Identify the means and the extremes in each of the following proportions.
2 answers:
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From trigonometry we know that:
if
then, (where
is an integer)
This can be rewritten in degrees as:
.............(Equation 1)
Now, in our case,
Therefore, (Equation 1) can be written as:
..........(Equation 2)
Now, to find the correct options all that we have to do is replace n by relevant integers and find the values of that match.
For n=2, (Equation 2) gives us: .
Thus,
Now, we know that:
Let n=-1, then:
Thus,
Likewise,
Only the last option will never match
because no integral value of
will ever give
Thus the last option is the correct option.
Answer:
(a) There are asymptotes at x=3/2 and x=-1/3
Step-by-step explanation:
The denominator zeros can be found by factoring:
f(x) = (x +1)/((2x -3)(3x +1))
Neither of the denominator factors is cancelled by the numerator factor, so each represents a vertical asyptote, not a function hole.
The asymptotes are at the values of x where the denominator is zero:
2x -3 = 0 ⇒ x = 3/2
3x +1 = 0 ⇒ x = -1/3
