anonymous
  • anonymous
Help! (k^2+8k+12) / (k^2 + 15k +54) * (k^2 +9k) / (k^2 +7k +10)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
multiplying rational expressions
Nnesha
  • Nnesha
1st) factor all quadratic equations
Nnesha
  • Nnesha
\[\large\rm \frac{ k^2+8k+12 } { 9k^2 + 15k +54} * \frac{k^2 +9k} { k^2 +7k +10}\] factor all qudratic equation so you can cancel out some stuff do you know how to factor quadratic equations ?

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anonymous
  • anonymous
I got k/k+5 is that right?
anonymous
  • anonymous
wait can you explain what happens to this
anonymous
  • anonymous
|dw:1433974987628:dw|
anonymous
  • anonymous
@Nnesha @ganeshie8
Nnesha
  • Nnesha
what's the common factor k^2 + 9k ???
Nnesha
  • Nnesha
what is common in those both terms ?
anonymous
  • anonymous
They cross out once factored
Nnesha
  • Nnesha
alright k is common k^2 can be written as k times k \[\huge\rm k \times \color{ReD}{k} + 9\color{Red}{k}\] k is common so it out \[\large\rm \color{ReD}{k}(k+9)\]
Nnesha
  • Nnesha
\[\large\rm \frac{ k^2+8k+12 } { 9k^2 + 15k +54} * \frac{\color{Red}{k(k+9)}} { k^2 +7k +10}\] now can you factor quadratic equations k^2 +8k +12 ?
anonymous
  • anonymous
yes give me a sec
Nnesha
  • Nnesha
alright :-)
anonymous
  • anonymous
(k+6) (k+2) /(k+6) (k+9) * k (x+9) / (k+5) (k+2)
Nnesha
  • Nnesha
GREAT job!
Nnesha
  • Nnesha
i made a typo .-. :P \[\large\rm \frac{ k^2+8k+12 } { k^2 + 15k +54} * \frac{\color{Red}{k(k+9)}} { k^2 +7k +10}\] here is it k^2 +15k +54 ?? denominator of first fraction ?
Nnesha
  • Nnesha
yes it's i typed 9 instead of ( <-- parentheses sorry about that
Nnesha
  • Nnesha
\[ \huge\rm \frac{(k+6) (k+2)}{(k+6) (k+9) }* \frac{ k (x+9) }{ (k+5) (k+2)} \] cancel!
anonymous
  • anonymous
ok so k/x=+5
anonymous
  • anonymous
k/x+5
anonymous
  • anonymous
thanlks!!!
Nnesha
  • Nnesha
\[ \huge\rm \frac{\cancel{(k+6)} \cancel{ (k+2)}}{\cancel{(k+6)} \cancel{\cancel{(k+9)}} }* \frac{ k \cancel{(x+9)} }{ (k+5) \cancel{(k+2)}} \] \[\large\rm \frac{ k }{ k+2 }\] yep right!
Nnesha
  • Nnesha
my pleasure!! :-)
anonymous
  • anonymous
yay! thanks! super helpful
Nnesha
  • Nnesha
thanks great work!

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