Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 1 - Problem Solving and Critical Thinking - 1.1 Inductive and Deductive Reasoning - Exercise Set 1.1 - Page 14: 75

Answer

given below.

Work Step by Step

(a) The expression of \[3367\times 3,\ 3367\times 6,\ 3367\times 9\], and \[3367\times 12\]. Solve this expression by using a calculator: \[\begin{align} & 3367\times 3=10101 \\ & 3367\times 6=20202 \\ & 3367\times 9=30303 \\ & 3367\times 12=40404 \end{align}\] (b) In the expression, the first multiplier is always 3367 and the second multiplier is a successive multiple of 3. Then, these products are 10101, 20202, 30303, 40404. The pattern in the products is that the first answer is \[10101\times 1=10101\], second is \[10101\times 2=20202\],and so on. Therefore, theexpression of the value is\[3367\times 3=10101\], \[3367\times 6=20202\], \[3367\times 9=30303\], and\[3367\times 12=40404\]. (c) The expression of \[3367\times 3,3367\times 6,3367\times 9\], and \[3367\times 12\]. Solve these two-next multiplication and product expression by usinga calculator: \[\begin{align} & 3367\times 3=10101 \\ & 3367\times 6=20202 \\ & 3367\times 9=30303 \\ & 3367\times 12=40404 \end{align}\] That is, \[3367\times 15=50505\] \[3367\times 18=60606\] Verify this result: \[3367\times 15=50505\] \[10101\times 5=50505\] And, \[\begin{align} & 3367\times 18=60606 \\ & 10101\times 6=60606 \end{align}\] Thus, theseresultsare correct and verified. (d) The expression of \[3367\times 3,\ 3367\times 6,\ 3367\times 9\], and \[3367\times 12\]. Solve this expression by using a calculator: \[\begin{align} & 3367\times 3=10101 \\ & 3367\times 6=20202 \\ & 3367\times 9=30303 \\ & 3367\times 12=40404 \end{align}\] This is an inductive reasoning, it uses an observed pattern and draws a conclusion from that pattern. In the expression, the first multiplier is always 3367 and the second multiplier isa successive multiple of 3. Then, these products are 10101, 20202, 30303, 40404.
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