Round
321.786 is about 322
240.23 is about 240
322+240=562
about 562
Answer:
-I haven't done this, but I have copied and pasted from another question and answer from the user syed514:
Graphing is one way to do the problem.But sometimes, graphing it is hard to do.So here’s an algebraic method.
If M(m1, m2) is the midpoint of two points A(x1, y1) and B(x2, y2),then m1 = (x1 + x2)/2 and m2 = (y1 + y2)/2.In other words, the x-coordinate of the midpointis the average of the x-coordinates of the two points,and the y-coordinate of the midpointis the average of the y-coordinates of the two points.
Let B have coordinates (x2, y2) in our problem.Then we have that 6 = (2 + x2)/2 and 8 = (3 + y2)/2.
Solving for the coordinates gives x2 = 10, y2 = 13
As there are 81 tiles total, that makes your area 81. Being a square, that makes your area equation A=s^2. As you need a side, this makes your equation s=√A. Plugging in 81 as A, we get s=√81=9, so an edge has 9 tiles.
F(x) = 18-x^2 is a parabola having vertex at (0, 18) and opening downwards.
g(x) = 2x^2-9 is a parabola having vertex at (0, -9) and opening upwards.
By symmetry, let the x-coordinates of the vertices of rectangle be x and -x => its width is 2x.
Height of the rectangle is y1 + y2, where y1 is the y-coordinate of the vertex on the parabola f and y2 is that of g.
=> Area, A
= 2x (y1 - y2)
= 2x (18 - x^2 - 2x^2 + 9)
= 2x (27 - 3x^2)
= 54x - 6x^3
For area to be maximum, dA/dx = 0 and d²A/dx² < 0
=> 54 - 18x^2 = 0
=> x = √3 (note: x = - √3 gives the x-coordinate of vertex in second and third quadrants)
d²A/dx² = - 36x < 0 for x = √3
=> maximum area
= 54(√3) - 6(√3)^3
= 54√3 - 18√3
= 36√3.
