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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).
I found the corresponding image. Pls. see attachment.
<span>The minimum number of rigid transformations required to show that polygon ABCDE is congruent to polygon FGHIJ is 2 (translation and rotation).
A rotation translation must be used to make the two polygons coincide.
A sequence of transformations of polygon ABCDE such that ABCDE does not coincide with polygon FGHIJ is a translation 2 units down and a 90° counterclockwise rotation about point D </span>
<span>The minimum number of rigid transformations required to show that polygon ABCDE is congruent to polygon FGHIJ is 2 (translation and rotation).
A rotation translation must be used to make the two polygons coincide.
A sequence of transformations of polygon ABCDE such that ABCDE does not coincide with polygon FGHIJ is a translation 2 units down and a 90° counterclockwise rotation about point D </span>