PLEASEEE HELP♥️♥️♥️ (I only need #12)

Step-by-step explanation:

solution

given KL is perpendicular bisector of PQ

1. in triangle KPL and KQL

a.∠KLP= ∠KLQ                                       1. a. KL is perpendicular bisector of PQ

b. KL is common in both triangle              b. common side

c. PL=LQ                                                   c. KL bisects the PQ

2 therefore triangle KPL and KQL is   2.by R.H.S axiom

congurent triangle

3.∠P=∠Q                                                3. corresponding angle of triangle are  

                                                                       equal

proved

- 2 + 10x³ - 21x² + 19x - 4

step-by-step explanation:

each term in the second factor is multiplied by each term in the first factor

x²(- 2x² + 4x - 1) - 3x(- 2x² + 4x - 1) + 4(- 2x² + 4x - 1) ← distribute

= - 2 + 4x³ - x² + 6x³ - 12x² + 3x -8x² + 16x - 4

collect like terms

= - 2 + 10x³ - 21x² + 19x - 4 ← in standard form