The answer is
![r= \sqrt[3]{GMT^{2}/4 \pi^{2}}](https://tex.z-dn.net/?f=r%3D%20%5Csqrt%5B3%5D%7BGMT%5E%7B2%7D%2F4%20%5Cpi%5E%7B2%7D%7D%20)
Move
to the other side of the equation:
Rearrange:
Since
![x^{3}= \sqrt[3]{x}](https://tex.z-dn.net/?f=%20x%5E%7B3%7D%3D%20%5Csqrt%5B3%5D%7Bx%7D%20%20)
, then
![r^{3} = GMT^{2}/4 \pi^{2} \\ r= \sqrt[3]{GMT^{2}/4 \pi^{2}}](https://tex.z-dn.net/?f=r%5E%7B3%7D%20%3D%20GMT%5E%7B2%7D%2F4%20%5Cpi%5E%7B2%7D%20%5C%5C%20%0Ar%3D%20%5Csqrt%5B3%5D%7BGMT%5E%7B2%7D%2F4%20%5Cpi%5E%7B2%7D%7D%20)
A-1/12 because if you write down h1, h2, h3, etc.-same for tail….tail and four would show up once out of the twelve options.
Step-by-step explanation:
(1 + cos θ + sin θ) / (1 + cos θ − sin θ)
Multiply by the reciprocal:
(1 + cos θ + sin θ) / (1 + cos θ − sin θ) × (1 + cos θ + sin θ) / (1 + cos θ + sin θ)
(1 + cos θ + sin θ)² / [ (1 + cos θ − sin θ) (1 + cos θ + sin θ) ]
(1 + cos θ + sin θ)² / [ (1 + cos θ)² − sin² θ) ]
Distribute and simplify:
(1 + cos θ + sin θ)² / (1 + 2 cos θ + cos² θ − sin² θ)
[ 1 + 2 (cos θ + sin θ) + (cos θ + sin θ)² ] / (1 + 2 cos θ + cos² θ − sin² θ)
(1 + 2 cos θ + 2 sin θ + cos² θ + 2 sin θ cos θ + sin² θ) / (1 + 2 cos θ + cos² θ − sin² θ)
Use Pythagorean identity:
(2 + 2 cos θ + 2 sin θ + 2 sin θ cos θ) / (sin² θ + cos² θ + 2 cos θ + cos² θ − sin² θ)
(2 + 2 cos θ + 2 sin θ + 2 sin θ cos θ) / (2 cos² θ + 2 cos θ)
(1 + cos θ + sin θ + sin θ cos θ) / (cos² θ + cos θ)
Factor:
(1 + cos θ + sin θ (1 + cos θ)) / (cos θ (1 + cos θ))
(1 + cos θ)(1 + sin θ) / (cos θ (1 + cos θ))
(1 + sin θ) / cos θ
If i were going by guesses i’d say y=x. sorry if i’m wrong.
Answer:
y−3=2(x+3)
y=2x+9
that's the answer hope this helps u
