Answer:
B. 6 units
Step-by-step explanation:
Given equations:
[tex]y = -2x^2 + 8x[/tex]
[tex]x-2.23y + 10.34 = 0[/tex]
The points at which the rod is attached to the archway are the points of intersection of the two equations.
Rearrange the second equation to make x the subject:
[tex]\implies x=2.23y-10.34[/tex]
Substitute this into the first equation to create a quadratic:
[tex]y = -2(2.23y-10.34)^2 + 8(2.23y-10.34)[/tex]
[tex]y = -2(4.9729y^2-46.1164y+106.9156) + 17.84y-82.72[/tex]
[tex]y=-9.9458y^2+92.2328y-213.8312+17.84y-82.72[/tex]
[tex]y=-9.9458y^2+110.0728y-296.5512[/tex]
[tex]-9.9458y^2+109.0728y-296.5512=0[/tex]
[tex]\boxed{\begin{minipage}{3.6 cm}\underline{Quadratic Formula}\\\\$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$\\\\when $ax^2+bx+c=0$ \\\end{minipage}}[/tex]
Solve the quadratic using the quadratic formula:
[tex]\implies y=\dfrac{-(109.0728) \pm \sqrt{(109.0728)^2-4(-9.9458)(-296.5512)}}{2(-9.9458)}[/tex]
[tex]\implies y=\dfrac{-109.0728 \pm \sqrt{99.12}}{-19.8916}[/tex]
[tex]\implies y=\dfrac{109.0728 \pm \sqrt{99.12}}{19.8916}[/tex]
[tex]\implies y=5.983867702, \quad y=4.982851918[/tex]
As point B is at a higher level than point A, the y-value of point B is approximately 6 units.
Therefore, point B is 6 units from ground level.
