Rewrite the equation x^2-4x+9=5 in the form (x-p)^2=q

(x - 2)² = 0

Step-by-step explanation:

given x² - 4x + 9 = 5 ( subtract 5 from both sides )

x² - 4x + 4 = 0

This is a perfect square of the form

(x - a)² = x² - 2a + a²

compare coefficients of the x-term

- 2a = - 4 ⇒ a = 2

⇒ (x - 2)² = 0 ← is the required form

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< 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⭐