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  1. #1
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    Exclamation Maths Questions - HELPA parabola

    Find the equation of the curve  y=x^2 - 7x^2 + 5 at the point (1, -1)


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    One card is selected at random from a pack of 20 numbered 1 to 20. What is the probability that it is either a multiple of 5 or 3. I had 1/2, but apparently its wrong. Please tell where I went wrong.


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  2. #2
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    Quote Originally Posted by mibamars View Post
    Find the equation of the curve  y=x^2 - 7x^2 + 5 at the point (1, -1)

    ...
    Hello,

    1. I assume that there is a typo and your function reads:
    y = x^2-7x+5

    2. do you want to find the equation of the tangent line at the graph of your function with the tangent point (1, -1)? If so:

    The slope of the tangent has the same value as the gradient of the function in the tangent point.
    Calculate the 1st derivation:

    EDIT: As angel.white pointed out I've made one of my usual silly mistakes. The following calculations are corrected - by me! Therefore check my calculations!

    y'=2x-7 . Now plug in x = 1. You'll get

    y'= -5 That means the tangent has a slope of m = -5.

    Now use point-slope-formula of a straight line:

    (y-(-1))=-5(x-1)~\iff~\boxed{y=-5x+4}
    Last edited by earboth; November 13th 2007 at 02:28 AM. Reason: used wrong sign
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  3. #3
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    Quote Originally Posted by mibamars View Post
    One card is selected at random from a pack of 20 numbered 1 to 20. What is the probability that it is either a multiple of 5 or 3. I had 1/2, but apparently its wrong. Please tell where I went wrong....
    Hi,

    obviously you counted th 15 twice because 15 is a multiple of 5 and a multiple of 3. But you can count the 15 only once!
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  4. #4
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    Quote Originally Posted by earboth View Post
    Hello,

    1. I assume that there is a typo and your function reads:
    y = x^2-7x+5

    2. do you want to find the equation of the tangent line at the graph of your function with the tangent point (1, -1)? If so:

    The slope of the tangent has the same value as the gradient of the function in the tangent point.
    Calculate the 1st derivation:

    y'=2x+7 . Now plug in x = 1. You'll get

    y'= 9 That means the tangent has a slope of m = 9.

    Now use point-slope-formula of a straight line:

    (y-(-1))=9(x-1)~\iff~\boxed{y=9x-10}
    Looks like a negative sign got misplaced in y'
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  5. #5
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    Hello, mibamars!

    One card is selected at random from a pack of 20 numbered 1 to 20.
    What is the probability that it is either a multiple of 5 or 3.

    As earboth pointed out, you overcounted.

    The cards with a multiple of 3 or 5 are: . \{3,\,5,\,6,\,9,\,10,\,12,\,15,\,18,\,20\}
    . . There are only nine of them.


    They expected you count carefully ... or use this formula:

    . . P(\text{mult.of 3 } \vee \text{ mult. of 5}) \;\;=\;\;P(\text{mult.of 3}) \;+\; P(\text{mult.of 5})  \;-\; P(\text{mult.of 3 } \wedge \text{ mult.of 5})

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