<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:
1/2
Step-by-step explanation:
(1/16)^x
Let x = 1/4
(1/16)^ 1/4
Rewriting 16 as 2^4
(1/2^4)^ 1/4
We know that 1 / a^b = a^-b
(2 ^ -4)^ 1/4
We know that a^b^c = a^(b*c)
2^(-4*1/4)
2^-1
We know that a^-b = 1/ a^b
2^-1 = 1/2^1 = 1/2
The domain and range of the given function are equal to (0, 3.85) and (0, 18.75) respectively.
<h3>How to calculate the domain of the function?</h3>
In this exercise, you're given the following function h(t) = -4.87t² + 18.75t. Next, we would equate the function to zero (0) to determine its domain as follows:
0= -4.87t² + 18.75t.
4.87t(-t + 3.85) = 0
t = 0 or t = 3.85.
Therefore, the domain is 0 ≤ t ≤ 3.85 or (0, 3.85).
<h3>How to calculate the range of the function?</h3>
h(t) = -4.87t² + 18.75t
h(t) = -4.87(t² - 3.85t + 3.85 - 3.85)
h(t) = -4.87(t - 1.925)² + 18.05
Therefore, the range is 0 ≤ h ≤ 18.05 or (0, 18.75).
Read more on domain here: brainly.com/question/17003159
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The answer is B because they never pass through each other therefore it has infinite possibilities
Answer:
C = 1.41n
Step-by-step explanation:
you have to multiply 1.41 by the weight(n)
