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Math Help - Can Anyone help me with these questions?

  1. #1
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    Can Anyone help me with these questions?

    Hi I was wondering if anyone can help me with these questions so I know I got them right.

    So can anyone go through these questions with step by step answers

    I got these on the internet with no mark schemes and I need answers for all the questions since I self teach.

    Thank You
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  2. #2
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    Re:

    #1.
    a.
    f(x)=x^3+3x^2+5
    f'(x)=3x^2+6x
    f''(x)=6x+6

    #2.

    a. (1-2x)^5=1-10x+40x^2-80x^3+80x^4-32x^5


    #4.

    5^x=17
    xln(5)=ln(17)

    Divide ln(17) by ln(5)

    Thus

    x=ln(17)/ln(5)

    #5.

    b. f(x)=x^3+4x^2+x-6=(x-1)(x+2)(x+3)
    c. f(x)=x^3+4x^2+x-6=(x-1)(x+2)(x+3)

    Hence

    x=1
    x=-2
    x=-3

    This worksheet is a piece of cake. Hopefully this will get you started.


    -qbkr21
    Attached Thumbnails Attached Thumbnails Can Anyone help me with these questions?-integrate.gif  
    Last edited by qbkr21; May 9th 2007 at 08:40 AM.
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  3. #3
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    #1.(a)
    f'(x)=3x^2-6x
    f"(x)=6x-6
    (b)see attachment
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  4. #4
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    Re:

    Quote Originally Posted by alinailiescu View Post
    #1.(a)
    f'(x)=3x^2-6x
    f"(x)=6x-6
    (b)see attachment
    Alinaliescu your differentiation and integration for part A & B are wrong

    you were asked to integrate

    f(x)=x^3+3x^2+5 NOT f(x)=x^3-3x^2+5
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  5. #5
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    Yes, I realized that after I posted the replay. I tried to print out the problems, and the printer didn't work properly.
    Anyway, you posted the right solutions.
    Thanks
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  6. #6
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    Re:

    I can see how you misread it. The vast majority of these PDF files people attach are very hard to read because of their clarity.
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  7. #7
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    #3
    Since the points given are the end points of the diameter, the midpoint would be the center of the circle.
    ((-1+3)/2, (4+6)/2)
    thus, (1, 5) is the center.
    You also need the radius,
    r=sqrt[(3-1)^2+(6-5)^2]=sqrt(5)
    So, the equation is:
    (x-1)^2+(y-5)^2=5
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  8. #8
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    Thanks so is anyone able to do questions 6-10
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  9. #9
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    #6

    first ,use cos^2(x)=1-sin^2(x) to eliminate cosine function
    That would give you:
    2(1-sin^2(x))+1=5sinx
    2-2sin^2(x)+1-5sinx=0
    -2sin^2(x)-5sinx+3=0
    This is a quadratic equation in sinx, so using the quadratic formula you have:
    sinx=[5+sqrt(25-4*-2*3)]/[-2*2]=[5+7]/(-4)=12/-4=-3(which has no solution for x, since sinx should be between -1 and 1)
    or
    sinx==[5-sqrt(25-4*-2*3)]/[-2*2]=[5-7]/(-4)=1/2
    so x=sin^-1(1/2)
    x=pi/6
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  10. #10
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    This is a full past paper you are asking others to do.
    Surely you can do at least one or two of them.
    England exams are not far away.

    @alinailiescu

    You missed out 5(pi)/6
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  11. #11
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    Sorry if it seems your helping me do my past paper so far I have done must of them but I was checking if I got them right.
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  12. #12
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    Get someone to check this.
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  13. #13
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    Pay attention,
    6x^3/3=2x^3 and not 3x^3
    So, the answer is -2. (which makes sense since the shaded area is under the x axis,)
    Your area is 2.
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  14. #14
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    Anyone for 8 and 9, I finished 10 so don't worry about that.
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  15. #15
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    Re:

    8.a

    v=-70
    v=70

    Do you want to see the work?

    8.b

    C''=(2800/v^3)
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