Solve the following initial value problem. StartFraction d squared s Over dt squared EndFraction d2s dt2equals=negative 16 Bold sine (4 t minus StartFraction pi Over 2 EndFraction )−16sin4t− π 2, s prime (0 )equals 500s′(0)=500, s(0)equals=0

Given: ad¯¯¯¯¯ is an altitude. prove: ab2+ac2=cb2 right triangle a b c with right angle a. point d lies on side b c and segment a d is drawn. angle a d c is a right angle. drag and drop a reason into each box to correctly complete the two-column proof. statement reason ad¯¯¯¯¯ is an altitude, and ∠bac is a right angle. given ∠adb and ∠adc are right angles. definition of altitude ∠bac≅∠bda ? ∠bac≅∠adc ? ∠b≅∠b ? ∠c≅∠c reflexive property of congruence △abc∼△dba ? △abc∼△dac aa similarity postulate abbd=cbab ? ab2=(cb)(bd) cross multiply and simplify. acdc=cbac polygon similarity postulate ac2=(cb)(dc) cross multiply and simplify. ab2+ac2=ab2+(cb)(dc) addition property of equality ab2+ac2=(cb)(bd)+(cb)(dc) substitution property of equality ab2+ac2=(cb)(bd+dc) ? bd+dc=cb segment addition postulate ab2+ac2=cb2 substitution property of equality