Maths year 11

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    (Original post by z_o_e)
    Do I not multiply them?

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    Yes, you do multiply them, but you had multiplied them incorrectly

     2 \times -2\sqrt5 = -4\sqrt5

    You had written  2 - 2\sqrt5 instead.
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    (Original post by K-Man_PhysCheM)
    Yes, you do multiply them, but you had multiplied them incorrectly

     2 \times -2\sqrt5 = -4\sqrt5

    You had written  2 - 2\sqrt5 instead.


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    (Original post by z_o_e)
    What do you mean like terms? Do I add all the surds up from each bracket?

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    You multiply the brackets out, as I had done, and then you collect the terms which can be collected.

     18-4\sqrt5+9\sqrt5-10 is what you get from multiplying out the brackets (you had already done this, bar one term which was incorrectly multiplied).

    You can collect the single numbers together (18 - 10) and collect the terms with the same surds together 99\sqrt5 - 4\sqrt5, and then simplify.
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    (Original post by K-Man_PhysCheM)
    Yes, you do multiply them, but you had multiplied them incorrectly

     2 \times -2\sqrt5 = -4\sqrt5

    You had written  2 - 2\sqrt5 instead.
    You told me I multiplied them all wrong?

    So is above right?
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    (Original post by z_o_e)
    You told me I multiplied them all wrong!

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    You had only multiplied that one incorrectly. The rest were correct.

    I had said:

    " The is not correct, because , not



    Now you need to collect like terms. "
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    (Original post by K-Man_PhysCheM)
    You had only multiplied that one incorrectly. The rest were correct.

    I had said:

    " The is not correct, because , not



    Now you need to collect like terms. "
    Ergh

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    Past papers !
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    (Original post by K-Man_PhysCheM)
    You had only multiplied that one incorrectly. The rest were correct.

    I had said:

    " The is not correct, because , not



    Now you need to collect like terms. "


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    Correct. Now remember that RDKGames' question was about a triangle, so how do you find the area of a triangle?
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     Area of triangle = \frac{1}{2} \times base \times height.

    You've already calculated base x height, so what must you do to calculate the triangle's area?
    Then you need to find 4.1% of that area to answer the question.
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    (Original post by K-Man_PhysCheM)
    Correct. Now remember that RDKGames' question was about a triangle, so how do you find the area of a triangle?
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     Area of triangle = \frac{1}{2} \times base \times height.

    You've already calculated base x height, so what must you do to calculate the triangle's area?
    Then you need to find 4.1% of that area to answer the question.
    Divide it by 2

    And then divide it by 4.1 and times it by 100?

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    (Original post by z_o_e)
    Divide it by 2

    And then divide it by 4.1 and times it by 100?

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    Divide by 2 is correct, but are you sure that you would calculate 4.1% by dividing by 4.1 and multiplying by 100? Wouldn't that give a much larger number?

    E.g  \dfrac {8.2}{4.1} \times 100 = 200 .

    200 couldn't be 4.1% of 8.2... 200 is much larger than 8.2 !!

    But if you try the other way:

    \dfrac {8.2}{100} \times 4.1 = 0.3362 , which looks about right.

    Hence, to get a percentage of something, first DIVIDE by 100, then multiply by the percentage you need to find.

    However, in that question, do NOT use a calculator: you need to leave it in the form:
     \frac{A}{B}(c^3 + 5\sqrt5)

    Note: you already have the c^3 + 5\sqrt5 bit:
    8+5\sqrt5 = 2^3+5\sqrt5.

    Hence all you need to do is to find the \frac{A}{B} bit.
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    (Original post by K-Man_PhysCheM)
    Hence, to get a percentage of something, first DIVIDE by 100, then multiply by the percentage you need to find.
    Or just turn 4.1% into \frac{4.1}{100}=\frac{41}{1000} straight away which makes the working out smoother.
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    (Original post by RDKGames)
    Or just turn 4.1% into \frac{4.1}{100}=\frac{41}{1000} straight away which makes the working out smoother.
    \frac{x}{100}\times{y} \equiv \frac{y}{100} (x) in my head, and I would just immediately convert between the two, but yeah, I agree the latter would be much neater and easier for the answer format.
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    (Original post by K-Man_PhysCheM)
    \frac{x}{100}\times{y} \equiv \frac{y}{100} (x) in my head, and I would just immediately convert between the two, but yeah, I agree the latter would be much neater and easier for the answer format.


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    Don't use a calculator! Remember, you have to leave the answer in the form:

    \frac{A}{B}(c^3+5\sqrt5)

    Moreover, that is NOT how you work out percentage. Multiply by 4.1 and divide by 100, or in other words:

    ans\times\dfrac{4.1}{100} \equiv \dfrac{41}{1000}\times\dfrac{c^3  +5\sqrt5}{2}
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    (Original post by RDKGames)
    Nope. Don't tell you forgot how to expand brackets with surds...
    How are these?


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    (Original post by z_o_e)
    How are these?


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    All correct. On the last one there shouldnt be a 2 next to \sqrt9 though
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    (Original post by RDKGames)
    All correct. On the last one there shouldnt be a 2 next to \sqrt9 though
    Oh alright x



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    (Original post by RDKGames)
    All correct. On the last one there shouldnt be a 2 next to \sqrt9 though
    How would I do this one?


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