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Area of the figure = 806.5 in²

Solution:

Length of the rectangle = 16 in

Breadth of the rectangle = 9 in

Area of the rectangle = length × breadth

                                    = 16 × 9

Area of the rectangle = 144 in²

Base of the triangle = 31 in

Height of the triangle = 20 in

Area of the triangle =

Area of the triangle = 310 in²

Parallel sides of the trapezium = 16 in and 31 in

Height of the trapezium = 35 – 20 = 15 in

Area of the trapezium =

Area of the trapezium = 352.5 in²

Area of the figure = Area of rectangle + Area of triangle + Area of trapezium

                              = 144 in² + 310 in² + 352.5 in²

Area of the figure = 806.5 in²

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

what is the information provided?

step-by-step explanation: