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Solution Manual For Business Mathematics In Canada 11th Edition By F. Ernest Jerome

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Review and Applications of Basic Mathematics Exercise 1.1 a. 10 + 10 x 0 = 10 + 0 = 10 b. 2 x 2 + 4 – 8 = 4 + 4 – 8 = 0 c. (10 + 10) x 0 = 20 x 0 = 0 d. 2 x (2 + 4) – 8 = 2 x 6 – 8 = 12 ... – 8 = 4 e. 0 + 3 x 3 – 3 2 + 10 = 0 + 9 – 9 + 10 = 10 f. 12 – 2 x 5 + 2 2 x 0 = 12 – 10 + 4 x 0 = 12 – 10 + 0 = 2 g. 0 + 3 x 3 – (32 + 10) = 0 + 9 – 19 = -10 h. (12 – 2) x (5 + 2 2 ) x 0 = 10 x 9 x 0 = 0 2 2 −4 = 4−4 = 0 = i. (4−2)2 2 2 4 0 (2−4)2 = (−2)2 = 4 = j. 5−2 2 5−4 1 1. 20 − 4 2 − 8 = 20 − 8 − 8 = 4 2. 18  3 + 6  2 = 6 + 12 = 18 3. (20 − 4)  2 − 8 = 16  2 − 8 = 32 − 8 = 24 4. 18  (3 + 6)  2 = 18  9  2 = 2  2 = 4 5. 20 − (4  2 − 8) = 20 − (8 − 8) = 20 6. (18  3 + 6)  2 = (6 + 6)  2 = 24 7. 54 − 36  4 + 2 2 = 54 − 9 + 4 = 49 8. (5 + 3) 2 − 3 2  9 + 3 = 8 2 − 9  9 + 3 = 64 −1+ 3 = 66 ) = 365 12 9. (54 − 36)  (4 + 2) 2 = 18  6 2 = 18  36 = 0.5 10. 5 + (3 2 − 3) 2  (9 + 3) = 5 + (9 − 3) 2 12 = 5 + 36 12 = 5 + 3 = 8 11. 8 2 − 4 2 (4 − 2) 3 = 64 −16 2 3 = 48 = 6 8 12. (8 − 4) 2 4 − 2 3 = 4 2 = 4 − 8 16 = − 4 − 4 13. 3(6+ 4) 2 −5(17−20) 2 = 3102 −5(−3) 2 = 3100−59 = 300−45 = 255 14. (4  3 − 2) 2  (4 − 3  2 2 )= (12 − 2) 2  (4 − 3  4) = 102  (4 −12) = 100  (− 8) = −12.5 15. (20 + 85)− 7 (− 3) 9 = (20 + 40 + 21) 9 = 81 9 = 9 16. 5  19  + (5 2 −16 2  2  519 + (2 [Show More]

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