<h3>2
Answers: Choice C and choice D</h3>
y = csc(x) and y = sec(x)
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Explanation:
The term "zeroes" in this case is the same as "roots" and "x intercepts". Any root is of the form (k, 0), where k is some real number. A root always occurs when y = 0.
Use GeoGebra, Desmos, or any graphing tool you prefer. If you graphed y = cos(x), you'll see that the curve crosses the x axis infinitely many times. Therefore, it has infinitely many roots. We can cross choice A off the list.
The same applies to...
- y = cot(x)
- y = sin(x)
- y = tan(x)
So we can rule out choices B, E and F.
Only choice C and D have graphs that do not have any x intercepts at all.
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If you're curious why csc doesn't have any roots, consider the fact that
csc(x) = 1/sin(x)
and ask yourself "when is that fraction equal to zero?". The answer is "never" because the numerator is always 1, and the denominator cannot be zero. If the denominator were zero, then we'd have a division by zero error. So that's why csc(x) can't ever be zero. The same applies to sec(x) as well.
sec(x) = 1/cos(x)
Part B seems to be missing, but I think I have enough information to be able to answer.
Let's say we had two numbers x and y. Let x be rational and y be irrational.
If x is some nonzero number, then x*y is irrational. The proof for this is a bit lengthy so I'll leave it out.
For instance,
x = 2 is rational, y = sqrt(3) is irrational, x*y = 2*sqrt(3) is irrational.
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If we made x = 0, then
x*y = 0*y = 0
This is true for any value of y that we want. The y value doesnt even have to be irrational. It can be any real number.
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So the distinction is that if x = 0, then x*y = 0 is rational since 0 is rational. Otherwise, x*y is irrational.
Answer:
2f+9
Step-by-step explanation: