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amy0799

  • one year ago

Find k such that the line y = 2x + 18 is tangent to the graph of the function below. y=k√x k=?

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  1. Loser66
    • one year ago
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    I told you, take derivative of the curve \(y = k\sqrt x\), what do you get?

  2. amy0799
    • one year ago
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    1/2k^-1/2

  3. Loser66
    • one year ago
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    |dw:1440973881338:dw|

  4. freckles
    • one year ago
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    |dw:1440974118718:dw| you want y=2x+8 and y=ksqrt(x) to have a common point (a,f(a)) and you also want the derivative of 2x+8 to be equal to the derivative of ksqrt(x) at (a,f(a))

  5. amy0799
    • one year ago
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    right, I understand that, I just don't know how to find it.

  6. freckles
    • one year ago
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    you have 2a+8=k sqrt(a) <--since we want 2a+8 to be equal to k sqrt(a) and you also have 2=k/(2sqrt(a))

  7. Loser66
    • one year ago
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    Thanks @freckles. Shame on me, I guided him wrong. @amy0799 I am sorry.

  8. freckles
    • one year ago
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    it is just a system of equations to solve

  9. amy0799
    • one year ago
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    \[\frac{ 2x+8 }{ \sqrt{x}} = k\]

  10. amy0799
    • one year ago
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    what do I do next?

  11. freckles
    • one year ago
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    \[2a+8=k \sqrt{a} \\ 2=\frac{k}{2 \sqrt{a}} \implies \sqrt{a}=\frac{k}{4} \implies a=\frac{k^2}{16} \text{ insert into first equation }\]

  12. freckles
    • one year ago
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    doing that will give you an equation just in terms of k

  13. amy0799
    • one year ago
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    insert it into which equation?

  14. freckles
    • one year ago
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    the First equation

  15. freckles
    • one year ago
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    the first equation I wrote above the second equation

  16. freckles
    • one year ago
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    \[2a+8=k \sqrt{a} \\ 2=\frac{k}{2 \sqrt{a}} \implies \sqrt{a}=\frac{k}{4} \implies a=\frac{k^2}{16} \text{ insert into first equation } \] there are equations I'm playing with here the first equation which is on the first line and the second equation which on the second line the second line I used the second equation to express sqrt(a) in terms o k and I also expressed a in terms of k using the second equation we can now write the first equation which I put on the first line in terms of k

  17. amy0799
    • one year ago
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    so I plug a=k^2/16 into 2a+8=ksqrt(a)?

  18. freckles
    • one year ago
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    yes you also plug in k/4 for sqrt(a)

  19. amy0799
    • one year ago
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    \[2(\frac{ k^{2} }{ 16 })+8=k \sqrt{k \sqrt{a}}\] this doesn't make sense

  20. freckles
    • one year ago
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    you are write that equation doesn't make sense you should have gotten something else entirely different

  21. freckles
    • one year ago
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    \[2a+8=k \sqrt{a} \\ \text{ we determined } \sqrt{a}=\frac{k}{4} \text{ then we squared both sides \to get } a=\frac{k^2}{16} \\ \text{ plug \in both of these results we obtained from the second equation } \\ \text{ into the first } \\ 2(\frac{k^2}{16})+8=k(\frac{k}{4})\]

  22. freckles
    • one year ago
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    now solve the equation for k (you will see you have to discard a value that you get for k when solving this equation because of the equation sqrt(a)=k/4 which tells us that sqrt(a) is positive and so k has to be positive )

  23. freckles
    • one year ago
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    do you know how to solve the above quadratic

  24. amy0799
    • one year ago
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    not really

  25. freckles
    • one year ago
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    k*k=k^2 and 2/16=1/8 you have: \[\frac{k^2}{8}+8=\frac{k^2}{4} \\ \text{ you can multiply both sides by 8 \to get } \\ k^2+64=2k^2 \] can you solve from here?

  26. amy0799
    • one year ago
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    k=8?

  27. freckles
    • one year ago
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    \[64=k^2 \\ k=8 \text{ or } -8 \\ \text{ but we did mentioned that } k>0 \\ \text{ so yes } k=8\]

  28. amy0799
    • one year ago
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    thank you! can u help me with another one?

  29. freckles
    • one year ago
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    i can try yes

  30. freckles
    • one year ago
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    but

  31. freckles
    • one year ago
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    before we go on you need to understand this one

  32. freckles
    • one year ago
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    like do you have any questions? do you understand how I solved the system of equations?

  33. amy0799
    • one year ago
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    yea I understood that

  34. amy0799
    • one year ago
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    Consider the following function. f(t) = 2t^2 − 3 (a) Find the average rate of change of the function below over the interval [3, 3.1]. (b) Compare this average rate of change with the instantaneous rates of change at the endpoints of the interval. (at t = 3) (at t = 3.1)

  35. freckles
    • one year ago
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    \[\text{average rate of change of } f \text{ from }t=a \text{ to } t=b \\ \text{ is given by the formula } \frac{f(b)-f(a)}{b-a}\]

  36. freckles
    • one year ago
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    \[\text{ instantaneous rate of change of} f \text{ at } x=c \text{ is given by } f'(c)\]

  37. amy0799
    • one year ago
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    so what's the first step I need to do?

  38. freckles
    • one year ago
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    you used the formula above to find the average rate of change from t=3 to t=3.1...

  39. freckles
    • one year ago
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    replace b with 3.1 and a with 3

  40. amy0799
    • one year ago
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    \[\frac{ f(3.1)-f(3) }{ 3.1-3 }\]

  41. freckles
    • one year ago
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    yes now evaluate f(3.1) and f(3) and do the difference on top and bottom

  42. freckles
    • one year ago
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    and then the division of the top by the bottom

  43. amy0799
    • one year ago
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    I don't understand what you mean

  44. freckles
    • one year ago
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    f(t) is given as 2t^2-3 you are to find f(3.1) and f(3) by using f(t)=2t^2-3 replace t with 3.1 and get a result for f(3.1) replace t with 3 and get a result for f(3)

  45. amy0799
    • one year ago
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    f(3.1)=16.22 f(3)=15

  46. freckles
    • one year ago
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    \[\frac{f(3.1)-f(3)}{3.1-3}=\frac{16.22-15}{3.1-3}\] now find the difference for both top and bottom

  47. freckles
    • one year ago
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    the difference of 16.22 and 15 is another way to say 16.22-15 if you are confused by the word difference

  48. amy0799
    • one year ago
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    12.2

  49. amy0799
    • one year ago
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    is that the average rate of change?

  50. freckles
    • one year ago
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    yes that is the average rate of change of f(t)=2t^2-3 from t=3 to t=3.1

  51. amy0799
    • one year ago
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    I got part b wrong

  52. freckles
    • one year ago
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    did you find the f'(3) and f'(3.1)

  53. amy0799
    • one year ago
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    yea isn't it 16.22 and 15?

  54. freckles
    • one year ago
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    no

  55. freckles
    • one year ago
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    f' means to find the derivative of f

  56. freckles
    • one year ago
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    and you just plugged into original function

  57. freckles
    • one year ago
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    did you not look earlier at how I said to find the instantaneous rate of change?

  58. freckles
    • one year ago
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    I posted the following two comments: \[\text{average rate of change of } f \text{ from }t=a \text{ to } t=b \\ \text{ is given by the formula } \frac{f(b)-f(a)}{b-a}\] \[\text{ instantaneous rate of change of} f \text{ at } x=c \text{ is given by } f'(c)\]

  59. amy0799
    • one year ago
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    I don't understand it

  60. freckles
    • one year ago
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    you don't know how to find the derivative of 2t^2-3?

  61. freckles
    • one year ago
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    do you know power rule and constant rule? if not use the definition of derivative

  62. amy0799
    • one year ago
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    4t.

  63. freckles
    • one year ago
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    yes so what do you not understand?

  64. freckles
    • one year ago
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    you have f'(t)=4t and you are asked to evaluate f'(t) at the endpoints

  65. freckles
    • one year ago
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    the endpoints are 3 and 3.1

  66. freckles
    • one year ago
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    f'(3)=? f'(3.1)=?

  67. amy0799
    • one year ago
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    12 and 12.4

  68. amy0799
    • one year ago
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    my very first question. I got k=8 wrong

  69. freckles
    • one year ago
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    did you look at what we wrote and what you asked?

  70. freckles
    • one year ago
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    you should see we changed 18 to 8

  71. freckles
    • one year ago
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    you know the line being y=2x+18 instead of y=2x+8

  72. amy0799
    • one year ago
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    is the answer not 8?

  73. freckles
    • one year ago
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    the answer is k=8 if the question involved y=2x+8 but the equation was y=2x+18 please look at our work and your question our answer was for if the line was y=2x+8 which I started over and over but the equation was y=2x+18 so just redo the work for the equation being y=2x+18 instead of it being y=2x+8

  74. freckles
    • one year ago
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    which I stated over and over in the answer*

  75. freckles
    • one year ago
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    do you understand?

  76. freckles
    • one year ago
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    the question was: Find k such that the line y = 2x + 18 is tangent to the graph of the function below. y=k√x k=? however our work was for: Find k such that the line y = 2x + 8 is tangent to the graph of the function below. y=k√x k=?

  77. freckles
    • one year ago
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    you should be able to do your question now if you actually understand the work we did to find k for this other question

  78. amy0799
    • one year ago
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    can you help me with it? @freckles

  79. freckles
    • one year ago
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    I thought you said you understand the process?

  80. freckles
    • one year ago
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    \[\text{ solve the system of equations } \\ (2x+18)'|_{x=a}=(k \sqrt{x})'|_{x=a} \\ 2a+18=k \sqrt{a}\]

  81. freckles
    • one year ago
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    it is the same process we did above

  82. freckles
    • one year ago
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    except there is just a little number change

  83. amy0799
    • one year ago
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    I understood how you solved the quadratic, but not how you got a

  84. freckles
    • one year ago
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    \[2a+8=k \sqrt{a} \\ 2=\frac{k}{2 \sqrt{a}} \implies \sqrt{a}=\frac{k}{4} \implies a=\frac{k^2}{16} \text{ insert into first equation } \] is this what you are talking about? you don't know how I solved the second equation for sqrt(a)?

  85. freckles
    • one year ago
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    \[2=\frac{k}{2 \sqrt{a}} \\ \text{ I just multiplied } 2 \sqrt{a} \text{ on both sides } 2(2 \sqrt{a})=k \\ 4 \sqrt{a}=k \\ \text{ and then divided both sides by } 4\]

  86. freckles
    • one year ago
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    and the only different and this set of equations is 8 is supposed to be 18

  87. freckles
    • one year ago
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    \[2a+18=k \sqrt{a} \\ \sqrt{a}=\frac{k}{4} \text{ squaring both sides gives } a=\frac{k^2}{16} \text{ replace the items in the first equation }\]

  88. amy0799
    • one year ago
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    k=18?

  89. freckles
    • one year ago
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    how did you get that?

  90. freckles
    • one year ago
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    did you replace a with k^2/16 and sqrt(a) with k/4?

  91. freckles
    • one year ago
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    \[2a+18=k \sqrt{a} \\ 2(\frac{k^2}{16})+18=k(\frac{k}{4})\]

  92. freckles
    • one year ago
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    solve the equation for k

  93. amy0799
    • one year ago
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    \[\frac{ 18k^{2} }{ 8 }+324=\frac{ 18k ^{2} }{ 4 }\] is this right so far?

  94. freckles
    • one year ago
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    why did you choose to multiply both sides by 18

  95. amy0799
    • one year ago
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    that's how u did it before

  96. freckles
    • one year ago
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    no I multiplied both sides by something to clear the fractions not to make the problem more fractionie (I know that isn't a word) \[\frac{2}{16}=\frac{1}{8} \\ k \cdot k=k^2 \\ \text{ you have } \\ 2(\frac{k^2}{16})+18=k(\frac{k}{4}) \\ \text{ so this is equivlant to } \\ \frac{k^2}{8}+18=\frac{k^2}{4} \\ \text{ if you multiply both sides by 8 } \text{ you can clear the fractions }\]

  97. freckles
    • one year ago
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    the denominators were same in previous previous problem and I chose to multiply both sides by 8 then so it shouldn't change here

  98. freckles
    • one year ago
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    unless you want to make the problem more complicated looking :p

  99. amy0799
    • one year ago
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    k=12

  100. freckles
    • one year ago
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    yes \[k^2+8(18)=2k^2 \\ 8(18)=k^2 \\ 16(9)=k^2 \\ 4(3)=k \\ k=12\] and again we chose 12 instead of -12 because we wanted k to be positive @amy0799 you might want to go and review how to solve algebraic equations practice, practice and you can get better at this I promise

  101. amy0799
    • one year ago
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    yeaa I know >.< I have one more I need help with

  102. freckles
    • one year ago
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    you can post it but brb

  103. freckles
    • one year ago
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    ok i'm here

  104. amy0799
    • one year ago
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    Consider the following function. \[f(x)=\sqrt{9-x^{2}}\] Find the derivative from the left at x = 3. If it does not exist, enter NONE. Find the derivative from the right at x = 3. If it does not exist, enter NONE.

  105. freckles
    • one year ago
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    do you know what the function looks like?

  106. freckles
    • one year ago
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    |dw:1440979836143:dw| does the function exist to the right of x=3?

  107. amy0799
    • one year ago
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    no

  108. freckles
    • one year ago
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    so we already know one answer is does not exist... now let's look at the left side: \[\text{ \left derivative of } f \text{ at } x=3 \text{ is given by } \lim_{x \rightarrow 3^-}\frac{f(x)-f(3)}{x-3} \\ \]

  109. amy0799
    • one year ago
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    it doesn't exist either

  110. freckles
    • one year ago
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    correct

  111. amy0799
    • one year ago
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    thank you so so much and I'm really sorry if I took up ur time!!

  112. freckles
    • one year ago
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    It is only fine if you take up my time if you have learned something

  113. freckles
    • one year ago
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    do you feel like you learned anything

  114. amy0799
    • one year ago
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    I learned a lot. thank you!

  115. freckles
    • one year ago
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    np :)

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