We are interested in the dimensions of a certain square a rectangle had length triple the side of this square and width two units less than the side of this square. if the two areas are equal, what are the squares dimensions (w x h)

Let s be the side length of the square. The dimensions of the rectangle are three times the side of the square (i.e. 3s), and two less than the side of the square (i.e. s-2).

So, the area of this rectangle is

The area of the square is , and we know that the two areas are the same, so we have

The solution s=0 would lead to the extreme case where the rectangle and the square are actually a point, so we accept the solution s=3.

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

whatvare the choices

step-by-step explanation: