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).
Because a line always goes past the y-intercept and x-intercept. It's not always both, it can sometimes just be the x-intercept or the y-intercept.
When a line intersects these points, for example if a line was to intersect the x-axis then y would be equal to 0, and the opposite for the y-axis. If a line was to intersect the y-axis x would be equal to 0.
Therefore by using that knowledge, and the equation of the line [ y=mx+c or y-y1=m(x-x1) ], we can find the equation of our line. Of course you would need the gradient of that line (the value "m").
Answer:
Explanation:
Using the PEMDAS rule:
P = parentheses
E = exponent
M = multiplication
D = division
A = Addition
S = subtraction
Step 1: 6 + 4 x 3 - 10/5
Step 2: 6 + 12 - 2
Step 3: 18 - 2
Step 4: 16
P= 500. I = 0.04. t=5. plug those numbers into the formula. So 1.04^5 and multiplying it time 500 and the final answer would be 608.33
Answer:
Observe that if the vector is in W then it satisfies:
This means that each vector in W can be expressed as a linear combination of the vectors
Also we can see that those vectors are linear independent. Then the set
is a basis for W and the dimension of W is 2.