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Math Help - common maths problems.

  1. #1
    Junior Member samsum's Avatar
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    common maths problems.

    what is the numerical values of the x & y intercepts of straight line are the same nonzero number. What's the slope of the line?

    a fraction is choosen at random from all positive unreduced proper fractions with denominators less than 6. Find the probability that fraction's decimal representation terminates.

    two adjacent faces of rectangular box have areas 36 and 63. If all three dimensions are positive intergers, find the ratio of the largest possible volume of the box to the smallest possible volume.

    a basketball player has a constant probability of 80% of making a free throw. Find the probability that her next successful free throw is the third or fourth one he attempts.

    in year 2006 is the product of exactly three distinct primes p, q, r. How many other years are also the product of three distinct primes with sum equal to p+q+r ?

    h(x) = 2x +2, k(x) = 2x^3 - 7x^2 - 11x +6, find the sum of all the irrational zeros of h(k(x)) and k(h(x)).
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    Quote Originally Posted by samsum View Post
    what is the numerical values of the x & y intercepts of straight line are the same nonzero number. What's the slope of the line?
    C\not =0

    I shall answer thee in the way I gaze.
    ---
    Any line and be expressed (though not uniquely):
    Ax+By+C=0 where both A,B are non-zero.

    You are told that the x and y intercepts are the same.

    When x=0 we have,
    By+C=0.
    Thus,
    y=-C/B (divison by zero did not occur based on the wording of the problem).
    ----> y-intercept.

    When y=0 we have,
    Ax+C=0.
    Thus,
    x=-C/A (divison by zero did not occur based on the wording of the problem).
    ----> x-intercept.

    They are equal.

    -C/B=-C/A
    Thus,
    A=B.
    Hence,
    Ax+Ay+C=0
    Divide by A,
    x+y+C/A=0,
    Thus,
    y=-x+C/A
    Thus,
    m=-1.
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  3. #3
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    Quote Originally Posted by samsum View Post
    two adjacent faces of rectangular box have areas 36 and 63. If all three dimensions are positive intergers, find the ratio of the largest possible volume of the box to the smallest possible volume.
    Look below at a hand drawn diagram I produced.

    Since,
    xy=36
    The possibilities since they are positive intergers are:
    (x,y)=(1,36),(2,18),(3,12),(4,9),(6,6),(9,4),(12,3  ),(18,2),(36,1)
    Since,
    yz=63
    The possibilities are:
    (y,z)=(1,63),(7,9),(9,7),(63,1)
    Since,
    y has to belong to both enumeration we have,
    y=1 \mbox{ or} 9
    Thus, the two possobilities are,
    (x,y,z)=(36,1,63)
    (x,y,z)=(4,9,7)
    Those art the two possible volumes.
    Now all you need to do if find their ratio.
    Attached Thumbnails Attached Thumbnails common maths problems.-picture10.gif  
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