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 )
The answer is B.
If you find a common denominator, the process becomes much easier. I hope this was helpful.
Answer: a
reason:
if you take any of the numbers and plug it in it works
______________________________
<h3>A = m ÷ t × 100</h3><h3>= $100 ÷ $350 × 100</h3><h3>= <u>28.6%</u></h3><h3>THE GOAL WAS SURPASSED BY 28.6%</h3>
______________________________
0=0
Step-by-step explanation:
4(×+3)=4x+12
4x+12=4x+12
Subtract 12 from both sides of the equation
4x+12-12=4x+12-12
simplify
4x=4x
Subtract 4x from both sides of the equation
4x-4x= 4x-4x
Simplify and combine like terms
0=0
