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lopez_hatesmath
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A construction company will be fined for each day it is late completing its current project. The daily fine will be $4000 for the first day and will increase by $1000 each day. Based on its budget, the company can only afford $60,000 in total fines. What is the maximum number of days it can be late?
 one year ago
 one year ago
lopez_hatesmath Group Title
A construction company will be fined for each day it is late completing its current project. The daily fine will be $4000 for the first day and will increase by $1000 each day. Based on its budget, the company can only afford $60,000 in total fines. What is the maximum number of days it can be late?
 one year ago
 one year ago

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theEric Group TitleBest ResponseYou've already chosen the best response.1
Can we make an equation to help us?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
There can be a few ways to go about this, it seems.
 one year ago

lopez_hatesmath Group TitleBest ResponseYou've already chosen the best response.0
yeahh anything that will help
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
This is a problem you can describe with a graph.dw:1345342500517:dw Do you like dealing with graphs? Or would you prefer just math? Are you required to use the pointslope formula?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
\[y=mx+b\]
 one year ago

lopez_hatesmath Group TitleBest ResponseYou've already chosen the best response.0
okay gotcha so far
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
\[totalCost = (\frac{$1000}{dayLate})(daysLate)+b\] where "b" is the y intercept.
 one year ago

lopez_hatesmath Group TitleBest ResponseYou've already chosen the best response.0
okay
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
I have to leave soon, sorry, but I'll provide some steps here! You're essentially saying you want to find the number of days late, x, when the total fine is $60,000, y. So you need to solve for the number of days late. With this equation, you know: y the total fine we have to consider ($60,000) m the $1000 per day slope but you don't know: x the number of days to get to that total fine But you want to know x, so you'd solve for it. SOLVE FOR X HERE You need to know "b" to go on! You know the slope, and a point on the graph (1 day, $4,000), so you can use \[y=mx+b\] and solve for "b" to get \[b = y  mx\] SUBSTITUTE AND CALCULATE "b" HERE In this way, you've found "b", only because you knew a point on the line and it's slope. "b" is going to be the same "b" as before, it's the same line. Now you know your "b", the $60,000 y, the $1000/day m, and you can solve for the days late, x. \[y=mx+b \rightarrow x=(yb)/m\]
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Good luck, and take care. I'll look at this question again later tonight, so feel free to ask questions!
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Thanks! Now I say two things: 1. I thought of a line because the change to the fine, per day, was unchanging, It was an unchanging $1000/day. This is a "linear relation" that can be explained with the pointslope formula, \[y=mx+b\]. 2. Picture! dw:1345398667899:dw The points from left to right, are: (0 days, ?) (1 day, $4,000) (? days, $60,000) That's sort of the overall picture here. You can look at it at any point while solving the problem. Except you find the the point at the yintercept is (0 days, $3000). And then you finish by finding the days it takes to accumulate $60,000 in fines...
 one year ago
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