Which of the following is an asymptote of y = sec(x)?
2 answers:
Answer: option d. x = 3π/2
Solution:
function y = sec(x)
1) y = 1 / cos(x)
2) When cos(x) = 0, 1 / cos(x) is not defined
3) cos(x) = 0 when x = π/2, 3π/2, 5π/2, 7π/2, ...
4) limit of sec(x) = lim of 1 / cos(x).
When x approaches π/2, 3π/2, 5π/2, 7π/2, ... the limit →+/- ∞.
So, x = π/2, x = 3π/2, x = 5π/2, ... are vertical asymptotes of sec(x).
Answer: 3π/2
The figures attached will help you to understand the graph and the existence of multiple asymptotes for y = sec(x).
Solution:
function y = sec(x)
1) y = 1 / cos(x)
2) When cos(x) = 0, 1 / cos(x) is not defined
3) cos(x) = 0 when x = π/2, 3π/2, 5π/2, 7π/2, ...
4) limit of sec(x) = lim of 1 / cos(x).
When x approaches π/2, 3π/2, 5π/2, 7π/2, ... the limit →+/- ∞.
So, x = π/2, x = 3π/2, x = 5π/2, ... are vertical asymptotes of sec(x).
Answer: 3π/2
The figures attached will help you to understand the graph and the existence of multiple asymptotes for y = sec(x).
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Answer:
Jennifer would complete the job on her own in 5 hours.
Step-by-step explanation:
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
.
This polynomial has roots such that
, given by the following formulas:
Rates
The together rate is the sum of their separates rate.
We have that:
Jennifer takes x hours to complete the job on her own, so her rate is 1/x.
Carl needs 15 hours longer than Jennifer, that is, 15 + x hours, so his rate is 1/(15+x).
The together rate is 1/4. So
Applying cross multiplication.
Quadratic equation with . So
Since the time to complete the job has to be a positive value.
Jennifer would complete the job on her own in 5 hours.