + 36 = 38

11. 2 ⋅ 4 2 + 1 = 2 × 16 + 1 = 32 + 1 = 33

12. 1 − 3 ⋅ (−2) 2 × 2 = 1 − 3 × 4 × 2 = 1 − 24 = −23

13. 3^2+2^2+1 = 3 2 + 2 2 + 1 = 9 + 4 + 1 = 14

14. 2^(2^2-2)

9. 2 ⋅ (−1) 2∕2 =

15.

3 − 2(−3) 2 −6(4 − 1) 2 3 − 2 × 9 3 − 18 = = −6 × 9 −6(3) 2 5 −15 = = −54 18

17. 10*(1+1/10)^3 1 3 = 10(1 + ) = 10(1.1) 3 10 = 10 × 1.331 = 13.31

2

= 2 (2 −2) = 2 4−2 = 2 2 = 4

16.

1 − 2(1 − 4) 2 2(5 − 1) 2 ⋅ 2 1 − 2(−3) 2 1−2×9 = = 2 × 16 × 2 2(4) 2 ⋅ 2 1 − 18 17 = =− 64 64

18. 121/(1+1/10)^2 121 121 = = 2 1 1.1 2 !1 + " 10

121 = = 100 1.21