Answer:
1. 3(9K-2)
2. 5(x+12y)
Step-by-step explanation:
Answer:
The change in volume is estimated to be 17.20 
Step-by-step explanation:
The linearization or linear approximation of a function
is given by:
where
is the total differential of the function evaluated in the given point.
For the given function, the linearization is:

Taking
inches and
inches and evaluating the partial derivatives we obtain:

substituting the values and taking
and
inches we have:

Therefore the change in volume is estimated to be 17.20 
Answer:
15.6
Step-by-step explanation:
40% × 39 = 15.6
Hope this helps
i think C but i dont know forsure
Answer:
The answer is A so put it