Sunday, December 6, 2009

Review Questions

Integrate the following DO NOT FORGET THE +c
1. dy/dx = 5x^3 + 2x^2 + 5
2. dy/dx = 6x + 1
3. dy/dx = 2
4. dy/dx = 8x^5 - 5x^3 + 4x
5. dy/dx = 7x^2 + 3x + 4

6. A particle moves in a straight line and at point P, it's velocity is given as v = 7t^2 - 5t +3. The particle comes to rest at point Q.1. What is the acceleration at Q if it arrives at Q when t=7?2. How far does the particle travel in t=1 to t=4?

7. Find dy/dx of the following:

1. y = 6x^3 + 4x^2 - 5x
2. y = x^-3 + 16
3. y = 4x + 4
4. y = 3x^4 + 3(x^2 + 5)
5. y + 6 = 4x + 9y - 3x^2


8. A curve has equation y = 4/(2)^.5, find dy/dx
9. A curve is such that dy/dx = 16/x^3 and (1,4) is a point on the curve, find the equation of the curve.
10. y = 6theta - 2sin theta, find dy/dx
11. The equation of a curve is y = 2x + 8/x^2, find dy/dx and d^2ydx^2
12. A curve is such that dy/dx = 2x^2 -5. Given that the point (3,8) lies on the curve, find the equation of the curve.
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212 comments:

  1. Sexy DucklinDecember 7, 2009 at 8:19 PM

    1. dy/dx = 5x^3 + 2x^2 + 5
    d^2y/dx^2=15x^2+4x

    ReplyDelete
  2. Sexy DucklinDecember 7, 2009 at 8:29 PM

    2. dy/dx = 6x + 1
    d^2y/dx^2=6

    3. dy/dx = 2
    d^2y/dx^2=0

    4. dy/dx = 8x^5 - 5x^3 + 4x
    d^2y/dx^2=40x^4-15x^2+4

    5. dy/dx = 7x^2 + 3x + 4
    d^2y/dx^2=14x+3

    ReplyDelete
  3. Sexy DucklinDecember 7, 2009 at 8:33 PM

    7. Find dy/dx of the following:

    1. y = 6x^3 + 4x^2 - 5x
    dy/dx=18x^2+8x-5

    2. y = x^-3 + 16
    dy/dx=-3x^-4

    3. y = 4x + 4
    dy/dx=4

    4. y = 3x^4 + 3(x^2 + 5)
    dy/dx=12x^3+6x

    5. y + 6 = 4x + 9y - 3x^2
    dy/dx=4+9-6x
    dy/dx=13-6x

    ReplyDelete
  4. Sexy DucklinDecember 7, 2009 at 8:43 PM

    11. The equation of a curve is y = 2x + 8/x^2, find dy/dx and d^2ydx^2
    y= 2x+8x^-2
    dy/dx= 2-16x
    d^2y/dx^2=-16

    ReplyDelete
  5. MATHS™December 8, 2009 at 6:06 AM

    2.

    dy/dx = 6x + 1

    y= 3x^2+x+c

    ReplyDelete
  6. Fadda AnonymousDecember 8, 2009 at 6:09 AM

    This comment has been removed by the author.

    ReplyDelete
  7. Fadda AnonymousDecember 8, 2009 at 6:09 AM

    2. dy/dx = 6x + 1
    S = (6x^2)/2 + c

    ReplyDelete
  8. Fadda AnonymousDecember 8, 2009 at 6:10 AM

    3. dy/dx = 2
    S = c

    ReplyDelete
  9. MATHS™December 8, 2009 at 6:12 AM

    4.

    dy/dx = 8x^5 - 5x^3 + 4x

    y=(8x^6/6)-(5x^4/4)+(4x^2/2)+c

    y=(4x^6/3)-(5x^4/4)+(2x^2)+c

    ReplyDelete
  10. Fadda AnonymousDecember 8, 2009 at 6:12 AM

    Alright....i'm not sure if i'm doing this correct anymore..... What is the integral of a constant????

    ReplyDelete
  11. Fadda AnonymousDecember 8, 2009 at 6:13 AM

    Is the integral of a consatnt 'c'......?? Sombody please clarify this.....

    ReplyDelete
  12. Fadda AnonymousDecember 8, 2009 at 6:17 AM

    4. dy/dx = 8x^5 - 5x^3 + 4x
    S = 8x^6/6 - 5x^4/4 + 4x^2/2 + c

    ReplyDelete
  13. MATHS™December 8, 2009 at 6:17 AM

    1.

    dy/dx = 5x^3 + 2x^2 + 5

    y=(5x^4)+(2x^3/3)+(5x)+c

    ReplyDelete
  14. Fadda AnonymousDecember 8, 2009 at 6:18 AM

    5. dy/dx = 7x^2 + 3x + 4
    S = 7x^3/3 + 3x^2/2 +c

    ReplyDelete
  15. MATHS™December 8, 2009 at 6:19 AM

    This comment has been removed by the author.

    ReplyDelete
  16. MATHS™December 8, 2009 at 6:20 AM

    3.

    dy/dx = 2

    y= 2x+c

    ReplyDelete
  17. Fadda AnonymousDecember 8, 2009 at 6:21 AM

    Ques. 7
    1. y = 6x^3 + 4x^2 - 5x
    dy/dx = 18x^2 + 8x - 5

    ReplyDelete
  18. MATHS™December 8, 2009 at 6:22 AM

    5.

    dy/dx = 7x^2 + 3x + 4

    y=(7x^3/3)+(3x^2/2)+(4x)+c

    ReplyDelete
  19. Fadda AnonymousDecember 8, 2009 at 6:22 AM

    Ques 7
    2. y = x^-3 + 16
    dy/dx = -3x^-4 + 0

    ReplyDelete
  20. Fadda AnonymousDecember 8, 2009 at 6:23 AM

    Ques. 7
    3. y = 4x + 4
    dy/dx = 4 + 0

    ReplyDelete
  21. MATHS™December 8, 2009 at 6:24 AM

    7

    1.

    y = 6x^3 + 4x^2 - 5x

    dy/dx =(18x^2)+(8x)-5

    ReplyDelete
  22. MATHS™December 8, 2009 at 6:25 AM

    7.
    2.
    y = x^-3 + 16

    dy/dx = (-3x^-4)

    ReplyDelete
  23. MATHS™December 8, 2009 at 6:26 AM

    7.
    3.
    y = 4x + 4

    dy/dx = 4

    ReplyDelete
  24. Fadda AnonymousDecember 8, 2009 at 6:28 AM

    Ques. 7
    5. y + 6 = 4x + 9y - 3x^2
    9y -y = 3x^2 - 4x + 6
    8y = 3x^2 - 4x + 6
    y = (3x^2 - 4x + 6) / 8
    y = 3x^2 - 4x + 6 + 8^-1
    dy/dx = 6x - 4 + 0 + 0
    dy/dx = 6x - 4

    ReplyDelete
  25. MATHS™December 8, 2009 at 6:29 AM

    7.
    4.
    y = 3x^4 + 3(x^2 + 5)

    y=(3x^4)+(3x^2)+(15)

    dy/dx =(12x^3)+(6x)

    ReplyDelete
  26. Fadda AnonymousDecember 8, 2009 at 6:31 AM

    Alright... 7 part 5 was very tricky....
    Could someone tell me if what i did is correct. especiall the part where i carried across the 8 and brought it up to the power of -1???????

    ReplyDelete
  27. MATHS™December 8, 2009 at 6:35 AM

    7.
    5.
    y + 6 = 4x + 9y - 3x^2

    y=[(3/8)x^2]-[(4/8)x]+[6/8]

    dy/dx =[(3/4)x]-[1/2]

    you made a mistake with no.5

    ReplyDelete
  28. MATHS™December 8, 2009 at 6:39 AM

    you were supposed to multiply the 8^(-1) by everything in the brackets

    ReplyDelete
  29. MATHS™December 8, 2009 at 6:47 AM

    for ex

    (8+y)/x

    =[x^(-1)]*(8+y)

    = [8x^(-1)] + [yx^(-1)]

    ReplyDelete
  30. doghousekrewDecember 8, 2009 at 7:10 AM

    #7:
    (1) y= 6x³ + 4x²- 5x
    dy/dx = 18x² + 8x -5
    (2) y= x-3 +16
    dy/dx= -3x-4
    (3) y= 4x +4
    dy/dx= 4
    (4) y=3x4 + 3(x²+5)
    y= 3x4 +3x² + 15
    dy/dx= 12x³ + 6x
    (5) y + 6 = 4x +9y – 3x²
    6+3x²-4x= 9y – y
    6 + 3x² - 4x = 8y
    (6 + 3x² - 4x) + 8-1 = y
    6x – 4 = dy/dx

    ReplyDelete
  31. doghousekrewDecember 8, 2009 at 7:37 AM

    #12:
    dy/dx = 2x² - 5
    in order to obtain the curve, the gradient must be integrated:
    y = ∫ (dy/dx) dx
    y = ∫ 2x² - 5 + c
    y = (2/3) x³ - 5x + c
    substitute point ( 3,8) to find c:
    8 = (2/3) 3³ - 5 (3) + c
    8 = -9 +c
    17 = c
    Therefore equation of the curve:
    y = (2/3) x³ - 5x + 17

    ReplyDelete
  32. doghousekrewDecember 8, 2009 at 7:48 AM

    #1:
    dy/dx = 5x³ + 2x² +5
    y = (5/4) x4 + (2/3) x³ + 5x + c

    ReplyDelete
  33. doghousekrewDecember 8, 2009 at 7:49 AM

    This comment has been removed by the author.

    ReplyDelete
  34. doghousekrewDecember 8, 2009 at 7:51 AM

    #2:
    dy/dx = 6x + 1
    y = 3x² + x + c

    ReplyDelete
  35. doghousekrewDecember 8, 2009 at 7:53 AM

    #3:
    dy/dx = 2
    y = 2x + c

    ReplyDelete
  36. "grapes"@@@@@@@@December 8, 2009 at 7:54 AM

    dy/dx = 5x^3 + 2x^2 + 5
    then
    d^2y/dx^2=15x^2+4x

    ReplyDelete
  37. doghousekrewDecember 8, 2009 at 7:55 AM

    #4:
    dy/dx = 8x5 – 5x³ + 4x
    y = (8/6) x6 – (5/4)x4 + 2x² + c

    ReplyDelete
  38. doghousekrewDecember 8, 2009 at 7:58 AM

    #5:
    dy/dx = 7x² + 3x + 4
    y = (7/3) x³ + (3/2) x² + 4x + c

    ReplyDelete
  39. "grapes"@@@@@@@@December 8, 2009 at 7:59 AM

    dy/dx = 6x + 1

    (6x^2)/2 + c

    ReplyDelete
  40. "grapes"@@@@@@@@December 8, 2009 at 8:00 AM

    3)dy/dx = 2
    =2x+c

    ReplyDelete
  41. "grapes"@@@@@@@@December 8, 2009 at 8:02 AM

    4. dy/dx = 8x^5 - 5x^3 + 4x
    =(8/6)x^6-(5/4)x^4- 2x^2

    ReplyDelete
  42. "grapes"@@@@@@@@December 8, 2009 at 8:04 AM

    5. dy/dx = 7x^2 + 3x + 4
    y=(7/3)x^3+(3/2)x^2 + 4x

    ReplyDelete
  43. "grapes"@@@@@@@@December 8, 2009 at 8:18 AM

    v = 7t^2 - 5t +3.
    dv/dt=14t-5
    when t=7
    subs into equ.
    14(7)-5= 93
    so acc. is 93
    intergrate v
    (7/3)t^3-(5/2)t^2 +3t
    when t=1
    subs into v
    (7/3)(1)^3-(5/2)t^2+3(1)
    =(7/3)-(5/2)+3
    =2/5/6

    when t=4
    subs into v
    (7/3)(4)^3-(5/2)(4)^2+3(4)
    okk i doh no if i doing this correct

    ReplyDelete
  44. "grapes"@@@@@@@@December 8, 2009 at 8:19 AM

    1. y = 6x^3 + 4x^2 - 5x
    dy/dx=18x^2+8x-5

    ReplyDelete
  45. "grapes"@@@@@@@@December 8, 2009 at 8:20 AM

    2. y = x^-3 + 16
    dy/dx=-3x^-4

    ReplyDelete
  46. "grapes"@@@@@@@@December 8, 2009 at 8:20 AM

    3. y = 4x + 4
    dy/dx=4

    ReplyDelete
  47. "grapes"@@@@@@@@December 8, 2009 at 8:21 AM

    4. y = 3x^4 + 3(x^2 + 5)
    dy/dx=12x^3+6x

    ReplyDelete
  48. "grapes"@@@@@@@@December 8, 2009 at 8:21 AM

    5. y + 6 = 4x + 9y - 3x^2
    dy/dx=4+9-6x
    dy/dx=13-6x

    ReplyDelete
  49. montana parkDecember 8, 2009 at 8:58 AM

    This comment has been removed by the author.

    ReplyDelete
  50. montana parkDecember 8, 2009 at 8:58 AM

    number (2)
    dy/dx = 6x + 1
    y= 3x^2+x+c

    ReplyDelete
  51. montana parkDecember 8, 2009 at 8:59 AM

    number (3)

    dy/dx = 2

    y= 2x+c

    ReplyDelete
  52. montana parkDecember 8, 2009 at 9:00 AM

    number (4)

    dy/dx = 8x^5 - 5x^3 + 4x
    S = 8x^6/6 - 5x^4/4 + 4x^2/2 + c

    ReplyDelete
  53. montana parkDecember 8, 2009 at 9:01 AM

    number (5)

    dy/dx = 7x^2 + 3x + 4

    y=(7x^3/3)+(3x^2/2)+(4x)+c

    ReplyDelete
  54. montana parkDecember 8, 2009 at 9:02 AM

    number (7)
    part 1

    y = 6x^3 + 4x^2 - 5x

    dy/dx =(18x^2)+(8x)-5

    ReplyDelete
  55. montana parkDecember 8, 2009 at 9:02 AM

    number (7)
    part 2

    y = x^-3 + 16
    dy/dx = (-3x^-4)

    ReplyDelete
  56. montana parkDecember 8, 2009 at 9:03 AM

    number (7)
    part 3

    y = 4x + 4
    dy/dx = 4

    ReplyDelete
  57. montana parkDecember 8, 2009 at 9:04 AM

    number (7)
    part 4

    y = 3x^4 + 3(x^2 + 5)
    y=(3x^4)+(3x^2)+(15)
    dy/dx =(12x^3)+(6x)

    ReplyDelete
  58. montana parkDecember 8, 2009 at 9:05 AM

    number (7)
    part 5

    y + 6 = 4x + 9y - 3x^2
    y=[(3/8)x^2]-[(4/8)x]+[6/8]
    dy/dx =[(3/4)x]-[1/2]

    ReplyDelete
  59. montana parkDecember 8, 2009 at 9:06 AM

    number (11)

    eq'n of curve is y = 2x + 8/x^2, find dy/dx and d^2ydx^2
    y= 2x+8x^-2
    dy/dx= 2-16x
    d^2y/dx^2=-16

    ReplyDelete
  60. montana parkDecember 8, 2009 at 9:07 AM

    number (12)

    dy/dx = 2x² - 5
    to get a curve, the gradient must be integrated:
    y = ∫ (dy/dx) dx
    y = ∫ 2x² - 5 + c
    y = (2/3) x³ - 5x + c
    substitute point ( 3,8) to find c:
    8 = (2/3) 3³ - 5 (3) + c
    8 = -9 +c
    17 = c
    Therefore equation of the curve:
    y = (2/3) x³ - 5x + 17

    ReplyDelete
  61. montana parkDecember 8, 2009 at 9:08 AM

    number (6)

    v = 7t^2 - 5t +3
    dv/dt=14t-5
    when t=7
    subs into equ.
    14(7)-5= 93
    so acc. is 93
    intergrate v
    (7/3)t^3-(5/2)t^2 +3t
    when t=1
    subs into v
    (7/3)(1)^3-(5/2)t^2+3(1)
    =(7/3)-(5/2)+3
    =2/5/6

    when t=4
    subs into v
    (7/3)(4)^3-(5/2)(4)^2+3(4)

    ReplyDelete
  62. doghousekrewDecember 8, 2009 at 10:32 AM

    #9:
    dy/dx = 16/x³
    in order to obtain the curve, the gradient must be integrated:
    y = ∫ (dy/dx) dx
    y = ∫ 16/x³ + c
    y= ∫ 16x-³
    y = 8x-² + c OR y = 8/x² + c
    substitute point ( 1,4) to find c:
    4 = 8/1² + c
    4 = 8 +c
    -4 = c
    Therefore equation of the curve:
    y = 8/x² - 4

    ReplyDelete
  63. SnowwhiteDecember 8, 2009 at 12:08 PM

    #1
    dy/dx = 5x^3 + 2x^2 + 5
    i.e y=5x^4/4 + 2x^3/3 +5x +c

    ReplyDelete
  64. montana parkDecember 8, 2009 at 12:09 PM

    number (1)

    dy/dx = 5x^3 + 2x^2 + 5
    d^2y/dx^2=15x^2+4x+c

    ReplyDelete
  65. SnowwhiteDecember 8, 2009 at 12:10 PM

    #2
    dy/dx = 6x + 1
    y=6x^2/2 + 1x +c

    ReplyDelete
  66. SnowwhiteDecember 8, 2009 at 12:11 PM

    miss i dont understand why u have to add +c

    ReplyDelete
  67. SnowwhiteDecember 8, 2009 at 12:12 PM

    #3
    dy/dx = 2
    y=2x +c

    ReplyDelete
  68. SnowwhiteDecember 8, 2009 at 12:13 PM

    miss are my ans correct?

    ReplyDelete
  69. SnowwhiteDecember 8, 2009 at 12:19 PM

    #4
    dy/dx = 8x^5 - 5x^3 + 4x
    y=8x^6/6 - 5x^4/4 + 4x +c

    ReplyDelete
  70. SnowwhiteDecember 8, 2009 at 12:22 PM

    #5
    dy/dx = 7x^2 + 3x + 4
    y=7x^3/3 + 3x^2/2 + 4x +c

    ReplyDelete
  71. SnowwhiteDecember 8, 2009 at 12:51 PM

    #6
    v = 7t^2 - 5t +3
    ueing loop
    dv/dt=14t - 5
    sud when t=7
    dv/dt=14(7)- 5
    dv/dt=98-5
    dv/dt=93

    ReplyDelete
  72. SnowwhiteDecember 8, 2009 at 12:55 PM

    #6
    v = 7t^2 - 5t +3
    v=7t^3/3-5t^2/2+3x+c

    IS IT 3X +C OR 3t +C ???????????

    ReplyDelete
  73. SnowwhiteDecember 8, 2009 at 1:01 PM

    sub when t=1
    v=7(1)^3/3-5(1)^2/2+3(1)+c
    v=7/3 -5/2 +3 +c

    ReplyDelete
  74. SnowwhiteDecember 8, 2009 at 1:06 PM

    sub when t=4
    v=7(4)^3/3-5(4)^2/2+3(4)+c
    v=448/3 -80/2 +12 +c

    ReplyDelete
  75. God Of WarDecember 8, 2009 at 2:21 PM

    This comment has been removed by the author.

    ReplyDelete
  76. God Of WarDecember 8, 2009 at 2:22 PM

    2. dy/dx = 6x + 1
    integrated
    y=3x^2+x+c

    ReplyDelete
  77. God Of WarDecember 8, 2009 at 2:25 PM

    1. dy/dx = 5x^3 + 2x^2 + 5
    integrated
    y=5x^4/4 + 2x^3/3 + 2x^3/3 + 5x + c

    ReplyDelete
  78. God Of WarDecember 8, 2009 at 2:26 PM

    3. dy/dx = 2
    integrated
    y=2x+ c

    ReplyDelete
  79. God Of WarDecember 8, 2009 at 2:27 PM

    4. dy/dx = 8x^5 - 5x^3 + 4x
    integrated
    y= 8x^6/6 - 5x^4/4 + 4x^2/2 + c

    ReplyDelete
  80. God Of WarDecember 8, 2009 at 2:28 PM

    5. dy/dx = 7x^2 + 3x + 4
    integrated
    y= 7x^3/3 + 3x^2/2 + 4x + c

    ReplyDelete
  81. God Of WarDecember 8, 2009 at 2:31 PM

    No.7
    part 1

    1. y = 6x^3 + 4x^2 - 5x
    dy/dx= 18x^2 + 8x - 5

    ReplyDelete
  82. God Of WarDecember 8, 2009 at 2:32 PM

    No.7

    2. y = x^-3 + 16
    dy/dx=-3x^-4

    ReplyDelete
  83. God Of WarDecember 8, 2009 at 2:33 PM

    No.7
    3. y = 4x + 4
    dy/dx=4

    ReplyDelete
  84. God Of WarDecember 8, 2009 at 2:34 PM

    No.7
    4. y = 3x^4 + 3(x^2 + 5)
    y = 3x^4 + 3x^2 + 15
    dy/dx = 12x^3 + 6x

    ReplyDelete
  85. God Of WarDecember 8, 2009 at 2:42 PM

    No.7
    5. y + 6 = 4x + 9y - 3x^2
    -8y = -3x^2 + 4x - 6
    y = -3x^2/-8 +4x/-8 -6/-8
    y = 3/8x^2 + 1/4x + 3/4
    dy/dx = 3/4x + 1/4

    ReplyDelete
  86. God Of WarDecember 8, 2009 at 2:45 PM

    8.
    y = 4/(2)^.5
    y = 4/32
    y = 1/8
    dy/dx = 0

    ReplyDelete
  87. God Of WarDecember 8, 2009 at 2:53 PM

    9.
    dy/dx = 16/x^3
    integrated
    y = 16x^-2/-2 + c
    y = -8x^-2 + c
    when(1,4)
    4 = -8(1)^-2 + c
    4 = -8 + c
    c = 12
    i.e. eq'n
    y = -8x^-2 + 12

    ReplyDelete
  88. God Of WarDecember 8, 2009 at 2:56 PM

    10. y = 6theta - 2sin theta
    dy/dx = 6 - 2 cos theta

    ReplyDelete
  89. God Of WarDecember 8, 2009 at 2:58 PM

    11.
    y = 2x + 8/x^2
    dy/dx = 2 + -16x^-3
    d2y/dx^2 = 48x^-4

    ReplyDelete
  90. God Of WarDecember 8, 2009 at 3:04 PM

    12.
    dy/dx = 2x^2 -5
    y = 2x^3/3 - 5x + c
    when (3,8)
    8 = 2(3)^3/3 - 5(3) + c
    8 = 18 - 15 + c
    c = 5
    i.e. eq'n
    y = 2x^3/3 - 5x + 5

    ReplyDelete
  91. *iheartseanpaul*December 8, 2009 at 3:27 PM

    Fadda Anonymous i saw that no one answered your question about the integral of a constant. *sigh*
    the integral of a constsnt is not c it is:
    (the constant)*(what you're integrating wrt)

    So if you're integrating 5 with respect to x, the answer is 5x + C, of course.

    ReplyDelete
  92. *iheartseanpaul*December 8, 2009 at 3:32 PM

    1. dy/dx = 5x^3 + 2x^2 + 5
    integral of(5x^3 + 2x^2 + 5)
    =[(5x^(3+1)/4]+[2x^(2+1)/3]+ 5x + c
    =5x^4/4 + 2x^3/3 + 5x + c

    ReplyDelete
  93. *iheartseanpaul*December 8, 2009 at 3:35 PM

    2. dy/dx = 6x + 1
    integral of (6x + 1)
    =[6x^(1+1)/2] + 1x
    =6x^2/2 + 1x
    =3x^2 + 1x

    ReplyDelete
  94. *iheartseanpaul*December 8, 2009 at 3:37 PM

    3. dy/dx = 2
    integral of 2
    =2x + c

    ReplyDelete
  95. *iheartseanpaul*December 8, 2009 at 3:42 PM

    4. dy/dx = 8x^5 - 5x^3 + 4x
    integral of (8x^5 - 5x^3 + 4x
    =[8x^(5+1)/6]-[5x^(3+1)/4]+[4x^(1+1)/2 + c
    =4x^6/3 - 5x^4/4 + 2x^2 + c

    ReplyDelete
  96. *iheartseanpaul*December 8, 2009 at 3:45 PM

    5. dy/dx = 7x^2 + 3x + 4
    integral of (7x^2 + 3x + 4)
    =[7x^(2+1)/3] + 3x^(1+1)/2 + 4x + c
    = 7x^3/3 + 3x^2/2 + 4x + c

    ReplyDelete
  97. SnowwhiteDecember 8, 2009 at 3:49 PM

    7)
    #1
    y = 6x^3 + 4x^2 - 5x
    dy/dx=18x^2 + 8x - 5

    ReplyDelete
  98. SnowwhiteDecember 8, 2009 at 3:52 PM

    #2
    y = x^-3 + 16
    dy/dx=-3x^-4

    ReplyDelete
  99. SnowwhiteDecember 8, 2009 at 3:53 PM

    #3
    y = 4x + 4
    dy/dx=4

    ReplyDelete
  100. SnowwhiteDecember 8, 2009 at 3:59 PM

    #4
    y = 3x^4 + 3(x^2 +5)
    y=3x^4+3x^2+15
    dy/dx=12x^3+6x

    ReplyDelete
  101. SnowwhiteDecember 8, 2009 at 4:08 PM

    #5
    y + 6 = 4x + 9y - 3x^2
    9y-y=-3x^2+4x-6
    8y=-3x^2+4x-6
    y=-3x^2/8+4x/8-6/8
    y=-3x^2/8+2x-6/8
    dy/dx=-6x/8+2

    ReplyDelete
  102. *iheartseanpaul*December 8, 2009 at 4:12 PM

    question #7
    part 1. y = 6x^3 + 4x^2 - 5x
    dy/dx = [(6*3)x^(3-1)] + [(4*2)x^(2-1) - 5
    dy/dx = 18x^2 + 8x - 5

    ReplyDelete
  103. SnowwhiteDecember 8, 2009 at 4:13 PM

    #5
    y + 6 = 4x + 9y - 3x^2
    dy/dx=4+9-6x
    dy/dx=13-6x

    not sure which is correct

    ReplyDelete
  104. SnowwhiteDecember 8, 2009 at 4:18 PM

    #8
    miss i need help on this one

    ReplyDelete
  105. *iheartseanpaul*December 8, 2009 at 4:19 PM

    part 2. y = x^-3 + 16
    dy/dx = [(1*-3)x^(-3-1)
    dy/dx = -3x^-4

    ReplyDelete
  106. SnowwhiteDecember 8, 2009 at 4:21 PM

    #9
    dy/dx = 16/x^3
    dy/dx=16x^-3

    ReplyDelete
  107. *iheartseanpaul*December 8, 2009 at 4:22 PM

    3. y = 4x + 4
    dy/dx= 4

    ReplyDelete
  108. SnowwhiteDecember 8, 2009 at 4:23 PM

    #9
    can someone tell me what to do from there

    ReplyDelete
  109. SnowwhiteDecember 8, 2009 at 4:24 PM

    *iheartseanpaul* like u timeing me lol

    ReplyDelete
  110. *iheartseanpaul*December 8, 2009 at 4:26 PM

    4. y = 3x^4 + 3(x^2 + 5)
    y = 3x^4 + 3x^2 + 15
    dy/dx = [(3*4)x^(4-1) + [(3*2)x^(2-1)
    dy/dx = 12x^3 + 6x

    ReplyDelete
  111. *iheartseanpaul*December 8, 2009 at 4:28 PM

    lol haha no snowhite lol i thought we were racing lol but i think u like way ahead of me lol but i'll catch up soon lol

    u ready for examz???? lol
    *sigh*

    ReplyDelete
  112. SnowwhiteDecember 8, 2009 at 4:31 PM

    #10
    y = 6theta - 2sin theta
    dy/dx=6-2cos theta

    ReplyDelete
  113. SnowwhiteDecember 8, 2009 at 4:36 PM

    lol ok yes i was racing lol

    no am not realy feelin prepared r u

    ReplyDelete
  114. SnowwhiteDecember 8, 2009 at 4:40 PM

    #11
    y = 2x + 8/x^2
    y=2x+8x^-2
    dy/dx=2-16x^-3

    ReplyDelete
  115. SnowwhiteDecember 8, 2009 at 4:42 PM

    #11
    dy/dx=2-16x^-3
    d^2y/dx^2=48x-4

    ReplyDelete
  116. SnowwhiteDecember 8, 2009 at 4:54 PM

    #12
    dy/dx = 2x^2 -5
    y=2x^3/3-5x+c
    sub when x=3 & y=8
    8=2(3)^3/3-5(3)+c
    8=18-15+c
    8=3+c
    3+c=8
    c=8-3
    c=5

    y=2x^3/3-5x+5

    ReplyDelete
  117. *iheartseanpaul*December 8, 2009 at 6:26 PM

    This comment has been removed by the author.

    ReplyDelete
  118. 7NewStarLight7December 8, 2009 at 6:29 PM

    1.
    dy/dx = 5x^3 + 2x^2 + 5
    y = ∫5x^3 + 2x^2 + 5
    y = 5x^4/4 +2x^3/4 + 5x + c

    ReplyDelete
  119. 7NewStarLight7December 8, 2009 at 6:31 PM

    2.
    dy/dx = 6x + 1
    y = ∫6x + 1
    y = 3x^2 + x + c

    ReplyDelete
  120. 7NewStarLight7December 8, 2009 at 6:33 PM

    3.
    dy/dx = 2
    y = ∫2
    y = 2x + c

    ReplyDelete
  121. 7NewStarLight7December 8, 2009 at 6:40 PM

    4.
    dy/dx = 8x^5 - 5x^3 + 4x
    y = ∫8x^5 - 5x^3 + 4x
    y = 4x^6/3 - 5x^4/4 + 2x^2 + c

    ReplyDelete
  122. 7NewStarLight7December 8, 2009 at 6:43 PM

    5.
    dy/dx = 7x^2 + 3x + 4
    y = ∫7x^2 + 3x + 4
    y = 7x^3/3 + 3x^2/2 + 4x + c

    ReplyDelete
  123. bornagain16December 8, 2009 at 7:46 PM

    fadda annonymous, the integral of zero is 'c'. the integral of a constant is the same constant by 'x'. eg. the integral of 3 would be 3x...

    ReplyDelete
  124. bornagain16December 8, 2009 at 7:48 PM

    the reason that the integral of zero is 'c' is that when you differentiate a constant you get zero and since integration is the opposite of differentiation, the integral of zero is 'c'.

    ReplyDelete
  125. bornagain16December 8, 2009 at 8:02 PM

    hey, fadda anonymous, for question 7 part 5, you have to divide each term by 8...then differentiate. if all the terms were multiplied to each other then it would have been possible to state it like that but since all the terms are added, each term has to be evaluated separately...understand what i mean?

    ReplyDelete
  126. *iheartseanpaul*December 8, 2009 at 8:11 PM

    here's how to do #9 snowwhite

    since you already have the gradient ie.dy/dx=16/x^3 you can integrate this to find the equation of the curve so it will be:
    y =∫16/x^3
    y =∫16x^-3
    y =[16x^(-3+1)]/-2 + c
    y = -8x^-2 + c OR -8/x^2 + c
    now that you have the integral u can substitute the points (1,4) into the equation to solve for c..that will give:

    4 = -8/1^2 +c
    4 = -8/1 + c
    4 = -8 + c
    c = 4 + 8
    c = 12

    so the equation of the curve is y = -8/x^2 + 12

    ReplyDelete
  127. bornagain16December 8, 2009 at 8:12 PM

    hey grapes, you did the differentiation part of question 6 correctly. but you did not have to integrate again as they already gave you the equation which is v = 7t^2 - 5t +3. all you have to do now is to use the limits that were given (t=4 and t=1) to find the area which, in this case, will be the distance travelled...

    ReplyDelete
  128. bornagain16December 8, 2009 at 8:16 PM

    snowwhite,see above for explanation on the reason for 'c'.

    ReplyDelete
  129. bornagain16December 8, 2009 at 8:18 PM

    also snow, see above for question six. however, if you had to integrate in this case, you would get 3t + c as the varibles are in 't'...

    ReplyDelete
  130. St. Anns graduateDecember 9, 2009 at 6:00 AM

    1).
    dy/dx=5x^3+2x^2+5

    the integral of that is 5/4x^4+2/3x^3+5x+c

    ReplyDelete
  131. St. Anns graduateDecember 9, 2009 at 6:02 AM

    2).
    dy/dx=6x+1

    the integral of that is 3x^2+x+c

    ReplyDelete
  132. St. Anns graduateDecember 9, 2009 at 6:02 AM

    3).
    dy/dx=2

    the integral of that is 2x+c

    ReplyDelete
  133. St. Anns graduateDecember 9, 2009 at 6:04 AM

    4).
    dy/dx=8x^5-5x^3+4x

    the integral of that is 4/3x^6-5/4x^4+2x^2+c

    ReplyDelete
  134. St. Anns graduateDecember 9, 2009 at 6:06 AM

    5).
    dy/dx=7x^2+3x+4

    the integral of that is 7/3x^3+3/2x^2+4x+c

    ReplyDelete
  135. St. Anns graduateDecember 9, 2009 at 6:21 AM

    6).(i) V=7t^2-5t+3

    dv/dt=14t-5

    when t=7

    dv/dt=14(7)-5
    = 98-5
    = 93

    therefore the acc. is 93

    (ii) V=7t^2-5t+3

    the integral of that is 7/3t^3-5/2t^2+3t

    using bounds t=1 and t=4

    [7(4)^3/3-5(4)^2/2+3(4)]-[7(1)^3/5-5(1)^2/2+3(1)]
    [149.33-40+12]-[1.4-2.5+3]
    121.33-1.9
    119.43

    the particle moves 119.43

    ReplyDelete
  136. St. Anns graduateDecember 9, 2009 at 6:27 AM

    11). y=2x+8/x^2
    y=2x+8x^-2

    dy/dx=2-16x^-3

    d2y/dx2= -48x^-4

    ReplyDelete
  137. St. Anns graduateDecember 9, 2009 at 6:31 AM

    12). dy/dx= 2x^2-5

    the integral of that is y= 2/3x^3-5x+c

    using the point (3,8) subst in (y)

    8=2(3)^3/3-5(3)+c
    8=18-15+c
    8=3+c
    c=8-3
    c=5

    equation the curve is y=2/3x^3-5x+5

    ReplyDelete
  138. *iheartseanpaul*December 9, 2009 at 6:33 AM

    8) A curve has equation y = 4/(2)^.5, find dy/dx

    ??????????????
    i don't understand how you're suppose to find the differential of this equation wrt x if there is no x in the entire equation...is this a trick question or am i missing something important????
    *sigh* :(

    ReplyDelete
  139. St. Anns graduateDecember 9, 2009 at 6:50 AM

    8). y=4/2^.5
    y= 4

    dy/dx= 0

    ReplyDelete
  140. St. Anns graduateDecember 9, 2009 at 6:53 AM

    9). dy/dx=16/x^3
    dy/dx=16x^-3

    the integral of that is y= 24x/x^2+c

    using the point (1,4) subst in (y)

    4=24/(1)^2+c
    4=24+c
    c=4-24
    c= -20

    so therefore the equation of the curve is
    y=24/x^2-20

    ReplyDelete
  141. *iheartseanpaul*December 9, 2009 at 7:00 AM

    10)

    y = 6theta - 2sin theta, find dy/dx
    let x = theta
    y = 6x - 2sinx
    dy/dx = 6 - 2cosx
    therefore dy/dx = 6 - 2cos theta

    ReplyDelete
  142. badman_nerdDecember 9, 2009 at 7:01 AM

    (8) y=4/(2)^.5
    y=4(2)^-1/2
    dy/dx= -2(2)^-3/2
    miss a try a thing tell me if that correct

    ReplyDelete
  143. *iheartseanpaul*December 9, 2009 at 7:11 AM

    The equation of a curve is y = 2x + 8/x^2, find dy/dx and d^2ydx^2

    y = 2x + 8/x^2
    y = 2x + 8x-2
    dy/dx = 2 + (8*-2)x^(-2-1)
    dy/dx = 2 - 16x^-3

    when dy/dx = 2 - 16x^-3
    d^2ydx^2 = (-16*-3)x^(-3-1)
    d^2ydx^2 = 48x^-4

    ReplyDelete
  144. *iheartseanpaul*December 9, 2009 at 7:13 AM

    badman_nerd how can u find dy/dx if there's is no mention of x in the entire equation????? :(

    ReplyDelete
  145. badman_nerdDecember 9, 2009 at 7:19 AM

    #9:
    dy/dx = 16/x³
    in order to obtain the curve, the gradient must be integrated:
    y = ∫ (dy/dx) dx
    y = ∫ 16/x³ + c
    y= ∫ 16x-³
    y = 8x-² + c OR y = 8/x² + c
    substitute point ( 1,4) to find c:
    4 = -8/x^2+c
    4= -8/1+c
    4+8=c
    c=12
    there the equation of the curve is y=-8/x^2+12

    ReplyDelete
  146. *iheartseanpaul*December 9, 2009 at 7:25 AM

    12. A curve is such that dy/dx = 2x^2 -5. Given that the point (3,8) lies on the curve, find the equation of the curve.

    y = ∫2x^2 - 5
    y = [2x^(2+1)]/3 - 5x + c
    y = 2x^3/3 - 5x + c

    substitute the points (3,8) into the equation to find c

    8 = 2(3)^3/3 - 5(3) + c
    8 = 18 - 15 + c
    8 = 3 + c
    c = 8 - 3
    c = 5

    therefore the equation of the cirve is:
    y = 2x^3/3 - 5x + 5

    ReplyDelete
  147. swagDecember 9, 2009 at 8:36 AM

    5. dy/dx = 7x^2 + 3x + 4
    y=(7/3)x^3+(3/2)x^2 + 4x

    ReplyDelete
  148. swagDecember 9, 2009 at 8:37 AM

    3)dy/dx=2

    the integral of that = 2x+c

    ReplyDelete
  149. swagDecember 9, 2009 at 8:38 AM

    8). y=4/2^.5
    y= 4

    dy/dx= 0

    ReplyDelete
  150. swagDecember 9, 2009 at 8:38 AM

    2)dy/dx=6x+1

    the integral of that = 3x^2+x+c

    ReplyDelete
  151. swagDecember 9, 2009 at 8:39 AM

    11). y=2x+8/x^2
    y=2x+8x^-2

    dy/dx=2-16x^-3

    d2y/dx2= -48x^-4

    ReplyDelete
  152. swagDecember 9, 2009 at 8:40 AM

    1. y = 6x^3 + 4x^2 - 5x
    dy/dx=18x^2+8x-5

    ReplyDelete
  153. swagDecember 9, 2009 at 8:41 AM

    need some help for number 12!!!!!!

    ReplyDelete
  154. sunflowerDecember 9, 2009 at 8:59 AM

    1.
    y = 6x^3 + 4x^2 - 5x
    dy/dx=18x^2+8x-5

    ReplyDelete
  155. sunflowerDecember 9, 2009 at 9:00 AM

    4.
    y=3x^4+3(x^2 + 5)
    dy/dx=12x^3+6x

    ReplyDelete
  156. d last man standingDecember 9, 2009 at 9:17 AM

    question 1
    y = 6x^3 + 4x^2 - 5x
    dy/dx=18x^2+8x-5

    ReplyDelete
  157. d last man standingDecember 9, 2009 at 9:19 AM

    question 5
    dy/dx = 7x^2 + 3x + 4
    y=(7/3)x^3+(3/2)x^2 + 4x

    ReplyDelete
  158. d last man standingDecember 9, 2009 at 9:22 AM

    question 3
    dy/dx=2

    integral= 2x+c

    ReplyDelete
  159. d last man standingDecember 9, 2009 at 9:23 AM

    question 8
    y=4/2^.5
    y= 4

    dy/dx= 0

    ReplyDelete
  160. d last man standingDecember 9, 2009 at 9:24 AM

    question 11
    y=2x+8/x^2
    y=2x+8x^-2

    dy/dx=2-16x^-3

    d2y/dx2= -48x^-4

    ReplyDelete
  161. d last man standingDecember 9, 2009 at 9:25 AM

    question 12
    dy/dx= 2x^2-5

    the integral of that is y= 2/3x^3-5x+c

    using the point (3,8) subst in (y)

    8=2(3)^3/3-5(3)+c
    8=18-15+c
    8=3+c
    c=8-3
    c=5

    equation the curve is y=2/3x^3-5x+5

    ReplyDelete
  162. d last man standingDecember 9, 2009 at 9:26 AM

    question 11
    y=2x+8/x^2
    y=2x+8x^-2

    dy/dx=2-16x^-3

    d2y/dx2= -48x^-4

    ReplyDelete
  163. d last man standingDecember 9, 2009 at 9:27 AM

    qeustion 4

    y = 3x^4 + 3(x^2 +5)
    y=3x^4+3x^2+15
    dy/dx=12x^3+6x

    ReplyDelete
  164. d last man standingDecember 9, 2009 at 9:29 AM

    question 2
    2. dy/dx = 6x + 1
    integral of (6x + 1)
    =[6x^(1+1)/2] + 1x
    =6x^2/2 + 1x
    =3x^2 + 1x

    ReplyDelete
  165. d last man standingDecember 9, 2009 at 9:30 AM

    question 6
    v = 7t^2 - 5t +3
    v=7t^3/3-5t^2/2+3x+c

    ReplyDelete
  166. d last man standingDecember 9, 2009 at 9:31 AM

    i stand to be corrected for the above answers!!!!!

    ReplyDelete
  167. UnknownDecember 9, 2009 at 10:14 AM

    for ques 4.

    dy/dx = 8x^5 - 5x^3 + 4x
    y=(8x^6/6)-(5x^4/4)+(4x^2/2)+c
    y=(4x^6/3)-(5x^4/4)+(2x^2)+c

    ReplyDelete
  168. UnknownDecember 9, 2009 at 10:18 AM

    for ques 7 #4

    y = 3x^4 + 3(x^2 + 5)
    y=(3x^4)+(3x^2)+(15)
    dy/dx =(12x^3)+(6x)

    ReplyDelete
  169. freakazoidDecember 9, 2009 at 10:20 AM

    2 INTEGRATION

    dy/dx = 6x + 1
    y = 6x^2/2 + x
    y = 3x^2 + x

    ReplyDelete
  170. freakazoidDecember 9, 2009 at 10:26 AM

    7 1)

    y = 6x^3 + 4x^2 - 5x
    dy/dx = 3(6x^2) + 2(4x) - 5
    dy/dx = 18x^2 + 8x - 5

    ReplyDelete
  171. UnknownDecember 9, 2009 at 10:28 AM

    for ques 1.

    dy/dx = 5x^3 + 2x^2 + 5
    y=(5x^4)+(2x^3/3)+(5x)+c

    ReplyDelete
  172. UnknownDecember 9, 2009 at 10:30 AM

    for ques 5
    dy/dx = 7x^2 + 3x + 4
    y=(7/3)x^3+(3/2)x^2 + 4x

    ReplyDelete
  173. freakazoidDecember 9, 2009 at 10:31 AM

    ques 12

    dy/dx = 2x^2 - 5
    y = 2x^2+1/3 - 5x
    y = 2x^3/3 - 5x

    the eq'n of the curve is
    y=2x^3/3 - 5x

    ReplyDelete
  174. "math works"December 9, 2009 at 11:44 AM

    The integral of dy/dx = 5x^3 + 2x^2 + 5
    = 5x^4/4 + 2x^3/3 + 5x + c

    ReplyDelete
  175. "math works"December 9, 2009 at 11:48 AM

    The integral of dy/dx = 6x + 1
    = 6x^2/2 + 1x +c

    ReplyDelete
  176. "math works"December 9, 2009 at 11:51 AM

    The integral of dy/dx = 2
    = 2x + c

    ReplyDelete
  177. "math works"December 9, 2009 at 11:55 AM

    The integral of dy/dx = 8x^5 - 5x^3 + 4x
    = 8x^6/6 - 5x^4/4 + 4X^2/2 + c

    ReplyDelete
  178. "math works"December 9, 2009 at 11:57 AM

    The integral of dy/dx = 7x^2 + 3x + 4
    = 7x^3/3 + 3x^2/2 + 4x +c

    ReplyDelete
  179. "math works"December 9, 2009 at 12:15 PM

    y = 6x^3 + 4x^2 - 5x
    using y =anx^n-1
    dy/dx = 18x^2 + 8x - 5

    ReplyDelete
  180. 7NewStarLight7December 9, 2009 at 1:31 PM

    7)
    1. y = 6x^3 + 4x^2 - 5x
    dy/dx = 18x^2 + 8x - 5

    ReplyDelete
  181. 7NewStarLight7December 9, 2009 at 1:33 PM

    7)
    2. y = x^-3 + 16
    dy/dx = -3x^-4
    dy/dx = -3/x^4

    ReplyDelete
  182. 7NewStarLight7December 9, 2009 at 1:34 PM

    7)
    3. y = 4x + 4
    dy/dx = 4

    ReplyDelete
  183. 7NewStarLight7December 9, 2009 at 1:36 PM

    7)
    4. y = 3x^4 + 3(x^2 + 5)
    y = 3x^4 + 3x^2 + 15
    dy/dx = 12x^3 + 6x

    ReplyDelete
  184. 7NewStarLight7December 9, 2009 at 1:45 PM

    7)
    5. y + 6 = 4x + 9y - 3x^2
    y - 9y = 4x - 3x^2 - 6
    -8y = 4x - 3x^2 - 6
    y = (4x - 3x^2 - 6)/-8
    y = 4x/-8 + 3x^2/8 + 3/4
    y = x*-2^-1 + 3x^2*8^-1 + 3*4^-1
    dy/dx = -x/(-2^-2) - 3x^2/(8^2) - 3/(4^2)
    dy/dx = -x/4 -3x^2/64 - 3/16

    ReplyDelete
  185. d blackmanDecember 9, 2009 at 3:56 PM

    1. dy/dx = 5x^3 + 2x^2 + 5
    int dy/dx = int 5x^3 + 2x^2 + 5
    = 5x^4/4 + 2x^3/3 +5x +c
    y = 5/4x^4 + 2/3x^3 +5x +c
    where c is the constant of proportionality

    ReplyDelete
  186. d blackmanDecember 9, 2009 at 4:04 PM

    2. dy/dx = 6x + 1
    int dy/dx = int 6x+1
    = 6x^2/2 +x +c
    y = 3x^2 + x +c

    ReplyDelete
  187. AnonymousDecember 9, 2009 at 4:04 PM

    question 1

    the integral of
    dy/dx=5x^3+2x^2+5

    y=(5/4)x^4+(2/3)x^3+5x+c

    ReplyDelete
  188. AnonymousDecember 9, 2009 at 4:07 PM

    question 2

    the integral of dy/dx=6x+1

    y=(6/2)x^2+x

    =3x^2+x+c

    ReplyDelete
  189. AnonymousDecember 9, 2009 at 4:08 PM

    question 3

    the integral of dy/dx=2

    y=2x+c

    ReplyDelete
  190. AnonymousDecember 9, 2009 at 4:11 PM

    question 4

    the integral of dy/dx=7x^2+3x+4

    y=(7/3)x^3+(3/2)x^2+4x+c

    ReplyDelete
  191. AnonymousDecember 9, 2009 at 4:22 PM

    question 5

    the integral of dy/dx=8x^5-5x^3+4x

    y=(8/6)x^6-(5/4)x^4+(4/2)x^2+c

    =(4/3)x^6-(5/4)x^4+2x^2=c

    ReplyDelete
  192. AnonymousDecember 9, 2009 at 4:27 PM

    question 12

    the integral of dy/dx=2x^2-5

    y=(2/3)x^3-5x+c

    at the point (3,8)

    8=(2/3)(3)^3-5(3)+c
    8=18-15+c
    8=3+c
    8-3=c
    c=5

    equation of the curve y=(2/3)x^3-5x+5

    ReplyDelete
  193. AnonymousDecember 9, 2009 at 4:33 PM

    question 10

    y=6 theta-2sin theta

    dy/dtheta=6-2cos theta

    ReplyDelete
  194. $$Empress$$December 9, 2009 at 4:37 PM

    #7 (1)
    Y=6X^3+4X^2-2x
    dy/dx=18x^2+8x-5

    ReplyDelete
  195. AnonymousDecember 9, 2009 at 4:37 PM

    question 11

    y=2x+8/x^2

    y=2x+8x^-2

    dy/dx=2-16x^-3

    =2-16/x^3

    d^2y/dx^2=48x^-4

    48/x^4

    ReplyDelete
  196. $$Empress$$December 9, 2009 at 4:38 PM

    #7 (2)
    y=x^-3 +16
    dy/dx= -3x^-4

    ReplyDelete
  197. $$Empress$$December 9, 2009 at 4:40 PM

    #7(3)
    y=4x+4
    dy/dx= 4

    ReplyDelete
  198. AnonymousDecember 9, 2009 at 4:40 PM

    question 7

    part 1

    y=6x^3+4x^2-5x

    dy/dx=18x^2+8x-5

    ReplyDelete
  199. $$Empress$$December 9, 2009 at 4:43 PM

    #7(4)
    y= 3x^4+3(x^2+5)
    dy/dx= 3x^4+3x^2+15
    dy/dx= 12x^3+6x

    ReplyDelete
  200. $$Empress$$December 9, 2009 at 4:46 PM

    #7(5)
    y+6=4x+9y-3x^2
    dy/dx=4x+9y-3x^2-6
    dy/dx=4+9-6x
    =13-6x

    ReplyDelete

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