Finding trigonometric expression
sec (arctan x/2)
Hello Lolaaaaaa!
f(x)=sec(arctan x2)=1cos (arctan x2)
x : 1 2 3 4 5 6 7 10
f(x) : 1.118 1.414 1.803 2.236 2.693 3.162 3.640 5.100
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What are the coordinates of the hole in the function?
f(x)=2x2−5x−123x2−11x−4
Is anyone able to help me with this? Would greatly appreciate it!
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