If the period of a sine function is the number of units before the cycle begins to repeat, why would the period of the regressio
1 answer:
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Answer:
Not a function.
Step-by-step explanation:
The given function is;
Let
Interchange x and y;
Solve for y;
Square both sides
The inverse is
is not a function because one x-value maps onto to different y-values.
Answer:
<u>The true statement is D</u>
Step-by-step explanation:
The rest of the question is the attached figure and the statement options.
- A. Over the interval [–4, –2], the local minimum is 0.
- B. Over the interval [–2, –1], the local minimum is 25.
- C. Over the interval [–1, 4], the local minimum is 0.
- D. Over the interval [4, 7], the local minimum is -7.
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According to the graph, we will check the options:
A. Over the interval [–4, –2], the local minimum is 0. (<u>Wrong</u>)
Because the minimum is -12
B. Over the interval [–2, –1], the local minimum is 25. (<u>Wrong</u>)
Because the minimum is 18
C. Over the interval [–1, 4], the local minimum is 0. (<u>Wrong</u>)
Because the minimum is at x = 4 less than zero
D. Over the interval [4, 7], the local minimum is -7. (<u>True</u>)
