Answer:
84 is the highest possible course average
Step-by-step explanation:
Total number of examinations = 5
Average = sum of scores in each examination/total number of examinations
Let the score for the last examination be x.
Average = (66+78+94+83+x)/5 = y
5y = 321+x
x = 5y -321
If y = 6, x = 5×6 -321 =-291.the student cannot score -291
If y = 80, x = 5×80 -321 =79.he can still score higher
If If y = 84, x = 5×84 -321 =99.This would be the highest possible course average after the last examination.
If y= 100
The average cannot be 100 as student cannot score 179(maximum score is 100)
Answer:
18
Step-by-step explanation:
1. 6x6=36
2. 8+36=44
3. 44/11=4
4. 7x2=14
5. 4+14=18
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).
Answer: I don’t know the answer to it but I will try and if I got it will tell u
Step-by-step explanation: