Find the first 3 Iterations of the function here: g(x)=1/3x+1 if you have an initial value of 2.

An example on how to complete it below.

1st it: g(2)=1/3(2)+1=0.67+1=1.67

2nd it: g^2(2)=1/3(1.67)+1=0.56+1=1.56

3rd it: g^3(2)=1/3(1.56)+1=0.52+1=1.52

The first three iterations are 1.67, 1.56 and 1.52

Step-by-step explanation:

Given the function g(x)=1/3x+1

To get the first threw iteration with initial value of x = 2

First iteration at x= 2:

g(2) = 2/3+1

g(2) = (2+3)/3

g(2) = 5/3 = 1.67

Second iteration will be when x = g(2) = 5/3

g²(2) = g(5/3) = 1/3(5/3) + 1

g²(2) = g(5/3) = 5/9 + 1

g²(2) = g(5/3) = 14/9 = 1.56

Third iteration will be at when

x = g²(2) = 14/9

g³(2) = g(14/9) = 1/3(14/9) + 1

g³(2) = g(14/9) = 14/27 + 1

g³(2) = g(14/9) = 41/27 = 1.52

The first three iterations are 1.67, 1.56 and 1.52

< 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: