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    O-Level Additional Math

    Scheduled Pinned Locked Moved Secondary Schools - Academic Support
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    • J Offline
      jieheng
      last edited by

      Acsian:




      Thanks Alot jieheng Thanks :thankyou:
      Hi Acsian ,

      You are welcome.

      Regards,

      1 Reply Last reply Reply Quote 0
      • H Offline
        Herbie
        last edited by

        When to use HCF and LCM to solve problem qn? Can someone advise?

        1 Reply Last reply Reply Quote 0
        • A Offline
          Acsian
          last edited by

          when


          10x^2+40xy-24x-96y is factorised the answer is (2x+8y)(5x-12) *not fully factorised*


          But when 2ax-4ay+3bx-6by is factorised, why is the answer is
          (2ab-3b)(x-2y)and not

          (2ab+3b)(x-2y)

          please help[/b]

          1 Reply Last reply Reply Quote 0
          • A Offline
            Acsian
            last edited by

            Herbie:
            When to use HCF and LCM to solve problem qn? Can someone advise?


            LCM can be used for questions like \" Bus A leaves the terminal every 10 minutes. Bus B leaves the terminal every 15 minutes. If the two buses leave the terminal at 08 00 when will they next leave together?\"

            HCF can be used for questions like \" There are 48 studens from Mexico 96 students from Canada and 72 students from Canada attending a education conference. What is the greatest number of groups that can be formed so that students from each country can be distributed equally among all groups?\"

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            • J Offline
              jieheng
              last edited by

              Acsian:
              when


              10x^2+40xy-24x-96y is factorised the answer is (2x+8y)(5x-12) *not fully factorised*


              But when 2ax-4ay+3bx-6by is factorised, why is the answer is
              (2ab-3b)(x-2y)and not

              (2ab+3b)(x-2y)

              please help[/b]
              2ax-4ay+3bx-6by

              =2ax+3bx-4ay-6by

              =x(2a+3b)-2y(2a+3b)

              =(2a+3b)(x-2y)

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              • K Offline
                k1ndan
                last edited by

                Please help to solve these 2 Sec 2 Maths problem.


                1) Given that 3a + e = 7 and 4ae/3 = 2, find the value of e-3a.



                2) Given that m^2 + 8m = v^2 + 8v and that m is not equal to v, find the value of 5(m + v).



                Thanks.

                1 Reply Last reply Reply Quote 0
                • S Offline
                  Sun_2010
                  last edited by

                  Hi k1ndan,


                  Let me give it a try and see if my rusty grey sells can work
                  1 ) Given that 3a + e = 7 and 4ae/3 = 2, find the value of e-3a.
                  3a+e=7
                  (3a+e)^2 = 7^2
                  9a^2 + e^2 + 6ae= 49 -------(i)

                  4ae/3=2 so 6ae=9 -------------(ii)

                  substituting in (i)
                  9a^2 + e^2 +9= 49
                  9a^2 + e^2 =40 -------------(iii)

                  (e-3a)^2
                  = 9a^2 + e^2 - 6ae
                  = 40-9
                  =31

                  hence (e-3a) = square root of 31

                  gtg, will be back later for 2nd question

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                  • S Offline
                    Sun_2010
                    last edited by

                    2) Given that m^2 + 8m = v^2 + 8v and that m is not equal to v, find the value of 5(m + v).


                    m^2 + 8m = v^2 + 8v
                    m^2 - v^2 + 8v -8M = 0
                    (m+v)(m-v) + 8(m-v) = 0
                    (m-v) (m+v+8)=0
                    since mis not equal to v, m+v= -8
                    5(m+v) = -40

                    1 Reply Last reply Reply Quote 0
                    • CoffeeCatC Offline
                      CoffeeCat
                      last edited by

                      Sun_2010:
                      Hi k1ndan,


                      Let me give it a try and see if my rusty grey sells can work
                      1 ) Given that 3a + e = 7 and 4ae/3 = 2, find the value of e-3a.
                      3a+e=7
                      (3a+e)^2 = 7^2
                      9a^2 + e^2 + 6ae= 49 -------(i)

                      4ae/3=2 so 6ae=9 -------------(ii)

                      substituting in (i)
                      9a^2 + e^2 +9= 49
                      9a^2 + e^2 =40 -------------(iii)

                      (e-3a)^2
                      = 9a^2 + e^2 - 6ae
                      = 40-9
                      =31

                      hence (e-3a) = square root of 31

                      gtg, will be back later for 2nd question
                      by the way, without any other restrictions on the values it is possible that the answer can be -sqrt(31) as well.

                      1 Reply Last reply Reply Quote 0
                      • J Offline
                        JadeDry
                        last edited by

                        I currently use "New Syllabus Mathematics" (8th Grade) books, and would appreciate if you could recommend additional good quality publications.


                        Thanks in advance.

                        1 Reply Last reply Reply Quote 0

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