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Math Help - function machines

  1. #1
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    Question function machines

    this is my question function machines-1.bmp


    i need to know what will be in the first box so all the answers = 3,6,10 if anyone can help i will be very greatfull thanks
    Last edited by andyboy179; February 14th 2009 at 10:55 AM. Reason: wrote something wrong
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  2. #2
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    Quote Originally Posted by andyboy179 View Post
    this is my question Click image for larger version. 

Name:	1.bmp 
Views:	11 
Size:	947.9 KB 
ID:	10116


    i need to know what will be in the first box so all the answers = 3,6,10 if anyone can help i will be very greatfull thanks
    Let "f" represent the function in that first box. Since you then add 2 you are looking for a function so that
    f(4)+ 2= 3, f(12)+ 2= 6, and f(28)+ 2= 10.

    That is the same as saying that f(4)= 1, f(12)= 4, and f(28)= 8. You can, at worst, fit a parabola through three points so try f(x)= ax^2+ bx+ c. Putting in the given values, f(4)= 16a+ 4b+ c= 1, f(12)= 144a+ 12b+ c= 10, and f(28)= 784a+ 28b+ 1= 10. That gives you three equations to solve for a, b, c.

    There are, of course, an infinite number of correct solutions to this problem.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Let "f" represent the function in that first box. Since you then add 2 you are looking for a function so that
    f(4)+ 2= 3, f(12)+ 2= 6, and f(28)+ 2= 10.

    That is the same as saying that f(4)= 1, f(12)= 4, and f(28)= 8. You can, at worst, fit a parabola through three points so try f(x)= ax^2+ bx+ c. Putting in the given values, f(4)= 16a+ 4b+ c= 1, f(12)= 144a+ 12b+ c= 10, and f(28)= 784a+ 28b+ 1= 10. That gives you three equations to solve for a, b, c.

    There are, of course, an infinite number of correct solutions to this problem.

    thanks for what you just posted but i forgot to say the number in the first box can only be 1 number i can't have a load of different numbers
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  4. #4
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    Quote Originally Posted by andyboy179 View Post
    i forgot to say the number in the first box can only be 1 number i can't have a load of different numbers
    So the box can contain only one operation...? Then:

    You have:

    . . . . .x = 4: [something done with 4] + 2 = 3
    . . . . .x = 12: [something done with 12] + 2 = 6
    . . . . .x = 28: [something done with 28] + 2 = 10

    Getting rid of the "plus two" after the "something done", and labelling the black-box output as "y", you have:

    . . . . .x = 4: y = 1
    . . . . .x = 12: y = 4
    . . . . .x = 28: y = 8

    For the first one, the "something done" might be "subtract three", but this won't work with the other two. And no subtraction that works for the either of the other two will work for the first. Clearly, addition won't work for any of them.

    You would multiply in the first one by 1/4 (or, which is the same thing, divide by 4), but this won't work for the others, and nothing that works for either of the others will work for the first.

    This shows that the only way to find a rule that works is to use the more complicated stuff that the previous replies explained. So either the instructions (to do one simple step using only one other number) were wrong, or else there is something wrong with the inputs or outputs.

    Please consult with your instructor regarding corrections and clarification. Thank you!
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