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anonymous

  • one year ago

PLEASE HELP WILL MEDAL AND FAN In the diagram below <ACB=6y-7 and M<DCA=8y-1 Write an equation to solve for y, then solve for y. Then give the measure of <ACB and <DCA in degrees. Diagram: https://imgur.com/KycHbxy

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  1. parker.goodbar
    • one year ago
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    I can't access the diagram

  2. michaelac959
    • one year ago
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    ok the equation would be 6y-7=8y-1

  3. anonymous
    • one year ago
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    Okay , how do I solve it?

  4. michaelac959
    • one year ago
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    now you combine like terms 14y-8

  5. michaelac959
    • one year ago
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    no 6 14y-6

  6. anonymous
    • one year ago
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    okay what next

  7. michaelac959
    • one year ago
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    can i get the answers please?

  8. michaelac959
    • one year ago
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    im gonna go back in my book because im doing this math im just gonna see how this goes one sec

  9. anonymous
    • one year ago
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    Um, how do I solve for Y, cause I didn't take Algebra.........

  10. michaelac959
    • one year ago
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    neither did i im on geometry now but i know how to do this

  11. anonymous
    • one year ago
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    yeah they just expect me to know this in my geometry class. How di I solve for Y from this.

  12. michaelac959
    • one year ago
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    r u in connections academy or k-12?

  13. anonymous
    • one year ago
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    k12 WAVA, u?

  14. michaelac959
    • one year ago
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    i just need to know cuz i went to both and have the books to help

  15. anonymous
    • one year ago
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    k12 WAVA OMAK HS

  16. anonymous
    • one year ago
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    ahh thats cool!

  17. michaelac959
    • one year ago
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    im in connections academy but i still have the book

  18. michaelac959
    • one year ago
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    my older sister had one

  19. anonymous
    • one year ago
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    I heard connections is better wish my parents lemme go

  20. anonymous
    • one year ago
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    Thanks!

  21. anonymous
    • one year ago
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    So would I do this |dw:1444322028651:dw|

  22. michaelac959
    • one year ago
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    is that how your book showed you how to do it?

  23. anonymous
    • one year ago
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    My book didn't tell me anything. The teacher just put this on the test without any explanation, I looked at everything

  24. anonymous
    • one year ago
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    I don't know any of this

  25. michaelac959
    • one year ago
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    ok so i will look this up because you cant do either of these without en equal sight the problem is acd=6y-7 but it need to be like 6y-7= a number

  26. michaelac959
    • one year ago
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    giv me the abcd answers if its multiple choice

  27. michaelac959
    • one year ago
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    if i have the answer i can work backward and show you

  28. anonymous
    • one year ago
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    I know! Thats how I need it to be to! It's not mulitple choice it's open ended.

  29. michaelac959
    • one year ago
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    how many questions r there?

  30. michaelac959
    • one year ago
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    cuz i can help you with the rest but i know i will look it up gimme 3 seconds

  31. anonymous
    • one year ago
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    its just this one I figured out all the rest....sorry

  32. michaelac959
    • one year ago
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    its fine i will get you the answer promice

  33. anonymous
    • one year ago
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    thanks alot!

  34. michaelac959
    • one year ago
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    are you adding the 2 numbers?

  35. anonymous
    • one year ago
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    yeah I calclend out the -6 with a +6 and added +6 to the 14

  36. michaelac959
    • one year ago
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    yes you are because you need a problem to solve for y

  37. michaelac959
    • one year ago
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    so it would be (6y-7) + (8y-1)

  38. michaelac959
    • one year ago
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    now combine like terms (14y-7)

  39. michaelac959
    • one year ago
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    minus 6

  40. michaelac959
    • one year ago
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    subtract and get y=8

  41. michaelac959
    • one year ago
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    ok i may be wrong but that is how the book im holding showed me to do it try that and if it is wrong i am sorry if it is right you now know how to do it so 50/50 chance im sorry i couldn't help you better

  42. anonymous
    • one year ago
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    thanks! ANy help is better than none! I like a 50/50 as versues a 0/100

  43. michaelac959
    • one year ago
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    any time just ask and i will try and help

  44. anonymous
    • one year ago
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    alrighty! :D

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