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jaylelile

  • one year ago

I'm having a hard time remembering how to go about this problem. help? Problem below

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  1. jaylelile
    • one year ago
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    \[\frac{ 4}{ 8 }^{2}b+\frac{ 40 }{ 32 }^{3}b\]

  2. jaylelile
    • one year ago
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    @Nnesha ?

  3. Nnesha
    • one year ago
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    well the variables are the same so they are like terms combine them

  4. Nnesha
    • one year ago
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    here is an example when the variables and the exponents are the same we can add/subtract their coefficients \[\rm 2x^2+5x^2 = (2+5)x^2=7x^2\]

  5. jaylelile
    • one year ago
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    \[\frac{ 44}{ 40 }^{3}b ?????\]

  6. Nnesha
    • one year ago
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    hmm we didn't add them correctly :=) \[\rm \frac{ 4}{ 8 }^{2}b+\frac{ 40 }{ 32 }^{3}b=(\frac{4^2}{8}+\frac{40^3}{32})b\]

  7. jaylelile
    • one year ago
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    that seems too simple though.. lol

  8. Nnesha
    • one year ago
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    \[\rm \frac{ 4}{ 8 }^{2}b+\frac{ 40 }{ 32 }^{3}b=(\frac{4^2}{8}+\frac{40^3}{32})b\] \[\rm (\color{ReD}{\frac{4^2}{8}+\frac{40^3}{32}})b\] solve paretheses what's the common denominator ??

  9. jaylelile
    • one year ago
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    do we add the exponents?

  10. jaylelile
    • one year ago
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    I'm so confused

  11. Nnesha
    • one year ago
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    no when we `multiply` same bases then we should add their exponents when we combine like terms base stay the same we just have to add/subtract their `coefficients `

  12. jaylelile
    • one year ago
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    I'm still getting \[\frac{ 44 }{ 40 }^{3}b\] ..........

  13. Nnesha
    • one year ago
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    alright show some work i mean what was your step how did you get 40 at the denominaotr ?

  14. Nnesha
    • one year ago
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    i see what you did there common mistake

  15. jaylelile
    • one year ago
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    I just added 8 and 32... is that wrong?

  16. Nnesha
    • one year ago
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    \[\frac{ 3 }{ 2} + \frac{4}{6}\] |dw:1444530347581:dw|wrong that's not how we should add

  17. jaylelile
    • one year ago
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    then how do we add? I'm lost

  18. Nnesha
    • one year ago
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    find common denominator then multiply the numerator of first fraction by the denominator of 2nd fraction multiply the numerator of 2nd fraction by the denominator of first fraction here is an example \[\huge\rm \frac{ a } {\color{Red}{ b} }+\frac{ c }{\color{blue}{ d} } =\frac{ a\color{blue}{d}+c\color{Red}{b} }{ bd}\]

  19. Nnesha
    • one year ago
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    to make it easy write factors of 8 and 32 8 = 1 ,2 , 4 , 8 32=1 ,2 ,4 ,8 ,16 ,32 what is the common number ??

  20. jaylelile
    • one year ago
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    1,2, 4 and 8

  21. Nnesha
    • one year ago
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    brb i need to refresh the page =.=

  22. Nnesha
    • one year ago
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    alright sorry i made a mistake there

  23. Nnesha
    • one year ago
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    when denominators aren't the same we should multiply them so multiply 8 times 32 that would be the common denominator

  24. Nnesha
    • one year ago
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    we don't need factors sorry about that

  25. Nnesha
    • one year ago
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    here is an example \[\huge\rm \frac{ a } {\color{Red}{ b} }+\frac{ c }{\color{blue}{ d} } =\frac{ a\color{blue}{d}+c\color{Red}{b} }{ bd}\] use this example \[\frac{ 4^2 } {\color{Red}{ 8} }+\frac{ 40^3}{\color{blue}{ 32} } \]

  26. jaylelile
    • one year ago
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    so \[\frac{ 44 }{ 256 }^{3}b\] ?????

  27. Nnesha
    • one year ago
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    hmm no use the example you can't just add their numerator if the denominators are the same THEN you can just add the numerators \[\frac{3}{\color{Red}{4}}+\frac{5}{\color{blue}{4}} = \frac{3+5}{4}\] in this example both denominators are the same so u can add their exponents

  28. Nnesha
    • one year ago
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    but when the denominators arn't the same multiply the numerator of first fraction by the denominator of 2nd fraction multiply the numerator of 2nd fraction by the denominator of first fraction here is an example \[\huge\rm \frac{ a } {\color{Red}{ b} }+\frac{ c }{\color{blue}{ d} } =\frac{ a\color{blue}{d}+c\color{Red}{b} }{ bd}\] same like cross multiplications |dw:1444531460579:dw|

  29. Nnesha
    • one year ago
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    make sense hmm ?

  30. Nnesha
    • one year ago
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    btw you can reduce the fraction before adding them but well let's stik with it

  31. anonymous
    • one year ago
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    \[\frac{ 4^2 }{ 8 }+\frac{ 40^3 }{ 32 }=\frac{ 4^3+40^3 }{ 32 }=\frac{ 4^3\left( 1^3+10^3 \right) }{ 32 }\] \[=\frac{ 64\left( 1+1000 \right) }{ 32 }=2\left( 1+1000 \right)=?\]

  32. Nnesha
    • one year ago
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    gtg in few mints so i'll just post this here just like the example i gave u \[\huge\rm \frac{4^2} {\color{Red}{ 8} }+\frac{ 40^3}{\color{blue}{ 32} } =\frac{ 4^2\color{blue}{(32)}+40^3\color{Red}{(8)} }{ 256}\] now you can simplify it

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