Answer:
Step-by-step explanation:
ST = w + 6,
PR = w
From the diagram given, we can deduce that PR is the midsegment of ∆QST. Therefore, according to the midsegment theorem:
PR = ½ of ST
Plug in the values into the equation and solve for w.
(distributive property of equality)
(subtraction property of equality)
(multiplication property of equality)
(subtraction property of equality)
Divide both sides by -1
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:
that's all about my answer
