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anonymous

  • one year ago

find the center and radius of the circle x^2+2x-4y+6=0 will FAN and MEDAL

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  1. mathstudent55
    • one year ago
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    This is not a circle. Are you sure you copied the equation correctly? You need a y^2 term.

  2. anonymous
    • one year ago
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    @mathstudent55 That is literally whats on my test. The answer options are: center (-1,2) radius :11 center (-1,2) radius: sq root of 11 center (1,-2) radius: 11 center (1,-2) radius sq root of 11

  3. Mertsj
    • one year ago
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    Well then let's work backwards from the answers and see which answer works.

  4. Mertsj
    • one year ago
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    \[(x+1)^2+(y-2)^2=11^2\]

  5. Mertsj
    • one year ago
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    \[x^2+2x+1+y^2-4y+4=121\]

  6. Mertsj
    • one year ago
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    \[x^2+2x+y^2-4y=116\]

  7. Mertsj
    • one year ago
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    Doesn't seem right, does it?

  8. anonymous
    • one year ago
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    no

  9. anonymous
    • one year ago
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    I am confused I kind of get what you did but I don't know how to continue it

  10. Mertsj
    • one year ago
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    The next choice has the same center but a different radius. But the radius should be sqrt116. So it doesn't seem to work either.

  11. Mertsj
    • one year ago
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    So try the third choice.

  12. Mertsj
    • one year ago
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    No. It couldn't be (x-1)^2 because that would be -2x

  13. Mertsj
    • one year ago
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    So of the choices given it has to be the first or the second one. And we know it isn't the first one.

  14. Mertsj
    • one year ago
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    Let's look at choice two more closely. I bet it will work.

  15. anonymous
    • one year ago
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    ok! after could you help me with another one? it's the same but the equation is x^2+8x-6y+9=0 options are center (4,-3) radius sq root of 34 center (-4,3) radius sq root of 34 center (4,-3) radius 34 center (-4,3) radius 34

  16. Mertsj
    • one year ago
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    \[(x-1)^2+(y+2)^2=(\sqrt{11})^2\] \[x^2-2x+1+y^2+4y+4=11\] \[x^2-2x+y^2+4y+5=11\] \[x^2-2x+y^2+4y=6\]

  17. Mertsj
    • one year ago
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    The question is obviously flawed if you have posted it correctly but since it is multiple choice, b is the best shot you have.

  18. anonymous
    • one year ago
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    thankyou! I've had so much trouble with that question!

  19. Mertsj
    • one year ago
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    Something is very wrong. Are you sure you posted the question EXACTLY as stated?

  20. anonymous
    • one year ago
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    yes! I can post a screenshot if you'd like. I just don't know how to upload a picture

  21. Mertsj
    • one year ago
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    That would be good.

  22. anonymous
    • one year ago
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    do you know how I can upload a pic?

  23. Mertsj
    • one year ago
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    All circles have x^2 and y^2 in their equations.

  24. anonymous
    • one year ago
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    yeah that's what I thought but this question is really weird and theres a lot similiar to these on my test

  25. Mertsj
    • one year ago
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    I have never done that on this site.

  26. anonymous
    • one year ago
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    I've seen other people do it...hmmm let me upload the pic to a site and then share the link

  27. Mertsj
    • one year ago
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    ok

  28. Mertsj
    • one year ago
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    It hasn't finished loading yet. I should warn you that my internet often goes down at 11 for some strange reason the Frontier can't seem to figure out. But the way the problem is posted, it has to be either choice b or .

  29. anonymous
    • one year ago
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    Oh aw alright! What time is it rn where ur at? I chose answer b but I haven't submitted it yet

  30. Mertsj
    • one year ago
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    I see what you mean. For the second problem, let's try choice b. 10:52

  31. mathstudent55
    • one year ago
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    The equation of a circle must have both the x-term and the y-term squared. The equation you were given is not a circle. I think they left out the y squared term.

  32. anonymous
    • one year ago
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    aw man thats soon :( oh well...... wait for the second problem? so not the first one right? @mathstudent55 there are many questions like this on my test :(

  33. Mertsj
    • one year ago
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    \[(x+4)^2+(y-3)^2=(\sqrt{34})^2\] \[x^2+8x+16+y^2-6y+9=34\] \[x^2+8x+y^2-6y=9\]

  34. anonymous
    • one year ago
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    so b for that one?

  35. Mertsj
    • one year ago
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    I would choose b and then I would point out to my teacher that these equations cannot be circles.

  36. anonymous
    • one year ago
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    Oh okay! Thankyou so much for your time and help!

  37. Mertsj
    • one year ago
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    yw

  38. anonymous
    • one year ago
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    do you know how to award a medal ? I'm not really sure how to haha

  39. anonymous
    • one year ago
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    nvm I think i got it

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