Answer:
False 8^0 is not greater than 8^1
Step-by-step explanation:
8^0 = 1
8^1= 8
8>1
First, tan(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>), so if cos(<em>θ</em>) = 3/5 > 0 and tan(<em>θ</em>) < 0, then it follows that sin(<em>θ</em>) < 0.
Recall the Pythagorean identity:
sin²(<em>θ</em>) + cos²(<em>θ</em>) = 1
Then
sin(<em>θ</em>) = -√(1 - cos²(<em>θ</em>)) = -4/5
and so
tan(<em>θ</em>) = (-4/5) / (3/5) = -4/3
The remaining trig ratios are just reciprocals of the ones found already:
sec(<em>θ</em>) = 1/cos(<em>θ</em>) = 5/3
csc(<em>θ</em>) = 1/sin(<em>θ</em>) = -5/4
cot(<em>θ</em>) = 1/tan(<em>θ</em>) = -3/4
yes you are correct evreything is correct
Answer:
Solve for x by simplifying both sides of the equation, then isolating the variable.
x=−845
Step-by-step explanation:
