Question

1. the height h of a balloon, in feet, t seconds after it is released is given by the function h(t)=2t + 6. a. what is the value of h(20), and what does it mean in the context of the situation? b. explain how to use the function to find the height of the balloon 2 minutes after it is released. c. are there any restrictions on the values of t that can be used as inputs for the function? if so, how would this affect the graph of the function? explain. 27. how can you determine whether a relation represents a function?

Explanation:

26.

Step1: Substitute t = 20 into the function

Given \(h(t)=2t + 6\), substitute \(t = 20\).
\[h(20)=2\times20+6\]

Step2: Calculate the value

\[h(20)=40 + 6=46\]
In the context of the situation, \(t\) represents the number of seconds after the balloon is released. So \(h(20)=46\) means that 20 seconds after the balloon is released, the height of the balloon is 46 feet.

Step3: Convert 2 minutes to seconds

Since 1 minute = 60 seconds, 2 minutes \(=2\times60 = 120\) seconds.

Step4: Find the height at 2 minutes

Substitute \(t = 120\) into \(h(t)=2t+6\).
\[h(120)=2\times120 + 6=240+6=246\]
The height of the balloon 2 minutes (120 seconds) after it is released is 246 feet.

Step1: Recall the definition of a function

A relation is a function if for every input value (in this case, value of \(t\)) there is exactly one output value (value of \(h(t)\)).
For the linear - function \(h(t)=2t + 6\), for any given real - number value of \(t\), we can calculate a unique value of \(h(t)\) by using the formula \(h(t)=2t + 6\).

Step2: Consider the graph

The graph of \(h(t)=2t + 6\) is a straight line. We can use the vertical - line test on the graph of the relation. If we draw any vertical line on the graph of \(y = 2t+6\), the vertical line will intersect the graph at most once. So the relation \(h(t)=2t + 6\) represents a function.

Step3: Analyze the domain restrictions

Theoretically, in the context of time, \(t\geq0\) because time cannot be negative in the context of the balloon being released. Also, there may be a practical upper - bound on \(t\) depending on factors like how long the balloon can stay in the air (e.g., if it pops or reaches a maximum altitude). But in terms of the mathematical function \(h(t)=2t + 6\) without considering physical limitations too strictly, the domain of the function (values of \(t\) that can be used as inputs) is all real numbers \(t\geq0\) in the context of the balloon - release situation. If we consider the physical limitations more strictly, there will be an upper - bound \(t_{max}\) such that \(0\leq t\leq t_{max}\).
The restrictions on \(t\) (the input values) affect the graph in that the graph of \(h(t)=2t + 6\) will only be drawn for the valid values of \(t\). For example, if \(t\geq0\), the graph will start at the point \((0,6)\) (when \(t = 0\), \(h(0)=2\times0+6 = 6\)) and will extend to the right. If there is an upper - bound \(t_{max}\), the graph will end at the point \((t_{max},2t_{max}+6)\).

Answer:

The value of \(h(20)\) is 46, which means the height of the balloon 20 seconds after it is released is 46 feet. The height of the balloon 2 minutes after it is released is 246 feet.

27.