Answer: Choice B. sqrt(2)
Draw out a right triangle in quadrant IV as you see in the attached image below. The horizontal and vertical legs are both 1 unit long. To ensure that the signs are properly set up, I am making the vertical leg BC have a label "-1" to mean this is below the x axis. Note how
tan(theta) = opposite/adjacent = BC/AB = -1/1 = -1
Use the pythagorean theorem to find that the hypotenuse AC is sqrt(2) units long
a^2 + b^2 = c^2
(1)^2 + (1)^2 = c^2
2 = c^2
c^2 = 2
c = sqrt(2)
The secant of theta is the ratio of the hypotenuse over the adjacent side, so we end up with
sec(theta) = hypotenuse/adjacent
sec(theta) = AC/AB
sec(theta) = sqrt(2)/1
sec(theta) = sqrt(2) which is why choice B is the answer
Answer:
CBenson1031
Step-by-step explanation:
p = 5 ÷ (12.37 - 1.12)
p = 5 ÷ 11.25
p = 2.25
Step-by-step explanation:
Answer:
Step-by-step explanation:
A(-6,2) =>A'(-5, -1)
B(-5,4) =>B'( -4,1)
C(-2,4)=> C'(-1,1)
D( 1,2) => D'(2,-1)
we move all these points to (x+1, y-3)
Answer:We need to see the net.
Step-by-step explanation:
Answer:
- Number of times that Noah volunteered at pet shelter.
- Number of times that Keith volunteered at pet shelter.
Step-by-step explanation:
After reading the statement of the problem, the following variables are described below:
- Number of times that Jason volunteered at pet shelter.
- Number of times that Noah volunteered at pet shelter.
- Number of times that Keith volunteered at pet shelter.
The following two relations are presented herein:
Noah and Jason
Jason and Keith
The relationship between the number of times Noah and Keith volunteered is:
