The company discovered that it costs $20.50 to produce 3 widgets, $60.50 to produce 7 widgets, and $133 to produce 12 widgets. Using quadratic functions, find the cost of producing 9 widgets.

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The company discovered that it costs $20.50 to produce 3 widgets, $60.50 to produce 7 widgets, and $133 to produce 12 widgets. Using quadratic functions, find the cost of producing 9 widgets.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Do you have access to a graphing calculator... Is it allowed for this problem...
no. all i have is a scientific calculator.
& yes it is allowed. i just don't have one.

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Just a minute.
okay
General formula y = ax^2 + bx + c put in the x,y for each point. (3,20.50) (7,60.50) (12,133.0) 20.50 = a(3^2) + b(3) + c 60.50 = a(7^2) + b(7) + c 133.00 = a(12^2) + b(12) + c 20.50 = 9a + 3b + c 60.50 = 49a + 7b + c 133.00 = 144a + 12b + c Subtract equation 1 - equation 2 -40 = -40a - 4b Subtract equation2 - equation 3 -72.50 = -95a - 5b -40 = -40a - 4b -72.50 = -95a - 5b Multiply by 5 and (-4) to get b's the same 5(-40 = -40a - 4b) -4(-72.50 = -95a - 5b) -200 = -200a - 20b 290 = 380a + 20b add these 90 = 180a a = 1/2 plug that into -200 = -200a - 20b -200 = -200(1/2) - 20b -200 = -100 - 20b -100 = -20b b = 5 plug those into 20.50 = 9a + 3b + c 20.50 = 9(1/2) + 3(5) + c 20.50 = 4.5 + 15 + c 20.50 = 19.5 + c 1 = c y = 1/2x^2 + 5x + 1 so put in x = 9 y = 1/2(9^2) + 5(9) + 1 y = 1/2*81 + 45 + 1 y = 40.5 + 45 + 1 y = 86.50
If you had a graphing calculator... you put in the three points... find regression and pick quadratic and it will give you the a,b, and c MUCH QUICKER...
yeah i've realized that. & i don't know anyone who has a graphing calculator which sucks.
but thanks.
If you executed the following expression in Mathematica v8 cost==Fit[{{3,20.50},{7,60.5},{12,133}}, {1,x,x^2}, x] /.x->9 Mathematica 8 running on a mid 2010 IMac with 8Gb ram would spit back with no perceptible delay: cost==86.5 Right now you could browse over to WolframAlpha.com and paste the following into their "=" box: Fit[{{3,20.50},{7,60.5},{12,133}}, {1,x,x^2}, x] Their result would be the cost expression: \[0.5x^2 + 5.x + 1.\] When evaluated at x=9 the cost expression would be equal to 86.5 as shown by blexting in his solution.

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