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Math Help - Advanced Functions (dont run away) need help badly

  1. #1
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    Advanced Functions (dont run away) need help badly

    1. Find the number of digits in the expansion of (2^120)(5^125) without any calculator or computer.

    2. Find the coordinates of the two points that trisect the line segment with the endpoints (A(2,3) B(8,2).

    l l = Absolute value
    3. Give the function f(x) = x^3 - 2x, sketch y = f(lxl).
    Sketch g(x) = lx^2 - 1l - lx^2 - 4l
    Sketch the region in the plane to show all points (x,y)
    such that lxl + lyl <= 2
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  2. #2
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    Hello, shane99!

    1. Find the number of digits in the expansion of: (2^{120})(5^{125}) without any calculator or computer.

    We have: . 2^{120}\cdot 5^{125} \;=\;2^{120}\cdot5^{120}\cdot5^5 \;=\;(2^{120}\cdot5^{120})\cdot5^5 \;=\;(2\cdot5)^{120}\cdot5^5 \;=\;10^{120}\cdot5^5

    This is: . 5^5 = 3125 followed by 120 zeros.

    Therefore, the product has 124 digits.

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  3. #3
    Junior Member qspeechc's Avatar
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    Quote Originally Posted by shane99 View Post
    2. Find the coordinates of the two points that trisect the line segment with the endpoints A(2,3) and B(8,2).

    l l = Absolute value
    3. Give the function f(x) = x^3 - 2x, sketch y = f(lxl).
    Sketch g(x) = lx^2 - 1l - lx^2 - 4l
    Sketch the region in the plane to show all points (x,y)
    such that lxl + lyl <= 2
    2. Here's one way I can think of, but someone else could probably find a simpler way.
    Find the equation of the straight line joining A and B.
    You want to trisect the line AB, so find 1/3 of the distance between A and B, call it r.
    Find the equations of the circles centred at A and B with radius r.
    Find where the circles intersect the straight line joining A and B. Take the coordinates of the relevant points (the straight line will intersect the two circles at four points, but you're only looking for two points).

    3. f(lxl) = |x|^3 - 2|x| = |x^3| - 2|x|
    Draw x^3 and then draw |x^3|, that is, where x^3 is negative you 'flip' it about the x-axis (y=0) so it becomes positive.
    Draw 2|x|.
    Subtract the two graphs (you have to have drawn the two graphs on the same set of axes).

    Try sketching g(x) yourself but doing similar reasoning as above.

    For |x|+|y|\leq 2 use the definition of the absolute value and look at the different cases, i.e.
    x<0 and y<0
    x<0 and y>0
    x>0 and y<0
    x>0 and y>0
    For instance, in the region x<0 and y>0 we have |x|=-x and |y|=y, and so |x|+|y|\leq 2 becomes -x+y\leq 2 and sketch this only in the region x<0 and y>0.
    Last edited by qspeechc; October 3rd 2009 at 12:29 PM.
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  4. #4
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    1. Find the number of digits in the expansion of (2^120)(5^125) without any calculator or computer.
    If I may give my 2 cents. A way to find the number of digits in a huge number is use logs.

    We have log(2^{120}\cdot 5^{125})=log(2^{120})+log(5^{125})=120log(2)+125lo  g(5)\approx 123.494850022

    Round up and we see there are 124 digits.

    To find the first few digits of this number. Take 10^{\text{the decimal part of the above}}

    We get 10^{.494850022}=3.12500000

    As Soroban said, it is 3125 and then 120 0's.

    Though you may need a calculator, it is a cool way to know and I thought I would show it anyway.
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  5. #5
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    Quote Originally Posted by Soroban View Post


    This is: .[tex]5^5 = 3125

    How do you know that 5^5= 3125 without using a calculator?
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  6. #6
    Eater of Worlds
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    5*5=25, 25*5=125, 125*5=625, 625*5=3125.

    Do it the old-fashioned way sans calculator.
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  7. #7
    Junior Member qspeechc's Avatar
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    Quote Originally Posted by shane99 View Post
    3. Give the function f(x) = x^3 - 2x, sketch y = f(lxl).
    Sketch g(x) = lx^2 - 1l - lx^2 - 4l
    Sketch the region in the plane to show all points (x,y)
    such that lxl + lyl <= 2
    Alternatively,
    f(|x|) = |x^3| - 2|x|
    use the definition of the absolute value and look at the different regions:
    |x^3| = -x^3 when -x^3<0, i.e. when x<0
    |x^3| = x^3 when x^3>0, i.e. when x>0

    |x| = -x when x<0
    |x| = x when x>0

    So for x<0
    f(|x|) = -x^3 +2x
    and for x>0
    f(|x|) = x^3-2x

    g(x) is done similarly, but is trickier because the regions you have to consider will be more complicated.
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  8. #8
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    Need more help with 3c)
    Last edited by shane99; October 4th 2009 at 11:22 AM.
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