Answer:
Let's solve the expression!
Step-by-step explanation:
Your answer is 75.
The fraction shows 150 ÷ 2 and 2x ÷ 2.
I hope this helps!
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Answer:
NO amount of hour passed between two consecutive times when the water in the tank is at its maximum height
Step-by-step explanation:
Given the water tank level modelled by the function h(t)=8cos(pi t /7)+11.5. At maximum height, the velocity of the water tank is zero
Velocity is the change in distance with respect to time.
V = {d(h(t)}/dt = -8π/7sin(πt/7)
At maximum height, -8π/7sin(πt/7) = 0
-Sin(πt/7) = 0
sin(πt/7) = 0
Taking the arcsin of both sides
arcsin(sin(πt/7)) = arcsin0
πt/7 = 0
t = 0
This shows that NO hour passed between two consecutive times when the water in the tank is at its maximum height
(tan²(<em>θ</em>) cos²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>))
Recall that
tan(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>)
so cos²(<em>θ</em>) cancels with the cos²(<em>θ</em>) in the tan²(<em>θ</em>) term:
(sin²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>))
Recall the double angle identity for cosine,
cos(2<em>θ</em>) = 2 cos²(<em>θ</em>) - 1
so the 1 in the denominator also vanishes:
(sin²(<em>θ</em>) - 1) / (2 cos²(<em>θ</em>))
Recall the Pythagorean identity,
cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1
which means
sin²(<em>θ</em>) - 1 = -cos²(<em>θ</em>):
-cos²(<em>θ</em>) / (2 cos²(<em>θ</em>))
Cancel the cos²(<em>θ</em>) terms to end up with
(tan²(<em>θ</em>) cos²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>)) = -1/2
Answer:
Step-by-step explanation:
The angle marks mean those angles have the same measure. That means the sides opposite them have the same measure, so ...
2x -24 = x -2
x = 22 . . . . . . . . . add 24-x
The length of BC is x, so is 22.
BC = 22
