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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:
      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)

      1 Reply Last reply Reply Quote 0
      • 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
                • PiggyLalalaP Offline
                  PiggyLalala
                  last edited by

                  What kind of assessment books are you looking at? Topical or paper by paper? Are you looking for more challenging questions? Or are you looking for one that focus more on drill and practice.

                  1 Reply Last reply Reply Quote 0
                  • H Offline
                    hometutors.018849sg
                    last edited by

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


                    Thanks in advance.
                    A good tutor / mentor would be able to select what is good for the student.

                    1 Reply Last reply Reply Quote 0
                    • E Offline
                      elkniwt
                      last edited by

                      Hi,


                      For compound interest. If the principal amt is invested for a non-integral no. of the compounding period, how is it computed?

                      Amt = $1000
                      Interest = 1% per annum compounded yearly
                      Invested for 2yr and 3mths
                      Final amt = (1.01)^2 x 1000
                      OR
                      Final amt = (1.01)^2.25 x 1000 ??

                      Thanks.

                      1 Reply Last reply Reply Quote 0
                      • F Offline
                        FrekiWang
                        last edited by

                        Final amt = (1.01)^2.25 x 1000 is the correct answer at secondary level.

                        1 Reply Last reply Reply Quote 0

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