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
You have not provided the diagram/coordinates for point Q, therefore, I cannot provide an exact answer.
However, I can help you with the concept.
When rotating a point 90° counter clock-wise, the following happens:
coordinates of the original point: (x,y)
coordinates of the image point: (-y,x)
Examples:
point (2,5) when rotated 90° counter clock-wise, the coordinates of the image would be (-5,2)
point (1,9) when rotated 90° counter clock-wise, the coordinates of the image would be (-9,1)
point (7,4) when rotated 90° counter clock-wise, the coordinates of the image would be (-4,7)
Therefore, for the given point Q, all you have to do to get the coordinates of the image is apply the transformation:
(x,y) .............> are changed into.............> (-y,x)
Hope this helps :)
Answer:
Step-by-step explanation:
big 4l what lil savage tooh say
Answer: 300*0.73^x
Step-by-step explanation:
300 is the initial value and every time x goes up by 1, the value get multiplied by 0.73
when x=3
then the value would be 300*0.73*0.73*0.73 or 300*0.73^3
then for x amount of times the equation would be 300*(0.73 x times ) or 300*0.73^x
