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chevygirl

  • 2 years ago

and again

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  1. chevygirl
    • 2 years ago
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  2. Harkirat
    • 2 years ago
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    Why don't u learn how to do them instead of asking for answers???

  3. phi
    • 2 years ago
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    This is the same problem as before. Just different numbers.

  4. chevygirl
    • 2 years ago
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    Well if you knew me good enough you would now that I don't learn real easy. It isn't that I am choosing not to learn I just don't get how to do it.

  5. skullpatrol
    • 2 years ago
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    |dw:1356563774476:dw|

  6. skullpatrol
    • 2 years ago
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    Multiply both sides by 7.

  7. chevygirl
    • 2 years ago
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    -35 and -7?

  8. skullpatrol
    • 2 years ago
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    |dw:1356563975056:dw|

  9. chevygirl
    • 2 years ago
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    Am I right?

  10. skullpatrol
    • 2 years ago
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    Add 5 to both sides.

  11. chevygirl
    • 2 years ago
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    -30 and -2?

  12. skullpatrol
    • 2 years ago
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    |dw:1356564176650:dw|

  13. phi
    • 2 years ago
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    no, when you multiply 7 times (z-5)/7 \[ \frac{7}{1}\cdot \frac{(z-5)}{7} \] you get \[\frac{7(z-5)}{7} \] or \[ \frac{7}{7}\cdot \frac{(z-5)}{1} \] 7/7 is 1 and (z-5)/1 is (z-5) multiplying by the 7 makes the denominator "go away"

  14. chevygirl
    • 2 years ago
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    OK????

  15. phi
    • 2 years ago
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    see skull's first post

  16. chevygirl
    • 2 years ago
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    what about it?

  17. phi
    • 2 years ago
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    you do not get -35 and -7? you get z-5= -7

  18. chevygirl
    • 2 years ago
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    ok. but never mind

  19. skullpatrol
    • 2 years ago
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    The answer is {-2} because -2 is the only number that will make z = -2 a true statement.

  20. skullpatrol
    • 2 years ago
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    -2 = -2

  21. chevygirl
    • 2 years ago
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    ok but the only reason I am asking these questions is because I dont get how to do them @Harkirat

  22. Harkirat
    • 2 years ago
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    @chevygirl Well ask your teacher to explain,in detail, how to solve such sums....u cannot go through life asking others to solve them for you......being able to do themselves will boost your confidence...

  23. chevygirl
    • 2 years ago
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    well if my teachers were here i would ask them but they arent here so I cant AND i got put in a different class because i got behind because i am not good at math

  24. skullpatrol
    • 2 years ago
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    |dw:1356564567196:dw|

  25. chevygirl
    • 2 years ago
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    ok. i will try to REMEMBER this

  26. skullpatrol
    • 2 years ago
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    Would you like me to explain how I solved the original equation?

  27. chevygirl
    • 2 years ago
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    um no thanks cause i have to do another test

  28. skullpatrol
    • 2 years ago
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    Is it a math test?

  29. chevygirl
    • 2 years ago
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    yeah

  30. skullpatrol
    • 2 years ago
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    If I explain this one you can use the method on most of your other questions.

  31. chevygirl
    • 2 years ago
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    yeah but i dont know if it is these type of questions or not

  32. skullpatrol
    • 2 years ago
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    There are a lot of these types of questions in math.

  33. skullpatrol
    • 2 years ago
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    What grade level?

  34. chevygirl
    • 2 years ago
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    7

  35. skullpatrol
    • 2 years ago
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    Do you know what an inverse operation is?

  36. chevygirl
    • 2 years ago
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    yeah

  37. skullpatrol
    • 2 years ago
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    inverse operations: Operations that "undo" each other for example multiplication and division.

  38. skullpatrol
    • 2 years ago
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    If the side containing the variable involves a certain order of operations, apply the inverse operations in the opposite order.

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