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anonymous

  • one year ago

hello i need help in math can anyone help me.

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  1. anonymous
    • one year ago
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    sure

  2. anonymous
    • one year ago
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    thanks

  3. anonymous
    • one year ago
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    ok here it is

  4. anonymous
    • one year ago
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    also i have a few is that ok

  5. anonymous
    • one year ago
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    I need to see it to see if i am able to help

  6. anonymous
    • one year ago
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    ok

  7. anonymous
    • one year ago
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    Line M is represented by the following equation: x + y = −1 What is most likely the equation for line P so the set of equations has infinitely many solutions? 2x + 2y = 2 2x + 2y = 4 2x + 2y = −2 x − y = 1

  8. anonymous
    • one year ago
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    also i only have 5 questions ok

  9. anonymous
    • one year ago
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    i will give medal after

  10. anonymous
    • one year ago
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    even if i get some wrong just because you helped me i will give medal at the end

  11. anonymous
    • one year ago
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    hello are you still there

  12. zepdrix
    • one year ago
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    You want to find a line that is a parallel to the given line. Mechanically what we'll be doing is, Taking our line \(\large\rm x + y = −1 \) Adding each of the options to it, and seeing which one gives us a result of 0=0. Example: I'll add our line and the 4th option:|dw:1440538891099:dw|This gave us 2x=0 which will lead to a `unique solution` of x=0. This is a single solution.

  13. zepdrix
    • one year ago
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    So this is no bueno, we want infinitely many solutions, this only happens when both variables disappear in the operation.

  14. anonymous
    • one year ago
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    so this is a single soulution

  15. zepdrix
    • one year ago
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    I'm saying that the "solution" or "intersection" of these two lines is at a single place, so it is clearly the wrong answer for us.

  16. anonymous
    • one year ago
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    but thats not what we want we want infinate soulutions correct

  17. zepdrix
    • one year ago
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    yes, correct :)

  18. anonymous
    • one year ago
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    ok got it

  19. zepdrix
    • one year ago
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    How bout we try the first one:|dw:1440539063251:dw|

  20. anonymous
    • one year ago
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    ok

  21. zepdrix
    • one year ago
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    We need the coefficients to match up, so what can we do to the first equation?

  22. anonymous
    • one year ago
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    im not sure im realy bad at math do you think you can give me like a hint to understand how to find the coefficients

  23. zepdrix
    • one year ago
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    Let's just pay attention to the x's. I want the x's to disappear through some type of addition. I have 2x's in the bottom so I'm going to need 2x's in the top equation as well. So let's multiply the entire top equation by 2.

  24. anonymous
    • one year ago
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    ok i think i understand were your going with this

  25. zepdrix
    • one year ago
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    |dw:1440539255066:dw|Gives us something like this, ya?

  26. zepdrix
    • one year ago
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    So um

  27. anonymous
    • one year ago
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    ok so we added the 2 right

  28. anonymous
    • one year ago
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    and then mutyplye

  29. zepdrix
    • one year ago
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    We `multiplied` by 2! :O not added. I guess in this case, in order to make the x's disappear, we'll need to `subtract` the bottom equation, not add it.

  30. anonymous
    • one year ago
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    ok

  31. zepdrix
    • one year ago
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    |dw:1440539392434:dw|Careful though, you have to subtract everything.

  32. zepdrix
    • one year ago
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    So what dare you left with? Try to deal with each column. 2x - 2x?

  33. anonymous
    • one year ago
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    would that make everything 0

  34. zepdrix
    • one year ago
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    2x - 2x = 0 2y - 2y = 0 The left side is clearly 0, good. -2 - 2 = ?

  35. anonymous
    • one year ago
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    so then the answer would be 0

  36. zepdrix
    • one year ago
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    Noooo :O Check that last line again carefully!

  37. anonymous
    • one year ago
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    ok sorry

  38. zepdrix
    • one year ago
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    \(\large\rm -2-2\ne0\)

  39. anonymous
    • one year ago
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    ok so 0

  40. anonymous
    • one year ago
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    lol im sorry im super confused

  41. zepdrix
    • one year ago
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    If you're at negative 2, and you take 2 away from that, you become even more negative.

  42. anonymous
    • one year ago
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    wait would it be 4

  43. anonymous
    • one year ago
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    -4

  44. zepdrix
    • one year ago
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    Ok good! So we end up with this result: \(\large\rm 0=-4\) Which is `false`. Therefore the system containing these two lines has NO solution. So again, this is not the one we're looking for. Your princess is in another castle :o

  45. anonymous
    • one year ago
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    ok lol so now on to the other aswer right

  46. zepdrix
    • one year ago
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    Let's check out the 3rd option: x + y = -1 2x + 2y = −2

  47. anonymous
    • one year ago
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    also im sorry math is the only thing i realy suck at

  48. anonymous
    • one year ago
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    ok

  49. anonymous
    • one year ago
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    so would we do the same thing we did last time

  50. zepdrix
    • one year ago
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    Yah, that seems like a good idea.

  51. anonymous
    • one year ago
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    ok

  52. zepdrix
    • one year ago
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    Again, we multiply by 2 to match up the x's. 2(x + y = -1) 2x + 2y = −2 Giving us: 2x + 2y = -2 2x + 2y = −2

  53. zepdrix
    • one year ago
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    And then do some subtraction or addition to combine them. What are we gonna get this time? :o

  54. anonymous
    • one year ago
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    umm the end parts on both sides are nagative so those be 0s

  55. zepdrix
    • one year ago
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    Well notice that both equations are exactly the same now, ya? What happens when you subtract something from itself? You're gonna get 0. So good, in this case we end up with 0=0. So this system of two lines gives us infinitely many solutions!

  56. anonymous
    • one year ago
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    so thats the answer

  57. zepdrix
    • one year ago
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    yay team \c:/ we got it!

  58. anonymous
    • one year ago
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    yayaya

  59. anonymous
    • one year ago
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    also wich question was this

  60. zepdrix
    • one year ago
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    The only one that you asked :D lol I dunno, you didn't number it :d

  61. anonymous
    • one year ago
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    no i mean the question to the question we just worked on is it x + y = -1 2x + 2y = −2

  62. zepdrix
    • one year ago
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    Line M was `given` to be x+y=-1. And our `answer` that we obtained for line P is the third one, 2x+2y=-2

  63. anonymous
    • one year ago
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    ok cool next question! :)

  64. anonymous
    • one year ago
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    Line Q is represented by the following equation: 2x + y = 11 Which equation completes the system that is satisfied by the solution (3, 5)? x + 2y = 15 x + y = 15 x + y = 8 x − y = 2

  65. anonymous
    • one year ago
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    i think its B or A

  66. zepdrix
    • one year ago
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    I gotta go make some foods. If you need further help, close this and open up another thread. This one is getting too long and messy. You can use the @ symbol to page someone for help. Example @zepdrix The people at the top of the lobby are usually a bunch of smarty pants. Include your question in title, it will help to get it answered faster. Don't just call it "need help with math". Just a little tip :) I'll try to come by and help when I can :) But the belly is rumblin at the moment!

  67. anonymous
    • one year ago
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    ok thaks

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