Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 14 Trigonometric Graphs, Identities, and Equations - Standardized Test Practice - Short Response - Page 973: 11

Answer

All possible answers: 31 17

Work Step by Step

$7k+3 \lt 50$ $7k \lt 47$ $k \lt 6 r 5$ Therefore, we are looking for a positive integer k that is 6 or below. Therefore, the possible values of k are 6, 5, 4, 3, 2, and 1. Let's try k = 6: $7k+3 = 7(6)+3 = 45$ Since 45 is not a prime number less than 50, let's try k=5. Let's try k = 5: $7k+3 = 7(5)+3 = 38$ Since 38 is not a prime number less than 50, let's try k=4. Let's try k = 4: $7k+3 = 7(4)+3 = 31$ *Since 31 is a prime number less than 50, it is a possible answer.* Since we have found a possible answer, we can stop there. However, to see other possible answers, let's try k = 2 and k = 1. If k = 2: $7k+3 = 7(2)+3 = 17$ Since 17 is also a prime number less than 50, it is a possible answer. If k = 1: $7k+3 = 7(1)+3 = 10$ 10 is not a prime number less than 50. So, 31 and 17 are two possible answers for this question.
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