Answer: option d. x = 3π/2Solution: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).
Step-by-step explanation:
a =45°
b=45°
c=30°
d=30°
e=30°
f=30°
total of the angle
90×4=360
C. Above the dashed line.
It becomes:
y > 1/2x +3/2
(Dashed, shaded above)
The small number is 2.
The large number is 3.
<u>Step-by-step explanation:</u>
Let the two consecutive numbers be x and x+1.
- x be the small integer.
- x+1 be the large integer.
The sum of these two consecutive integers = small integer + large integer
The sum of these two consecutive integers is x+x+1 = (2x+1)
It is given that,
- The sum of two consecutive integers is one less than three times the smaller integer.
- This means that, (2x+1) is one less than three times the smaller integer.
- Here, the small integer is represented as x.
<u>Therefore, it can determined that :</u>
(2x+1) = 3x-1
Keeping x term on one side and constants on other side,
3x-2x = 1+1
x = 2
Therefore, the small number is 2 and the large number is x+1 = 3.
I tried my best at it and my guess is that in standard form is 0.00086