Why do Taylor polynomials improve in their approximations as you continue to iterate them? What does the error (remainder) term have to do with this?

use this formula:

p(a\b) = p(a and b)/p(b).

take it from here.

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.

bfgbh bhf gfdzf bhgfrdfgbn cvdgsytrghvn ncvhgfdtrseawa hgdrsxcvbgfrewq

step-by-step explanation:

answer: entity is prob the answer

step-by-step explanation: