Answer:
The width is 7 cm
Step-by-step explanation:
The volume of a cereal box is
V = l*w*h
6860 = 35 * w * 28
6860 =980 w
Divide each side by 980
6860/980 = w
7 = w
Answer:
Required series is:

Step-by-step explanation:
Given that
---(1)
We know that:
---(2)
Comparing (1) and (2)
---- (3)
Using power series expansion for 


![=-[c+\sum\limits^{ \infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}]](https://tex.z-dn.net/?f=%3D-%5Bc%2B%5Csum%5Climits%5E%7B%20%5Cinfty%7D_%7Bn%3D0%7D%20%28-1%29%5E%7Bn%7D%5Cfrac%7Bx%5E%7B2n%2B1%7D%7D%7B2n%2B1%7D%5D)


as

Hence,

.394= 197/500
you can't make it a mixed number since there isn't a whole number as with an improper fraction.
Please ask some questions and I will assist to the best of my abilities. <span />