Consider these three changes in the possible distributions of six gaseous particles within three interconnected boxes. Which change has a positive ÎS?

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

the answer is 150.6

step-by-step explanation:

you multiply .02 by 30 and get .6.

and then you multiply 5.00 by 30 and get 150.

hope this !