## lopez_hatesmath 2 years ago 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? • This Question is Closed 1. theEric Can we make an equation to help us? 2. theEric There can be a few ways to go about this, it seems. 3. lopez_hatesmath yeahh anything that will help 4. theEric 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? 5. theEric $y=mx+b$ 6. lopez_hatesmath okay gotcha so far 7. theEric $totalCost = (\frac{1000}{dayLate})(daysLate)+b$ where "b" is the y intercept. 8. lopez_hatesmath okay 9. theEric 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. $y=mx+b \rightarrow x=(y-b)/m$ 10. theEric Good luck, and take care. I'll look at this question again later tonight, so feel free to ask questions! 11. theEric 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...

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