Problem 51 - Algebra and Trigonometry Real Mathematics, Real People

Question

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

Step-by-Step Solution

Answer

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 1: Calculate value for t = -5

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

Step 2: Calculate value for t = -4

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

Step 3: Calculate value for t = -3

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

Step 4: Calculate value for t = -2

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

Step 5: Calculate value for t = -1

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