Find the equation of the tangent line to the graph of y=x^3-3x^2+x at the point (2,-2)

< 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

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step-by-step explanation:

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x> 6

step-by-step explanation:

8x+8> 7x+14

8x-7x> 14-8

x> 6

answer is the third choice:

3x^5+4x^4+x^3+2x^2-10x-7=0

step-by-step explanation: