A pentagon ABCDE has sides AE and CD parallel and the line EC is parallel to side AB. Sides ED and BC, when extended, meet at a point F. ∠ABC is equal to ∠CDE. Show that ∠AEC = ∠CFD Pls hel

< 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⭐

the correct answers i think are b. and d.

give me brainliest if it is correct!

i only know the sequence of arithmetic it adds the first is 5+4 then it goes up as it adds to then it is 9+5 then 14+6 then to 20+7 and so on. 5,9,14,20,27,35.

step-by-step explanation:

answer: the answer is one

step-by-step explanation:

pemdas, in this situation there is no parenthesis or exponents so you go to multiplication, 1x1=1, then division 1/1= 1 then addition 1+1=2 then subtraction 2-1=1