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anonymous

  • one year ago

Piecewise function question

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  1. misty1212
    • one year ago
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    \[f(x) = |x+4| = \left\{\begin{array}{rcc} x + 4 & \text{if} & x \geq -4 \\ - x - 4& \text{if} & x < -4 \end{array} \right. \]

  2. anonymous
    • one year ago
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    yes i have a graph that i really need help on

  3. anonymous
    • one year ago
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    could you please hhelp me out

  4. misty1212
    • one year ago
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    you need to graph something?

  5. anonymous
    • one year ago
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    no ive got a graph and have some questions about it

  6. misty1212
    • one year ago
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    ok i can try to help, post the picture

  7. anonymous
    • one year ago
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    i can post a picture of the worksheet so it'll be easier to see the graph

  8. anonymous
    • one year ago
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    okay pleasegive me a sec

  9. anonymous
    • one year ago
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  10. anonymous
    • one year ago
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    this is the graph and the first question is f(1)=

  11. anonymous
    • one year ago
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    and i got 3 as the answer

  12. misty1212
    • one year ago
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    oh i thought you had to find the function

  13. anonymous
    • one year ago
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    i do later on

  14. misty1212
    • one year ago
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    yeah \(f(1)=3\)

  15. anonymous
    • one year ago
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    and the next one is f(-11)

  16. anonymous
    • one year ago
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    but on the graph its only till -10

  17. misty1212
    • one year ago
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    then we better find the function for \(x\leq -5\)

  18. misty1212
    • one year ago
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    you got the slope right, it is \(\frac{1}{2}\)

  19. anonymous
    • one year ago
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    yes that is the slope

  20. misty1212
    • one year ago
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    goes through the point \((-5,3)\) so it is \[y+5=\frac{1}{2}(x-3)\]

  21. misty1212
    • one year ago
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    you can plug 11 in to that if you like

  22. anonymous
    • one year ago
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    oh alright that makes sense

  23. misty1212
    • one year ago
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    or else write it as \[y=\frac{1}{2}(x-3)+5\]

  24. anonymous
    • one year ago
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    is it -2

  25. misty1212
    • one year ago
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    whoah did i screw that up!!

  26. anonymous
    • one year ago
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    im not sure haha

  27. misty1212
    • one year ago
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    slope is \(\frac{1}{2}\) point is \((-5,3)\) equation is \[y-3=\frac{1}{2}(x+5)\] doe

  28. anonymous
    • one year ago
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    so its y= 1/2(x+5)+3?

  29. misty1212
    • one year ago
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    or \[y=\frac{x}{2}+\frac{11}{2}\]

  30. misty1212
    • one year ago
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    whatever you like

  31. anonymous
    • one year ago
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    alright let me solve one sec

  32. misty1212
    • one year ago
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    cool henna btw

  33. anonymous
    • one year ago
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    thank you :)

  34. misty1212
    • one year ago
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    did you get zero yet?

  35. anonymous
    • one year ago
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    yepp

  36. anonymous
    • one year ago
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    okay so the next one says find x when f(x)=0

  37. anonymous
    • one year ago
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    is it 5/2

  38. misty1212
    • one year ago
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    lol you just did it

  39. anonymous
    • one year ago
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    im not supposed to use decimals

  40. misty1212
    • one year ago
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    you found \(f(-11)=0\) right?

  41. anonymous
    • one year ago
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    yes i did

  42. misty1212
    • one year ago
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    so that makes "find x when f(x)=0" rather obvious !!

  43. anonymous
    • one year ago
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    lol idk why i put 5/2

  44. misty1212
    • one year ago
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    plug in \(-11\) get out zero !!

  45. misty1212
    • one year ago
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    me neither lol

  46. anonymous
    • one year ago
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    i got it!

  47. misty1212
    • one year ago
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    ok next...

  48. anonymous
    • one year ago
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    alright then lol on to the next one

  49. anonymous
    • one year ago
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    i have to find the x intercepts

  50. misty1212
    • one year ago
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    ok we got one right? one is \(-11\)

  51. anonymous
    • one year ago
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    and i got (5/2,0) and (-11,0)

  52. misty1212
    • one year ago
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    there is another as well, because it is coming back up at the end and that little arrow indicates it continues up

  53. misty1212
    • one year ago
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    lets just find the damn function and be done with it k?

  54. anonymous
    • one year ago
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    lol alright

  55. anonymous
    • one year ago
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    it actually says to write in point slope form and restricted domain in inequality form

  56. anonymous
    • one year ago
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    so basically its asking me to write a fuction for each piece

  57. misty1212
    • one year ago
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    right so far we are at \[f(x) = \left\{\begin{array}{rcc} y-3=\frac{1}{2}x+5 & \text{if} & x <-5 \\ 3& \text{if} & -5\leq x\leq 1 \end{array} \right.\]

  58. misty1212
    • one year ago
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    what is the slope of the next piece? it is a little hard for me to see it

  59. misty1212
    • one year ago
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    maybe \(-2\)?

  60. anonymous
    • one year ago
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    you said in the first function theres a 3 ?

  61. anonymous
    • one year ago
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    under where you write the 5 in the first equation

  62. misty1212
    • one year ago
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    it is a constant (horizontal line) between -5 and 1 right?

  63. anonymous
    • one year ago
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    yes thats correct

  64. misty1212
    • one year ago
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    ooh damn i made a typo there forgot the parentheses

  65. misty1212
    • one year ago
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    \[f(x) = \left\{\begin{array}{rcc} y-3=\frac{1}{2}(x+5) & \text{if} & x <-5 \\ 3& \text{if} & -5\leq x\leq 1 \end{array} \right.\]

  66. anonymous
    • one year ago
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    im a little confused

  67. misty1212
    • one year ago
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    ok lets go slow and make sure you get it

  68. anonymous
    • one year ago
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    under the first equation you put a 3

  69. anonymous
    • one year ago
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    but im not sure why

  70. misty1212
    • one year ago
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    yeah we are not done you see how you have 4 long lines in your answer space?

  71. misty1212
    • one year ago
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    because this piecewise function has four distinct parts

  72. anonymous
    • one year ago
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    yes i do

  73. misty1212
    • one year ago
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    4 different lines

  74. anonymous
    • one year ago
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    okay yes i understand now thank you :)

  75. misty1212
    • one year ago
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    on the interval from \(-5\) to \(1\) the function is a constant, it is \(f(x)=3\) a horizontal line

  76. anonymous
    • one year ago
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    yes

  77. misty1212
    • one year ago
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    so the second line down on the left you should have \(3,-5\leq x\leq 1\) because on that interval it is 3

  78. anonymous
    • one year ago
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    i understand

  79. anonymous
    • one year ago
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    oh okay

  80. misty1212
    • one year ago
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    the next line has slope \(-2\) and goes through \((1,3)\) so its equation is \[y=y-3=-2(x-1)\] in point slope form

  81. anonymous
    • one year ago
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    would the slope not be a 0?

  82. misty1212
    • one year ago
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    yes, the slope is zero, the equation is \(y=3\)

  83. anonymous
    • one year ago
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    okay i got it

  84. anonymous
    • one year ago
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    okay i got it

  85. misty1212
    • one year ago
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    \[f(x) = \left\{\begin{array}{rcc} \frac{1}{2}(x+5)+3 & \text{if} & x <-5 \\ 3& \text{if} & -5\leq x\leq 1\\ -2(x-1)+3 &\text{if}&1<x<6 \end{array} \right.\]

  86. misty1212
    • one year ago
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    one more to go

  87. anonymous
    • one year ago
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    okay i got it

  88. misty1212
    • one year ago
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    last one also has slope \(\frac{1}{2}\) right?

  89. anonymous
    • one year ago
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    yes it does

  90. misty1212
    • one year ago
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    and goes through the point \((6,-7\)?

  91. anonymous
    • one year ago
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    yes i think so

  92. misty1212
    • one year ago
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    \[f(x) = \left\{\begin{array}{rcc} \frac{1}{2}(x+5)+3 & \text{if} & x <-5 \\ 3& \text{if} & -5\leq x\leq 1\\ -2(x-1)+3 &\text{if}&1<x<6\\ \frac{1}{2}(x-6)-7&\text{if}&x\geq6 \end{array} \right.\]

  93. misty1212
    • one year ago
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    that should do it

  94. anonymous
    • one year ago
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    yes i got it i just have a couple of small questions that i wanted to confirm

  95. misty1212
    • one year ago
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    ok

  96. anonymous
    • one year ago
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    are the x intercepts (5/2,0) and (-11,0)?

  97. misty1212
    • one year ago
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    yes, but there is one more

  98. misty1212
    • one year ago
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    set \[\frac{1}{2}(x-7)-7=0\] for to find the last one

  99. misty1212
    • one year ago
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    aggh!!

  100. misty1212
    • one year ago
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    \[\frac{1}{2}(x-6)-7=0\]

  101. misty1212
    • one year ago
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    that is the part on the right think you get \(20\) if i did it right

  102. anonymous
    • one year ago
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    i was solving it one sec ill confirm

  103. anonymous
    • one year ago
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    yes i got 20

  104. anonymous
    • one year ago
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    so itll be (20,0)

  105. misty1212
    • one year ago
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    yeah

  106. anonymous
    • one year ago
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    okay and the range is (-infinity, infintiy)?

  107. misty1212
    • one year ago
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    yeah that is what the arrows indicate

  108. anonymous
    • one year ago
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    and the positives are (-11, 5/2)?

  109. anonymous
    • one year ago
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    thast the last one haha

  110. misty1212
    • one year ago
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    i don't know what that means

  111. misty1212
    • one year ago
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    oh you mean the interval over which it is positive ?

  112. anonymous
    • one year ago
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    the positives as in the parts in the upper part of the graph

  113. anonymous
    • one year ago
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    yes

  114. misty1212
    • one year ago
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    \[(-11,\frac{5}{2})\] is one interval but there is another

  115. misty1212
    • one year ago
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    since it is zero at 20, and it keeps going up, the other one is \((20,\infty)\)

  116. anonymous
    • one year ago
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    alrighty thank you so much

  117. anonymous
    • one year ago
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    im just putting the functions in slope intercept form now and the doamins in interval notation

  118. anonymous
    • one year ago
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    ill be done in a sec if you have time to check those

  119. misty1212
    • one year ago
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    ok good sure

  120. anonymous
    • one year ago
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    y= 1/2x+11/2

  121. anonymous
    • one year ago
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    y=3 y=-2x+5 y=1/2x-10

  122. misty1212
    • one year ago
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    slope intercept for the first one yes

  123. anonymous
    • one year ago
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    yes i am doing slope intercept for all of them

  124. misty1212
    • one year ago
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    ok let me scroll way up and see i know the first two are right

  125. misty1212
    • one year ago
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    third is right too

  126. anonymous
    • one year ago
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    okay annddd the last one

  127. misty1212
    • one year ago
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    yeah that also looks good

  128. anonymous
    • one year ago
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    okay and the interval notation is (infinity, -5)? for the first one?

  129. anonymous
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    [-5,-1] (1,6) (infinity, 6]

  130. misty1212
    • one year ago
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    ooh mistake there

  131. misty1212
    • one year ago
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    second interval should be \([-5,1]\)

  132. misty1212
    • one year ago
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    and also do not put infinity on the left on the last interval!!

  133. anonymous
    • one year ago
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    i think thats what i said for the second one

  134. misty1212
    • one year ago
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    should be on the right, go from least to greatest i.e. \((6,\infty)\)

  135. anonymous
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    is the first interbal correct?

  136. misty1212
    • one year ago
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    \((-\infty,-5)\) yes

  137. anonymous
    • one year ago
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    and for the last one do i use a bracket before 6 since its greater than or equal to

  138. misty1212
    • one year ago
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    since it meets up there it is your choice

  139. misty1212
    • one year ago
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    btw usually these piecewise functions are written with inequalities not intervals but if you are supposed to use intervals that is fine too

  140. anonymous
    • one year ago
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    yeah it asked us to write it nboth wwyas

  141. anonymous
    • one year ago
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    do you think i should use a bracket?

  142. misty1212
    • one year ago
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    kk makes no difference

  143. anonymous
    • one year ago
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    alrigh thank you :)

  144. misty1212
    • one year ago
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    yw ms kumar lol \[\huge \color\magenta\heartsuit\]

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