Both companies charge the same amount when the total duration of monthly calls is

[eddy2017]

Yes

Cool. I'll get on it and report back

 

[eddy2017]

okay, this is my work,
for y = 25 + 0.05x,
or y=0.05x+25
i set up a relation table and assign values to x to find the value of y
i came up with these two points.
(0,25) x=0, y=25
(20,1) x=20, y=1
i graphed the points.
for y= 50, well, here the slope is 0, m=0, i have only the y-intercept which i plotted as you can see in the graph.
okay my question is this.
how does graphing these equations help with the solution to the problem?

 

[lev888]

okay, this is my work,
for y = 25 + 0.05x,
or y=0.05x+25
i set up a relation table and assign values to x to find the value of y
i came up with these two points.
(0,25) x=0, y=25
(20,1) x=20, y=1
i graphed the points.
for y= 50, well, here the slope is 0, m=0, i have only the y-intercept which i plotted as you can see in the graph.
okay my question is this.
how does graphing these equations help with the solution to the problem?

Ok, y=50 makes sense. The other function doesn't look right. It looks like the longer you talk the less you pay. And at some point you start getting paid for calls (the charge is negative). Please check your work.

 

[lev888]

it is not a pretty graph but gives an idea. the slope is pretty steep. i did not assign values to x.
i just use the y intercept and the slope given in the equation.
give your opinion.

Y intersect is 25, however, you marked it below the x-axis, why? Below x-axis y values are negative. Please fix it. Why are there 4 arrows? They should indicate positive direction for each axis, so you should have only 2 of them.
What's the other point used to construct the line?

 

[eddy2017]

yes, i was careless cos i know that. so i'm plotting 0.05 as 1/20. so one up and 20 to the right and connecting the points.
check it pls and tell me. i still have a couple of questions.

 

[eddy2017]

i was going to ask you why we change 0.005 into a fraction before plotting it, but this short piece cleared it up for me

 

[lev888]

yes, i was careless cos i know that. so i'm plotting 0.05 as 1/20. so one up and 20 to the right and connecting the points.
check it pls and tell me. i still have a couple of questions.

Ok, the left point is (0, 25). To find the second point you are using the slope. No problem, but I don't understand the diagram. The point is 20 units to the right and one unit up. x is 20. But what is its y coordinate? What is 0.05 on the graph? Please mark the the coordinates of these points.
Please also graph the y=50 line.
Back to my original question - which point on the graph corresponds to equal charges?

 

[eddy2017]

i think i got it this time
i was plotting it the wrong way. i needed to plot (0,25) and then from that point, the y intercept then i plot the other point, the slope in the equation in this case 0.05 that i turned into 1/20. i rise 1 and run 20.

 

[eddy2017]

Back to my original question - which point on the graph corresponds to equal charges?
i'm stumped there.

 

[lev888]

i think i got it this time
i was plotting it the wrong way. i needed to plot (0,25) and then from that point, the y intercept then i plot the other point, the slope in the equation in this case 0.05 that i turned into 1/20. i rise 1 and run 20.

I don't see the y=50 line. The whole point of the exercise is to graph both lines in the same coordinate system to see how they relate to each other.
Here's the graph from desmos. Same question, at which point the lines show the same charge? Charge is y. So, at which point the lines have the same y coordinates?

 

[eddy2017]

I don't see the y=50 line. The whole point of the exercise is to graph both lines in the same coordinate system to see how they relate to each other.
Here's the graph from desmos. Same question, at which point the lines show the same charge? Charge is y. So, at which point the lines have the same y coordinates?

at 500 but i do not get it. if i were to graph it on a coordinate plane drawn by me, how would i chart 500?. there is no space!

 

[eddy2017]

I don't see the y=50 line. The whole point of the exercise is to graph both lines in the same coordinate system to see how they relate to each other.
Here's the graph from desmos. Same question, at which point the lines show the same charge? Charge is y. So, at which point the lines have the same y coordinates?

i understand now. where they meet. that is where they have the same charge.

 

[lev888]

i understand now. where they meet. that is where they have the same charge.

Yes. What does it means algebraically?
y is the same means we should equate the 2 right side expressions and we get 50 = 0.05x + 25.
Solve for x => x = 500min.
It's up to you to decide whether graphing in this particular case was a helpful step.

 

[Otis]

… how would i chart 500 [on the horizontal axis]?. there is no space!

Hi Eddy. You could graph it as shown in post #30, with the horizontal axis drawn parallel to the longest side of your paper.

Note: There are no negative times or charges, so you don't need to draw Quadrants II, III and IV.

Alternatively, if you prefer to not rotate your paper, reduce the scale on the horizontal axis. The horizontal and vertical axes' tick marks don't need to match, in this type of graph. (Identical scales are necessary for things like graphing circles or trying to show symmetry.)

?

 

[eddy2017]

Hi Eddy. You could graph it as shown in post #30, with the horizontal axis drawn parallel to the longest side of your paper.

Note: There are no negative times or charges, so you don't need to draw Quadrants II, III and IV.

Alternatively, if you prefer to not rotate your paper, then reduce the scale on the horizontal axis. The horizontal and vertical axes' tick marks don't need to match, in this type of graph. (Identical scales are necessary for things like graphing circles or trying to show symmetry.)

?

interesting, thank you!.

 

[eddy2017]

Yes. What does it means algebraically?
y is the same means we should equate the 2 right side expressions and we get 50 = 0.05x + 25.
Solve for x => x = 500min.
It's up to you to decide whether graphing in this particular case was a helpful step.

yes, it was. for me it was. even if i see now that HallsofIvy was right. there is no need to graph it to solve it, but for me it is good being a newbie at this anything visual speaks volumes to me.

 

[eddy2017]

Note: There are no negative times or charges, so you don't need to draw Quadrants II, III and IV.

yes, i see it. all the action is on Quadrant 1.

 

[eddy2017]

Thanks for the collective advice and help on solving this. You do a world of good to people like me who struggle to learn. Thanks.