A good letter of recommendation would provide personal insight to the student. Option C.
Whos up for giving each other brainiest, I need to level up....
Answer:
I'm up to it because I also need brainliest
determine the ........ of y= 4sin2x
o Amplitude
o Period
o Intercepts
o Maximum
o Minimum
Answer:
Amplitude: [tex]A = 4[/tex]
Period: [tex]\tau = \pi[/tex]
Minimum: -4
Maximum: 4
Intercepts: [tex]x = \frac{1}{2}\cdot [0 \pm \pi \cdot n][/tex], [tex]\forall \,n\in \mathbb{N}_{O}[/tex]
Explanation:
The expression described on statement is a sinusoidal formula, whose expression is of the form:
[tex]y = A\cdot \sin \left(\frac{2\pi\cdot x}{\tau}\right)[/tex] (1)
Where:
[tex]x[/tex] - Independent variable.
[tex]y[/tex] - Dependent variable.
[tex]A[/tex] - Amplitude.
[tex]\tau[/tex] - Period.
By direct comparison, we calculate the amplitude and period:
Amplitude
[tex]A = 4[/tex]
Period
[tex]\frac{2\pi}{\tau} = 2[/tex]
[tex]\tau = \pi[/tex]
Minimum and Maximum
The sine is a bounded function between -1 and 1, meaning that sinusoidal formula is bounded between [tex]-A[/tex] and [tex]A[/tex]. Hence, the minimum and maximum are -4 and 4, respectively.
Intercepts
The intercepts are set of points of the sinusoidal formula such that [tex]y = 0[/tex]. The sine function is a periodic function which equals 0 each [tex]\pi[/tex] radians.
[tex]4\cdot \sin 2x = 0[/tex]
[tex]\sin 2x = 0[/tex]
[tex]2\cdot x = \sin^{-1} 0[/tex]
[tex]x = \frac{1}{2}\cdot \sin^{-1} 0[/tex]
[tex]x = \frac{1}{2}\cdot [0 \pm \pi \cdot n][/tex], [tex]\forall \,n\in \mathbb{N}_{O}[/tex]
Show an example of how to evaluate a trigonometric function using reference
angle (without a calculator/ you may use left hand rule or unit circle)
Answer:
[tex]\cos(150) = -\frac{\sqrt 3}{2}[/tex]
Explanation:
An example is as follows;
Using reference angle, evaluate [tex]\cos(150)[/tex]
Solution:
We have: [tex]\cos(150)[/tex]
150 degrees is in the second quadrant, and it makes 30 degrees (i.e. 180 - 150) with the x-axis.
This implies that, cos(150) has the same magnitude as cos(30);
The only difference in the values is the sign; because
30 degrees is in the first quadrant where all functions are positive150 degrees is in the third quadrant where all cosine is negativeSo, we have:
[tex]\cos(150) = -\cos(30)[/tex]
-----------------------------------------------
[tex]\cos(30) = \frac{\sqrt 3}{2}[/tex]
-----------------------------------------------
[tex]\cos(150) = -\frac{\sqrt 3}{2}[/tex]
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Answer:
americans be like:
Explanation: