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vredina [299]
3 years ago
13

A rational function can have infinitely many x-values at which it is not continuous. True or False. If false, explain

Mathematics
1 answer:
sattari [20]3 years ago
3 0
<span>It is false since the rational function is discontinuous when the denominator is zero. But the denominator is a polynomial and a polynomial has only finitely many zeros. So the discontinuity points of a rational function is finite. </span>
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Licemer1 [7]

Answer:

9(x-11) = 9

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x = 108/9

thus,

x = 12

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I need help answering this question!
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Step-by-step explanation:

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Let p(x) = x3 and q(x) = x + 5. If p(q(x)) and q(p(x)) are shown on the graph, which statements are true? Check all that apply.
liraira [26]

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Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
I dont understand this, can i get some help please
Ray Of Light [21]

When we want to find the roots of a one-variable function, we look for where its graph intersects the x-axis. In this case, the graph intersects the x-axis at x=1.

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For the y-intercept, just look for where the graph intersects the y-axis; in this case, that point is (0,1).

Using this information, the vertex-form equation of the parabola is y=1\cdot(x-1)^2+0, so the factors are two copies of x-1. In this case, the value of a in the equation y=a(x-h)^2+k was conveniently 1; if that's not the case, you'll want to plug in x=0 to solve for the value of a that gives the correct y-intercept.

Does that help clear things up?

3 0
3 years ago
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