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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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  1. theEric Group Title
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    Can we make an equation to help us?

    • one year ago
  2. theEric Group Title
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    There can be a few ways to go about this, it seems.

    • one year ago
  3. lopez_hatesmath Group Title
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    yeahh anything that will help

    • one year ago
  4. theEric Group Title
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    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 point-slope formula?

    • one year ago
  5. theEric Group Title
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    \[y=mx+b\]

    • one year ago
  6. lopez_hatesmath Group Title
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    okay gotcha so far

    • one year ago
  7. theEric Group Title
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    \[totalCost = (\frac{$1000}{dayLate})(daysLate)+b\] where "b" is the y intercept.

    • one year ago
  8. lopez_hatesmath Group Title
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    okay

    • one year ago
  9. theEric Group Title
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    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. \[y=mx+b \rightarrow x=(y-b)/m\]

    • one year ago
  10. theEric Group Title
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    Good luck, and take care. I'll look at this question again later tonight, so feel free to ask questions!

    • one year ago
  11. theEric Group Title
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    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 point-slope 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 y-intercept 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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