The inequality for sec (x) < cot (x) is; π/2 < x < π
<h3>How to express trigonometric inequality?</h3>
We are given that;
We want to find the intervals that the trigonometric inequality sec (x) < cot (x) always hold true.
This can also be expressed as;
1/cos (x) < 1/tan (x)
Now, this can happen only in the quadrant where tan (x) is negative and cos x is positive which is in fourth quadrant where;
π/2 < x < π
The inequality for sec (x) < cot (x) is; π/2 < x < π
Read more about Trigonometric Inequality at; brainly.com/question/12094532
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Y - y1 = m(x - x1)
slope(m) = 13
(-2,5)...x1 = -2 and y1 = 5
sub...pay attention to ur signs
y - 5 = 13(x - (-2)...not done yet
y - 5 = 13(x + 2) <== point slope form
A^2 + b^2 = c^
a^2 + 7.8^2 = 8^2
a^2 + 2.79= 2.83
a^2 = 0.04
a = 0.2m
i did that in my head so like it might be off or wrong but that’s the basic idea of it
Answer:
m= -1 4/11
Step-by-step explanation: all you're doing is dividing the side with the variable (11/5m) by the side with the number by itself (-3)