Determine if the sequence below is arithmetic or geometric and determine the common difference / ratio in simplest form.
2, comma, 5, comma, 8, comma, point, point, .
2,5,8,...
The sequence is arithmetic because there is a common difference between each term. To find the common difference, we can subtract any two consecutive terms. Subtracting the second term (5) from the first term (2), we get 5 - 2 = 3. Therefore, the common difference is 3 in simplest form.
Determine if the sequence below is arithmetic or geometric and determine the common difference / ratio in simplest form.
2, comma, 5, comma, 8, comma, point, point, .
2,5,8,...
Answer
Attempt 1 out of 3
This is ______ sequence and the ______ is equal to _____
fill in the blankets correctly
.
This is an arithmetic sequence and the common difference is equal to 3.
Determine if the sequence below is arithmetic or geometric and determine the common difference / ratio in simplest form.
3, comma, 15, comma, 75, comma, point, point, .
3,15,75,...
Answer
Attempt 1 out of 3
This is _____ sequence and the _____ is equal to _____
fill the blanks correctly
.
This is a geometric sequence and the common ratio is equal to 5.
One spring day, Harper noted the time of day and the temperature, in degrees Fahrenheit. Her findings are as follows: At 6 a.m., the temperature was 58° F. For the next 3 hours, the temperature rose 4° per hour. For the next 5 hours, it rose 1° per hour. The temperature then stayed steady until 6 p.m. For the next 2 hours, the temperature dropped 1° per hour. The temperature then dropped steadily until the temperature was 70° at midnight. On the set of axes below, graph Harper's data.
I'm sorry, but as a text-based AI, I am unable to create or display graphical images. However, I can help you understand how to graph Harper's data.
To graph Harper's data on a set of axes, you will plot the time of day (hours) on the x-axis, and the temperature (degrees Fahrenheit) on the y-axis.
Here is a breakdown of Harper's data:
- At 6 a.m. (6 hours), the temperature was 58° F.
- For the next 3 hours, the temperature rose 4° per hour. This means that at 9 a.m. (9 hours), the temperature would be 58° F + (4° per hour * 3 hours) = 58° F + 12° = 70° F.
- For the next 5 hours, the temperature rose 1° per hour. So at 2 p.m. (14 hours), the temperature would be 70° F + (1° per hour * 5 hours) = 70° F + 5° = 75° F.
- The temperature then stayed steady until 6 p.m. (18 hours). So the temperature would remain 75° F from 2 p.m. to 6 p.m.
- For the next 2 hours, the temperature dropped 1° per hour. So at 8 p.m. (20 hours), the temperature would be 75° F - (1° per hour * 2 hours) = 75° F - 2° = 73° F.
- The temperature then dropped steadily until it was 70° at midnight (24 hours).
Using this information, you can plot the points on the graph. Start with the point (6, 58) and connect it with a line to the point (9, 70). Then, continue to plot and connect the rest of the points until midnight (24, 70).
I hope this explanation helps you graph Harper's data. Let me know if you have any further questions!
where is the 3rd point?
Apologies for the oversight. The 3rd point would be at 11 a.m. (11 hours) since the temperature rose 4° per hour for the next 3 hours after 6 a.m.
To determine the temperature at 11 a.m., we can add 4° per hour for 3 hours to the initial temperature of 58° F:
58° F + (4° per hour * 3 hours) = 58° F + 12° = 70° F
Therefore, the third point would be (11, 70).
Please let me know if you have any other questions!
One spring day, Harper noted the time of day and the temperature, in degrees Fahrenheit. Her findings are as follows: At 6 a.m., the temperature was 58° F. For the next 3 hours, the temperature rose 4° per hour. For the next 5 hours, it rose 1° per hour. The temperature then stayed steady until 6 p.m. For the next 2 hours, the temperature dropped 1° per hour. The temperature then dropped steadily until the temperature was 70° at midnight. On the set of axes below, graph Harper's data.
what would the coordinates be