Answer:The numbers are already in standard form
Step-by-step explanation:
Answer:
√(4/5)
Step-by-step explanation:
First, let's use reflection property to find tan θ.
tan(-θ) = 1/2
-tan θ = 1/2
tan θ = -1/2
Since tan θ < 0 and sec θ > 0, θ must be in the fourth quadrant.
Now let's look at the problem we need to solve:
sin(5π/2 + θ)
Use angle sum formula:
sin(5π/2) cos θ + sin θ cos(5π/2)
Sine and cosine have periods of 2π, so:
sin(π/2) cos θ + sin θ cos(π/2)
Evaluate:
(1) cos θ + sin θ (0)
cos θ
We need to write this in terms of tan θ. We can use Pythagorean identity:
1 + tan² θ = sec² θ
1 + tan² θ = (1 / cos θ)²
±√(1 + tan² θ) = 1 / cos θ
cos θ = ±1 / √(1 + tan² θ)
Plugging in:
cos θ = ±1 / √(1 + (-1/2)²)
cos θ = ±1 / √(1 + 1/4)
cos θ = ±1 / √(5/4)
cos θ = ±√(4/5)
Since θ is in the fourth quadrant, cos θ > 0. So:
cos θ = √(4/5)
Or, written in proper form:
cos θ = (2√5) / 5
Y = mx + b
slope(m) = 3/2
(5,9)...x = 5 and y = 9
now we sub and find b, the y int
9 = 3/2(5) + b
9 = 15/2 + b
9 - 15/2 = b
18/2 - 15/2 = b
3/2 = b
so ur equation is : y = 3/2x + 3/2.........(3,a)....so x = 3
y = 3/2(3) + 3/2
y = 9/2 + 3/2
y = 12/2
y = 6
so ur missing value is 6...a (or y) = 6
Answer:
11 units i think
Step-by-step explanation:
