Use the formula ln(1 + x) = ∞ n = 1 (−1)n − 1xn n = x − x2 2 + x3 3 − x4 4 + for |x| < 1 and your knowledge of alternating series to find the smallest n such that the partial sum sn approximates ln 3 2 to within an error of at most 10−3. confirm this using a calculator to compute both sn and ln 3 2 . (round your answers to six decimal places.)

10010010001010010101001001010010100101010010101010010010101010110010001011010010010001

step-by-step explanation:

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl