(a)

Tim charted his distance from home. Click Here for Help Video

Plot the point (10, 16).
(b)
How many total minutes did Tim stop to rest?

How many times did Tim get closer to his home?

The speed from the 14th minute to the 18th minute is miles_____per minute. (Remember that speed is never negative.)

The information provided is not sufficient to accurately answer the questions without additional context or a graph/chart to analyze. Here's what we can infer or if given the proper data, how it could be approached:

To plot the point (10, 16):

(a) On a Cartesian coordinate system, typically the x-axis represents time (in minutes, if that's the case here) and the y-axis represents distance (in miles, perhaps). The point (10, 16) would indicate that at the 10-minute mark, Tim was 16 miles from home. To plot this:
1. Move 10 units along the horizontal (x) axis.
2. From that point, move 16 units up the vertical (y) axis.
3. Mark that position with a point. This would be Tim's location at the 10-minute mark.

To determine how many total minutes Tim stopped to rest:

(b) Normally, to find the resting time on a distance vs. time chart, you would look for segments where the distance remains constant over time, indicating no movement. You would add up the lengths of these time segments.

To calculate how many times Tim got closer to his home:

You would look for segments on the graph that trend downward, meaning the distance from home is decreasing.

To find the speed from the 14th minute to the 18th minute:

You would need to know the difference in the distance from minute 14 to minute 18 and divide it by the time interval. If the distance remained the same, the speed would be 0 indicating rest. If the distance decreased, you would have:

(speed) = (distance at minute 14 - distance at minute 18) / (18 minutes - 14 minutes)

Speed is indeed always a positive value or zero when it's a scalar quantity, as negative speed doesn't make practical sense; speed denotes magnitude only, while velocity is the term used for speed with direction, which can be positive or negative depending on the defined direction. If Tim's distance from home decreased, he was getting closer to home, which would be reflected in a negative velocity, indicating direction, but the speed itself would be positive.

Again, actual calculations are not possible without the specific data from the distance-time chart, which is not provided here.

(b)

How many total minutes did Tim stop to rest? ______

How many times did Tim get closer to his home? ______

The speed from the 14th minute to the 18th minute is miles _______ per minute. (Remember that speed is never negative.)

Without the graph or data chart that details Tim's distance from home over time, I cannot provide specific answers to the questions in part (b). To answer these questions, one would typically look at the graph to interpret Tim's movements:

- **How many total minutes did Tim stop to rest?** To answer this, one would sum the lengths of time segments where the graph shows a constant distance, which indicates that Tim wasn't moving and therefore was resting.

- **How many times did Tim get closer to his home?** To determine this, one would count the number of segments on the graph where the slope is negative, meaning the distance is decreasing over time, which indicates movement towards home.

- **The speed from the 14th minute to the 18th minute:** To calculate this, one would need to know the distance Tim was from home at the 14th minute and the distance at the 18th minute. Assuming this is a straight line segment on the graph, the speed would be calculated by the change in distance over the change in time:

\[ \text{Speed} = \frac{\text{Distance at minute 18} - \text{Distance at minute 14}}{18 - 14} \]

This would provide the average speed in miles per minute over that 4-minute interval.

If the graph or data is made available, these questions can be answered with precise values. Without that information, it is impossible to give a definite answer.

How many total minutes did Tim stop to rest?



How many times did Tim get closer to his home?

As mentioned earlier, to answer these questions, I would need the graph or distance-time data for Tim's journey. With that graph or data, we could then analyze the information to determine the rest times and the instances when Tim got closer to his home.

To reiterate the approach:

1. **How many total minutes did Tim stop to rest?** This would involve looking at the graph or data for time periods where the distance from home does not change, as constant distance over a period indicates rest. Adding the duration of all such periods would give the total rest time.

2. **How many times did Tim get closer to his home?** This would involve observing the graph or data for portions where the distance from home is decreasing over time. Each separate decreasing segment would count as a time when Tim got closer to his home.

It is not possible to provide specific numbers without the actual graph or chart. If you can provide the graph or chart detailing Tim's journey, I would be able to analyze it and give you the answers to your questions.

Graph the following information.

Andrew started at home. For the first 15 minutes he traveled at a constant speed and traveled a total of 4 miles.
For the next 10 minutes Andrew rested.
Andrew then rode his bike home which took 20 minutes.
(b)
What was Andrew's average speed on his way home? (Note that speed cannot be negative.)
Andrew's speed was_____miles per minute.

Based on the information provided, we can analyze Andrew's journey and calculate his average speed on the return trip home.

Let's break down the journey:

1. **First 15 minutes:** Andrew traveled at a constant speed and covered 4 miles.

2. **Next 10 minutes:** Andrew rested (didn't travel any distance).

3. **Last 20 minutes:** Andrew rode his bike home.

We do not know the exact distance he traveled on his way home since it might not necessarily be the same 4 miles (due to different paths or speeds). But if we assume that he returned the same way he came, then we can use the 4 miles as the distance for his return trip home.

To calculate his average speed on the way home:

\[ \text{Average speed} = \frac{\text{Total distance traveled}}{\text{Total time spent traveling}} \]

If he traveled 4 miles back home in 20 minutes, the speed would be:

\[ \text{Average speed} = \frac{4 \text{ miles}}{20 \text{ minutes}} \]
\[ \text{Average speed} = 0.2 \text{ miles per minute} \]

Therefore, Andrew's average speed on his way home was **0.2 miles per minute**.

Please note, in reality, we can't assume that the distance back is the same as the distance he went out unless it's explicitly stated. However, for the purpose of this calculation, we are using the data provided under that assumption. If the journey back home is a different distance, we would need that exact value to calculate the accurate average speed.

Graph the following information.

Andrew started at home. For the first 20 minutes he traveled at a constant speed and traveled a total of 5 miles.
For the next 10 minutes Andrew rested.
Andrew then rode his bike home which took 20 m
(b)
What was Andrew's average speed on his way home? (Note that speed cannot be negative.)
Andrew's speed was_____miles per minute. nutes.

Now let's analyze the given information and calculate Andrew's average speed for the return journey home.

1. **First 20 minutes:** Andrew traveled at a constant speed for a total of 5 miles.

2. **Next 10 minutes:** Andrew rested (didn't travel any distance).

3. **Final 20 minutes:** Andrew rode his bike home.

For simplicity, let's assume Andrew traveled the same 5 miles back home. This is a reasonable assumption if he returned exactly the way he came. However, it's important to note that this is an assumption, and in reality, more information would be required to state this definitively.

Now, to find Andrew's average speed on the return trip, we can use the formula for speed:

\[ \text{Average speed} = \frac{\text{Total distance traveled}}{\text{Total time spent traveling}} \]

Assuming a distance of 5 miles for the return trip and a travel time of 20 minutes, his speed would be:

\[ \text{Average speed} = \frac{5 \text{ miles}}{20 \text{ minutes}} \]
\[ \text{Average speed} = 0.25 \text{ miles per minute} \]

So, Andrew's average speed on his way home was **0.25 miles per minute**.

Write an equation from slope-intercept form (y=mx+b ) from the table below. Click Here for Help Video.

x - y
-3 - -2
-6 - -5
-9 - -8

y=____x+_____

To write an equation in slope-intercept form (y = mx + b), we need to first determine the slope (m) and then the y-intercept (b).

The slope (m) is calculated by taking the change in y-values divided by the corresponding change in x-values between any two points on the table. Let's use the first two points to calculate the slope:

Point 1: (x1, y1) = (-3, -2)
Point 2: (x2, y2) = (-6, -5)

Using the slope formula:

\[ m = \frac{y2 - y1}{x2 - x1} \]

\[ m = \frac{-5 - (-2)}{-6 - (-3)} \]
\[ m = \frac{-5 + 2}{-6 + 3} \]
\[ m = \frac{-3}{-3} \]
\[ m = 1 \]

So the slope of the line is 1.

Now, to find the y-intercept (b), we can substitute the slope and the coordinates of any point on the line into the equation y = mx + b. Let's use Point 1 for this:

\[ -2 = (1)(-3) + b \]
\[ -2 = -3 + b \]
\[ b = -2 + 3 \]
\[ b = 1 \]

Now we have both the slope (m = 1) and the y-intercept (b = 1). Therefore, the equation in slope-intercept form using the given points is:

\[ y = x + 1 \]

Write an equation from slope-intercept form (y=mx+b ) from the table below.

x - y
6 - 10
12 - 14
18 -18

y=___x+____

To write an equation in slope-intercept form (y = mx + b) using the provided points, let's start by finding the slope (m):

First, we calculate the slope using the first two points:

Point 1: (x1, y1) = (6, 10)
Point 2: (x2, y2) = (12, 14)

Slope (m) is the change in y divided by the change in x:

\[ m = \frac{y2 - y1}{x2 - x1} \]

\[ m = \frac{14 - 10}{12 - 6} \]

\[ m = \frac{4}{6} \]

\[ m = \frac{2}{3} \]

Now that we have the slope, we can use it along with any of the provided points to find the y-intercept (b). Let's use Point 1:

\[ 10 = \left(\frac{2}{3}\right)(6) + b \]

\[ 10 = 4 + b \]

\[ b = 10 - 4 \]

\[ b = 6 \]

So, the slope (m) is \( \frac{2}{3} \) and the y-intercept (b) is 6. The equation in slope-intercept form is:

\[ y = \frac{2}{3}x + 6 \]

I don't understand

\[ y = \frac{2}{3}x + 6 \]

Certainly, let's break down the equation to make it clearer. The equation

\[ y = \frac{2}{3}x + 6 \]

is written in slope-intercept form, which is:

\[ y = mx + b \]

where:
- \( y \) is the dependent variable (the output of the equation or the value you calculate),
- \( x \) is the independent variable (the input or the value you know),
- \( m \) is the slope of the line, and
- \( b \) is the y-intercept of the line (the point where the line crosses the y-axis).

In your equation:

- The slope \( m \) is \( \frac{2}{3} \). This tells you that for each unit increase in \( x \), \( y \) increases by \( \frac{2}{3} \) units.
- The y-intercept \( b \) is 6. This means that when \( x = 0 \), \( y \) is 6. So the point (0, 6) is where the line crosses the y-axis.

To understand it visually, if you were to graph this line on a coordinate plane, you would start by plotting the y-intercept at point (0, 6) on the y-axis. Then, to continue drawing the line, you would use the slope. A slope of \( \frac{2}{3} \) means that for every 3 units you move horizontally to the right on the x-axis, you move up 2 units vertically on the y-axis. So, starting from the y-intercept (0, 6), you would move 3 units to the right and 2 units up, plotting another point at (3, 8). You can repeat this process to plot additional points if needed.

Connecting these points would give you a straight line, which represents the equation. Every point (x, y) on this line would satisfy the equation \( y = \frac{2}{3}x + 6 \).

If you want to calculate the value of \( y \) for a specific value of \( x \), you would simply substitute the value of \( x \) into the equation and solve for \( y \). For example, if \( x = 9 \):

\[ y = \frac{2}{3} \cdot 9 + 6 \]
\[ y = 6 + 6 \]
\[ y = 12 \]

So when \( x = 9 \), \( y = 12 \), and the point (9, 12) would lie on the graphed line.

Write an equation from slope-intercept form (



=


+

y=mx+b ) from the table below. Click Here for Help Video.


x

y
6
6
10
10
12
12
14
14
18
18
18
18




=
y =



+
x +

Based on the table provided, it appears that the value of y is always equal to the value of x for each pair given. This means that the slope (m) of the line is 1 since y increases by the same amount as x does.

To confirm this, let's calculate the slope just like we would for any two points, although it's quite clear from the pattern:

Using any two points from the table:

Point 1: (x1, y1) = (6, 6)
Point 2: (x2, y2) = (10, 10)

Slope (m) = (y2 - y1) / (x2 - x1)

Substituting the values, we get:

m = (10 - 6) / (10 - 6)
m = 4 / 4
m = 1

Now, since the slope is 1, the slope-intercept form equation we're looking for would be y = mx + b.

We know that m = 1, and now we need to determine b (the y-intercept). However, since for each data point y equals x, when x = 0, y would also be 0. Therefore, b = 0 (because the line passes through the origin).

So the equation with m = 1 and b = 0 is:

y = 1x + 0

In the simplest form, the equation can be written as:

y = x

Every (x, y) point given in your table satisfies this equation, confirming that our equation is correct.

Write an equation from slope-intercept form (y=mx+b ) from the table below.

y -x
0 - 7
6 - 2
12 - -3
18 - -8

To determine the equation of a line in slope-intercept form (y = mx + b) from the table, we need to calculate the slope (m) using two points from the table, and then use one of those points to find the y-intercept (b).

Let's choose the first two points to calculate the slope:
Point 1 (x1, y1) = (7, 0)
Point 2 (x2, y2) = (2, 6)

The formula for the slope (m) is:
\[ m = \frac{y2 - y1}{x2 - x1} \]

Substitute the known values:
\[ m = \frac{6 - 0}{2 - 7} \]
\[ m = \frac{6}{-5} \]
\[ m = -\frac{6}{5} \]

Now that we have the slope, we can find the y-intercept (b). Let's use the original Point 1 for this:
\[ y1 = m \cdot x1 + b \]

Substitute the known values (including m and the coordinates of Point 1):
\[ 0 = -\frac{6}{5} \cdot 7 + b \]
\[ 0 = -\frac{42}{5} + b \]
\[ b = \frac{42}{5} \]

The y-intercept (b) is \( \frac{42}{5} \), which can also be written as 8.4 if you prefer decimal form.

Therefore, the equation in slope-intercept form is:
\[ y = -\frac{6}{5}x + \frac{42}{5} \]
or
\[ y = -1.2x + 8.4 \]

Both equations are equivalent, just expressed in different forms (fraction and decimal, respectively).

Find the slope between the points

(10, 8)(10, 8) and (1, 8)(1, 8) .
m=

The slope (m) between two points is determined by the change in y-coordinates divided by the change in x-coordinates. The formula for the slope between two points (x1, y1) and (x2, y2) is:

\[ m = \frac{y2 - y1}{x2 - x1} \]

Using the points provided:

Point 1: (10, 8)
Point 2: (1, 8)

\[ m = \frac{8 - 8}{1 - 10} \]
\[ m = \frac{0}{-9} \]
\[ m = 0 \]

The slope (m) between the points (10, 8) and (1, 8) is 0. This indicates that there is no vertical change between the two points; they are at the same height on the graph, meaning the line segment connecting them is horizontal.

Find the slope between the points

(−9,1)(−9, 1) and (1, 6)(1, 6) .
m=

To find the slope (m) between two points, we use the formula:

\[ m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \]

Given the points:

Point 1: (−9, 1)
Point 2: (1, 6)

We can determine the slope as follows:

\[ m = \frac{6 - 1}{1 - (-9)} \]

\[ m = \frac{5}{1 + 9} \]

\[ m = \frac{5}{10} \]

\[ m = \frac{1}{2} \]

So, the slope (m) between the points (−9, 1) and (1, 6) is \( \frac{1}{2} \).

Find the slope between the points

(−9,1) and (1, 6) .
m=

To find the slope (m) between two points (x1, y1) and (x2, y2), you can use the slope formula:

\[ m = \frac{y2 - y1}{x2 - x1} \]

Let's apply the formula to the points (−9, 1) and (1, 6):

Point 1: (x1, y1) = (−9, 1)
Point 2: (x2, y2) = (1, 6)

Now, plug the values into the slope formula:

\[ m = \frac{6 - 1}{1 - (−9)} \]
\[ m = \frac{5}{1 + 9} \]
\[ m = \frac{5}{10} \]
\[ m = \frac{1}{2} \]

The slope (m) between the points (−9, 1) and (1, 6) is \( \frac{1}{2} \) or 0.5.

Find the slope between the points

(5, 3) and (11,10).
m=

To find the slope (m) of the line that passes through the points (5, 3) and (11, 10), use the slope formula:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Given the points:

Point 1: (x1, y1) = (5, 3)
Point 2: (x2, y2) = (11, 10)

Substitute the values into the slope formula:

\[ m = \frac{10 - 3}{11 - 5} \]
\[ m = \frac{7}{6} \]

The slope (m) between the points (5, 3) and (11, 10) is \( \frac{7}{6} \).

7/6

Correct, the slope (m) between the points (5, 3) and (11, 10) is \( \frac{7}{6} \).