parking in a student lot cost $1 for the first hour and 1.75 for each hour there after a partial hour is changed the same as a full hour what is the longest time that a student can park in this lot for $7 ? the longest time is ? have to show my work.

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parking in a student lot cost $1 for the first hour and 1.75 for each hour there after a partial hour is changed the same as a full hour what is the longest time that a student can park in this lot for $7 ? the longest time is ? have to show my work.

Linear Algebra
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the expression for how much money it costs is 1+1.75t where t is time in hours. If the cost has to be less than or equal to 7, set the expression less than or equal to 7 1+1.75t<=7 subtract 1 from both sides and divide by 1.75 this will tell you how many hours you can have remember to round down to the nearest integer
1+1.75t<=7 1.75t<=6 calculate t now..
okay

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the answer is 6
@hitten101 i think we both made a mistake...shouldnt it be 1+1.75(t-1)<=7 because if you plug in 1 hour into the original, you get charged 2.75 not 1.00
okay
oh yes you are right @etemplin
the answer is 6 then?
no offends you a very intelligent handsome guy.
no. using the new equation (the old one was wrong...sorry about that) you get t<=4.42 so it should be 4. ill check by taking 1 for the first hour, adding 1.75 for the second hour (total = 2.75) add 1.75 for the third hour (total = 4.5) and 1.75 for the 4th hour (total = 6.25) since you cant go over 7, the hours = 4
No @etemplin the time can go in fractions also. it does not need to be a natural number. it can be a fraction.
it says if you only use a partial hour it counts for a whole one
@Emilyggarza whom u referring to?
not me ;)
hmmm... ur right @etemplin i ignored that part of the question
and thank you @Emilyggarza
\[1+\text{Floor}\left[\frac{7-1}{1.75}\right]=4 \text{ hours}\]Floor[x] gives the greatest integer less than or equal to x.

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