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Liv1234

  • one year ago

Can someone please help me with a question?

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  1. Liv1234
    • one year ago
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  2. Michele_Laino
    • one year ago
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    the new function g(x) has to be equal to the old function f at point x+1

  3. Liv1234
    • one year ago
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    What does that mean?

  4. Michele_Laino
    • one year ago
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    in formula, my statements means: g(x)=f(x+1)

  5. Liv1234
    • one year ago
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    So kind of like for example y= mx+b?

  6. Michele_Laino
    • one year ago
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    in that case, we can write this: g(x9=m(x+1)+b

  7. Michele_Laino
    • one year ago
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    oops.. g(x)=m(x+1)+b

  8. Liv1234
    • one year ago
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    By the way, sorry if I seem slow with this, I'm just really bad with math.

  9. Michele_Laino
    • one year ago
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    no worries! :)

  10. Liv1234
    • one year ago
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    So, g(x)=m(x+1)+b

  11. Michele_Laino
    • one year ago
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    yes!

  12. Liv1234
    • one year ago
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    How would I solve the answer with that?

  13. Michele_Laino
    • one year ago
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    I think you have to do the multiplication, like below: g(x)=mx+m+b

  14. Liv1234
    • one year ago
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    How do I do the multiplication with this equation?

  15. Michele_Laino
    • one year ago
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    we can do the multiplication above using the distributive property of multiplication over addition

  16. Liv1234
    • one year ago
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    I think I remember how to do that, is there a formula?

  17. Liv1234
    • one year ago
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    So, (sorry to ask again) what numbers would be put in?

  18. Michele_Laino
    • one year ago
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    the general formula, is: \[\left( {a + b} \right)c = ac + bc,\quad \forall a,b,c \in \mathbb{R}\]

  19. Michele_Laino
    • one year ago
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    you have to know the values of the coefficients m and b

  20. Liv1234
    • one year ago
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    I feel really stupid to keep asking for help, but how would I find the coefficients?

  21. Michele_Laino
    • one year ago
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    you can retrieve the value of those coefficients, from the text of your exercise

  22. Liv1234
    • one year ago
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    The only number I had from the exercise is "1"

  23. Michele_Laino
    • one year ago
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    in that case your exercise is not a numeric exercise

  24. Liv1234
    • one year ago
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    Hang on, let me repost the question

  25. Michele_Laino
    • one year ago
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    ok!

  26. Liv1234
    • one year ago
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  27. Liv1234
    • one year ago
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    Okay, that's the question

  28. Michele_Laino
    • one year ago
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    your answer, is: \[g\left( x \right) = m\left( {x + 1} \right) + b = mx + m + b\] and we have completed your exercise

  29. Liv1234
    • one year ago
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    Wait, so what option would I choose? I'm confused.

  30. Michele_Laino
    • one year ago
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    the first option

  31. Liv1234
    • one year ago
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    Can you help me with another question?

  32. Michele_Laino
    • one year ago
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    ok!

  33. Liv1234
    • one year ago
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  34. Michele_Laino
    • one year ago
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    when we make a reflection with respect to the x-axis, then we can write the new function g(x) as below: \[g\left( x \right) = - f\left( x \right)\]

  35. Liv1234
    • one year ago
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    So, how would I solve?

  36. Michele_Laino
    • one year ago
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    it is simple, you have to replace f(x) with its formula: \[g\left( x \right) = - f\left( x \right) = - \left( {{x^2} + 5} \right) = ...?\]

  37. Liv1234
    • one year ago
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    Would the answer be g(x)= -x^2+5?

  38. Michele_Laino
    • one year ago
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    not exactly, since when you cancel the parentheses, you get this: \[g\left( x \right) = - f\left( x \right) = - \left( {{x^2} + 5} \right) = - {x^2} - 5\]

  39. Liv1234
    • one year ago
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    I'm confused D:

  40. Michele_Laino
    • one year ago
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    every term inside the parentheses change its sign

  41. Michele_Laino
    • one year ago
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    so we can write this: \[ - \left( {{x^2} + 5} \right) = - {x^2} - 5\]

  42. Liv1234
    • one year ago
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    So, the first option?

  43. Michele_Laino
    • one year ago
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    yes!

  44. Liv1234
    • one year ago
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    Can you help me with another one?

  45. Michele_Laino
    • one year ago
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    ok!

  46. Liv1234
    • one year ago
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  47. Michele_Laino
    • one year ago
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    here we have to consider all points whose y-coordinate is less the y coordinate of the points which belong to the straight line

  48. Liv1234
    • one year ago
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    Let me guess, it's the first option right?

  49. Liv1234
    • one year ago
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    Or is it another one?

  50. Michele_Laino
    • one year ago
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    please wait, first we have to write the equation of the straight line

  51. Liv1234
    • one year ago
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    Okay, so how would we do that?

  52. Liv1234
    • one year ago
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    I know that the line is at -6 and 5

  53. Michele_Laino
    • one year ago
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    your line passes at points (-5,0) and (0,6), so its equation is: \[\frac{{y - 0}}{{6 - 0}} = \frac{{x + 5}}{{0 + 5}}\]

  54. Liv1234
    • one year ago
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    It actually passes at -6 and 5 though.

  55. Michele_Laino
    • one year ago
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    which can be simplified to this form: \[6x - 5y = - 30\]

  56. Liv1234
    • one year ago
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    So, what option would I choose?

  57. Michele_Laino
    • one year ago
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    please wait, I think that your line passes at points (-6,0) and (0,5), so its equation is: \[5x - 6y = - 30\]

  58. Liv1234
    • one year ago
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    So, would it be C or D?

  59. Michele_Laino
    • one year ago
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    so the shaded part of cartesian plane is represented by this inequality: \[5x - 6y \geqslant - 30\]

  60. Liv1234
    • one year ago
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    So, D?

  61. Liv1234
    • one year ago
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    Can you help me with another one please?

  62. Michele_Laino
    • one year ago
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    more explanation: the y-coordinate of the points of the shaded region has to be less than the y-coordinate of the point of the straight line, so we can write this: \[y \leqslant \frac{{5x - 30}}{6}\]

  63. Liv1234
    • one year ago
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    Can you help me with this question?

  64. Liv1234
    • one year ago
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    Please?

  65. Michele_Laino
    • one year ago
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    we can write this: \[g\left( x \right) = 5\left( { - 7{x^2}} \right) + 5 = 5f\left( x \right) + 5\]

  66. Liv1234
    • one year ago
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    I think the answer is D

  67. Michele_Laino
    • one year ago
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    yes! correct!

  68. Liv1234
    • one year ago
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    Yay! Can you help me with another one please?

  69. Michele_Laino
    • one year ago
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    ok!

  70. Liv1234
    • one year ago
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  71. Liv1234
    • one year ago
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    Okay, that's the question.

  72. Michele_Laino
    • one year ago
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    the y-coordinates of the new function g(x) have to be six time greater than the respectively y-coordinate of the old function f(x)

  73. Liv1234
    • one year ago
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    So, would the answer be C?

  74. Michele_Laino
    • one year ago
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    no, I meant six as a factor

  75. Liv1234
    • one year ago
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    Oh, so would it be A? Because it has multiplication?

  76. Michele_Laino
    • one year ago
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    correct!

  77. Liv1234
    • one year ago
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    Can you help me with another one?

  78. Michele_Laino
    • one year ago
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    please wait, since another student has requested my help

  79. Liv1234
    • one year ago
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    It's a quick question though

  80. Michele_Laino
    • one year ago
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    ok!

  81. Liv1234
    • one year ago
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  82. Michele_Laino
    • one year ago
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    -7 means shifting down by 7 units furthermore, as we can see in your first exercise, x+3 means shifting to left

  83. Liv1234
    • one year ago
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    So, would it be C?

  84. Liv1234
    • one year ago
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    Oh, wait, B?

  85. Michele_Laino
    • one year ago
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    no, since the vertical shift is by 7 units down and the horizontal shift is by 3 units left

  86. Liv1234
    • one year ago
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    So, B?

  87. Michele_Laino
    • one year ago
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    correct! it is option B

  88. Liv1234
    • one year ago
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    Do you think you could help me with another?

  89. Michele_Laino
    • one year ago
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    yes! please wait

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