Find the probability that a randomly selected point within the circle falls into the white area round to the nearest tenth of a
percent
1 answer:
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Answer:
The function y = sec(x) shifted 3 units left and 7 units down .
Step-by-step explanation:
Given the function: y = sec(x)
- If k is any positive real number, then the graph of f(x) - k is the graph of y = f(x) shifted downward k units.
- If p is a positive real number, then the graph of f(x+p) is the graph of y=f(x) shifted to the left p units.
The function comes from the base function y= sec(x).
Since 3 is added added on the inside, this is a horizontal shift Left 3 unit, and since 7 is subtracted on the outside, this is a vertical shift down 7 units.
Therefore, the transformation on the given function is shifted 3 units left and 7 units down
Answer:
B.
Step-by-step explanation:
Here the best method to solve is by substituting the end values of the set in each option , otherwise it will a time consuming problem.
Now substitute x=-4 in all the options
A.
16+8-8=16>0
so out of option A and C option C is correct.
B.
16-8-8=0 which means for the values of x>-4 is less than 0
Now substitute x=2 in all the options
A.
4-4-8=-8<0 . so option A and C both are incorrect.
B.
4+4-8=0 which means for the values of x<2 is less than 0
Therefore the correct option is B