ASAP PLEASE!!! Start with the following conditional statement: If it is October 31, then it is Halloween.
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This problem is half understanding the question and half just plugging in numbers.
Understanding the Question:
So its saying that n = whole number bigger than one, making n = 2, 3, 4, 5, etc.
Now try plugging any number n into the
equation. Let's use 2 to make it simple. If you plug in 2, you get:
Remember that the subscripts (the small numbers in the corner) are simply the term number.
is term 1, 
is term 2, etc. The equation is saying that "term 2 = 4 times term 1."
If you keep plugging in random numbers for n, you'll see that the equation basically says, "a term = 4 times the term right before it." That makes it easy to plug and chug to get our answer.
Plugging and Chugging
Now we understand the question is asking for the first 4 terms, and each term is 4 times the last term. You're told that term 1, = 6. That makes the next four terms:
Term 2
Term 3
Term 4
Your first 4 terms are 4, 24, 96, and 384.
--------
Answer: A) F
Understanding the Question:
So its saying that n = whole number bigger than one, making n = 2, 3, 4, 5, etc.
Now try plugging any number n into the
Remember that the subscripts (the small numbers in the corner) are simply the term number.
If you keep plugging in random numbers for n, you'll see that the equation basically says, "a term = 4 times the term right before it." That makes it easy to plug and chug to get our answer.
Plugging and Chugging
Now we understand the question is asking for the first 4 terms, and each term is 4 times the last term. You're told that term 1,
Term 2
Term 3
Term 4
Your first 4 terms are 4, 24, 96, and 384.
--------
Answer: A) F
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).