• Use whichever method you prefer (the area model or the distributive property)

2a^2+22a+−1

Step-by-step explanation:

Let's simplify step-by-step.

2a^2−5a+(9)(3)a−1

=2a^2+−5a+27a+−1

Combine Like Terms:

=2a^2+−5a+27a+−1

=(2a^2)+(−5a+27a)+(−1)

=2a^2+22a+−1

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐