Find all points on the x-axis that are 14 units from the point (3,-7).

  (3±√147, 0) ≈ (-9.124, 0) and (15.124, 0)

Step-by-step explanation:

The points (x, 0) satisfy the distance formula:

  d = 14 = √((x -3)² +(0+7)²)

  196 = (x -3)² +49 . . . . . square both sides

  147 = (x -3)² . . . . . . . . . subtract 49

  3 ±√147 = x . . . . . . . . . take the square root and add 3

The points at distance 14 from (3, -7) on the x-axis are (3±√147, 0).

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step-by-step explanation:

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