Applications of Slope and Intercept

Linear equations word problems: tables - Questions

Answers

1. $7.5$ degrees Celsius per minute.

Since the pizza was heated at a constant rate, the table describes a linear relationship.

Moreover, the rate of change of this relationship corresponds to the rate at which the pizza was heated.

The table of values shows that for each increase of $6$ minutes in Time, Temperature increased by $45$ degrees Celsius. The rate at which the pizza was heated is the ratio of those corresponding differences:

$\frac{\Delta \text { Temperature }}{\Delta \text { Time }}=\frac{45}{6}=7.5$

In conclusion, the pizza was heated at a rate of $7.5$ degrees Celsius per minute.

2. $14$ minutes.

Since the pizza was heated at a constant rate, the table describes a linear relationship.

Moreover, the time it took the pizza to reach $100$ degrees Celsius corresponds to the case where the temperature was $100$ degrees Celsius.

The table of values shows that for each increase of $2$ minutes in the time, the temperature increased by $15$ degrees celsius.

Time (minutes)	Temperature (degrees Celsius)
$4$	$25$
$\stackrel{+2}{\longrightarrow} 6$	$40 \stackrel{+15}{\longleftarrow}$
$\stackrel{+2}{\longrightarrow} 8$	$55 \stackrel{+15}{\longleftarrow}$

Let's extend the table to get to $100$ degrees Celsius.

Time (minutes)	Temperature (degrees Celsius)
$8$	$55$
$\stackrel{+2}{\longrightarrow} 10$	$70 \stackrel{+15}{\longleftarrow}$
$\stackrel{+2}{\longrightarrow} 12$	$85 \stackrel{+15}{\longleftarrow}$

In conclusion, the pizza reached $100$ degrees Celsius after $14$ minutes.

3. $6$ minutes

Since Julia rode at a constant speed, the table describes a linear relationship.

Moreover, the rate of change of this relationship corresponds to the time it takes Julia to ride $1$ kilometer.

The table of values shows that for each increase of $2$ kilometers in Distance, Time increased by $12$ minutes. The time it takes Julia to ride $1$ kilometer is the ratio of those corresponding differences:

In conclusion, Julia rode $1$ kilometer in $6$ minutes.

4. $10$ minutes

Since Julia rode at a constant speed, the table describes a linear relationship.

Moreover, the time it took Julia to get ready for the delivery corresponds to the case where the distance $0$ kilometers.

The table of values shows that for each increase of $2$ kilometers in the distance, the time increases by $12$ minutes.

Distance (kilometers)	Time (minutes)
$6$	$46$
$\stackrel{+2}{\longrightarrow} 8$	$58 \stackrel{+12}{\longleftarrow}$
$\stackrel{+2}{\longrightarrow} 10$	$70 \stackrel{+12}{\longleftarrow}$

Let's extend the table backwards to get to $0$ kilometers.

Distance (kilometers)	Time (minutes)
$6$	$46$
$\stackrel{-(2)}{\longrightarrow} 4$	$34 \stackrel{-(12)}{\longleftarrow}$
$\stackrel{-(2)}{\longrightarrow} 2$	$22 \stackrel{-(12)}{\longleftarrow}$
$\stackrel{-(2)}{\longrightarrow} 0$	$10 \stackrel{-(12)}{\longleftarrow}$

In conclusion, it took Julia $10$ minutes to get ready for the delivery.