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Updated on: 17 Jun 2022, 11:48
Originally posted by EthanTheTutor on 17 Jun 2022, 00:30.
Last edited by EthanTheTutor on 17 Jun 2022, 11:48, edited 1 time in total.
Last edited by EthanTheTutor on 17 Jun 2022, 11:48, edited 1 time in total.
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B
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Difficulty:
55%
(hard)
Question Stats:
65% (02:17) correct
35%
(02:31)
wrong
based on 156
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If the length and width of a certain rectangle with positive area are both solutions to the equation \(-3x^3+13x^2-11x=0\), what is the area of that rectangle?
a) \(3.\overline{3}\)
b) \(3.\overline{6}\)
c) \(4.\overline{3}\)
d) \(4.\overline{6}\)
e) \(5.\overline{3}\)
a) \(3.\overline{3}\)
b) \(3.\overline{6}\)
c) \(4.\overline{3}\)
d) \(4.\overline{6}\)
e) \(5.\overline{3}\)
Updated on: 17 Jun 2022, 11:47
Originally posted by EthanTheTutor on 17 Jun 2022, 00:34.
Last edited by EthanTheTutor on 17 Jun 2022, 11:47, edited 1 time in total.
Last edited by EthanTheTutor on 17 Jun 2022, 11:47, edited 1 time in total.
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EthanTheTutor
\(x=0\) is one solution to the equation \(-3x^3+13x^2-11x=0\), but we are told that the area is positive; thus, none of the sides are equal to zero, and we can discard that possibility.
This allows us to divide by \(x\), giving: \(-3x^2+13x-11=0\). Then, the question is equivalent to asking, "what is the product of the solutions to this quadratic?"
We could use the quadratic formula to find the solutions to this equation, but it will be much easier to remember that the product of the solutions to a quadratic is given by \(\frac{c}{a}\).
In this case, that gives: \(\frac{c}{a}=\frac{-11}{-3}=3.\overline{6}\). Answer B.
17 Jun 2022, 11:30
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as per the standard form,
Here, -11 should be considered as c, right?
Here, -11 should be considered as c, right?
