The function represents a <em>cosine</em> graph with axis at y = - 1, period of 6, and amplitude of 2.5.
<h3>How to analyze sinusoidal functions</h3>
In this question we have a <em>sinusoidal</em> function, of which we are supposed to find the following variables based on given picture:
- Equation of the axis - Horizontal that represents the mean of the bounds of the function.
- Period - Horizontal distance needed between two maxima or two minima.
- Amplitude - Mean of the difference of the bounds of the function.
- Type of sinusoidal function - The function represents either a sine or a cosine if and only if trigonometric function is continuous and bounded between - 1 and 1.
Then, we have the following results:
- Equation of the axis: y = - 1
- Period: 6
- Amplitude: 2.5
- The graph may be represented by a cosine with no <em>angular</em> phase and a sine with <em>angular</em> phase, based on the following trigonometric expression:
cos θ = sin (θ + π/2)
To learn more on sinusoidal functions: brainly.com/question/12060967
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Answer: See below
Step-by-step explanation:
<u>Area of the square</u> = l*w = 11*11 = 121
<u>Area of the circle</u> = pi r^2 = pi*1^2 = 3.142
<u>Part A</u>
P(hitting the circle) = 3.142 / 121 = 0.026 which is closer to 0.
<u>Part B</u>
P(hitting the outside portion) = 1 - 0.026 = 0.974 which is closer to 1.
9514 1404 393
Answer:
(0, π/4] ∪ (π, 5π/4]
Step-by-step explanation:
Multiplying by 2 gives ...
cot(x) ≥ 1
The cotangent function decreases from ∞ to 1 in the domain (0, π/4], and again in the domain (π, 5π/4]. The solution is the union of these two intervals.
x ∈ (0, π/4] ∪ (π, 5π/4]
_____
(a, b] is interval notation for a < x ≤ b
<span>If the clock is held at a
constant 0.0ºc over a period of 24 hours, the clock will be exactly the same as
the perfect clock because it is at a
constant 0.0</span> <span>ºc for 24. Meaning there is
no deviation on its reading</span>
Answer:
Subtract the two diameters and multiply by 0.5
Step-by-step explanation:
Calculating the Radius of the hole gives distance to the inner edges WHILE the Radius of the coin gives us the distance to the outer edges.
By subtracting the output we can then get the difference between both distances.
If hole diameter = 5mm and coin diameter = 22mm
Difference = [(22 - 5)mm] ÷ 2
17mm / 2 = 8.5mm
