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Math Help - Caculus Homework Help!!

  1. #1
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    Exclamation Caculus Homework Help!!

    I start my Ap Cal class tomarrow and our summer homework included some problems I just cannot figure out, any help would be amazing!

    ~factor and simplify. express answer as a fraction withought neg. exponiets
    x(9x-10)^-1/2 + 2(x-1)^1/2


    ~Another one is this... I thnk you may have to use compleating the squares but couldnt figure it out)

    x^2-x=5
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  2. #2
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    Quote Originally Posted by Caycee7
    ~Another one is this... I thnk you may have to use compleating the squares but couldnt figure it out)

    x^2-x=5
    subtract 5 from both sides: x^2-x-5=0

    now use the quadratic formula: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

    substitute: x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-5)}}{2(1)}

    simplify: x=\frac{1\pm\sqrt{1+20}}{2}

    you can solve from there...
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  3. #3
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    Thanks so much for the help
    but I dont really understand why you use the quadratic,
    I havnt seen that formula in over a year and dont know how it relates...
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  4. #4
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    Quote Originally Posted by Caycee7
    I start my Ap Cal class tomarrow and our summer homework included some problems I just cannot figure out, any help would be amazing!

    ~factor and simplify. express answer as a fraction withought neg. exponiets
    x(9x-10)^-1/2 + 2(x-1)^1/2
    first thing you need to remember is (x^a)^b=x^{(a\times b)}

    second thing you need to remember is x^{\frac{1}{2}}=\sqrt{x}

    therefore: (9x-10)^{-\frac{1}{2}}=\left((9x-10)^{\frac{1}{2}}\right)^{-1} =(\sqrt{9x-10})^{-1}=\frac{1}{\sqrt{9x-10}}

    therefore you get:
    \frac{x}{\sqrt{9x-10}}+2\sqrt{x-1}

    and solve from there, (if you can't just ask me to help you further)
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  5. #5
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    Quote Originally Posted by Caycee7
    I start my Ap Cal class tomarrow and our summer homework included some problems I just cannot figure out, any help would be amazing!

    ~factor and simplify. express answer as a fraction withought neg. exponiets
    x(9x-10)^-1/2 + 2(x-1)^1/2
    <br />
x(9x-10)^{-1/2}  + 2(x-1)^{1/2}=\frac{x}{\sqrt{9x-10}}+ 2\sqrt{x-1}=<br />
\frac{x+2\sqrt{x-1}\sqrt{9x-10}}{\sqrt{9x-10}}<br />

    RonL
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  6. #6
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    Quote Originally Posted by Caycee7
    Thanks so much for the help
    but I dont really understand why you use the quadratic,
    I havnt seen that formula in over a year and dont know how it relates...
    when given a quadratic equation: ax^2+bx+c=0

    then you can use the quadratic formula to find the value(s) of x: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
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  7. #7
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    Quote Originally Posted by Quick
    when given a quadratic equation: ax^2+bx+c=0

    then you can use the quadratic equation to find the value(s) of x: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

    Oh okay that makes sense, wow it is amazing how rusty you get over the summer! Thanks so much you guys! I might have a few more in a bit, I have a few more pages to go hopefully I will be able to figure them out but if not you'll be hearing from me again
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  8. #8
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    Got another one!

    the question reads:

    find the domain of the function

    f(x)=ln(3x+2)


    (Ahh I dont remember what find the domain means... Am I suppose to solve for x instead of f(x)... That is all I recall domain being, x on a graph!)
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  9. #9
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    Quote Originally Posted by Caycee7
    the question reads:

    find the domain of the function

    f(x)=ln(3x+2)


    (Ahh I dont remember what find the domain means... Am I suppose to solve for x instead of f(x)... That is all I recall domain being, x on a graph!)
    Domain means what values of "x" can take, basically.

    Thus, since \ln (3x+2) can only make sense when the expression inside the parantheses is positive. Thus,
    3x+2>0
    Thus,
    x>-\frac{2}{3}
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  10. #10
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    Okay I feel bad to keep posting questions but I have one that is asking me for the general form of a line passing through pint (3,1) slope=5/2


    I thought this isnt bad and drew the graph but now I found out that that is a formula that I have never studied at all, I tried to read up on it on some other sites but got extreamly confused, and ideas?
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  11. #11
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    Quote Originally Posted by Caycee7
    Okay I feel bad to keep posting questions but I have one that is asking me for the general form of a line passing through pint (3,1) slope=5/2


    I thought this isnt bad and drew the graph but now I found out that that is a formula that I have never studied at all, I tried to read up on it on some other sites but got extreamly confused, and ideas?
    label your point as (x_1,y_1) then make an equation in point-slope form: y-y_1=m(x-x_1)

    now substitute: y-1=\frac{5}{2}(x-3)
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  12. #12
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    Quote Originally Posted by ThePerfectHacker
    Domain means what values of "x" can take, basically.

    Thus, since \ln (3x+2) can only make sense when the expression inside the parantheses is positive. Thus,
    3x+2>0
    Thus,
    x>-\frac{2}{3}

    So if that is true the domain of f(x)= the square root of 2x+3

    does this mean x> or equal to -4?

    I really need to figure out the math typing thing but you know what I mean...
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  13. #13
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    Quote Originally Posted by Quick
    label your point as (x_1,y_1) then make an equation in point-slope form: y-y_1=m(x-x_1)

    now substitute: y-1=\frac{5}{2}(x-3)

    So the point slope form is simply the gen. form of the line?
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  14. #14
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    Quote Originally Posted by Caycee7
    So the point slope form is simply the gen. form of the line?
    it's a general form of a line, I don't think there is one specified "general form" (although y=mx+b would be most likely)

    ---------------------------------------

    to answer your other question:
    you are given the equation f(x)=\sqrt{2x+3}

    therefore: 2x+3\geq0

    subtract 3 from both sides: 2x\geq -3

    divide 2 from both sides: x\geq \frac{-3}{2}
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  15. #15
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    Ahh Relief!

    Thanks everyone I finally finished all of my problems! Took a few hours but a relief to have it all finished. Now I am off to finish my English work, I have a feeling it is going to be a long night! Oh well thatís what I get for procrastinating. Thanks again, I'm sure I'll be around now that I found this site, I have AP both for Cal and Stats, and math isnít my best subject...

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