Given f(x)=x^2 +3 find h(x) h(x)=f(5x)-f(-2)

  h(x) = 25x^2 -4

Step-by-step explanation:

Fill in the given arguments and evaluate:

  h(x) = f(5x) -f(-2)

  = ((5x)^2 +3) - ((-2)^2 +3)

  h(x) = 25x^2 -4

< 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⭐