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Math Help - 3 short questions

  1. #1
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    3 short questions

    Question 1:
    ImageShack - Hosting :: quesiton92208ny9.jpg

    This one should be simple, I already have the answers to a, b and c. But for some reason I can't seem to remember how to calculate d. That is if the motion of an object (in this case a ball) can be modeled by a parabola, then what was the initial angle the ball was fired at (in radians)? Any pointers on this one would be vastly appreciated.


    Question 2:
    ImageShack - Hosting :: question42208lu9.jpg

    This is how I worked through the problem:

    ImageShack - Hosting :: rwer21234oq5.jpg


    But obviously my answer is incorrect, so if anyone could help explain where in my steps to solve the question I went wrong, that would be fantastic.

    Question 3:

    ImageShack - Hosting :: question102208cr5.jpg

    This question is flooring me at the moment, I have no clue where to start or even what the notation in the second line of the question is referring to. If someone could set me on the path to solving for a, b and c I would be most thankful
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  2. #2
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    bump
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by jpatrie View Post
    Question 2:
    ImageShack - Hosting :: question42208lu9.jpg

    This is how I worked through the problem:

    ImageShack - Hosting :: rwer21234oq5.jpg


    But obviously my answer is incorrect, so if anyone could help explain where in my steps to solve the question I went wrong, that would be fantastic.
    \sum_{i=1}^{2}\left[\left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^{\left(\sum\limits_{l=1}^i l\right)-1}\right)!\right]

    When i=1, we see that in the exponent we have \sum_{l=1}^1 l=1. Thus, 1-1=0

    We then end up with the first term in the outermost series: \left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^0\right)!=1!=1

    Now, when i=2:

    we see that in the exponent, \sum_{l=1}^2 j=3. Thus, 3-1=2

    Now, let's focus on \sum_{j=1}^2\prod_{k=1}^j k

    When j=1, we have \prod_{k=1}^1 k=1

    When j=2, we have \prod_{k=1}^2 k=2

    Thus, \sum_{j=1}^2\prod_{k=1}^j k= 1+2=3

    So we now see that (3^2)!=9!

    So, we then see that \sum_{i=1}^{2}\left[\left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^{\left(\sum\limits_{l=1}^i l\right)-1}\right)!\right]=1+9!=\dots

    Does this make sense?

    --Chris
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  4. #4
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    Quote Originally Posted by jpatrie View Post
    bump
    (1) a) Let the equation of the parapola is

    y = ax^2 + bx + c

    since it passes through (0, 0), put this in eqn, we got, c=0.

    So, the eqn becomes, y = ax^2 + bx

    since it passes through (10, 100),

    100 = a (10)^2 + b(10)

    \Rightarrow 100a + 10b = 100

    \Rightarrow 10a + b = 10   ...................eqn (1)

    Since it passes through (12, 112),

    \Rightarrow 112 = a(12)^2 + b(12)
    <br />
\Rightarrow 144a + 12b = 112

    \Rightarrow 36a + 3b = 28   .........................eqn (2)

    Now, you solve (1) and (2),to find a and b. we got a = \frac {-1}{3}, \;\;b= \frac{40}{3}

    Put the value of a and b to get the equation.

    equation is \boxed{\;\;y = \frac{-1}{3}x^2 + \frac{40}{3}x \;\;}

    b) y = \frac{-1}{3}x^2 + \frac{40}{3}x

    y = \frac{-1}{3} \left(x^2 - 40x \right)

    y = \frac{-1}{3} \left(x^2 - 40x + 20^2 - 20^2\right)

    y = \frac{-1}{3} \left(x - 20 \right)^2 + \frac{400}{3}

    so, the maximum height reached is \frac{400}{3} \;\;metres

    c) For horizontal distance, put y=0

    0 = \frac{-1}{3}x^2 + \frac{40}{3}x

    \Rightarrow x = 0, \;\; 40

    Horizontal distance = 40 -0 = 40 m

    Now, you try the last (d) part.
    Last edited by Shyam; September 22nd 2008 at 09:42 PM.
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  5. #5
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    Quote Originally Posted by Shyam View Post
    (1) a) Let the equation of the parapola is

    y = ax^2 + bx + c

    since it passes through (0, 0), put this in eqn, we got, c=0.

    So, the eqn becomes, y = ax^2 + bx

    since it passes through (10, 100),

    100 = a (10)^2 + b(10)

    \Rightarrow 100a + 10b = 100

    \Rightarrow 10a + b = 10   ...................eqn (1)

    Since it passes through (12, 112),

    \Rightarrow 112 = a(12)^2 + b(12)
    <br />
\Rightarrow 144a + 12b = 112

    \Rightarrow 36a + 3b = 28   .........................eqn (2)

    Now, you solve (1) and (2),to find a and b. we got a = \frac {-1}{3}, \;\;b= \frac{40}{3}

    Put the value of a and b to get the equation.

    equation is \boxed{\;\;y = \frac{-1}{3}x^2 + \frac{40}{3}x \;\;}

    b) y = \frac{-1}{3}x^2 + \frac{40}{3}x

    y = \frac{-1}{3} \left(x^2 - 40x \right)

    y = \frac{-1}{3} \left(x^2 - 40x + 20^2 - 20^2\right)

    y = \frac{-1}{3} \left(x - 20 \right)^2 + \frac{400}{3}

    so, the maximum height reached is \frac{400}{3} \;\;metres

    c) For horizontal distance, put y=0

    0 = \frac{-1}{3}x^2 + \frac{40}{3}x

    \Rightarrow x = 0, \;\; 40

    Horizontal distance = 40 -0 = 40 m

    Now, you try the last (d) part.
    well, you see part a,b and c are easy, part d was the only part I couldn't figure out, but thanks for confirming my answers for part a,b and c
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  6. #6
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    Quote Originally Posted by Chris L T521 View Post
    \sum_{i=1}^{2}\left[\left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^{\left(\sum\limits_{l=1}^i l\right)-1}\right)!\right]

    When i=1, we see that in the exponent we have \sum_{l=1}^1 l=1. Thus, 1-1=0

    We then end up with the first term in the outermost series: \left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^0\right)!=1!=1

    Now, when i=2:

    we see that in the exponent, \sum_{l=1}^2 j=3. Thus, 3-1=2

    Now, let's focus on \sum_{j=1}^2\prod_{k=1}^j k

    When j=1, we have \prod_{k=1}^1 k=1

    When j=2, we have \prod_{k=1}^2 k=2

    Thus, \sum_{j=1}^2\prod_{k=1}^j k= 1+2=3

    So we now see that (3^2)!=9!

    So, we then see that \sum_{i=1}^{2}\left[\left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^{\left(\sum\limits_{l=1}^i l\right)-1}\right)!\right]=1+9!=\dots

    Does this make sense?

    --Chris
    Actually, I believe in the question the exponent of (\sum_{l=1}^i l) is to the power of (-1) NOT subtracting 1
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by jpatrie View Post
    Actually, I believe in the question the exponent of (\sum_{l=1}^i l) is to the power of (-1) NOT subtracting 1
    If that was the case, then the second term in the outer most series would be (\sqrt{3})!, which is not possible. In order to evaluate n!, n\in\mathbb{Z^+}, where \mathbb{Z^+}=\left\{0,1,2,3,...\right\}. Here, n is irrational.

    So I'm pretty sure you are subtracting 1 from that sum in the exponent.

    Do you see what I'm trying to say?

    --Chris
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  8. #8
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    Quote Originally Posted by Chris L T521 View Post
    If that was the case, then the second term in the outer most series would be (\sqrt{3})!, which is not possible. In order to evaluate n!, n\in\mathbb{Z^+}, where \mathbb{Z^+}=\left\{0,1,2,3,...\right\}. Here, n is irrational.

    So I'm pretty sure you are subtracting 1 from that sum in the exponent.

    Do you see what I'm trying to say?

    --Chris
    Yea I do understand, I just thought that maybe I had made an error when I found root 3 factorial as the answer.

    Thanks
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  9. #9
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    Quote Originally Posted by Chris L T521 View Post
    \sum_{i=1}^{2}\left[\left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^{\left(\sum\limits_{l=1}^i l\right)-1}\right)!\right]

    When i=1, we see that in the exponent we have \sum_{l=1}^1 l=1. Thus, 1-1=0

    We then end up with the first term in the outermost series: \left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^0\right)!=1!=1

    Now, when i=2:

    we see that in the exponent, \sum_{l=1}^2 j=3. Thus, 3-1=2

    Now, let's focus on \sum_{j=1}^2\prod_{k=1}^j k

    When j=1, we have \prod_{k=1}^1 k=1

    When j=2, we have \prod_{k=1}^2 k=2

    Thus, \sum_{j=1}^2\prod_{k=1}^j k= 1+2=3

    So we now see that (3^2)!=9!

    So, we then see that \sum_{i=1}^{2}\left[\left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^{\left(\sum\limits_{l=1}^i l\right)-1}\right)!\right]=1+9!=\dots

    Does this make sense?

    --Chris
    Also in general note that \prod_{k=1}^{j}k=j! holds for j\in\mathbb{N} and \sum\limits_{l=1}^{i}l=\frac{i(i+1)}{2} for i\in\mathbb{N}.
    Hence
    \sum_{i=1}^{2}\left[\left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^{\left(\sum\limits_{l=1}^i l\right)-1}\right)!\right]
    =\sum_{i=1}^{2}\left[\left(\left(\sum_{j=1}^{2}j!\right)^{\left(\sum\li  mits_{l=1}^i l\right)-1}\right)!\right]
    =\sum_{i=1}^{2}((1!+2!)^{\frac{i(i+1)}{2}-1})!
    =\sum_{i=1}^{2}(3^{\frac{i(i+1)}{2}-1})!
    =1!+9!=362881
    This pathern works when 2 is replaced by any positive integer n.
    Last edited by bkarpuz; September 23rd 2008 at 08:51 AM. Reason: Explanation is expanded.
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