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Chapter 1: Problem 51

Complete the table. $$h(t)=\frac{1}{2}|t+3|$$ $$\begin{array}{|l|l|l|l|l|l|} \hline t & -5 & -4 & -3 & -2 & -1 \\ \hline h(t) & & & & & \\ \hline \end{array}$$

Short Answer

Expert verified
The completed table is: \[\begin{array}{|l|l|l|l|l|l|} \hline t & -5 & -4 & -3 & -2 & -1 \ \hline h(t) & 1 & 0.5 & 0 & 0.5 & 1 \ \hline \end{array}\]

Step by step solution

01

Calculate value for t = -5

Plug t = -5 into the function: \( h(t) = \frac{1}{2}|t+3| = \frac{1}{2}|-5+3| = \frac{1}{2}|-2| = 1. \)
02

Calculate value for t = -4

Plug t = -4 into the function: \( h(t) = \frac{1}{2}|t+3| = \frac{1}{2}|-4+3| = \frac{1}{2}|-1| = 0.5. \)
03

Calculate value for t = -3

Plug t = -3 into the function: \( h(t) = \frac{1}{2}|t+3| = \frac{1}{2}|-3+3| = \frac{1}{2}|0| = 0. \)
04

Calculate value for t = -2

Plug t = -2 into the function: \( h(t) = \frac{1}{2}|t+3| = \frac{1}{2}|-2+3| = \frac{1}{2}|1| = 0.5. \)
05

Calculate value for t = -1

Plug t = -1 into the function: \( h(t) = \frac{1}{2}|t+3| = \frac{1}{2}|-1+3| = \frac{1}{2}|2| = 1. \)

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Most popular questions from this chapter

Write the rational expression in simplest form. $$\frac{5 x^{2} y^{2}+25 x^{2} y}{x y+5 x}$$

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Perform the operation and simplify. $$12-\frac{4}{x+2}$$

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