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    did everyone get 8 + h as h tends to zero ?
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    (Original post by Crozzer24)
    The first I got 10+5+10 so 25 and the second one I got 0
    I got the same for the first one, but I wasn't sure about the second one. Wasn't sure what the remainer would be, whether it was zero or 3.
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    Oh my **** log(a^m)^3. I got 3mloga but that didn't work.

    Also, the find and simplify f(5+h)etc. Wtf am I meant to find. H?


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    (Original post by tom476zf)
    did everyone get 8 + h as h tends to zero ?
    Yes, however didnt get the as h tends to 0 bit?
    just 8+h
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    (Original post by WuMyster)
    Oh my **** log(a^m)^3. I got 3mloga but that didn't work.

    Also, the find and simplify f(5+h)etc. Wtf am I meant to find. H?


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    wasnt it just m3? as logaa is just 1 or is that wrong
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    (Original post by ScalyJake)
    I got the same for the first one, but I wasn't sure about the second one. Wasn't sure what the remainer would be, whether it was zero or 3.
    I took it as 6/3 = 2 so no remainder but I was unsure
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    for the simplifying log question i got -3m
    is this familiar to anyone?!
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    (Original post by Lucuniexet)
    for the simplifying log question i got -3m
    is this familiar to anyone?!
    I got -m^3
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    (Original post by Lucuniexet)
    Yes, however didnt get the as h tends to 0 bit?
    just 8+h
    don't think you get any extra marks for it its just how i was taught
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    (Original post by Lucuniexet)
    for the simplifying log question i got -3m
    is this familiar to anyone?!
    yes got this not sure is its right though ?
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    (Original post by Lucuniexet)
    This is exactly what i got
    Damn. I got 0.6 for intercept but 0.4 for gradient.

    You had to work out gradient from the graph right?

    I used (0,0.6) and (1,1)


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    And I thought S1 was difficult
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    (Original post by jaseho98)
    oh my **** same struggle, i was so confused by that part... liKE WTF
    You were meant to find the height at the lowest point of the graph using the width of 3.6, so you do the centre - half the width of the vehicle and work out that height

    I think thats how you do it anyway.
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    to draw sinx-3 between 0 and 450
    You do just draw a normal sin graph and move it downwards by 3 units yeah?
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    (Original post by tom476zf)
    wasnt it just m3? as logaa is just 1 or is that wrong
    it is -3m yes.
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    (Original post by Crozzer24)
    I got -m^3
    I also got this. Because the cubed was for the entire log, so you didn't bring it down to the front.
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    (Original post by Dowel)
    it is just 3m yes.
    but it was log a1 which is 0.
    and then 0 minus the 3m is -3m?
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    (Original post by Lucuniexet)
    but it was log a1 which is 0.
    and then 0 minus the 3m is -3m?
    it was logaa i think
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    (Original post by Lucuniexet)
    but it was log a1 which is 0.
    and then 0 minus the 3m is -3m?
    Yes it is, I just edited that quote :P
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    (Original post by Lucuniexet)
    but it was log a1 which is 0.
    and then 0 minus the 3m is -3m?
    Yeah there's a negative. I wrote -3mlogaa. But I forgot logaa is 1.


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