the value ax^2 + bx is 5 when x=1, and is equal to 0 when x=1.what are the values of a and b. explain how u arrived at your answer

a = 1.25, b = 2.5, c = 1.25

Step-by-step explanation:

Use vertex form and substitute a point into it (1, 5)

5 = a(1 + 1)^2 + 0

5 = a(2)^2 + 0

5 = 4a

a = 5/4

y = 5/4 (x + 1)^2

y = 5/4 (x^2 + 2x + 1)

1.25 x^2 + 2.5x + 5/4

Is there a "c" value otherwise it's not possible

< 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⭐

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