Let us have four distinct collinear points $a,$ $b,$ $c,$ and $d$ on the cartesian plane. the point $c$ is such that $\dfrac{ab}{cb} = \dfrac{1}{2}$ and the point $d$ is such that $\dfrac{da}{ba} = 3$ and $\dfrac{db}{ba} = 2.$ if $c = (0, 4),$ $d = (4, 0),$ and $a = (x, y),$ what is the value of $2x + y$?

Start with a line segment connecting two points, A and B. means DA is 3 times longer than BA. Clearly, D cannot fall between A and B because that would mean DA is shorter than BA. So there are two possible locations where D can be placed on the line relative to A and B.

But with , or the fact that DB is 2 times longer than BA, we can rule out one of these positions; referring to the attachment, if we place D to the left of A, then DB would be 4 times longer than BA.

Finally, , so that CB is 2 times longer than AB. Again we have two possible locations for point C (it cannot fall between A and B), but one of them forces C to occupy the same point as D. However, A, B, C, D are distinct, so C must fall to the left of A.

Now let be the length of AB. Then the length of CD in terms of is . We have the coordinates of C and D, and the distance between them is . So

The slope of the line through C and D is

and so the equation of the line through these points is

So the coordinates of A are . The distance between C and A is , so we have

Since A falls to the right of C (in the plane, not just in the sketch), we know to take the positive value . Then the coordinate is .

All this to say that A is the point (1, 3), so

incomplete question. how do you expect anyone to you when you do not post the question in its entirety?

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