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:
B. y - 35 = 2(x - 10)
Step-by-step explanation:
The height of the plant, y, after x days could be modeled by the equation
<h3>y-y0=k(x-xo) (1)
,</h3><h3>
where y0 was the initial height at 'x0'th. day, and k is the constant of proportionality.</h3><h3>
From equation (1), k could be evaluated as follows:</h3><h3>
k=(y-y0)/(x-x0) </h3><h3>
From the problem statement, we may determine k by plugging in the given values, e.g. y0= 35, x0=10, y=55, x=20.</h3><h3>
Thus,</h3><h3>
k=(55-35)/(20-10)=2</h3><h3>
Therefore, the model equation becomes</h3><h3>
y-35=2(x-10)</h3><h3>
</h3>
Answer:
I think it is non proportional
Answer:
X is 50
Step-by-step explanation:
Answer:
0.32608696
Step-by-step explanation:
