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anonymous

  • 5 years ago

\[{ 1 \over R} = {1 \over R1} + {1 \over R2} \]

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  1. anonymous
    • 5 years ago
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    Solve the equation for the indicated variable.

  2. anonymous
    • 5 years ago
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    guess it depends on the indicated variable.

  3. anonymous
    • 5 years ago
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    \[{1 \over R}\] ? lol

  4. anonymous
    • 5 years ago
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    are you solving for \[R\] \[R_1\] or \[R_2\]?

  5. anonymous
    • 5 years ago
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    no. i'm sorry. it's R1

  6. anonymous
    • 5 years ago
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    whew. ok

  7. anonymous
    • 5 years ago
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    haha. work with me satellite. :p

  8. anonymous
    • 5 years ago
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    subtract \[\frac{1}{R_2}\] from both sides, take the reciprocal

  9. anonymous
    • 5 years ago
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    \[\frac{1}{R}-\frac{1}{R_2}=\frac{1}{R_1}\]

  10. anonymous
    • 5 years ago
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    \[{1 \over R} - {1 \over R2} = {1 \over R1} ? \]

  11. anonymous
    • 5 years ago
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    \[\frac{R_2-R}{RR_2}=\frac{1}{R_1}\]

  12. anonymous
    • 5 years ago
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    haha. same thing! this is good!

  13. anonymous
    • 5 years ago
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    now flip everything to get \[\frac{RR_2}{R_2-R}=R_1\]

  14. anonymous
    • 5 years ago
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    where did you come up with the last one?

  15. anonymous
    • 5 years ago
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    the one before the last... not flipping but the previous one

  16. anonymous
    • 5 years ago
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    oh i had 1/R and wanted R. just flip it

  17. anonymous
    • 5 years ago
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    oh previous one. ok hold on

  18. anonymous
    • 5 years ago
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    \[\frac{1}{a}-\frac{1}{b}=\frac{b}{ab}-\frac{a}{ab}\] yes? common denominator is ab and i have to build up the fractions

  19. anonymous
    • 5 years ago
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    then you subtract to get \[\frac{b-a}{ab}\]

  20. anonymous
    • 5 years ago
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    clear?

  21. anonymous
    • 5 years ago
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    no. give me a second. i'm going to write it...

  22. anonymous
    • 5 years ago
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    ok if not clear let me know. think of \[\frac{1}{5}-\frac{1}{3}\]

  23. anonymous
    • 5 years ago
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    forgot to close your tag

  24. anonymous
    • 5 years ago
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    this: \[\frac{1}{a}-\frac{1}{b}=\frac{b}{ab}-\frac{a}{ab} \] is important for me to remember. i am taking note of it. i think it's just that i need to understand this before being able to do these kind of problems. does that sound right from what you are seeing with me?

  25. anonymous
    • 5 years ago
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    but i understood

  26. anonymous
    • 5 years ago
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    yes, but don't get married to that formula. to add and subtract fraction it is always \[\frac{a}{bb}\pm\frac{c}{d}=\frac{ad\pm bc}{bd}\]

  27. anonymous
    • 5 years ago
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    thanks for confusing me more :p

  28. anonymous
    • 5 years ago
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    just in this case the numerators happened to be 1!

  29. anonymous
    • 5 years ago
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    that should really not confuse you much. how do you add \[\frac{2}{3}+\frac{5}{7}\]? it is just \[\frac{2\times 7 + 3\times 5}{3\times 7}\]

  30. anonymous
    • 5 years ago
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    i would rather find the common denominator for the fraction with numbers lol

  31. anonymous
    • 5 years ago
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    but you are showing me a more advanced way. i believe i must learn this way to be succesful

  32. anonymous
    • 5 years ago
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    in your case it was \[\frac{1}{R}-\frac{1}{R_2}=\frac{1\times R_2-1\times R}{R\times R_2}\]

  33. anonymous
    • 5 years ago
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    so the 1's were eliminated.... then we are left with variables

  34. anonymous
    • 5 years ago
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    wtf. that's the easiest way to add fractions that i have seen.

  35. anonymous
    • 5 years ago
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    thank you again! so the: \[\frac{a}{bb}\pm\frac{c}{d}=\frac{ad\pm bc}{bd} \] applies to all fractions?

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