the area of a rhombus formed by the points of intersection of the parabolas y=x^2−b, y=−x^2+b, and their vertices is 54. find b.

So hmmm if you notice  y = x² - b is really just the parent function x² with a downward shift of "b".  And y = - x²  + b is really just x² upside-down shifted upwards by "b".  so, both parabolas, look like the picture below, and their vertices is at the origin and then they get shifted, one upwards and the other downwards.

So the rhombus lies in their vertices and intersection points.

now, the area of the rhombus is 54, we're taking the green triangle in the picture, which is half of the rhombus, so its area is 27 then, it has a base of "2b", and an altitude of whatever the value of "x" at the intersection happens to be.

so hmmm let's check what's "x" anyway.

1. 0.3 •4 is the same as 4 groups of 0.32. 7•0.8 is the same as 7 groups of 0.83. it increases by 0.5 every time, so 0.5x4 it increases by 0.9 every time, so 0.9x5. 0.9•12=10.86. 0.47•5=2.35i can finish the next ones in a separate post