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anonymous

  • 5 years ago

please help...... ((2z-8)/(z^2-4))/((z^2+6z+8)/(z-4)) simplify completly..... show work

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  1. anonymous
    • 5 years ago
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    do you know how to do this?

  2. dumbcow
    • 5 years ago
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    First Flip bottom fraction and Multiply Then factor and things will cancel \[2z-8 = 2(z-4)\] \[z^{2}-4 = (z+2)(z-2)\] \[z^{2}+6z+8=(z+4)(z+2)\]

  3. anonymous
    • 5 years ago
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    wow i forgot that the things cancelled.... okay what did you get for your answer?

  4. anonymous
    • 5 years ago
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    im not sure how you did that can you show me?

  5. dumbcow
    • 5 years ago
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    \[=\frac{2(z-4)(z-4)}{(z+2)(z+2)(z-2)(z+4)}\] hmm maybe they dont cancel in this case..haha

  6. anonymous
    • 5 years ago
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    yea and i think you factored wrong.....

  7. anonymous
    • 5 years ago
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    wait how did you factor that?

  8. dumbcow
    • 5 years ago
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    no everything is factored correctly

  9. dumbcow
    • 5 years ago
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    which one

  10. anonymous
    • 5 years ago
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    how did you do it then? i got something completly different but i thrust what you got....

  11. dumbcow
    • 5 years ago
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    are you sure its division and not multiplying the 2 fractions ??

  12. anonymous
    • 5 years ago
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    you flip the second one and change to multi.

  13. dumbcow
    • 5 years ago
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    correct

  14. anonymous
    • 5 years ago
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    and then can you show me what you did im completly lost

  15. dumbcow
    • 5 years ago
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    oh for the factoring part. well for 2z -8 i just pulled out a 2 from each term 2(z-4) = 2z - 8

  16. anonymous
    • 5 years ago
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    ok

  17. dumbcow
    • 5 years ago
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    For z^2 - 4, find factors of -4 that add up to 0 since there is no "z" term -2*2 = -4 and -2+2=0 (z-2)(z+2)

  18. anonymous
    • 5 years ago
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    wow thats sooo much easier now thanks :) can you help me with a couple more?

  19. dumbcow
    • 5 years ago
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    z^2 +6z +8 same thing, look for factors of 8 that add up to 6 4*2 =8 and 4+2 =6 (z+4)(z+2)

  20. dumbcow
    • 5 years ago
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    ok

  21. anonymous
    • 5 years ago
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    okay....... ((4)/(2x+1))-((3)/(2x))=

  22. anonymous
    • 5 years ago
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    can you show me step by step please?

  23. dumbcow
    • 5 years ago
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    adding/subtracting fractions you need to get common denominator 2x+1 cannot be factored 2x cannot be factored common denominator is 2x(2x+1) just like adding 1/2 + 1/3, common denominator is 2*3=6

  24. dumbcow
    • 5 years ago
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    now change numerators \[\frac{4}{2x+1} = \frac{(2x)(4)}{2x(2x+1)}\] \[\frac{3}{2x} = \frac{3(2x+1)}{2x(2x+1)}\] combine into 1 fraction \[\frac{(2x)(4) - 3(2x+1)}{2x(2x+1)}\]

  25. anonymous
    • 5 years ago
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    ok

  26. dumbcow
    • 5 years ago
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    multiply and add like terms on top \[\frac{8x -6x -3}{2x(2x+1)} = \frac{2x-3}{2x(2x+1)}\]

  27. anonymous
    • 5 years ago
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    can you help me with this one? ((m+5)/(2m^2-2))+((3)/(1-m))+((5)/(2m+2))

  28. dumbcow
    • 5 years ago
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    factor all the denominators 2m^2 -2 = 2(m-1)(m+1) 1-m = -(m-1) 2m+2 = 2(m+1) common denominator will include everything w/out repeats -2(m-1)(m+1)

  29. anonymous
    • 5 years ago
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    okay i can do it from there can you help me with this one? ((n+1)-(2/n))/((n+4)+(4/n))

  30. dumbcow
    • 5 years ago
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    \[=\frac{-(m+5)+(2(m+1))(3) + (-(m-1))(5)}{-2(m+1)(m-1)}\]

  31. anonymous
    • 5 years ago
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    can you help me with this one? please???? ((n+1)-(2/n))/((n+4)+(4/n))

  32. dumbcow
    • 5 years ago
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    ok combine fractions on top and bottom, then flip and multiply

  33. dumbcow
    • 5 years ago
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    common denominator will just be n

  34. anonymous
    • 5 years ago
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    ok

  35. dumbcow
    • 5 years ago
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    answer should be (n-1)/(n+2)

  36. anonymous
    • 5 years ago
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    yup that what i got are you good with summation notation and geometric means/sequences?

  37. dumbcow
    • 5 years ago
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    sure

  38. anonymous
    • 5 years ago
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    okay.... wellthe question is... insert four geometirc means between -7 and -224

  39. dumbcow
    • 5 years ago
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    hmm dont understand?

  40. anonymous
    • 5 years ago
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    nevermind i figured that one out......

  41. anonymous
    • 5 years ago
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    what about....

  42. anonymous
    • 5 years ago
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    \[\sum_{n=1}^{50}(1/4)(n+2)\]

  43. dumbcow
    • 5 years ago
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    \[=\frac{1}{4}\sum_{n=1}^{50}n+2 = \frac{1}{4}(\sum_{n=1}^{50}n +\sum_{n=1}^{50}2)\]

  44. anonymous
    • 5 years ago
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    im stuck on this one.. which term of the geometric sequence 243, -81, 27, . . . is (-1/9)?

  45. dumbcow
    • 5 years ago
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    8 th term

  46. anonymous
    • 5 years ago
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    how did you figure that out?

  47. dumbcow
    • 5 years ago
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    haha tried proving it using geometric sequence formula but got stuck when taking log anyway just continue the sequence of dividing by 3 and flipping the sign

  48. anonymous
    • 5 years ago
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    o so you flip the sign every other time? could you help me with this one? there are seven houses; in each are seven cats. each cat kills seven mice. each mouse would hvae eaten seven ears of wheat. each ear of wheat prodece seven measures of grain. how much grain is saved? i got 16807 grains saved

  49. dumbcow
    • 5 years ago
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    correct 7^3 mice 7^4 ears 7^5 grain

  50. anonymous
    • 5 years ago
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    ok that equals 16807 which is what i got:)

  51. anonymous
    • 5 years ago
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    okay change the repeating decimal .25 (sopposto have the bar above the 25) to an equivlant common fraction

  52. dumbcow
    • 5 years ago
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    ?? why do you need to know that some fraction close to proportional of 1/4

  53. anonymous
    • 5 years ago
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    i dontk know its on my homework though ydo you know what it'd be?

  54. dumbcow
    • 5 years ago
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    haha i just guessed and put it in my calculator 25/99

  55. anonymous
    • 5 years ago
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    wow your good.... do you know how i would show that? just say guess and check? we turn this in for a grade thats why im asking

  56. dumbcow
    • 5 years ago
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    actually any n/99 = .nnnnnn

  57. anonymous
    • 5 years ago
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    in a certain credit union, money left on deposit for one year earns 4% intrest at the end ot the year. if you invested $100 at the beginning of each year in this credit union and did NOT withdraw the intrest due at the end of the year, how much would you hvae on deposit at the end of the tenth year?

  58. dumbcow
    • 5 years ago
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    =100(1.04 + 1.04^2+... +1.04^10) need sum of geometric sequence sum = a(1-r^n)/(1-r) sum = 1.04(1-1.04^10)/(1-1.04)

  59. dumbcow
    • 5 years ago
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    i get 1248.64

  60. anonymous
    • 5 years ago
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    so when you get your answer are you sopposto multiply it by 100?

  61. dumbcow
    • 5 years ago
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    yes

  62. anonymous
    • 5 years ago
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    okay just stay right there i have a couple more i just have to go and get another pencil mine just broke

  63. anonymous
    • 5 years ago
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    ok im back i dont know what formula you used

  64. dumbcow
    • 5 years ago
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    for finding sum of geometric sequence

  65. anonymous
    • 5 years ago
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    yea but can you tell me it with all of the variable in it please? i need to look and see if i have that one

  66. dumbcow
    • 5 years ago
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    \[s _{n} = \frac{a _{1}(1-r ^{n})}{1-r}\]

  67. anonymous
    • 5 years ago
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    i only have SofN=(Asub1-Asub1R^n)/(1-R) and... SofN=(Asub1-AsubnR)/(1-R) sooo.. which one should i use, and what numbers?

  68. dumbcow
    • 5 years ago
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    ues the first one, only difference is mine has factored out the a1 on top

  69. dumbcow
    • 5 years ago
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    a1 = first term r = common ratio

  70. anonymous
    • 5 years ago
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    ummmm... i should get the same answer right? i think im doing someting wrong im getting 1200.610712

  71. dumbcow
    • 5 years ago
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    1-(1.04)^10 = -.4802 1.04(-.4802) = -.49945 -.49945/(1-1.04) = 12.48635

  72. anonymous
    • 5 years ago
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    wait shouldnt it be..... (100-100(1.04)^10)/(1-1.04)

  73. dumbcow
    • 5 years ago
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    no im leaving the 100 on outside of sequence then we multiply the sum by 100

  74. anonymous
    • 5 years ago
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    but when i do it shouldnt i be getting the same thing? and i dont understand how your doing it

  75. dumbcow
    • 5 years ago
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    ok if you include the 100 then sequence will look like this =100(1.04) + 100(1.04^2) +...+100(1.04^10) a1 = 100(1.04) r = 1.04 sn = 100(1.04)(1-1.04^10)/(1-1.04)

  76. anonymous
    • 5 years ago
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    but my equation is Sn=(A1-A1R^n)/(1-R)

  77. dumbcow
    • 5 years ago
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    same thing factor out the A1

  78. anonymous
    • 5 years ago
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    can you do it my way when you factor out the A1 it confuses me

  79. anonymous
    • 5 years ago
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    ....^because when you factor....

  80. dumbcow
    • 5 years ago
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    ok Sn = 100(1.04) -100(1.04)(1.04^10) / (1-1.04)

  81. anonymous
    • 5 years ago
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    ok ... a ball which rolls off a penthouse terrace falls 16 feet the first second, 48 feet the next second, and 80 feet the third second. if it continues to fall in this mannor, how far does it fall in the seventh second?

  82. dumbcow
    • 5 years ago
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    i believe this models a parabola of y=-16x^2 the change in y from 6 to 7 is how far it falls in 7th second 16(7^2) - 16(6^2) = 16(49-36) = 16(13) = 208

  83. anonymous
    • 5 years ago
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    a rubber ball dropped 40 feet rebounds on each bounce 2/5 of the distance from which it fell. how far will it travel before comming to a rest?

  84. dumbcow
    • 5 years ago
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    geometric sequence 40+40(2/5)+ 40(2/5)^2 +... looks like an infinite sequence before you get 0 lim n->inf (2/5)^n = 0 so in formula r^n = 0 a1=40 r=2/5 sum = 40 - 40(0) / 1-(2/5) sum = 40 /3/5 = 200/3 = 66.666

  85. anonymous
    • 5 years ago
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    what formula is that?

  86. dumbcow
    • 5 years ago
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    same as before sum of geometric sequence

  87. anonymous
    • 5 years ago
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    no but which one of mine because yours confuse me

  88. dumbcow
    • 5 years ago
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    \[s _{n}=\frac{a _{1}-a _{1}r^{n}}{1-r}\]

  89. anonymous
    • 5 years ago
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    what would i put for n?

  90. dumbcow
    • 5 years ago
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    infinity

  91. anonymous
    • 5 years ago
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    hu?

  92. dumbcow
    • 5 years ago
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    just substitute r^n=0

  93. anonymous
    • 5 years ago
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    would my answer be iin feet?

  94. dumbcow
    • 5 years ago
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    yes

  95. anonymous
    • 5 years ago
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    what does the graph for ..... look like? -cubic with one real solution and two complex solutions look like? -cubic with no real solutions? -quartic with no real solutions? -a quadratic with one real solution and one complex solution? -a quartic with two real solutions and two complex solutions? -a quadratic with no real solutions? -a quartic with no real solutions? -a cubic with 3 real solutions, but one is a double root? -a quartic with for real roots, but both are double roots?

  96. dumbcow
    • 5 years ago
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    umm number of real solutions represents number of x_intercepts cubic with no real solutions does not exist i believe

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