Thinking Mathematically (6th Edition)

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

Chapter 5 - Number Theory and the Real Number System - 5.4 The Irrational Numbers - Exercise Set 5.4 - Page 297: 81

Answer

Consider the provided formula. $h=2.9\sqrt{x}+36$ Substitute $0$ for $x$ and find $h$, that is., $\begin{align} & h=2.9\sqrt{0}+36 \\ & =0+36 \\ & =36 \end{align}$ So, the value of $h$ is $36\text{ cm}$ (b) Consider the given formula. $h=2.9\sqrt{x}+36$ Substitute $9$ for $x$ and find $h$, that is., $\begin{align} & h=2.9\sqrt{9}+36 \\ & =2.9\sqrt{{{3}^{2}}}+36 \\ & =2.9\left( 3 \right)+36 \\ & =44.7 \end{align}$ So, the value of $h$ is $44.7\text{ cm}$ (c) Consider the provided formula. $h=2.9\sqrt{x}+36$ Substitute $14$ for $x$ and find $h$, that is., $h=2.9\sqrt{14}+36$ Now, use a calculator to find the value. That is., $\begin{align} & h=46.85 \\ & \simeq 46.9 \end{align}$ So, the value of $h$ is $46.9\text{ cm}$ (d) From the above graph check the corresponding values on $y\text{-axis}$ for both curves for $x=0,9,14$. Also, write the values as calculated in above parts.
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