An archway is modeled by the equation y = -2x2 + 8x. A rod is to be placed across the archway at an angle defined by the equatio
n x − 2.23y + 10.34 = 0. If the rod is attached to the archway at points A and B, such that point B is at a higher level than point A, at what distance from the ground level is point B?
1) y= - 2x² + 8x. It's a parabola open downward (a<0) 2) x - 2.23.y + 10.34 = 0 . Re-write it : y = (x/2.23) + (10.34/2.23), a linear equation. To find the intersections between 1) & 2), let 1) = 2) -2x² + 8x = (x/2.23) + (10.34/2.23) -2x² + 8x - (x/2.23) - (10.34/2.23) =0 ; solve this quadratic for x values:
x' (that is A) = 0.772 & x" (that is B) = 3. (these are the values of x-intercept (parabola with line). To calculate the y-values, plug x' & x' in the equation: for x' = 0.772, y = 0.34 → B(0.772 , 0.34) for x" = 3, y = 0.016 → A(3 , 0.O16) So B IS AT 0.34 Unit from the ground
she has 17 slik roses already, and 6 friends give her 3 silk roses each, so that's 6 times 3, which makes 18, and 17 plus 18 is 35, so she has 35 silk roses in all.
