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Nuetrik [128]
3 years ago
6

Please help me! I'm spending 50 points on this question and my last two questions never even got answered!

Mathematics
1 answer:
Verdich [7]3 years ago
8 0

Answer:

C.

Step-by-step explanation:

Domain is only affected if certain values can make the result something that either doesn't make sense or is an error. In your situation, it wants you to find the domain of f/g, which is a fraction. The only rule we have for fractions regarding domain is that we simply can't divide by zero, to find the domain for any fraction with variables in the denominator, what I do is take the denominator, set it equal to 0, and solve for x.

So, if we're taking f/g, we have:

\frac{3x^2 + x^4 + 2}{4x - 3}

4x - 3 makes up our denominator, the bottom part of the fraction. So

4x - 3 = 0

4x = 3

x = 3/4

So the denominator is equal to zero when x is equal to 3/4. This would produce an error in any calculator because you can't divide by zero. The domain here is all real numbers except for 3/4.

(For future problems, the only thing to look out for is if your numerator, the top part of the fraction, factors into something that cancels with the bottom. In that case, it wouldn't affect domain because you aren't actually dividing by anything. For example:

\frac{x^2 - 4}{x-2}

If we factor the top, we see that x^2 - 4 = (x - 2)(x + 2). In this case, we have (x - 2) in both the top and bottom of the fraction, so it cancels out, and from there nothing else is restricting our domain. It would be all real numbers in a case like that.)

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jek_recluse [69]

Answer:

h(u)=-7/6u

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Find the roots of h(t) = (139kt)^2 − 69t + 80
Sonbull [250]

Answer:

The positive value of k will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80, roots are those values of t so that h(t) = 0. That is:

19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0 (1)

Roots are determined analytically by the Quadratic Formula:

t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}

t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }

The smaller root is t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }, and the larger root is t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }.

h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80 has one real root when \frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0. Then, we solve the discriminant for k:

\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}

k \approx \pm 0.028

The positive value of k will result in exactly one real root is approximately 0.028.

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2 years ago
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