Step-by-step explanation:
︻╦デ╤━╼︻╦デ╤━╼︻╦デ╤━╼︻╦デ╤━╼︻╦デ╤━╼︻╦デ╤━╼︻╦デ╤━╼
Answer:
Solution:
, 3.5, -15, 5
Not a solution: 129 and 
Step-by-step explanation:
20 ≥ 5 + 3x
3x + 5 ≤ 20
3x ≤ 20 - 5
3x ≤ 15
x ≤ 5
x must be LESS than or EQUAL to 5
Answer:
Commutative
Step-by-step explanation:
Now, the cosecant of θ is -6, or namely -6/1.
however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.
we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

recall that

therefore, let's just plug that on the remaining ones,

now, let's rationalize the denominator on tangent and secant,
Answer:
-650
Step-by-step explanation: