keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above
so we're really looking for the equation of a line with a slope of -2 and that passes through (1 , -1)
Can someone help me, im really stuck.
2 answers:
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<h3>Given</h3>
tan(x)²·sin(x) = tan(x)²
<h3>Find</h3>x on the interval [0, 2π)
<h3>Solution</h3>Subtract the right side and factor. Then make use of the zero-product rule.
... tan(x)²·sin(x) -tan(x)² = 0
... tan(x)²·(sin(x) -1) = 0
This is an indeterminate form at x = π/2 and undefined at x = 3π/2. We can resolve the indeterminate form by using an identity for tan(x)²:
... tan(x)² = sin(x)²/cos(x)² = sin(x)²/(1 -sin(x)²)
Then our equation becomes
... sin(x)²·(sin(x) -1)/((1 -sin(x))(1 +sin(x))) = 0
... -sin(x)²/(1 +sin(x)) = 0
Now, we know the only solutions are found where sin(x) = 0, at ...
... x ∈ {0, π}
Answer:
The inverse function of f(x) is ... f⁻¹(x) = log (x/9.16)/log (1.0054)
Step-by-step explanation:
Starting equation... f(x) = 9.16(1.0054)ˣ
Visually rewritten... y = 9.16(1.0054)ˣ
Flipping x and y... x = 9.16(1.0054)ʸ
Isolating (1.0054)ʸ... x/9.16 = (1.0054)ʸ
Taking the log of both sides... log (x/9.16) = y log (1.0054)
Isolating y... log (x/9.16)/log (1.0054) = y
Solution... y = log (x/9.16)/log (1.0054)
