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Thread: I have 1 hour please help!!

  1. #1
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    I have 1 hour please help!!

    I'm having some serious trouble trying to answer these questions could someone please help me, I don't understand how to come up with the final figure! Please include your working as it will benefit me in working out how the answer was found!I have 1 hour please help!!-image.jpg


    Any and all help would be much appreciated!
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  2. #2
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    Re: I have 1 hour please help!!

    What have you achieved so far?
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  3. #3
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    Re: I have 1 hour please help!!

    Nothing, I've done the questions on the other side of the page and skipped that one, I just don't understand how to do it in the slightest
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  4. #4
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    Re: I have 1 hour please help!!

    Well, the first thing you need to do is to put the two pairs of values into the given equation. That gives you two equations with variables $\displaystyle a$ and $\displaystyle b$ which you can solve using elimination and/or substitution.
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  5. #5
    MHF Contributor Matt Westwood's Avatar
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    Re: I have 1 hour please help!!

    Plug the values in.

    11.0 = 4a + b
    18.5 = 9a + b

    Solve the simultaneous equations.
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  6. #6
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    Re: I have 1 hour please help!!

    You are told to assume that $F = aL + b.$ Right? That has FOUR apparent unknowns, which leaves you helpless.

    BUT you are also given two pairs of values for F and L, correct?

    If you insert one pair into your equation, you get an equation in a and b.

    If you insert the other pair into your equation, you get a second equation in a and b.

    Two equations with two unknowns.

    What are the equations? Can you solve them?
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  7. #7
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    Re: I have 1 hour please help!!

    I'll try giving it a go
    Last edited by kiwi18; Oct 15th 2015 at 05:46 AM.
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  8. #8
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    Re: I have 1 hour please help!!

    Quote Originally Posted by Matt Westwood View Post
    Plug the values in.

    11.0 = 4a + b
    18.5 = 9a + b

    Solve the simultaneous equations.
    There are two ways to solve these equations. The simplest to understand is substitution. You can rearrange one of the equations (it doesn't matter which) so that $\displaystyle b$ is the subject*. Then substitute that expression into the second equation in place of $\displaystyle b$. Here's an example:
    Suppose we have
    $\displaystyle 10 = 5x + y$
    $\displaystyle 12 = 7x + y$
    We can take (for example) the first equation and write
    $$\begin{aligned}
    10 &= 5x + y \\
    10 - 5x &= y \\
    \end{aligned}$$
    which is an expression for $\displaystyle y$. We then use that in the other equation
    $$\begin{aligned}
    12 &= 7x + y \\
    12 &= 7x + (10 - 5x) \\
    \end{aligned}$$
    We can solve this equation for $\displaystyle x$, and when we have a value for $\displaystyle x$ we can put it into either of the original equations to find $\displaystyle y$.
    $$\begin{aligned}
    12 &= 7x + (10 - 5x) \\
    2 &= 2x \\
    x &= 1 \\[8pt]
    10 &= 5x + y \\
    10 &= 5(1) + y \\
    10 &= 5 + y \\
    5 &= y
    \end{aligned}$$

    The other approach is elimination. For this, we add (or subtract) multiples of the equations to make one of the variables disappear. In this case, since each equation has $\displaystyle 1y$, we can just subtract one equation from the other.
    $$\begin{aligned}
    12 &= 7x + y \\
    10 &= 5x + y \\
    12 - 10 &= (7x + y) - (5x + y) \\
    2 &= 7x - 5x + y - y \\
    2 &= 2x \\
    x &= 1
    \end{aligned}$$
    And then we substitute our value for $\displaystyle x$ into one of the original equations as before.

    * You can make $\displaystyle a$ the subject if you prefer, but it's a little more complicated.
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  9. #9
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    Re: I have 1 hour please help!!

    Quote Originally Posted by kiwi18 View Post
    I'll try giving it a go
    You'll TRY?
    Solving those 2 equations is quite basic:
    who's missing classes: you or your teacher?
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