F ` ( x ) = ( x² )` · e^(5x) + x² · ( e^(5x) )` =
= 2 x · e^(5x) + 5 e^(5x) · x² =
= x e^(5x) ( 2 + 5 x )
f `` ( x ) = ( 2 x e^(5x) + 5 x² e^(5x) ) ` =
= ( 2 x ) ˙e^(5x) + 2 x ( e^(5x) )` + ( 5 x² ) ` · e^(5x) + ( e^(5x)) ` · 5 x² =
= 2 · e^(5x) + 10 x · e^(5x) + 10 x · e^(5x) + 25 x² · e^(5x) =
= e^(5x) · ( 2 + 20 x + 25 x² )
Answer:
yes
Step-by-step explanation:
The cones are not concurrent because one cone is obliquewhich makes the length of these land height is different have the same radius same volumes and same height are the cones concurrent is the corresponding slant height sorry equal .concurrent counts must have concurrent corresponding slant height with all this corresponding being concurrent
the answer is C
the dependent variable is distance, equation is y = 22x
Answer:
Step-by-step explanation:
Domain of 5x^2 + 2x - 1 is all real numbers.
