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    (Original post by KaylaB)
    Well if we have the Opposite and the Adjacent of a right-angled triangle we can use Pythagoras to find the length of the Hypotenuse
    Thank you so much!! 😊😊
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    (Original post by Gemma_98)
    Thank you so much!! 😊😊
    No problem Happy to be of service :hat2:
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    (Original post by KaylaB)
    I still seem to to be getting the same value for a, and thus the final answer
    I'll work through it in a sec and let you know.
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    (Original post by Zacken)
    I'll work through it in a sec and let you know.
    Is it not like 2:40am where you are, do you not require sleep?
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    not that the help isn't appreciated
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    (Original post by KaylaB)
    Is it not like 2:40am where you are, do you not require sleep?
    Spoiler:
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    not that the help isn't appreciated
    Ssshhh.

    Okay, so resolving down the plane: 15g\sin \theta - T = 15a for the first particle and resolving vertically up for the second particle: T - 10g = 10a.

    Adding these: 25a = 15g\sin \theta - 10g \Rightarrow a \approx -3.8 - can you take it from here?
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    (Original post by Zacken)
    Ssshhh.

    Okay, so resolving down the plane: 15g\sin \theta - T = 15a for the first particle and resolving vertically up for the second particle: T - 10g = 10a.

    Adding these: 25a = 15g\sin \theta - 10g \Rightarrow a \approx -3.8 - can you take it from here?
    Yeah I can finish it, but I have a question. I made both of those equations but I substituted in 10g-10a for T instead of adding them, how come you can just add them like that or am I just really tired?
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    (Original post by KaylaB)
    Yeah I can finish it, but I have a question. I made both of those equations but I substituted in 10g-10a for T instead of adding them, how come you can just add them like that or am I just really tired?
    If you have two equations:

    a+b = c

    2a + 3b = 5c

    you are allowed to substitute things in to solve. Or you can add both of them or subtract both of them or divide both of them, etc...

    So if you add this is valid and you get 3a + 4b = 6c.

    You can test this out: 5 + 3 =8 and 7-5 = 2 so 5 +3 + 7  -5 = 10 = 8 + 2 which is true, etc...
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    (Original post by Zacken)
    If you have two equations:

    a+b = c

    2a + 3b = 5c

    you are allowed to substitute things in to solve. Or you can add both of them or subtract both of them or divide both of them, etc...

    So if you add this is valid and you get 3a + 4b = 6c.

    You can test this out: 5 + 3 =8 and 7-5 = 2 so 5 +3 + 7  -5 = 10 = 8 + 2 which is true, etc...
    Ah okay cheers :hat2:, now rest that giant brain of yours.
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    (Original post by KaylaB)
    Ah okay cheers :hat2:, now rest that giant brain of yours.
    You two, Mrs. "am I really tired? Yes you are.".
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    (Original post by Zacken)
    You two, Mrs. "am I really tired? Yes you are.".
    Touché
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    (Original post by KaylaB)
    Touché
    On the other hand, I just spelt "too" as "two" so...
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    (Original post by Zacken)
    On the other hand, I just spelt "too" as "two" so...
    I was being nice and wasn't going to mention it :laugh:, you just have numbers on the brain
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    (Original post by KaylaB)
    I was being nice and wasn't going to mention it :laugh:, you just have numbers on the brain
    I appreciate it. Let's not make fun of Zacken the illiterate dumbo. Lol jk let's totally make fun of Zacken the illiterate dumbo, it'll be hilarious. :lol:
 
 
 
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