Solve N=2rt^2-7rx for x

          N - 2 r t²

x =  -  

                7x

Step-by-step explanation:

N = 2 r t² - 7 r x

solve for x

N = 2 r t² - 7 r x

N + (-2 r t²) = (2 r t²  - 7 r x) + (-2 r t²)

N - 2 r t² = 2 r t² - 7 r x - 2 r t²

N - 2 r t²  - 2 r t² +  2 r t² = - 7 r x

N - 2 r t² = 7 r x

therefore,

          N - 2 r t²

x =  -  

                7x

< 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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step-by-step explanation: