PLEASEEE HELP ME FO THISSS

As C is the mid-point of BD.

Therefore, BC=DC ...(i)

As AB ⊥ BD and ED ⊥ BD

So, ∠ABC= 90° and ∠EDC= 90°

Therefore, ∠ABC=∠EDC= 90° ...(ii)

As two line segments, AE and BD intersect at point C, so the vertically opposite angles are equal.

Therefore, ∠BCA=∠DCE ...(iii)

Now, in ΔABC and ΔEDC,

Angle, ∠BCA=∠DCE [ from equation (iii)]

Side, BC=DC [ from equation (i)]

Angle, ∠ABC=∠EDC [ from equation (ii)]

So, by ASA property of congruency, ΔABC and ΔEDC are congruent

Hence, ΔABC ≅ ΔEDC.

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.