anonymous
  • anonymous
What is the simplified form of 4x^2-25/2x-5?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@Michele_Laino
Michele_Laino
  • Michele_Laino
hint: we can factorize the numerator as follows: \[\Large \frac{{4{x^2} - 25}}{{2x - 5}} = \frac{{\left( {2x - 5} \right)\left( {2x + 5} \right)}}{{2x - 5}} = ...?\]
anonymous
  • anonymous
2x+5

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Michele_Laino
  • Michele_Laino
since in general we have: \[\Large {a^2}{x^2} - {b^2} = \left( {ax - b} \right)\left( {ax + b} \right)\]
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
What would the restriction be?
Michele_Laino
  • Michele_Laino
since we can not divide by zero, then we have to exclude those values of x such that: \[\Large 2x - 5 = 0\]
anonymous
  • anonymous
So -5/2
Michele_Laino
  • Michele_Laino
we have to exclòude x=5/2
Michele_Laino
  • Michele_Laino
exclude*
anonymous
  • anonymous
So I was wrong when I said -5/2?
Michele_Laino
  • Michele_Laino
yes! since \[2x - 5 = 0\] when: x=5/2
Michele_Laino
  • Michele_Laino
please try to solve that equation: 2x-5=0
anonymous
  • anonymous
Ok! I see! So the answer would be 2x+5, with a restriction of 5/2
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
:)!!! May I ask another?
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
What is the simplifed form of |dw:1437058472772:dw|
Michele_Laino
  • Michele_Laino
here we can write your division, as a multiplication of the first fraction, times the inverse of the second fraction, like below: \[\Large \begin{gathered} \frac{{x + 1}}{{{x^2} + x - 6}}:\frac{{{x^2} + 5x + 4}}{{x - 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{{x^2} + x - 6}} \cdot \frac{{x - 2}}{{{x^2} + 5x + 4}} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
yes!
Michele_Laino
  • Michele_Laino
now we have to factorize the subsequent polynomials: \[\Large \begin{gathered} {x^2} + x - 6 \hfill \\ {x^2} + 5x + 4 \hfill \\ \end{gathered} \]
anonymous
  • anonymous
(x+2)(x-3) for the first one
Michele_Laino
  • Michele_Laino
ok! correct!
anonymous
  • anonymous
(x+1)(x+4) for the second one
Michele_Laino
  • Michele_Laino
sorry, for first trinomial I got this: \[{x^2} + x - 6 = \left( {x - 2} \right)\left( {x + 3} \right)\]
anonymous
  • anonymous
whoops yes:) you are right!
Michele_Laino
  • Michele_Laino
so, substituting our factorizations, in our computation, we get: \[\Large \begin{gathered} \frac{{x + 1}}{{{x^2} + x - 6}}:\frac{{{x^2} + 5x + 4}}{{x - 2}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{{x^2} + x - 6}} \cdot \frac{{x - 2}}{{{x^2} + 5x + 4}} = \hfill \\ \hfill \\ = \frac{{x + 1}}{{\left( {x - 2} \right)\left( {x + 3} \right)}} \cdot \frac{{x - 2}}{{\left( {x + 1} \right)\left( {x + 4} \right)}} = ...? \hfill \\ \hfill \\ \end{gathered} \]
anonymous
  • anonymous
1/(x+3)(x+4)
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
Last one?
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
Robin can mow a lawn in 3 hours, while Brady can mow the same lawn in 4 hours. How many hours would it take for them to mow the lawn together?
anonymous
  • anonymous
a. 7/12 b. 12/7 c. 12 d. 7
Michele_Laino
  • Michele_Laino
If I call with W the work to be done, then the working rate of Robin is W/3, whereas the working rate of Brady is W/4
anonymous
  • anonymous
Yes:)
Michele_Laino
  • Michele_Laino
now, when Robin and brady work together, then the working rate is: \[\Large \frac{W}{3} + \frac{W}{4}\]
Michele_Laino
  • Michele_Laino
Brady*
Michele_Laino
  • Michele_Laino
so, we can write this equation: \[\Large \left( {\frac{W}{3} + \frac{W}{4}} \right)t = W\] where t is the requested time
anonymous
  • anonymous
I understand! Continue:) And then find like denominators
Michele_Laino
  • Michele_Laino
after a simplification at the left side, we can write: \[\Large \frac{{4W + 3W}}{{12}}t = W\]
Michele_Laino
  • Michele_Laino
please continue
anonymous
  • anonymous
4x/12
Michele_Laino
  • Michele_Laino
hint: we can simplify the left side, so we get: \[\Large \frac{{7W}}{{12}}t = W\]
anonymous
  • anonymous
:) Would that be the answer?
Michele_Laino
  • Michele_Laino
no, please we can simplify further that equation, and we get this: \[\Large \frac{7}{{12}}t = 1\]
Michele_Laino
  • Michele_Laino
now I multiply both sides by 12: |dw:1437059913299:dw|
anonymous
  • anonymous
OH! Lol I'm sorry! I must have missed that.
anonymous
  • anonymous
12/7 ?
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
Yay!!
Michele_Laino
  • Michele_Laino
:)

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