By OpenStax

Algebra and Trigonometry presents a finished and multi-layered exploration of algebraic ideas. The textual content is acceptable for a customary introductory Algebra & Trigonometry direction, and used to be constructed for use flexibly. The modular method and the richness of content material guarantees that the ebook meets the desires of quite a few courses. Algebra and Trigonometry publications and helps scholars with differing degrees of education and adventure with arithmetic. rules are awarded as basically as attainable, and growth to extra advanced understandings with massive reinforcement alongside the best way. A wealth of examples - frequently a number of dozen in keeping with bankruptcy - provide particular, conceptual motives, as a way to construct in scholars a powerful, cumulative starting place within the fabric prior to asking them to use what they have discovered. this can be a full-color textbook.

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How To… Given an expression with a single square root radical term in the denominator, rationalize the denominator. 1. Multiply the numerator and denominator by the radical in the denominator. 2. Simplify. Example 8 Rationalizing a Denominator Containing a Single Term — 2√3 in simplest form. Write ______ — 3√10 Solution — — 10 √ The radical in the denominator is √ 10 . Then simplify. 10 √ — — 2 3 10 √ √ ______ — · _____ — 3√10 √ 10 — 2√30 ______ 30 — 30 √ _____ 15 36 CHAPTER 1 Prerequisites Try It #8 — 12√3 in simplest form.

A 23 38. 4x + x(13 − 7) 39. 2y − (4)2 y − 11 40. __ ( 64) − 12a ÷ 6 41. 8b − 4b(3) + 1 42. 5l ÷ 3l · (9 − 6) 43. 7z − 3 + z · 62 44. 4 · 3 + 18x ÷ 9 − 12 45. 9(y + 8) − 27 46. _ t − 4 2 47. 6 + 12b − 3 · 6b 48. 18y − 2(1 + 7y) 4 49. _ · 27x 50. 8(3 − m) + 1(−8) 51. 9x + 4x(2 + 3) − 4(2x + 3x) 69 9 2 52. 52 − 4(3x) 16 CHAPTER 1 Prerequisites REAL-WORLD APPLICATIONS For the following exercises, consider this scenario: Fred earns $40 mowing lawns. He spends $10 on mp3s, puts half of what is left in a savings account, and gets another $5 for washing his neighbor’s car.

U−1v 2 a. (−2a3b−1)(5a−2b2) −1 v 4 −4 (3w2)5 — — e. _ (6w−2)2 Solution a. (6m2n−1)3 = (6)3(m2)3(n−1)3 The power of a product rule The power rule =6m n 3 2 ∙ 3 −1 ∙ 3 = 216m n Simplify. 216m6 The negative exponent rule = _ n3 b. 175 ∙ 17−4 ∙ 17−3 = 175 − 4 − 3 The product rule 6 −3 = 17−2 Simplify. 1 1 The negative exponent rule = ___ 2 or ___ 17 289 2 (u−1v)2 u−1v c. _ = _ The power of a quotient rule −1 v (v−1)2 u−2v2 The power of a product rule = _ v−2 The quotient rule = u−2v2−(−2) = u−2v 4 Simplify.