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1 3 1 2 1 2 1 2 1 5 23 2 1 2 1 the correct

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Unformatted text preview: Chapter 15 4. The correct answer is $800. Solve for a in the general equation a · r (n – 1) = T. Let T = 2,700. The value at the date of the purchase is the first term in the sequence, and so the value three years later is the fourth term; accordingly, n = 4. Given that painting’s value increased by 3 50% (or 1 ) per year on average, r = 1.5 = 2 . 2 Solving for a: a× 3 2 Exercise 2 1. (E) The union of the two sets is the set that contains all integers — negative, positive, and zero (0). The correct answer is 4. The positive factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. The positive factors of 18 are 1, 2, 3, 6, 9, and 18.The two sets have in common four members: 1, 2, 3, and 6. (C) 19 is a prime number, and therefore has only one prime factor: 19. There are two prime factors of 38: 2 and 19. The union of the sets described in choice (C) is the set that contains two members: 2 and 19. (A) Through 10, the multiples of 2 , or 2 1 , 2 are 2 1 , 5, 2 1 , and 10. Through 10, the 2 2 multiples of 2 are 2, 4, 6, 8, and 10. As you can see, the two sets desribed in choice (A) intersect at, but only at, every multiple of 10. (D) You can express set R:{|x| ≤ 10} as R:{–10 ≤ x ≤ 10}. The three sets have only one real number in common: the integer 10. 5 2. a × 3 = 2, 700 2 a × 27 = 2, 700 8 a = 2, 700 × 8 27 a = 800 () () ( 4 −1) = 2, 700 3 3. At an increase of 50% per year, the collector must have paid $800 for the painting three years ago. 5. The correct answer is 21. First, find r: 3 × r (3−1) = 147 3 × r = 147 2 4. 5. r 2 = 49 r=7 To find the second term in the sequence, multiply the first term (3) by r : 3 · 7 = 21. www.petersons.com Numbers and Operations, Algebra, and Functions 267 Exercise 3 1. 2. (D) |7 – 2| – |2 – 7| = |5| – |–5| = 5 – 5 = 0 Exercise 4 1. (E) First, cancel common factors in each term. Then, multiply the first term by the reciprocal of the second term. You can now see that all terms cancel out: a 2 b ÷ a 2 c = a 2 ÷ a 2 = a 2 × bc = 1 bc bc bc a 2 b 2 c bc 2 (C) If b – a is a negative integer, then a > b, in which case a – b must be a positive integer. (When you subtract one integer from anot...
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This note was uploaded on 08/15/2010 for the course MATH a4d4 taught by Professor Colon during the Spring '10 term at Embry-Riddle FL/AZ.

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