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astraxan [27]
2 years ago
9

YOU DONT HAVE TO DO ALL IF YOU DONT WANT TO JUST DO WHAT YOU CAN

Mathematics
2 answers:
natima [27]2 years ago
8 0

Answer:

Part A

Does the data for Amit’s puppy show a function? Why or why not?

It does show a function because it passes the vertical line test (no two points have the same x value).

Part B

Is the relationship for Amit’s puppy’s weight in terms of time linear or nonlinear? Explain your response.

Nonlinear because the line isn’t straight

Part C

Is the relationship between Amit’s puppy’s weight in terms of time increasing or decreasing? Explain your response.

Increasing because it is gaining weight

Part D

Does the data for Camille’s puppy show a function? Why or why not?

Yes, it does, because each input value has a unique output value

Part E

Is the relationship for Camille’s puppy’s weight in terms of time linear or nonlinear? Explain your response.

It is a linear function because the line has no curve, and the line is constant.

Part F

Is the relationship between Camille’s puppy’s weight in terms of time increasing or decreasing? Explain your response.

Increasing because as the puppy gets older it gains weight.

Part G

Does the data for Olivia’s puppy show a function? Why or why not?

Yes, it does, because each input value has a unique output value. The graph attached ( which shows the data for Camille’s puppy), that each x-value (Weeks) has a unique y-value (Weight in pounds).

Therefore, based on this and keeping in mind the explanation before, you can conclude that the data for Camille’s puppy shows a function.

Part H

Is the relationship for Olivia’s puppy’s weight in terms of time linear or nonlinear? Explain your response

Yes, it linear because it’s a straight line.

Part I

Is the relationship for Olivia’s puppy’s weight in terms of time increasing or decreasing? Explain your response. Increasing. For every week that goes by, Olivia's puppy is gaining one pound. 6-5= 1  14-13= 1. Gaining a pound every week makes the puppy’s weight increase.

Part J

Which two relationships have a y-intercept and a constant rate of change?

They all have y-intercepts and only Olivia and camilles have a constant rate of change.

Part A

To compare the linear functions, you first need to find their equations. For each of the linear functions, write an equation to represent the puppy’s weight in terms of the number of weeks since the person got the puppy.

Linear equation, y=mx+b

Exponential equation, y=a(b)×

Part B

Now you can compare the functions. In each equation, what do the slope and y-intercept represent in terms of the situation?

The y-intercept in the situation is 2/6.

Part C

Whose puppy weighed the most when the person got it? How much did it weigh?

Olivia’s puppy, it weighed 5 pounds

Part D

Whose puppy gained weight the slowest? How much did it gain per week?

Olivia’s puppy gained weight the slowest because it started off with more weight but only gained around 1 pound every week.

Part E

You can also graph the functions to compare them. Using the Edmentum Graphing Tool, graph the two linear functions. Paste a screenshot of the two functions in the space provided. How could you find which puppy had a greater initial weight from the graph? How could you find which puppy gained weight the slowest?

The edmentum graphing tool is opening up I tried it more than once but, the linear graphs would be Camille puppy and olivia's. And I could tell which one had a greater weight by how much they had at week 1 and how much they gained the weeks later. I could find which puppy gained weight the slowest by looking at the weight gained and graphed.

Step-by-step explanation:

Vikki [24]2 years ago
6 0

Answer:

(Warning) Not sure this is completley correct but this is just what I did.

Part A

Does the data for Amit’s puppy show a function? Why or why not?

It does show a function because it passes the vertical line test (no two points have the same x value).

Part B

Is the relationship for Amit’s puppy’s weight in terms of time linear or nonlinear? Explain your response.

Nonlinear because the line isn’t straight

Part C

Is the relationship between Amit’s puppy’s weight in terms of time increasing or decreasing? Explain your response.

Increasing because it is gaining weight

Part D

Does the data for Camille’s puppy show a function? Why or why not?

Yes, it does, because each input value has a unique output value

Part E

Is the relationship for Camille’s puppy’s weight in terms of time linear or nonlinear? Explain your response.

It is a linear function because the line has no curve, and the line is constant.

Part F

Is the relationship between Camille’s puppy’s weight in terms of time increasing or decreasing? Explain your response.

Increasing because as the puppy gets older it gains weight.

Part G

Does the data for Olivia’s puppy show a function? Why or why not?

Yes, it does, because each input value has a unique output value. The graph attached ( which shows the data for Camille’s puppy), that each x-value (Weeks) has a unique y-value (Weight in pounds).

Therefore, based on this and keeping in mind the explanation before, you can conclude that the data for Camille’s puppy shows a function.

Part H

Is the relationship for Olivia’s puppy’s weight in terms of time linear or nonlinear? Explain your response

Yes, it linear because it’s a straight line.

Part I

Is the relationship for Olivia’s puppy’s weight in terms of time increasing or decreasing? Explain your response. Increasing. For every week that goes by, Olivia's puppy is gaining one pound. 6-5= 1  14-13= 1. Gaining a pound every week makes the puppy’s weight increase.

Part J

Which two relationships have a y-intercept and a constant rate of change?

They all have y-intercepts and only Olivia and camilles have a constant rate of change.

Part A

To compare the linear functions, you first need to find their equations. For each of the linear functions, write an equation to represent the puppy’s weight in terms of the number of weeks since the person got the puppy.

Linear equation, y=mx+b

Exponential equation, y=a(b)×

Part B

Now you can compare the functions. In each equation, what do the slope and y-intercept represent in terms of the situation?

The y-intercept in the situation is 2/6.

Part C

Whose puppy weighed the most when the person got it? How much did it weigh?

Olivia’s puppy, it weighed 5 pounds

Part D

Whose puppy gained weight the slowest? How much did it gain per week?

Olivia’s puppy gained weight the slowest because it started off with more weight but only gained around 1 pound every week.

Part E

You can also graph the functions to compare them. Using the Edmentum Graphing Tool, graph the two linear functions. Paste a screenshot of the two functions in the space provided. How could you find which puppy had a greater initial weight from the graph? How could you find which puppy gained weight the slowest?

The edmentum graphing tool is opening up I tried it more than once but, the linear graphs would be Camille puppy and olivia's. And I could tell which one had a greater weight by how much they had at week 1 and how much they gained the weeks later. I could find which puppy gained weight the slowest by looking at the weight gained and graphed.

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

1. You can add the 2 halves together making 1

2. Then add a 3/4 and one 1/4 together making another 1

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4. Convert 2 1/2 to eighths. 2 4/8

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Answer:

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2 years ago
Choose the number pair that has digits of 4 that differ in place by a factor of 1000. 4 A. 49,823 and 81,540 B. 83,419 and 46,85
Ne4ueva [31]

Answer:

The correct option is D) 47,063 and 54,722.

Step-by-step explanation:

Consider the provided information.

Here we need to find the number pair that has digits of 4 that differ in place by a factor of 1000.

Now consider the option A) 49,823 and 81,540

The place value of 4 in 49,823 and 81,540 is 40,000 and 40 respectively.

So the difference=40,000-40=39960 which is not a multiple of 1000.

Consider the option B) 83,419 and 46,859

The place value of 4 in 83,419 and 46,859 is 400 and 40,000 respectively.

So the difference=40,000-400=39600 which is not a multiple of 1000.

Consider the option C) 54,890 and 81,745

The place value of 4 in 54,890 and 81,745 is 4000 and 40 respectively.

So the difference=4000-40=3960 which is not a multiple of 1000.

Consider the option D) 47,063 and 54,722

The place value of 4 in 47,063 and 54,722 is 40,000 and 4000 respectively.

So the difference=40000-4000=36000 which is a multiple of 1000.

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