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Math Help - 10 questions that I need HELP ON!!! Please i can't do them!!!

  1. #1
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    10 questions that I need HELP ON!!! Please i can't do them!!!

    questions on attachment!!!!
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  2. #2
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    Use sigma notation to write the sum. -2.4-.4+1.6+3.6
    That do not make no sense to me.

    Find a formula for an for the arithmetic sequence. a1=7, a11=19
    By the formula's of arithmetical sequences:
    a_{11}=a_1+10k
    Substitute know values,
    19=7+10k
    Thus,
    12=10k
    Thus,
    k=1.2

    "An arithemtica sequnce with initial term of 7 and constant diffrence of 1.2"

    Find a formula for an for the arithmetic sequence. a1=11, a11=15
    Same idea.

    A rubber ball on a hard surface takes a sequence of vertical bounces. Each bounce is 1/6 as high as the proceeding one. If this ball is dropped from a height of 12 feet, find the total distance it has traveled when it hits the surface the fifth time.
    This is a geometric series because the situation is,
    12+12(1/6)+12(1/6)(1/6)+12(1/6)(1/6)(1/6)+...
    Thus,
    12+12(1/6)+12(1/6)^2+...+12(1/6)^4---> Because you only got 5 bounces (do not think the exponent is 4 a common mistake it should be 4)

    Thus, factor
    12(1+(1/6)+...+(1/6)^4)
    Use sum for geometric series,
    12\cdot \left( \frac{1-(1/6)^6}{1-1/6} \right)

    6. Write the first five terms of the sequence.
    a0=-2
    a1=4
    an=-2a n -1-4an-2
    -2,4,0,-16,32

    7. Find the coefficient of b6a2 in the expansion of (b+2a)8
    Which term is that? The third thus,
    {8 \choose 3}=56

    8. Use the binomial theorem to expand and simplify the expression.
    (s-u)5
    Tooooooooooooooooooooooooo long.


    9. A college has seven instructors qualified to teach a special computer lab course which requires two instructors to be present. How many different pairs of teachers could teach the class.
    Order no important thus,
    {7 \choose 2}=21

    10. Eight cards are drawn, without replacement, from a standard deck of 52 cards. How many sets of eight cards are possible?
    Same as before,
    {52 \choose 8}=752538150

    11.
    Not doing this one you promised only 10.
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  3. #3
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    Hello, Lane!

    Use sigma notation to write the sum: -2.4 - 0.4+1.6+3.6

    Here are two possible sigma-representations.

    It starts with -2.4 ... and 2 is added three times: . \sum^3_{n=0}(-2.4 + 2n)

    It starts with -4.4 and 2 is added four times: . \sum^4_{n=1}(-4.4 + 2n)



    2. Find a formula for a_n for the arithemtic sequence: a_1 = 7,\;a_{11}=19

    Recall the formula for the n^{th}term: . a_n\:=\:a_1 + d(n-1)
    . . where a_1 is the first term and d is the common difference.

    We are given a_1=7, so the formula becomes: . a_n\:=\:7 + d(n-1)

    We are given a_{11}=19 ... the 11^{th} term is 19.
    So we have: . 19\:=\:7 + d(11-1)\quad\Rightarrow\quad d = 1.2

    Therefore: . a_n\:=\:7 + 1.2(n-1)



    5. A rubber ball on a hard surface takes a sequence of vertical bounces.
    Each bounce is 1/6 as high as the preceeding one.
    If the ball is dropped from a height of 12 feet,
    find the total distance it has traveled when it hits the surface for the fifth time.

    Warning: "The Bouncing Ball" is a classic trick question.
    I'll baby-step through the explanation . . .

    The ball falls 12 feet. .(Contact #1)

    It bounces up . \frac{1}{6}(12) feet . . . and falls . \frac{1}{6}(12) feet. .(Contact #2)

    It bounces up \left(\frac{1}{6}\right)^2(12) feet . . . and falls \left(\frac{1}{6}\right)^2(12) feet. .(Contact #3)

    . . You get the idea . . .


    By the 5th contact, the total distance is:

    . . D \:=\:12 + 2\cdot12\left(\frac{1}{6}\right) + 2\cdot12\left(\frac{1}{6}\right)^2 +  2\cdot12\left(\frac{1}{6}\right)^3 + 2\cdot12\left(\frac{1}{6}\right)^4

    . . D\:=\:12 + 4\bigg[1 + \frac{1}{6} + \left(\frac{1}{6}\right)^2 +  \left(\frac{1}{6}\right)^3\bigg] \:=\:12 + 4\left(\frac{259}{216}\right)\:=\:\boxed{\frac{907  }{54}\text{ feet}}

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  4. #4
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    Hello again, Lane!

    7) Find the coefficient of b^6a^2 in the expansion of (b+2a)^8

    The expansion begins: . \binom{8}{0}b^8 + \binom{8}{1}b^7(2a) + \underbrace{\binom{8}{2}b^6(2a)^2} + \hdots

    The third term is: . \frac{8!}{2!\,6!}\,b^6(4a^2) \;=\;\boxed{112}\,b^6a^2



    8. Use the binomial theorem to expand and simplify: (s - u)^5

    If you don't know the Binomial Theorem, we can't be much help.

    (s - u)^5\;=\;\binom{5}{0}s^5 + \binom{5}{1}s^4(-u) + \binom{5}{2}s^3(-u)^2 + \binom{5}{3}s^2(-u)^3  + \binom{5}{4}s(-u)^4 + \binom{5}{5}(-u)^5

    . . I expanded it . . . you can simplify it.

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