chemistry1111

  1. reflected in the x-axis
  2. reflected in the y-axis

**I'm not quite sure if y = (5/x) -6 or if it's y = 5/(x-6) so I'll show the working for both

let f[x] = (5/x) -6

  1. x reflection: -f[x] = f1[x] = -[ (5/x) - 6 ] = -( 5/x ) + 6

  2. y reflection: f1[-x] = f2[x] = -(5/ [-x] ) +6 = -(5/ x ) +6

let g[x] = 5/(x-6)

  1. x reflection: -g[x] = f1[x] = -[ 5/(x-6) ] = -5/(x-6)

  1. y reflection: g1[-x] = f2[x] = -5/( [-x] - 6) = -5/( -x - 6)

hope that helps!

    Hi everyone,

    Is (2x-1) the same as (x-1/2)
    I mean they both give x= 1/2
    But let's say if we had to factorise 12x3 + 20x2 − x − 6
    and we get (2x-1)(2x+3)(3x+2)

    Can I also write this as:
    (x-1/2)(2x+3)(3x+2)?

    ALSO can some please help and find the roots of this: 12x3 + 20x2 − x − 6
    using rational root theorem

    much appreciated!!!

      PizzaMaster
      good question! to answer your first one, (2x-1) = 2 (x-1/2). they're not the same but the same x value will make both of them equal (0). I'm not home rn so I'll reply to the second one when I have access to my iPad.

      EDIT
      I think it's easier if you watch a video, here's a good one I found (watching 0:00-2:40 should be enough): https://www.youtube.com/watch?v=4XytYH35AP0

        So I can write (2x-1)(2x+3)(3x+2) as 2(x-1/2)(2x+3)(3x+2) NOT (x-1/2)(2x+3)(3x+2)
        Great, thanks crumblycupcakes!

        With rational root theorem, I find that it is super time consuming... after we find the possible roots, is there a shortcut to know which ones to try first?
        For example with 12x3 + 20x2 − x − 6:
        the possible roots are +- [1/2, 1/3, 1/4, 1/6, 1/12, 2/3, 3/4, 3/2]
        That is a lot! Is there a way to know which one we can try first?

        Also does simplifying work?
        for example with 12x3 + 20x2 − x − 6, we do like factors of 6/ factors of 12. Can we do factors of 1/ factors of 2 (since 6/12 = 1/2?)

        Billzene
        I think that would only be the case in Specialist maths with VCAA so 'no solutions' is usually accepted in methods because complex numbers aren't taught in it. BUT we can't say for sure so it's always good to say 'no real solutions'. 😅

        [ Edit ]
        istg VCAA is so weird. They changed the study design for spesh and they included proofs but didn't specify if it was matrix proofs, number proofs or circle proofs or all of them. Apparently my teacher emailed them a couple of times but they never replied.

          5 days later

          Hi, could someone please help me with this question. I am pretty sure the answer is 150-t but i'm not too sure how I got there and what the working out should look like.

          Towns A and B are 150 km apart along a straight road. A car leaves town A and heads towards town B at 60 km/h. The distance, in kilometres, of the car from town B after t minutes is...

          Thanks

            :) It sounds like you might have caught on to the right idea. We know the car travels at 60 km/h which equals 1 km/minute. Each minute, the car travels 1 km, therefore in t minutes the car travels t km. Therefore the distance from town B is 150-t km.
            I feel that the following working should be sufficient, but it might depend on mark allocation.
            distance = 150 - (60 / 60) * t = 150 - t

              2 months later

              hey guys! I was just wondering if any of you know if the methods checkpoints books are designed to be done using a calculator or without one? thank you!!

                chimichurri

                This depends on the question. Anything from exam 1 is non-CAS whereas exam 2 is CAS. Iirc there was normally a symbol next to the question showing a calculator if applicable but idk if that's still the case.

                  4 months later

                  Are events in tree diagrams always dependent and
                  independent in a two-way table?
                  What does that mean?

                  20 days later

                  Hi,
                  I always struggle to recognise which differentiating rule to use, the product/chain/quotient rule?
                  How would I figure this out?

                  You use product rule when you want to differentiating values that are being multiplied (hence it is named product).
                  For example: To find the derivative of 2xe3x you use product rule.
                  = (2)(e3x) + (2x)(3e3x) = 2e3x + 6xe3x

                  You use quotient rule when you want to differentiating values that are being divided (hence it is named quotient).
                  For example: To find the derivative of 2x/e3x you use quotient rule.
                  = (2)(e3x) - (2x)(3e3x)/(e3x)2

                  5 months later

                  Hi! Could someone please help me understand what this question means, what it is asking for? I tend to strruggle with approaching extended-response.
                  AB and CD are crossroads. A jogger runs along road AB at a speed of 8 km/h and
                  passes O at 1 p.m. Another runner is moving along road CD. The second runner is
                  moving at 10 km/h and passes O at 1:30 p.m.
                  a Let y km be their distance apart t hours
                  after 1 p.m.
                  i Find an expression for y in terms of t.

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