Graph the function in the interval from 0 to 2pi. y = sin(θ − 2)
2 answers:
Answer: It is the second graph. See the attached figure.
Explanation:
1) You can consider the function y = sin(θ − 2) as a transformation of the parent function <span>y = sin θ
2) Subtracting a constant from the argument of a function leads to shifting the graph as many units to the right as the constant.
3) So, subtracting 2 from the argument of y = sin </span><span>θ shifts its graph 2 units to the right.
4) Since the x-intercepts of </span><span>y = sin θ in the interval 0 to 2π are 0, π, and 2π, the x-intercepts of </span><span>y = sin(θ − 2) will be 2, π + 2, and 2π + 2.
5) The x-intercepts are not enough to differentiate between some graphs, so take into account the local maxima or minima.
The local maxima of the function </span><span>y = sin θ in the interval 0 to 2π are at x = π/2, and 3π/2, then the local maximum of <span>y = sin(θ − 2) is at π/2 + 2.</span>
That is the second graph.
I attach the graph for avoiding confussions.
</span>
Explanation:
1) You can consider the function y = sin(θ − 2) as a transformation of the parent function <span>y = sin θ
2) Subtracting a constant from the argument of a function leads to shifting the graph as many units to the right as the constant.
3) So, subtracting 2 from the argument of y = sin </span><span>θ shifts its graph 2 units to the right.
4) Since the x-intercepts of </span><span>y = sin θ in the interval 0 to 2π are 0, π, and 2π, the x-intercepts of </span><span>y = sin(θ − 2) will be 2, π + 2, and 2π + 2.
5) The x-intercepts are not enough to differentiate between some graphs, so take into account the local maxima or minima.
The local maxima of the function </span><span>y = sin θ in the interval 0 to 2π are at x = π/2, and 3π/2, then the local maximum of <span>y = sin(θ − 2) is at π/2 + 2.</span>
That is the second graph.
I attach the graph for avoiding confussions.
</span>
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Answer:
a) 0.00001
b) 0.001
Step-by-step explanation:
Given numbers,
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
a) If by these numbers we have to make six digit number,
So, the total ways of arranging 6 digit number =
Now, if all six digit in the number are same,
Then the possible ways = 10 ( eg : 000000, 111111, 222222, 333333, 444444... etc ),
Hence, the probability that all six digits are the same
b) The ways that five or more of the digits of the same = exactly 5 digits are same + Exactly 6 digits are same
∵ The ways that 5 digits are same = 10 × 9 = 90 (i.e. 111110, 111112, 111113,.....9 times, 222220, 222221,.....9 times .... etc..)
So, the number of ways that 5 or more than 5 digits are same = 90 + 10 = 100
Hence, the probability that five or more of the digits of the same