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Math Help - Finding integers? Point of tangency?

  1. #1
    Member realintegerz's Avatar
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    Finding integers? Point of tangency?

    Intro to Calc HELP!?????? Finding integrs?

    1)Find an integer, x, greater than 4 where the sum of x consecutive integers is divisible by x



    2) Show that the equation above gives the correct slope for the line tangent to a circle with a radius of 13 at the point (-5,12)



    So I'm thinking, the circle's center is at (0,0) because then the point of tangency will be (-5,12) exactly...But I don't know what to do


    For #1, I got 5n+10....is that right or wrong?
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  2. #2
    Eater of Worlds
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    Assuming the circle is centered at the origin, then its equation above the x-axis is y=\sqrt{169-x^{2}}

    We can use this and its derivative to find the slope at (-5,12) to find the equation.

    y'=\frac{-x}{\sqrt{169-x^{2}}}

    Using y=mx+b, we have 12=\frac{5}{12}(-5)+b

    b=\frac{169}{12}

    y=\frac{5}{12}x+\frac{169}{12} is the line equation.
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  3. #3
    Member realintegerz's Avatar
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    Ok now im lost, how do i do #1 then?

    I got 5n+10..i dont know how the two relate now.....
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  4. #4
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    You are correct, X=5.
    Note the sum is \sum\limits_{k = 0}^4 {\left( {5N + k} \right)}  = 5N + \frac{{4(5)}}{2} = 5N + 10 which is clearly divisible by 5.
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  5. #5
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    Hello, realintegerz!

    Could you state the original problem?
    I'm sure that this isn't what "they" said.


    1) Find an integer n > 4 where the sum of n consecutive integers
    is divisible by n.
    Let a be the first (smallest) integer.

    \text{Then: }\;S \;=\; \underbrace{a + (a + 1) + (a + 2) + (a + 3) + \hdots + (a + [n-1])}_{n\text{ terms}}

    . . . . S \;=\;\underbrace{(a + a + a + \hdots + a)}_{n\text{ terms}} + (1 + 2 + 3 + \hdots + [n-1])

    . . . . S\;=\;na + \frac{n(n-1)}{2} \;=\;n\left(a + \frac{n-1}{2}\right)


    S is divisible by n if \left(a + \frac{n-1}{2}\right) is an integer.
    . . This happens when n is odd.


    Therefore, n is any odd integer greater than 4.




    2) Show that the equation above gives the correct slope for the line tangent
    to a circle with a radius of 13 at the point (-5,12)
    I don't see what this question has to do with #1.
    . . There is no "equation above."

    I will assume that the circle is centered at the origin.

    The slope of the radius from (0, 0) to (-5, 12) is: . \frac{12-0}{-5-0} \:=\:-\frac{12}{5}

    The tangent at the point is perpendicular to that radius.

    Therefore, the slope of the tangent is: . \frac{5}{12}

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  6. #6
    Member realintegerz's Avatar
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    Well actually it's a series of 4 problems

    1) Write an expression in terms of an integer, a, for the sum of any 3 consecutive integers and show that this sum is divisible by 3

    a+(a+1)+(a+2)
    =3a+3

    3(a+1)

    3(a+1)/3= (a+1) x 3/3 = (a+1) x 1 = a+1

    2) Wrote an expression in terms of an integer, a, for the sum of any 4 consecutive integers and show that this sum isn't divisible by 4

    a+(a+1)+(a+2)+(a+3)
    =4a+6

    2(2a+3)

    2(2a+3)/4 = (2a+3) x 2/4 = (2a+3) x 1/2 = (2a+3)/2

    3) Find an integer, n, greater than 4 where the sum of n consecutive integers is divisible by n

    4) Show that the equation above gives the correct slope for the line tangent to a circle with radius of 13 at the point (-5,12)





    Yeah im lost on how 4 relates to 3....should i just ignore it?
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