Answer:
Yes
Domain: 1, -1, 0, -3, 3
Range: 1, -1, 6
Step-by-step explanation:
The domain is in the left oval, the range is in the right oval. To find if its a function you have to make sure the domain (left side) doesn't "cheat". One number in the domain cant have more than one number in the range (right side), BUT (for future reference) the range can cheat. As shown, the right side numbers 1 and 6 have two different arrow pointed at them. As long as the domain doesnt "cheat" it is a function
Answer:
b
Step-by-step explanation:
I DK I DK I DK I DK but maybe it’s already in that order
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
