16. Position and label the following triangle on the coordinate plane. right isosceles ABC with congruent sides AB and AC a units long C(0, 1) C(0.4) O A1,0) A0 B(1,0) |A(0,0) B(20) Option 1 O Option A C(0,0) A(0, 2) Cl2,2) 0 A(0,0) Ba, 0) B 2.0)

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

b. b < 1 might be the correct answer