Answer is <span>x ≥ 7
........................................</span>
Answer:
3 min per lap
Step-by-step explanation:
15 (minutes)/5 (lap)= 3 minutes
You are correct. The answer is choice B.
The base angles of an isosceles triangle are congruent. The base angles are opposite the congruent sides as shown by the tickmarks.
I will use the letter x instead of theta.
Then the problem is, given sec(x) + tan(x) = P, show that
sin(x) = [P^2 - 1] / [P^2 + 1]
I am going to take a non regular path.
First, develop a little the left side of the first equation:
sec(x) + tan(x) = 1 / cos(x) + sin(x) / cos(x) = [1 + sin(x)] / cos(x)
and that is equal to P.
Second, develop the rigth side of the second equation:
[p^2 - 1] / [p^2 + 1] =
= [ { [1 + sin(x)] / cos(x) }^2 - 1] / [ { [1 + sin(x)] / cos(x)}^2 +1 ] =
= { [1 + sin(x)]^2 - [cos(x)]^2 } / { [1 + sin(x)]^2 + [cos(x)]^2 } =
= {1 + 2sin(x) + [sin(x)^2] - [cos(x)^2] } / {1 + 2sin(x) + [sin(x)^2] + [cos(x)^2] }
= {2sin(x) + [sin(x)]^2 + [sin(x)]^2 } / { 1 + 2 sin(x) + 1} =
= {2sin(x) + 2 [sin(x)]^2 } / {2 + 2sin(x)} = {2sin(x) ( 1 + sin(x)} / {2(1+sin(x)} =
= sin(x)
Then, working with the first equation, we have proved that [p^2 - 1] / [p^2 + 1] = sin(x), the second equation.
