X = 89
41 + 50 = 91 which is 89 away from 180
You should have drawn1 - x-axis and y-axis in light pencil.2 - graphed a down-facing parabola with the top of the frown on the y-axis at y = 2. It should be crossing the x-axis at ±√2. This should be in dark pencil or another color.3 - In dark pencil or a completely new color, draw a rectangle with one of the horizontal sides sitting on top of the x-axis and the other horizontal side touching the parabola at each of the top corners of the rectangle. The rectangle will have half of its base in the positive x-axis and the other half on the negative x-axis. It should be split right down the middle by the y-axis. So each half of the base we will say is "x" units long. So the whole base is 2x units long (the x units to the right of the y-axis, and the x units to the left of the y-axis) I so wish I could draw you this picture... In the vertical direction, both vertical edges are the same length and we will call that y. The area that we want to maximize has a width 2x long, and a height of y tall. So A = 2xy This is the equation we want to maximize (take derivative and set it = 0), we call it the "primary equation", but we need it in one variable. This is where the "secondary equation" comes in. We need to find a way to change the area formula to all x's or all y's. Since it is constrained to having its height limited by the parabola, we could use the fact that y=2 - x2 to make the area formula in only x's. Substitute in place of the "y", "2 - x2" into the area formula. A = 2xy = 2x(2 - x2) then simplify A = 4x - 2x3 NOW you are ready to take the deriv and set it = 0 dA/dx = 4 - 6x2 0 = 4 - 6x2 6x2 = 4 x2 = 4/6 or 2/3 So x = ±√(2/3) Width remember was 2x. So the width is 2[√(2/3)]Height is y which is 2 - x2 = 2 - 2/3 =4/3
Answer:
y=3/10x+0.6
Step-by-step explanation:
3x-10y=-6
-3x -3x
-10y=-3x-6
Divide -10 from everything.
y=3/10x+0.6
Answer:
Parabolas are frequently used in physics and engineering for things such as the design of automobile headlight reflectors and the paths of ballistic missiles. Parabolas are frequently encountered as graphs of quadratic functions.
To fill the squares look for the one filled in column/row/diagonal and figure out its sum.
All other columns/rows/diagonals must be equal to that, so whenever there is a column/row/diagonal with a single missing value you can add the others up and calculate the missing value.
So you have to find those single missing value columns/rows/diagonals, fill them out and continue with the next one until the whole square is filled in.
