Question

Phone Company 1 charges a flat rate of $25.00 a month for local calling, and $0.05 per minute for each long distance call. Phone Company 2 charges a flat rate of $20.00 a month for local calling, and $0.10 per minute for long distance calling. After how many minutes of long distance calling do the phone companies charge an equal amount?

Answer 1

someone....

Answer 2

i'm thinking hold on

Answer 3

100 minutes of distance calls will have to be made for the phone companies to charge equal amounts.

Answer 4

how did you figure that out?

Answer 5

hello...

Answer 6

25 + .05m = 20 + .10m (subtract .05m from both sides)
25 = 20 + .10m - .05m (subtract 20 from both sides)
25 - 20 = .10m - .05m (combine like terms)
5 = .05m (divide by .05)
5/.05 = m
100 = m

They are equal at 100 minutes

Answer 7

any questions ?

Answer 8

no but thnx

Answer 9

anytime :)

Answer 10

wait...

Answer 11

Which of the following is the equation of the line perpendicular to 6x - y - 5 = 0 and passes through the point (-3, 2)?
Choose one answer.

Answer 12

first you have to find the slope by putting the equation in y = mx + b form and m will be your slope.
6x - y - 5 = 0
-y = -6x + 5 (multiply by -1 to make y positive)
y = 6x - 5
the slope for this equation is 6. However, we need a perpendicular line, therefore, the slope must be the negative reciprocal of the slope given. That just means " flip " the slope over and change the sign. So the negative reciprocal of slope 6 = -1/6.
Now we can sub this back into y = mx + b form and solve for b, the y intercept.
y = mx + b
slope(m) = -1/6
(-3,2) x = -3 and y = 2
now we sub
2 = -1/6(-3) + b
2 = 1/2 + b
2 - 1/2 = b
4/2 - 1/2 = b
3/2 = b
Therefore, your perpendicular line is : y = -1/6x + 3/2
any questions ?

Answer 13

no

Answer 14

glad I could help....that was an answer choice, right ?

Answer 15

yes

Answer 16

good to hear :)

Answer 17

<img src=" http://moodlemid.ucps.k12.nc.us/file.php/753/Weekly_Math_Images/WM3Q8.JPG " alt="q8"/>

Answer 18

what is x in that picture????

Answer 19

hello...

Answer 20

can someone help me????

Answer 21

do you have to log in or something ??

Answer 22

can u not click on the link?

Answer 23

r u still there?