By John Peterson, Robert D. Smith
With an emphasis on real-world math functions, the 6th version of INTRODUCTORY TECHNICAL arithmetic presents readers with present and sensible technical math functions for ultra-modern refined exchange and technical paintings environments. easy and simple to appreciate, this hands-on e-book is helping readers construct a superb figuring out of math innovations via step by step examples and difficulties drawn from quite a few occupations. up to date to incorporate the most up-tp-date details within the box, the 6th version comprises elevated insurance of issues comparable to estimation utilization, spreadsheets, and energy-efficient electric purposes.
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Example text
400 ϫ 8,000 ϭ 3,200,000 Compute the answer. Write the multiplier under the multiplicand, placing digits in 7812 proper place positions. ϫ 436 46 872 Multiply the multiplicand by the units of the multiplier, using 234 36 the procedure for short multiplication. This answer is called 3 124 8 the first partial product. 3 , 4 0 6 , 0 3 2 Ans 6 ϫ 7,812 ϭ 46,872 Write the first partial product, starting at the units position and going from right to left. Multiply the multiplicand by the tens of the multiplier to get the second partial product.
Evaluate. 28 ϩ 16 Ϭ 4 Solution. 28 16 4 32 Ans Because the calculator has algebraic logic, the division operation (16 Ϭ 4) was performed before the addition operation (adding 28) was performed. NOTE: If the calculator does not have algebraic logic, the answer to the expression if solved in the order as shown is incorrect: 28 16 4 11 INCORRECT ANSWER. The calculator merely performed the operations in the order entered, without assigning priorities to various operations, and gave an incorrect answer.
14 5 10 14. 3 63 15. 4 47 16. 32 79 16 96 18. 5 87 19. 8 133 20. 64 13. 318 32 217 22. 8 451 23. 64 412 24. 25 17. 21. Express each mixed number as an equivalent improper fraction as indicated. 1 25. 1 ϭ 2 3 26. 5 ϭ 4 2–8 ? 8 ? 16 5 27. 7 ϭ 8 3 28. 9 ϭ 4 ? 32 ? 64 3 ? 29. 81 ϭ 8 32 9 ? 30. 46 ϭ 10 50 Division of Whole Numbers; Quotients as Mixed Numbers In unit 1–9 the answer to a problem like 49 ÷ 6 was written as 8 R1 because 8 6ͤ 49 48 1 The answer to this type of problem is often written as a mixed number.