<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:
7 1/2+ 7 1/2=146 of 146 Kerry and Eric ran the same amount each
The graph of y = -3x+5
Step 1: Since the y - intercept is 5, plot the point (0,5).
Step 2: Since the slope is -3, move 1 unit to the right and 3 units down, so plot the point (1,2).
Step 3: Connect those points by a straight line.
Answer:
Thank you lol. I am so bored right now.
Step-by-step explanation: