2. pb and j array ken has made a selection of square jam sandwiches for party, which he arranges in a single layer on the table in perfect rectangle. sarah has brought along the same number of square peanut butter sandwiches, which she puts out, also as a single layer, to form a border va (one sandwich wide) around ken's given that all of the sandwiches are the same size and they are all used, how many sandwiches could each person possibly have made?

Let's assume Ken arranged his sandwiches in a rectangular shape with length 'L' and width 'W'. Now Sarah arranged her sandwiches to form a border around Ken's rectangle. The solution diagram is attached below.

Ken needs as much sandwiches as area of rectangle i.e. number of Ken's sandwiches would be = LW.

It is clear that Sarah needs as much sandwiches as perimeter of rectangle and extra four sandwiches to put on corners. It means number of Sarah's sandwiches would be = 2·(L+W) + 4.

Given that all of the sandwiches are the same size and they are all used.

Ken and Sarah bought same number of sandwiches. So we need to solve the following equation :-

Number of Ken's sandwiches = Number of Sarah's sandwiches

LW = 2·(L+W) + 4

Using trial and error approach, we get 6 x 4 = 2 x (6 + 4) + 4.

So number of Ken's sandwiches would be LW = 6 x 4 = 24 sandwiches.

Hence, final answer is "Each person could have 24 sandwiches".

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents

sine

step-by-step explanation: