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 12 − 1 = 5 3 − 1 = 2 3 4. 2 − (3 * 4)∕10 = 2 − 12 10 = 2 − 6 5 = 4 5 5. 3 + ([3 + (−5)]) 3 − 2 × 2 = 3 + (−2) 3 − 4 = 1 −1 = −1 6. 12 − (1 − 4) 2(5 − 1) ⋅ 2 − 1 = 12 − (−3) 16 − 1 = 15 15 = 1 7. (2 − 5 * (−1))∕1 − 2 * (−1) = 2 − 5 ⋅ (−1) 1 − 2 ⋅ (−1) = 2 + 5 1 + 2 = 7 + 2 = 9 8. 2 − 5 * (−1)∕(1 − 2 * (−1)) = 2 − 5 ⋅ (−1) 1 − 2 ⋅ (−1) = 2 − −5 1 + 2 + 2 = 2 + 5 3 = 11 5 9. 2 ⋅ (−1)2 ∕2 = 2 × (−1)2 2 = 2 × 1 2 = 2 2 = 1 10. 2 + 4 ⋅ 32 = 2 + 4 × 9 = 2 + 36 = 38 11. 2 ⋅ 42 + 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 = 32 + 22 + 1 = 9 + 4 + 1 = 14 14. 2^(2^2-2) = 2(22−2) = 24−2 = 22 =