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Finding trigonometric expression

(sec (arctan x/2) 

 Sep 18, 2020
 #1
avatar+15068 
+1

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 

laugh  !

 Sep 18, 2020
 #2
avatar+6 
0

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!

 Sep 18, 2020
 #3
avatar+2096 
-1

You can post your question as a separate question :) Go to forum and click the QUESTION button! :D 

 Sep 18, 2020

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