## Section 3.4 Slope

In SectionĀ 3, we observed that a constant rate of change between points produces a linear relationship, whose graph is a straight line. Such a constant rate of change has a special name,

*slope*, and weāll explore slope in more depth here.### Subsection 3.4.1 What is slope?

Given a graph with points, when the rate of change from one point to the next never changes, those points must fall on a straight line, as in FigureĀ 2. Rather than say āconstant rate of changeā in every such situation, there is a specific word for this.

#### Definition 3.4.3. Slope.

When \(x\) and \(y\) are two variables where the rate of change between any two points is always the same, we call this common rate of change the slope. Since having a constant rate of change means the graph will be a straight line, its also called the slope of the line.

Considering the definition for rate of change, this means that when \(x\) and \(y\) are two variables where the rate of change between any two points is always the same, then you can calculate slope, \(m\text{,}\) by finding two distinct data points \((x_1,y_1)\) and \((x_2,y_2)\text{,}\) and calculating

\begin{equation}
m=\frac{\text{change in $y$}}{\text{change in $x$}}=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}\text{.}\tag{3.4.1}
\end{equation}

A slope is a rate of change. So if there are units for the horizontal and vertical variables, then there will be units for the slope. The slope will be measured in \(\frac{\text{vertical units}}{\text{horizontal units}}\text{.}\) If the slope is nonzero, we say that there is a linear relationship between \(x\) and \(y\text{.}\) When the slope is \(0\text{,}\) we say that \(y\) is constant with respect to \(x\text{.}\)

Here are some scenarios with different slopes. As you read each scenario, note how a slope is more meaningful with units.

- If a tree grows \(2.5\) feet every year, its rate of change in height is the same from year to year. So the height and time have a linear relationship where the slope is 2.5
^{ft}ā_{yr}. - If a company loses \(2\) million dollars every year, its rate of change in reserve funds is the same from year to year. So the companyās reserve funds and time have a linear relationship where the slope is \(-2\) million dollars per year.
- If Sakura is an adult who has stopped growing, her rate of change in height is the same from year to yearāitās zero. So the slope is 0
^{in}ā_{yr}. Sakuraās height is constant with respect to time.

#### Remark 3.4.4.

A useful phrase for remembering the definition of slope is ārise over run.ā Here, āriseā refers to āchange in \(y\text{,}\)ā \(\Delta y\text{,}\) and ārunā refers to āchange in \(x\text{,}\)ā \(\Delta x\text{.}\) Be careful though. As we have learned, the horizontal direction comes

*first*in mathematics, followed by the vertical direction. The phrase ārise over runā reverses this. (Itās a bit awkward to say, but the phrase ārun under riseā puts the horizontal change first.)#### Example 3.4.5. Yaraās Savings.

On Dec.Ā 31, Yara had only \(\$50\) in her savings account. For the the new year, she resolved to deposit \(\$20\) into her savings account each week, without withdrawing any money from the account.

Yara keeps her resolution, and her account balance increases steadily by \(\$20\) each week. Thatās a constant rate of change, so her account balance and time have a linear relationship with slope \(20\, \frac{\text{dollars}}{\text{wk}}\text{.}\)

We can model the balance, \(y\text{,}\) in dollars, in Yaraās savings account \(x\) weeks after she started making deposits with an equation. Since Yara started with \(\$50\) and adds \(\$20\) each week, then \(x\) weeks after she started making deposits,

\begin{equation}
y = 50 + 20x\tag{3.4.2}
\end{equation}

where \(y\) is a dollar amount. Notice that the slope, \(20\,\frac{\text{dollars}}{\text{wk}}\text{,}\) serves as the multiplier for \(x\) weeks.

We can also consider Yaraās savings using a table as in FigureĀ 6.

\(x\text{,}\) weeks since Dec. 31 |
\(y\text{,}\) savings account balance (dollars) |
||

\(0\) | \(50\) | ||

\(x\) increases by \(1\)\(\longrightarrow\) | \(1\) | \(70\) | \(\longleftarrow\)\(y\) increases by \(20\) |

\(x\) increases by \(1\)\(\longrightarrow\) | \(2\) | \(90\) | \(\longleftarrow\)\(y\) increases by \(20\) |

\(x\) increases by \(2\)\(\longrightarrow\) | \(4\) | \(130\) | \(\longleftarrow\)\(y\) increases by \(40\) |

\(x\) increases by \(3\)\(\longrightarrow\) | \(7\) | \(190\) | \(\longleftarrow\)\(y\) increases by \(60\) |

\(x\) increases by \(5\)\(\longrightarrow\) | \(12\) | \(290\) | \(\longleftarrow\)\(y\) increases by \(100\) |

In first few rows of the table, we see that when the number of weeks \(x\) increases by \(1\text{,}\) the balance \(y\) increases by \(20\text{.}\) The row-to-row rate of change is \(\frac{20\,\text{dollars}}{1\,\text{wk}} = 20\,\frac{\text{dollars}}{\text{wk}}\text{,}\) the slope. In any table for a linear relationship, whenever \(x\) increases by \(1\) unit, \(y\) will increase by the slope.

In further rows, notice that as row-to-row change in \(x\) increases, row-to-row change in \(y\) increases proportionally to preserve the constant rate of change. Looking at the change in the last two rows of the table, we see \(x\) increases by \(5\) and \(y\) increases by \(100\text{,}\) which gives a rate of change of \(\frac{100\,\text{dollars}}{5\,\text{wk}} = 20\,\frac{\text{dollars}}{\text{wk}}\text{,}\) the value of the slope again.

On a graph of Yaraās savings, we can āseeā the rates of change between consecutive rows of the table by including slope triangles.

The large slope triangle indicates that when \(5\) weeks pass, Yara saves \(\$100\text{.}\) This is the rate of change between the last two rows of the table, \(\frac{100}{5} = 20 \, \frac{\text{dollars}}{\text{wk}}\text{.}\) The smaller slope triangles indicate, from left to right, the rates of change \(\frac{20\,\text{dollars}}{1\,\text{wk}}\text{,}\) \(\frac{20\,\text{dollars}}{1\,\text{wk}}\text{,}\) \(\frac{40\,\text{dollars}}{2\,\text{wk}}\text{,}\) and \(\frac{60\,\text{dollars}}{3\,\text{wk}}\) respectively. All of these rates simplify to the slope, \(20 \, \frac{\text{dollars}}{\text{wk}}\text{.}\)

Every slope triangle on the graph of Yaraās savings has the same shape (geometrically, they are called similar triangles) since the ratio of vertical change to horizontal change is always \(20 \, \frac{\text{dollars}}{\text{wk}}\text{.}\) On any graph of any line, we can draw a slope triangle and compute slope as ārise over run.ā

Of course, we could draw a slope triangle on the other side of the line:

This slope triangle works just as well for identifying āriseā and ārun,ā but it focuses on vertical change before horizontal change. For consistency with mathematical conventions, we will generally draw slope triangles showing horizontal change followed by vertical change, as in FigureĀ 7.

#### Example 3.4.9.

The following graph of a line models the amount of gas, in gallons, in Kiranās gas tank as they drive their car. Find the lineās slope, and interpret its meaning in this context.

## Explanation.

To find a lineās slope using its graph, we first identify two points on it, and then draw a slope triangle. Naturally, we would want to choose two points whose \(x\)- and \(y\)-coordinates are easy to identify exactly based on the graph. We will pick the two points where \(x=3\) and \(x=6\text{,}\) because they are right on grid line crossings:

Notice that the

*change*in \(y\) is negative, because the amount of gas is decreasing. Since we chose points with integer coordinates, we can easily calculate the slope:
\begin{equation*}
\text{slope}=\frac{-2\,\text{gallons}}{3\,\text{miles}}=-\frac{2}{3}\,\frac{\text{gal}}{\text{mi}}\text{.}
\end{equation*}

In the given context, this slope implies gas in the tank is

*decreasing*at the rate of \(\frac{2}{3}\)^{gal}ā_{h}. Since this slope is written as a fraction, another way to understand it is that Kiran is using up \(2\) gallons of gas every \(3\) hours.#### Checkpoint 3.4.12.

Find the slope of the line.

## Explanation.

To find the slope of a line from its graph, we first need to identify two points that the line passes through. It is wise to choose points with integer coordinates. For this problem, we choose \((0,-4)\) and \((3,1)\text{.}\)

Next, we sketch a slope triangle and find the

*rise*and*run*. In the sketch below, the rise is \(5\) and the run is \(3\text{.}\)
\begin{equation*}
\text{slope}=\frac{\text{rise}}{\text{run}}=\frac{5}{3}
\end{equation*}

This lineās slope is \({{\frac{5}{3}}}\text{.}\)

#### Checkpoint 3.4.13.

Make a table and plot the equation \(y=\frac{3}{4}x+2\text{,}\) which makes a straight line. Use the plot to determine the slope of this line.

## Explanation.

First, we choose some \(x\)-values to make a table, and compute the corresponding \(y\)-values.

\(x\) |
\(y=\frac{3}{4}x+2\) | Point |

\(-2\) | \(\frac{3}{4}(-2)+2=0.5\) | \((-2,0.5)\) |

\(-1\) | \(\frac{3}{4}(-1)+2=1.25\) | \((-1,1.25)\) |

\(0\) | \(\frac{3}{4}(0)+2=2\) | \((0,2)\) |

\(1\) | \(\frac{3}{4}(1)+2=2.75\) | \((1,2.75)\) |

\(2\) | \(\frac{3}{4}(2)+2=3.5\) | \((2,3.5)\) |

\(3\) | \(\frac{3}{4}(3)+2=4.25\) | \((3,4.25)\) |

\(4\) | \(\frac{3}{4}(4)+2=5\) | \((4,5)\) |

\(5\) | \(\frac{3}{4}(5)+2=5.75\) | \((5,5.75)\) |

This table lets us plot the graph and identify a slope triangle that is easy to work with.

Since the slope triangle runs \(4\) units and then rises \(3\) units, the slope is \(\frac{3}{4}\text{.}\)

### Subsection 3.4.2 Comparing Slopes

Itās useful to understand what it means for different slopes to appear on the same coordinate system.

#### Example 3.4.14.

Effie, Ivan and Cleo are in a foot race. FigureĀ 15 models the distance each has traveled in the first few seconds. Each runner takes a second to accelerate up to their running speed, but then runs at a constant speed. So they are then traveling with a constant rate of change, and the straight line portions of their graphs have a slope. Find each lineās slope, and interpret its meaning in this context. What comparisons can you make with these runners?

We will draw slope triangles to find each lineās slope.

Using the slope equation, we have:

- Effieās slope is \(\frac{8\,\text{m}}{3\,\text{s}}\approx2.666\,\frac{\text{m}}{\text{s}}\text{.}\)
- Ivanās slope is \(\frac{7\,\text{m}}{2\,\text{s}}=3.5\,\frac{\text{m}}{\text{s}}\text{.}\)
- Cleoās slope is \(\frac{8\,\text{m}}{2\,\text{s}}=4\,\frac{\text{m}}{\text{s}}\text{.}\)

In a time-distance graph, the slope of a line represents speed. The slopes in these examples and the running speeds of these runners are measured in

^{m}ā_{s}. A relationship we can see is that the more sharply a line is slanted, the bigger the slope is. This should make sense because for each passing second, the faster person travels longer, making a slope triangleās height taller. This means that, numerically, we can tell that Cleo is the fastest runner (and Effie is the slowest) just by comparing the slopes \(4>3.5>2.666\text{.}\)#### Checkpoint 3.4.17. Jogging on Mt.Ā Hood.

Kato is training for a race up the slope of Mt.Ā Hood, from Sandy to Government Camp, and then back. The graph below models his elevation from his starting point as time passes. Find the slopes of the three line segments, and interpret their meanings in this context.

- What is the slope of the first segment?
- What is the slope of the second segment?
- What is the slope of the third segment?

## Explanation.

The first segment started at \((0,0)\) and stopped at \((7,3500)\text{.}\) This implies, Kato started at the starting point, traveled \(7\) hours and reached a point \(3500\) feet higher in elevation from the starting point. The slope of the line is

\begin{equation*}
\frac{\Delta y}{\Delta x}=\frac{3500\,\text{ft}}{7\,\text{h}}=500\,\frac{\text{ft}}{\text{h}}
\text{.}
\end{equation*}

In context, Kato was running, gaining \(500\) feet in elevation per hour.

What happened in the second segment, which started at \((7,3500)\) and ended at \((19,3500)\text{?}\) This implies he started this portion \(3500\) feet higher in elevation from the starting point, and didnāt change elevation for \(19\) hours. The slope of the line is

\begin{equation*}
\frac{\Delta y}{\Delta x}=\frac{0\,\text{ft}}{12\,\text{h}}=0\,\frac{\text{ft}}{\text{h}}
\end{equation*}

In context, Kato was running but neither gaining nor losing elevation.

The third segment started from \((19,3500)\) and stopped at \((23,0)\text{.}\) This implies, Kato started this part of his trip from a spot \(3500\) feet higher in elevation from the starting point, traveled for \(4\) hours and returned to the starting elevation. The slope of the line is

\begin{equation*}
\frac{\Delta y}{\Delta x}=\frac{-3500\,\text{ft}}{4\,\text{h}}=-875\,\frac{\text{ft}}{\text{h}}
\end{equation*}

In context, Kato was running, dropping in elevation by \(875\) feet per hour.

Some important properties are demonstrated in ExerciseĀ 17.

#### Fact 3.4.18. The Relationship Between Slope and Increase/Decrease.

In a linear relationship, as the \(x\)-value increases (in other words as you read its graph from left to right):

- if the \(y\)-values increase (in other words, the line goes upward), its slope is positive.
- if the \(y\)-values decrease (in other words, the line goes downward), its slope is negative.
- if the \(y\)-values donāt change (in other words, the line is flat, or horizontal), its slope is \(0\text{.}\)

These properties are summarized graphically in FigureĀ 19.

### Subsection 3.4.3 Finding Slope by Two Given Points

Several times in this section we computed a slope by drawing a slope triangle. Thatās not really necessary if you have coordinates for two points that a line passes through. In fact, sometimes itās impractical to draw a slope triangle.

^{ā1ā}Here we will stress how to find a lineās slope without drawing a slope triangle.#### Example 3.4.20.

Your neighbor planted a sapling from a local nursery in his front yard. Ever since, for several years now, it has been growing at a constant rate. By the end of the third year, the tree was 15 ft tall; by the end of the sixth year, the tree was 27 ft tall. Whatās the treeās rate of growth (i.e.Ā the slope)?

We

*could*sketch a graph for this scenario, and include a slope triangle. If we did that, it would look like:By the slope triangle and EquationĀ (3.4.1) we have:

\begin{align*}
\text{slope}=m\amp =\frac{\Delta y}{\Delta x}\\
\amp =\frac{12\,\text{ft}}{3\,\text{yr}}\\
\amp =4\frac{\text{ft}}{\text{yr}}
\end{align*}

So the tree is growing at a rate of 4

^{ft}ā_{yr}.But hold on. Did we really

*need*this picture? The āriseā of \(12\) came from a subtraction of two \(y\)-values: \(27-15\text{.}\) And the ārunā of \(3\) came from a subtraction of two \(x\)-values: \(6-3\text{.}\)Here is a picture-free approach. We know that after 3 yr, the height is 15 ft. As an ordered pair, that information gives us the point \((3,15)\) which we can label as \((\overset{x_1}{3},\overset{y_1}{15})\text{.}\) Similarly, the background information tells us to consider \((6,27)\text{,}\) which we label as \((\overset{x_2}{6},\overset{y_2}{27})\text{.}\) Here, \(x_1\) and \(y_1\) represent the first pointās \(x\)-value and \(y\)-value, and \(x_2\) and \(y_2\) represent the second pointās \(x\)-value and \(y\)-value.

Now we can write an alternative to EquationĀ (3.4.1):

\begin{equation}
\text{slope}=m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}\tag{3.4.3}
\end{equation}

This is known as the slope formula. The following graphs help to understand why this formula works. Basically, we are still using a slope triangle to calculate the slope.

#### Remark 3.4.23.

Itās important to use subscript instead of superscript in the slope equation, because \(y^2\) means to take the number \(y\) and square it. Whereas \(y_2\) tells you that there are at least two \(y\)-values in the conversation, and \(y_2\) is the second of them.

The beauty of the slope formula is that to find a lineās slope, we donāt need to draw a slope triangle any more. Letās look at an example.

#### Example 3.4.24.

A line passes the points \((-5,25)\) and \((4,-2)\text{.}\) Find this lineās slope.

## Explanation.

If you are new to this formula, it may help to label each number before using the formula. The two given points are:

\begin{equation*}
(\overset{x_1}{-5},\overset{y_1}{25}) \text{ and } (\overset{x_2}{4},\overset{y_2}{-2})
\end{equation*}

Now apply the slope formula:

\begin{align*}
\text{slope}\amp=\frac{y_2-y_1}{x_2-x_1}\\
\amp=\frac{-2-25}{4-(-5)}\\
\amp=\frac{-27}{9}\\
\amp=-3
\end{align*}

Note that we used parentheses when substituting negative numbers in \(x_1\) and \(y_1\text{.}\) This is a good habit to protect yourself from making errors with subtraction and double negatives.

#### Checkpoint 3.4.25.

A line passes through the points \((-2,{-4})\) and \((4,{-13})\text{.}\) Find this lineās slope.

## Explanation.

To find a lineās slope, we can use the slope formula:

\begin{equation*}
\text{slope}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}
\end{equation*}

First, we mark which number corresponds to which variable in the formula:

\begin{equation*}
(-2,{-4}) \longrightarrow (x_{1},y_{1})
\end{equation*}

\begin{equation*}
(4,{-13}) \longrightarrow (x_{2},y_{2})
\end{equation*}

Now we substitute these numbers into the corresponding variables in the slope formula:

\begin{equation*}
\begin{aligned} \text{slope}\amp =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\
\amp =\frac{{-13}-((-4))}{4-(-2)}\\
\amp =\frac{{-9}}{6}\\
\amp =-\frac{3}{2} \end{aligned}
\end{equation*}

So the lineās slope is \(\displaystyle{-\frac{3}{2}}\text{.}\)

### Reading Questions 3.4.4 Reading Questions

#### 1.

Have you memorized a formula for finding the slope between two points using their coordinates?

#### 2.

What is an important thing to do with slope to make it more meaningful in an application problem?

#### 3.

Drawing a slope triangle can be helpful to think about slope. But what might happen that could make it impractical to draw a slope triangle?

### Exercises 3.4.5 Exercises

#### Review and Warmup

##### Exercise Group.

###### 1.

Reduce the fraction \(\displaystyle{ \frac{5}{25} }\text{.}\)

###### 2.

Reduce the fraction \(\displaystyle{ \frac{3}{15} }\text{.}\)

###### 3.

Reduce the fraction \(\displaystyle{ \frac{9}{30} }\text{.}\)

###### 4.

Reduce the fraction \(\displaystyle{ \frac{15}{50} }\text{.}\)

###### 5.

Reduce the fraction \(\displaystyle{ \frac{84}{147} }\text{.}\)

###### 6.

Reduce the fraction \(\displaystyle{ \frac{24}{54} }\text{.}\)

###### 7.

Reduce the fraction \(\displaystyle{ \frac{168}{70} }\text{.}\)

###### 8.

Reduce the fraction \(\displaystyle{ \frac{180}{75} }\text{.}\)

###### 9.

Reduce the fraction \(\displaystyle{ \frac{84}{21} }\text{.}\)

###### 10.

Reduce the fraction \(\displaystyle{ \frac{75}{15} }\text{.}\)

#### Skills Practice

##### Slope and Points.

###### 11.

A line passes through the points \((4,2)\) and \((9,12)\text{.}\) Find this lineās slope.

###### 12.

A line passes through the points \((1,9)\) and \((8,23)\text{.}\) Find this lineās slope.

###### 13.

A line passes through the points \((3,-9)\) and \((6,-18)\text{.}\) Find this lineās slope.

###### 14.

A line passes through the points \((1,-10)\) and \((10,-37)\text{.}\) Find this lineās slope.

###### 15.

A line passes through the points \((-3,-6)\) and \((-6,-18)\text{.}\) Find this lineās slope.

###### 16.

A line passes through the points \((-1,-6)\) and \((-7,-36)\text{.}\) Find this lineās slope.

###### 17.

A line passes through the points \((-3,-23)\) and \((1,-3)\text{.}\) Find this lineās slope.

###### 18.

A line passes through the points \((-2,16)\) and \((4,-14)\text{.}\) Find this lineās slope.

###### 19.

A line passes through the points \((-4,18)\) and \((-6,28)\text{.}\) Find this lineās slope.

###### 20.

A line passes through the points \((-2,-1)\) and \((-8,23)\text{.}\) Find this lineās slope.

###### 21.

A line passes through the points \((21,{21})\) and \((-14,{1})\text{.}\) Find this lineās slope.

###### 22.

A line passes through the points \((15,{9})\) and \((-5,{-7})\text{.}\) Find this lineās slope.

###### 23.

A line passes through the points \((-4,{19})\) and \((2,{-2})\text{.}\) Find this lineās slope.

###### 24.

A line passes through the points \((-18,{14})\) and \((6,{-14})\text{.}\) Find this lineās slope.

###### 25.

A line passes through the points \((2,7)\) and \((-3,7)\text{.}\) Find this lineās slope.

###### 26.

A line passes through the points \((5,9)\) and \((-5,9)\text{.}\) Find this lineās slope.

###### 27.

A line passes through the points \((-10,-2)\) and \((-10,3)\text{.}\) Find this lineās slope.

###### 28.

A line passes through the points \((-7,-4)\) and \((-7,5)\text{.}\) Find this lineās slope.

##### Slope and Graphs.

###### 29.

Below is a lineās graph.

The slope of this line is .

###### 30.

Below is a lineās graph.

The slope of this line is .

###### 31.

Below is a lineās graph.

The slope of this line is .

###### 32.

Below is a lineās graph.

The slope of this line is .

###### 33.

Find the slope of the line.

###### 34.

Find the slope of the line.

###### 35.

Below is a lineās graph.

The slope of this line is .

###### 36.

Below is a lineās graph.

The slope of this line is .

###### 37.

Below is a lineās graph.

The slope of this line is .

###### 38.

Below is a lineās graph.

The slope of this line is .

###### 39.

Below is a lineās graph.

The slope of this line is .

###### 40.

Below is a lineās graph.

The slope of this line is .

###### 41.

Below is a lineās graph.

The slope of this line is .

###### 42.

Below is a lineās graph.

The slope of this line is .

###### 43.

Below is a lineās graph.

The slope of this line is .

###### 44.

Below is a lineās graph.

The slope of this line is .

#### Slope in Context

##### 45.

By your cell phone contract, you pay a monthly fee plus some money for each minute you use the phone during the month. In one month, you spent \(300\) minutes on the phone, and paid \({\$18.50}\text{.}\) In another month, you spent \(330\) minutes on the phone, and paid \({\$18.95}\text{.}\) What is the rate (in dollars per minute) that the phone company is charging you? That is, what is the slope of the line if you plotted the bill versus the number of minutes spent on the phone?

The rate is per minute.

##### 46.

By your cell phone contract, you pay a monthly fee plus some money for each minute you use the phone during the month. In one month, you spent \(240\) minutes on the phone, and paid \({\$14.60}\text{.}\) In another month, you spent \(310\) minutes on the phone, and paid \({\$15.65}\text{.}\) What is the rate (in dollars per minute) that the phone company is charging you? That is, what is the slope of the line if you plotted the bill versus the number of minutes spent on the phone?

The rate is per minute.

##### 47.

A company set aside a certain amount of money in the year 2000. The company spent exactly the same amount from that fund each year on perks for its employees. In \(2004\text{,}\) there was still \({\$764{,}000}\) left in the fund. In \(2007\text{,}\) there was \({\$683{,}000}\) left. What is the rate (in dollars per year) at which this company is spending from this fund?

The company is spending per year on perks for its employees.

##### 48.

A company set aside a certain amount of money in the year 2000. The company spent exactly the same amount from that fund each year on perks for its employees. In \(2002\text{,}\) there was still \({\$671{,}000}\) left in the fund. In \(2006\text{,}\) there was \({\$547{,}000}\) left. What is the rate (in dollars per year) at which this company is spending from this fund?

The company is spending per year on perks for its employees.

##### 49.

A biologist has been observing a treeās height. Thirteen months into the observation, the tree was \(21.37\) feet tall. Sixteen months into the observation, the tree was \(21.94\) feet tall. What is the rate at which the tree is growing? In other words, what is the slope if you plotted height versus time?

##### 50.

A biologist has been observing a treeās height. Ten months into the observation, the tree was \(18.3\) feet tall. Twenty months into the observation, the tree was \(20.5\) feet tall. What is the rate at which the tree is growing? In other words, what is the slope if you plotted height versus time?

##### 51.

Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Seven minutes since the experiment started, the gas had a mass of \(144\) grams. Sixteen minutes since the experiment started, the gas had a mass of \(111.6\) grams. At what rate is the gas leaking?

##### 52.

Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Ten minutes since the experiment started, the gas had a mass of \(357.2\) grams. Fourteen minutes since the experiment started, the gas had a mass of \(319.6\) grams. At what rate is the gas leaking?

##### 53.

Grace is training for a race up the slope of Mt.Ā Hood, from Sandy to Government Camp, and then back. The graph below models their elevation from their starting point as time passes. Find the slopes of the four line segments, and interpret their meanings in this context.

##### 54.

Danny is learning to ski on the slopes of Mt.Ā Hood. The graph below models his elevation from the ski parkās base as time passes during one ski run on a small hill. Find the slopes of the four line segments, and interpret their meanings in this context.

##### 55.

The graph plots the number of invasive cancer diagnoses in Oregon over time, and a trend-line has been drawn.

Estimate the slope of the trend-line.

#### Challenge

##### 56.

True or False: A slope of \({{\frac{1}{7}}}\) is steeper than a slope of \({0.1}\text{.}\)

- true
- false

##### 57.

True or False: A slope of \({{\frac{5}{7}}}\) is steeper than a slope of \({0.7}\text{.}\)

- true
- false

You have attempted of activities on this page.