sandy is upgrading her internet service. Fast internet charges $60 for installation $50.45 per month. Quick internet has free in
stallation but charges $57.95 per month write an equation that can be used to find the number of months for which the internet service would cost the same
In order to know at what price the two services offered would be the same, we can actually use the process of trial and error starting from 1 onwards. The first one is 50.45 per month plus a standard fee of 60 for installation. So this will be 60+50.45x (number of days). After that. there is the other one that has free installation but charges 57.95. So it can be expressed as 57.95x. Now after trial and error with numbers as x, I came upon 8. If x is 8, then the first one will be 60+50.45 (8) = 463.6. For the second, 57.95(8) will also equal to 463.6. So the day in which the two services will charge the same is during the 8th day.
To find the number of even factors, we can multiply the number of odd factors by the power of 2 (not the power of 2 + 1!!!). For 540, we have (3 + 1)(1 + 1)(2) = 16 even factors.
