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  • Algebra.... CRAP, here we go again....

    I'm really sorta embarassed that I've forgotten more about Algebra than I ever probably knew in the first place, so helping Amy with her homework is frustrating when she brings a question to dear OLD dad (as she shakes her head yes in agreement :hissyfit:) and I can't remember how to help her. So once again, I turn to my intelligent forum friends to lend the quadman (and Amy) a hand here. :dizzy:

    Problem: The top row of a supermarket display contains 12 cans. Each successive row has 1 more can than the previous row. If the display has 18 rows, what is the total number of cans in the display? What they DON'T mention, is you'd have to be at least ten feet tall to even SEE that the top row had 12 cans! :crazy:

    Thanks in advance for the assist guys....
    John W.
    Indy

  • #2
    I used Excel and came up with 369.
    PhenomeNhan Audio Video

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    • #3
      Originally posted by PhenomeNhan
      I used Excel and came up with 369.
      Haha, with modern technology, why is algebra even relavent? :neener:

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      • #4
        I'm really rusty on this too, so I can't give you a definitive setup. You're looking at

        12 + (12 +1) + (12 +2) + ... (12+17)

        I'm not sure how advanced the math is. I'd probably do a summation of i = 0 to 17 with the sum of (12+i), but that might sound like it's beyond the basic algebra setup. I don't really know...

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        • #5
          Originally posted by PhenomeNhan
          I used Excel and came up with 369.
          Well, that's all well and good my friend, but I know they want her to show her work and just adding the rows by hand (or Excel) isn't what they are looking for. Especially considering it's Algebra homework.

          Thanks all the same though.
          John W.
          Indy

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          • #6
            I think Gaussian's formula is what you're looking for:

            http://everything2.com/title/Gaussian+formula

            If I'm reading the problem correctly, you need to solve for the entire amt (1 to 30) then subtract the amt from rows 1-11 since they don't exist in the example.

            So the summation of 30 rows (initial 12 + 18) = 30*31 / 2 = 435
            then you need to subtract the sum of rows 1 - 11 from this: 11*12 / 2 = 66

            435-66 = 369

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            • #7
              You can also view the problem as if calculating a trapezoid's area : (1/2) * height * (width_top + width_bottom).

              Height = 18, width_top = 12, width_bottom = (12 + 18 - 1). Gives you 369 just like all the other answers here.

              [EDIT] NVM if it's an algebra homework the approach/formula in above post should be used, although the numbers in the previous post are a bit off.

              ie., summation of numbers from 1->n = n * (n+1) / 2

              then you calculate sum(12->29) as sum(1->29) - sum(1->11) = (29*30/2) - (11*12/2) = 369

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              • #8
                Originally posted by JazzySmooth
                I think Gaussian's formula is what you're looking for:

                http://everything2.com/title/Gaussian+formula

                If I'm reading the problem correctly, you need to solve for the entire amt (1 to 30) then subtract the amt from rows 1-11 since they don't exist in the example.

                So the summation of 30 rows (initial 12 + 18) = 30*31 / 2 = 435
                then you need to subtract the sum of rows 1 - 11 from this: 11*12 / 2 = 66

                435-66 = 369
                Use the same formula but eliminate the extra step (just find the value of rows 12 thru 29):

                ((12+29)(18))/2=369

                Comment


                • #9
                  Originally posted by hqubic
                  You can also view the problem as if calculating a trapezoid's area : (1/2) * height * (width_top + width_bottom).

                  Height = 18, width_top = 12, width_bottom = (12 + 18 - 1). Gives you 369 just like all the other answers here.

                  [EDIT] NVM if it's an algebra homework the approach/formula in above post should be used, although the numbers in the previous post are a bit off.

                  ie., summation of numbers from 1->n = n * (n+1) / 2

                  then you calculate sum(12->29) as sum(1->29) - sum(1->11) = (29*30/2) - (11*12/2) = 369
                  Ok.... this is what I think they were looking for. And, I was sorta, kinda on the right track but got stuck on the +/- 1 for that bottom row equation. I don't understand why it's (12 + 18 - 1). Weren't we "adding" a can on each row? If so, I miss the logic of the - 1.
                  John W.
                  Indy

                  Comment


                  • #10
                    Originally posted by quadman
                    Ok.... this is what I think they were looking for. And, I was sorta, kinda on the right track but got stuck on the +/- 1 for that bottom row equation. I don't understand why it's (12 + 18 - 1). Weren't we "adding" a can on each row? If so, I miss the logic of the - 1.
                    In order to show her work JazzySmooth has it correctly she needs to show using the formula that she has been given, so you plug the summation of i = 1 to 30 into the formula sum = n(n+1)/2
                    30 corresponds to the 18th row of cans.
                    Then plug 18 into the formula which represent the first row of cans and then subtract that from the first total.

                    So what you have is the total of i=1 to 30 minus the total of i=1 to 18

                    You should be able to find a sample problem that demonstrates how she should show her work.

                    Originally posted by JazzySmooth
                    I think Gaussian's formula is what you're looking for:

                    http://everything2.com/title/Gaussian+formula

                    If I'm reading the problem correctly, you need to solve for the entire amt (1 to 30) then subtract the amt from rows 1-11 since they don't exist in the example.

                    So the summation of 30 rows (initial 12 + 18) = 30*31 / 2 = 435
                    then you need to subtract the sum of rows 1 - 11 from this: 11*12 / 2 = 66

                    435-66 = 369

                    Comment


                    • #11
                      Originally posted by quadman
                      Ok.... this is what I think they were looking for. And, I was sorta, kinda on the right track but got stuck on the +/- 1 for that bottom row equation. I don't understand why it's (12 + 18 - 1). Weren't we "adding" a can on each row? If so, I miss the logic of the - 1.
                      Looks like most of the questions got answered; maybe I can help with this one.

                      The answer for subtracting the one was that the bottom width is actually 12 plus one can for each of 17 more rows (the row with 12 cans being the first), so I'd actually calculate the bottom row width as 12 + 17 rather than 12+18-1. Its like counting from 1 to 10 has ten numbers, but the difference is only 9.

                      Beyond that, I like the elegance of solving the "area" of the trapezoid. But I'm also having trouble seeing this as "algebra" since there's no unknown to solve for. Curious what else is in this module.

                      take care,
                      Sent to my room. :smoke1:

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                      • #12
                        Originally posted by quadman
                        Ok.... this is what I think they were looking for. And, I was sorta, kinda on the right track but got stuck on the +/- 1 for that bottom row equation. I don't understand why it's (12 + 18 - 1). Weren't we "adding" a can on each row? If so, I miss the logic of the - 1.
                        That last row will only have 29 cans not 30. The first row is 12 cans, so the set of 18 would be row 1: 12 +0, row 2: 12 + 1, ... row 18: 12 + 17.

                        It is a simple take the average number of cans per row and times by 18.

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                        • #13
                          Umm, yeah, what Eric said.

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                          • #14
                            Or go to your local wearhouse store and buy her cases of tomato paste and visually show her. I'm sure you'll use up the paste. You could the do the calcs why how many cases were needed and how many cans were in excess for extra credit. :whoopie:

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                            • #15
                              But I'm also having trouble seeing this as "algebra" since there's no unknown to solve for. Curious what else is in this module.
                              Correct.... this is "supposed" to be Algebra II and for the LIFE of me, I can't understand why some of her homework has had some of the things it has. I know it's been a LONG time since I took Algebra II, but I don't remember stuff like what I've seen her bring home these last couple of nights.

                              I consider myself half way intelligent and did fairly well in school (graduated 91st in a class of 750 with a 94.8 GPA with 95.0 - 99.0 being an A). Obviously I have forgotten a lot of stuff that never gets used on a regular basis, but some of the things they seem to be doing now, I just don't understand. Am i really THAT old?!? :no clue:
                              John W.
                              Indy

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