Find an equation of the tangent line to the curve at the given point. y = 7x2 − x3, (1, 6)y = correct: your answer is correct. illustrate by graphing the curve and the tangent line on the same screen.

  y = 11(x -1) +6

Step-by-step explanation:

The slope at the given point is ...

  y' = 14x -3x^2 = 14(1) -3(1^2) = 11

Then the point-slope equation for the line can be written as ...

  y = m(x -h) +k . . . . . . for slope m at point (h, k)

  y = 11(x -1) +6

< 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⭐

the answer is a i belaeve atleast thats what my calculator just gfave me

step-by-step explanation: