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Seth and Karen are college students who take on different jobs for income. Seth needs $750 for expenses and savings each month; Karen needs $700.

Seth has a part-time job where he earns $7.50 per hour. He also mows lawns and earns $15 per lawn that he mows. His monthly earnings can be expressed with the equation 7.50x + 15y = 750, where x is the number of hours Seth works and y is the number of lawns he mows.

Karen has a part-time job babysitting, where she earns $6 per hour. She also walks dogs and earns $4 per dog that she walks. Her monthly earnings can be expressed with the equation 6x + 4y = 700, where x is the number of hours Karen works and y is the number of dogs she walks.

Part A: Each partner now has an equation, function, and graph that represent the combinations of activities resulting in the income needed for each student. Compare these items and answer the following questions:
Which student, Seth or Karen, has the higher slope when the graphs of their earnings are compared? How can you tell?

Part B: Whose graph, Seth's or Karen's, has a greater y-intercept? What does that mean for this person?

Part C: If both students work 80 hours for the month, is the number of lawns Seth has to mow greater than or less than the number of dogs Karen has to walk? Justify your answer.

Part D: Karen begins charging $8 per hour for babysitting and Seth begins earning $8 per hour. How does this change affect their graphs if everything else remains the same?

P.S. Those who answer wrong answers on purpose will be reported. (Has happened too many times by the same person. You know who you are.) This is very important, so the faster I get the correct answer for each part, the better. Thank you for those who help!!! I will be in your debt!

Respuesta :

Seth:

Equation 7.50x + 15y = 750, where x is the number of hours Seth works and y is the number of lawns he mows.

Karen:

Equation 6x + 4y = 700, where x is the number of hours Karen works and y is the number of dogs she walks.

Part A: Converting it in function for by solving for y first.

Solving 7.50x + 15y = 750, equation we get

7.50x + 15y = 750

15y = -7.50x + 750

y = -1/2x + 50

Solving 6x + 4y = 700 for y, we get

6x + 4y = 700

4y = -6x + 700

y= -3/2x + 175.

On comaring with slope-intercept form y=mx+b, we got slope for Seth equation is -1/2 and slope for Karen is -3/2.

If we take absolute of those -1/2 and -3/2 we get 1/2 and 3/2.

Therefore, Karen has the higher slope. We can see the graph that blue line has greater slope as it's increasing/decreasing with greater rate.

In function form we could write equation as

f(x) = -1/2x + 50 and

f(x) = -3/2x + 175.

Part B : y-intercept of Karen equation is 175 and y-intercept of Seth equation is 50.

Therefore, Karen has greater y-intercept.

This means Karen earns $175 by just walking dogs.

Part C: If both students work 80 hours for the month.

Let us plug x =80 in each of the equations.

y = -1/2x + 50  => y = -1/2(80) + 50 = -40 +50 = $10.

Seth would earn $10 from mowing lawn.

Therefore, the number of lawns Seth has to mow = 10/15 = 0.67 that is approximately 1 lawn.

y = -3/2x + 175  => y = -3/2(80) + 175 = -120 +175 = $55.

Karen would earn $50 from walking dog.

So, the number of dogs Karne has to walk = 55/4 = 13.75 that ia approximately 14 dogs.

Therefore, the number of lawns Seth has to mow is less than the number of dogs Karen has to walk.

Part D: If Karen begins charging $8 per hour for babysitting and Seth begins earning $8 per hour.

The equations would become : 8x + 15y = 750, 8x + 4y = 700.

It would not effect the graph much. Even the y-intercepts would remain same.

There would be a slightly change in slopes.

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Answer:

Seth:


Equation 7.50x + 15y = 750, where x is the number of hours Seth works and y is the number of lawns he mows.


Karen:


Equation 6x + 4y = 700, where x is the number of hours Karen works and y is the number of dogs she walks.


Part A: Converting it in function for by solving for y first.


Solving 7.50x + 15y = 750, equation we get


7.50x + 15y = 750


15y = -7.50x + 750


y = -1/2x + 50

Solving 6x + 4y = 700 for y, we get


6x + 4y = 700


4y = -6x + 700


y= -3/2x + 175.

On comaring with slope-intercept form y=mx+b, we got slope for Seth equation is -1/2 and slope for Karen is -3/2.


If we take absolute of those -1/2 and -3/2 we get 1/2 and 3/2.


Therefore, Karen has the higher slope. We can see the graph that blue line has greater slope as it's increasing/decreasing with greater rate.


In function form we could write equation as


f(x) = -1/2x + 50 and

f(x) = -3/2x + 175.

Part B : y-intercept of Karen equation is 175 and y-intercept of Seth equation is 50.


Therefore, Karen has greater y-intercept.

This means Karen earns $175 by just walking dogs.

Part C: If both students work 80 hours for the month.


Let us plug x =80 in each of the equations.


y = -1/2x + 50  => y = -1/2(80) + 50 = -40 +50 = $10.

Seth would earn $10 from mowing lawn.


Therefore, the number of lawns Seth has to mow = 10/15 = 0.67 that is approximately 1 lawn.


y = -3/2x + 175  => y = -3/2(80) + 175 = -120 +175 = $55.

Karen would earn $50 from walking dog.


So, the number of dogs Karne has to walk = 55/4 = 13.75 that ia approximately 14 dogs.


Therefore, the number of lawns Seth has to mow is less than the number of dogs Karen has to walk.

Part D: If Karen begins charging $8 per hour for babysitting and Seth begins earning $8 per hour.


The equations would become : 8x + 15y = 750, 8x + 4y = 700.


It would not effect the graph much. Even the y-intercepts would remain same.

There would be a slightly change in slopes.


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