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).
$29+$17+$13 = $59 weekly spending
So 1 month gonna be $59 times 4 = $236 per month
Answer:
x = (-5 ± 2√10) / 3
Step-by-step explanation:
5 − 10x − 3x² = 0
Write in standard form:
-3x² − 10x + 5 = 0
Solve with quadratic formula:
x = [ -b ± √(b² − 4ac) ] / 2a
x = [ -(-10) ± √((-10)² − 4(-3)(5)) ] / 2(-3)
x = [ 10 ± √(100 + 60) ] / -6
x = (10 ± 4√10) / -6
x = (-5 ± 2√10) / 3
First set the equation equal to y by adding x to each side, then divide everything by 8. then set x to zero and solve to find the y - intercept of 1/2 which is just 4/8 reduced. you get this answer because x/8 becoming 0/8 is just 0 leaving you with only 1/2. then reset the equation with x and set y to zero and solve for x to get the x- intercept of -4. you get this by subtracting 1/2 from both sides leaving you -1/2=x/8 and multiply by 8 to get -4=x
Polygon HJKL has vertices H(-3, 4), J(2, 6), K(2, 1), and L(-3, -1). What are the slopes of sides HJ, JK, KL, and LH and how wou
aleksklad [387]
<span> the slopes of sides HJ, JK, KL, and LH and how would you classify HJKL?</span>