SOLUTION: have one question left on a summer packet, not sure how to answer. need to sketch both equations on the same x-y plane..company A offers a cell phone fee of $90/mnth...company B co
Algebra.Com
Question 484136: have one question left on a summer packet, not sure how to answer. need to sketch both equations on the same x-y plane..company A offers a cell phone fee of $90/mnth...company B costs $15per month for service and $.10 per minute...need to compare the 2 plans ???????????????????
Found 2 solutions by ankor@dixie-net.com, Theo:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
sketch both equations on the same x-y plane.
.company A offers a cell phone fee of $90/mnth.
..company B costs $15per month for service and $.10 per minute.
.need to compare the 2 plans:
:
Plan A offers a flat $90 with unlimted calling, the equation for that
y = 90
:
Let t = no. of min used in Plan B
y = .10m + 15
:
Plot these two equation, should look like this,
x axis, no of minutes
y axis total cost per month
You can see that Plan A is just a horizontal line at y=90 (Red)
Plan B cost increases with the number of minutes used
To find how many minutes for them to cost the same write
.10m + 15 = 90
.10m = 90-15
.10m = 75
m =
m = 750 mins which is illustrated by the graph
:
To summarize, if you use more than 750 min Plan A is best if you use less Plan B is best, right?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
company A charges $90.00 per month flat fee.
company B charges $15.00 pr month plus $.10 per minute.
equation for company A is:
y = 90
equation for company B is:
y = .10*x + 15
you can graph both equations.
looks like this:
the plan with the flat rate is more expensive up to about x minutes.
you can solve for the exact value by solving the 2 equations simultaneously.
the 2 equations are:
y = 90
y = .10x + 15
since they both equal to y, then set them equal to each other to get:
90 = .10x + 15
subtract 15 from both sides of the equation to get:
75 = .10x
divide both sides of the equation by .10 to get:
x = 750
the break even point is at 750 minutes.
the graph has been modified to provide a horizontal line at y = 90 and a vertical line at x = 750.
the intersection of the horizontal line and the vertical line should be the break even point.
the horizontal line at y = 90 is the equation for Company A.
The slanted line is the equation for Company B.
It intercepts the y axis at x = 15 which is the cost to the customer with 0 minutes used.
The vertical line shows you that the intersection of the graph of the equation for Company A and the graph of the equation for Company B occurs at x = 750 which is the break even point.
the value of x in the graph is the number of minutes used.
the value of y in the graph is the cost for the number of minutes used.
you can see that the cost for company A customer is $90.00 regardless of the number of minutes they used, while the cost for company B customer has a fixed component ($15.00) and a variable component (10 cents a minute).
RELATED QUESTIONS
Here is the problem I received on a summer homework packet. I think there is not enough... (answered by ankor@dixie-net.com)
hi, my name is nia. i am having a problem with equations. i think it's called equation... (answered by Alan3354)
I need help to simplify this equation on my summer math packet. Thank you.... (answered by Alan3354)
I have a question that i need help with:
Sketch the graph of y=x^3-3x^2+1/x^3
1.... (answered by arallie)
I have no idea how to approach this, can you please help me.
The question is as... (answered by Alan3354,ikleyn)
i have to a packet of math work for summer as a review to see if we remember how to do... (answered by jim_thompson5910)
I need help on these two summer packet equations. It says state the axiom or property... (answered by rothauserc)
Hi! I have a question with inequalities and x and y on the same side. Can you help? It... (answered by Edwin McCravy)
This is like a disscussion question. Problem 51 on Page 512:
You now have had practice... (answered by stanbon)