anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Do you have access to a graphing calculator... Is it allowed for this problem...
anonymous
  • anonymous
no. all i have is a scientific calculator.
anonymous
  • anonymous
& yes it is allowed. i just don't have one.

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anonymous
  • anonymous
Just a minute.
anonymous
  • anonymous
okay
anonymous
  • anonymous
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
anonymous
  • anonymous
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...
anonymous
  • anonymous
yeah i've realized that. & i don't know anyone who has a graphing calculator which sucks.
anonymous
  • anonymous
but thanks.
anonymous
  • anonymous
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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