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Will give brainliest! 2017 amc 12a problem 24 quadrilateral [tex]abcd[/tex] is inscribed in circle [tex]o[/tex] and has side lengths [tex]ab=3, bc=2, cd=6[/tex], and [tex]da=8[/tex]. let [tex]x[/tex] and [tex]y[/tex] be points on [tex]\overline{bd}[/tex] such that [tex]\frac{dx}{bd} = \frac{1}{4}[/tex] and [tex]\frac{by}{bd} = \frac{11}{36}[/tex]. let [tex]e[/tex] be the intersection of line [tex]ax[/tex] and the line through [tex]y[/tex] parallel to [tex]\overline{ad}[/tex]. let [tex]f[/tex] be the intersection of line [tex]cx[/tex] and the line through [tex]e[/tex] parallel to [tex]\overline{ac}[/tex]. let [tex]g[/tex] be the point on circle [tex]o[/tex] other than [tex]c[/tex] that lies on line [tex]cx[/tex]. what is [tex]xf\cdot xg[/tex]? show all work!

Consider this equation. |y + 6| = 2 what can be concluded of the equation? check all that apply. there will be one solution. there will be two solutions. the solution to –(y + 6) = 2 will be also be a solution to the given absolute value equation. the solution(s) will be the number(s) on the number line 2 units away from –6. the value of y must be positive since the variable is inside absolute value signs.

20 solve 2(4 x + 3) < 5 x + 21. a) { x | x < 9} b) { x | x > -5} c) { x | x > -9} d) { x | x < 5}