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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:

    What?....No Calculators allowed! You would never see me signing up for this course...



    -qbkr21
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    Quote Originally Posted by qbkr21 View Post
    What?....No Calculators allowed! You would never see me signing up for this course...



    -qbkr21
    After a quick look a calculator is not needed for these.

    RonL
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    So can anyone help me with the questions please.
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  5. #5
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    Quote Originally Posted by qbkr21 View Post
    What?....No Calculators allowed! You would never see me signing up for this course...



    -qbkr21
    In my College the Math Department forbids the use of calculus, no matter in what course (except for many super easy ones for people who just need to take it).

    And I agree with that approach greatly.
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  6. #6
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    Question 1: 12x^2+x^(-1/2)

    Question 2a: sqrt(108) = sqrt(3*36) = sqrt(3)*sqrt(36) = 6*sqrt(3)

    2b: (2-rt(3))^2 = (2-rt(3))(2-rt(3)) = 4 - 4*rt(3) +3 = 7 - 4*rt(3)

    I'll answer the rest tomorrow if no-one else does, and if I have time.
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  7. #7
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    Quote Originally Posted by jl5000 View Post
    Question 1: 12x^2+x^(3/2)
    the power of the last x should be -1/2

    you are correct for question 2

    2(b)

    (2 - sqrt(3))^2 = 4 - 4sqrt(3) + 3 = 7 - 4sqrt(3)

    got it?

    3(a)

    You should know what the graph for f(x) = 1/x looks like. it has a horizontal asymptote at the x-axis and a vertical asymptote at the y-axis. as we approach the y-axis from the right we go to + infinity, as we approach it from the left we fo to -infinity

    y = f(x) + 3 is just this graph shifted up. the vertical asymptote remains the same, the horizontal asymptote becomes the line y = 3

    3(b)

    do we cross an axis? you bet we do, we cross the x-axis if we shift up. so let's find the x-interscept:

    for x-int, y = 0

    => 0 = 1/x + 3
    => 1/x = -3
    => x = -1/3
    Last edited by ThePerfectHacker; April 25th 2007 at 06:45 PM.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    3(a)

    You should know what the graph for f(x) = 1/x looks like. it has a horizontal asymptote at the x-axis and a vertical asymptote at the y-axis. as we approach the y-axis from the right we go to + infinity, as we approach it from the left we fo to -infinity

    y = f(x) + 3 is just this graph shifted up. the vertical asymptote remains the same, the horizontal asymptote becomes the line y = 3

    3(b)

    do we cross an axis? you bet we do, we cross the x-axis if we shift up. so let's find the x-interscept:

    for x-int, y = 0

    => 0 = 1/x + 3
    => 1/x = -3
    => x = -1/3
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  9. #9
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    4.

    y = x -2 .........................(1)
    y^2 + x^2 = 10 ...............(2)

    this system is just begging for substitution:
    From (1) we see y = x -2, substitute x - 2 for y in (2)

    => (x - 2)^2 + x^2 = 10
    => x^2 - 4x + 4 + x^2 = 10
    => 2x^2 - 4x - 6 = 0
    => x^2 - 2x - 3 = 0
    => (x - 3)(x + 1) = 0

    so x = 3 or x = -1

    but y = x - 2
    so when x = 3
    y = 3 - 2 = 1

    when x = -1
    y = -1 - 2 = -3

    so the solution is: x = 3, y = 1 or x = -1, y = -3
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  10. #10
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    5)

    2x^2 - 3x - (k + 1) = 0
    By the quadratic formula:
    x = [3 +/- sqrt(9 + 8(k + 1))]/4

    we have no real roots if what's under the sqrt is negative. so the set of k for which we have no real roots is given by:

    9 + 8(k + 1) < 0
    => 17 + 8k < 0
    => k < -17/8

    so the set of k for which this happens is:

    k = {x : x < -17/8} = (-infinity, -17/8)
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  11. #11
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    6(a)

    (4 + 3sqrt(x))^2 = 16 + 24sqrt(x) + 9x ...........so we can do it in one step, i hope you can follow the process, tell me if you can't

    6(b)

    int{(4 + 3sqrt(x))^2 }dx = int{16 + 24sqrt(x) + 9x}dx
    ..................................= int{16 + 24x^(1/2) + 9x}dx
    ..................................= 16x + 24(2/3)x^(3/2) + (9/2)x^2 + C
    ..................................= 16x + 16x^(3/2) + (9/2)x^2 + C
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  12. #12
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    7)

    (a) f ' (x) = 3x^2 - 6 - 8x^-2

    => f(x) = int{3x^2 - 6 - 8x^-2}dx
    ...........= x^3 - 6x + 8x^-1 + C

    since (2,1) is on f(x), we have f(x) = 1 when x = 2
    => 1 = 2^3 - 6(2) + 8(2)^-1 + C
    => C = 1 - 2^3 + 6(2) - 4 = 1

    so our curve is, f(x) = x^3 - 6x + 8x^-1 + 9

    (b) f ' (x) gives the slope of the tangent to the curve at any value of x. we want the one passing through (2,1). so we have

    f ' (2) = 3(2)^2 - 6 - 8(2)^-2 = 12 - 6 - 2 = 4 = m

    using m = 4, (x1,y1) = (2,1), we have by the point-slope form:
    y - y1 = m(x - x1)
    => y - 1 = 4(x - 2)
    => y = 4x - 8 + 1
    => y = 4x - 7 ..............the tangent line to f(x) at (2,1)
    Last edited by Jhevon; April 26th 2007 at 12:55 AM. Reason: Made a mistake in calculating c
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  13. #13
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    8)

    a)

    y = 4x + 3x^(3/2) - 2x^2 , x > 0

    => dy/dx = 4 + (9/2)x^(1/2) - 4x


    b) weird question. the simpliest way to do this i see, is plug in x = 4 into y and show that the output is y = 8

    y = 4x + 3x^(3/2) - 2x^2
    when x = 4, we have
    y = 4(4) + 3(4)^(3/2) - 2(4)^2 = 16 + 24 - 32 = 8

    so we see, when x = 4, y = 8, therefore, (4,8) lies on the curve C


    c)
    the normal is the line that is perpendicular to the tangent line. so let's find the slope of the tangent line, and take it's negative inverse to find the slope of the normal, the rest is precalculus.

    at (4,8)
    dy/dx = 4 + (9/2)(4)^(1/2) - 4(4) = 4 + 9 - 16 = -3 ....slope of the tangent line

    => for the normal, m = 1/3
    using m = 1/3, (x1,y1) = (4,8), we have by the point slope form:
    y - y1 = m(x - x1)
    => y - 8 = (1/3)(x - 4)
    => y = (1/3)x - 4/3 + 8
    => y = (1/3)x + 20/3 .............equation of the normal. multiplying through by 3 we get

    3y = x + 20 ...........as desired


    d)
    where does the normal cut the x-axis? where y = 0 of course, so set y = 0
    => x + 20 = 0
    => x = -20

    so Q(-20,0)

    using the length of line formula:
    P(4,8) Q(-20,0)

    PQ = sqrt[(4 + 20)^2 + (8 - 0)^2] = sqrt(24^2 + 8^2) = sqrt(640) = 4sqrt(40) .........you can break this down even simpler, i can't bother
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  14. #14
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    9)
    a) let's call the rows n, you will see why

    when:
    rows.....................sticks
    n=1 ........................4
    n=2 ........................7
    n=3 ........................10

    notice a pattern? we add three going on to each new term. that is a common difference, thus we can think of this as an arithmetic sequence with the formula:
    a_n = 1 + 3n for n = 1,2,3,4 .... a_n gives the number of sticks needed to make n squares in the nth row


    b) a_10 = number of sticks needed to make 10 squares in the 10th row

    a_10 = 1 + 3(10) = 31 ......number of sticks in the 10th row

    now to find the total number of sticks to make 10 rows, we need to find the sum of the first 10 terms of the sequence a_n, that is, S_n, in particular, S_10

    Now, S_n = n(a_1 + a_n)/2
    => S_10 = 10(4 + 31)/2 = 175

    she uses 175 sticks to make 10 rows


    c)
    Not too sure about this one, i'll have to think about it


    d) Ann completes k rows with 1750 sticks, lets find what row she went up to:
    a_k = 1 + 3k
    => 1750 = 1 + 3k
    => 1749 = 3k
    => k = 583
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  15. #15
    is up to his old tricks again! Jhevon's Avatar
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    10)
    a) There is a graph of the curves below, obviosly the blue curve is graph (ii)

    so just putting the curve up like this doesn't help you, since sketching by hand is a lot more difficult than just imputting functions in a graphing utility. but think of this as a way to check if the sketch you came up with is correct. (if you have trouble with the sketching itself, just say so)


    b)
    We find where the graphs intersect by equating the two functions:
    they intersect where:

    (x - 2)x^2 = x(6 - x)
    => (x - 2)x^2 - x(6 - x) = 0
    => x^3 - 2x^2 - 6x + x^2 = 0
    => x^3 - x^2 - 6x = 0
    => x(x^2 - x - 6) = 0
    => x(x - 3)(x + 2) = 0
    => x = 0, 3, -2
    these are the x-values for which they intersect, but we need the coordinates

    plug in the value into either function--it makes sense to do it for the easier one. using graph (ii)

    when x = 0
    y = 0(6 - 0) = 0
    so one point is (0,0)

    when x = 3
    y = 3(6 - 3) = 9
    another point is (3,9)

    when x = -2
    y = -2(6 + 2) = -16
    yet another is (-2,-16)

    so the coordinates for interscetion are: (0,0), (3,9), and (-2,-16). these are shown on the graph i posted
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