A whale swims due east for a distance of 7.299x10^1 km, turns around and goes due west for 3.58x10^0 km, and finally turns around again and heads 2.64x10^1 km due east. What is the displacement (referenced to the East) of the whale, to 2 significant figures?

72.99 - 3.58 + 26.4 = _____

Add and round off to two sig. figs.

To find the displacement of the whale, we need to add up all the distances traveled in the eastward direction and subtract the distances traveled in the westward direction.

The whale swims east for a distance of 7.299x10^1 km and then turns around and swims west for 3.58x10^0 km. The total distance traveled in the eastward direction is 7.299x10^1 km - 3.58x10^0 km.

Next, the whale turns around again and swims east for 2.64x10^1 km. We need to add this distance to the previous total.

Adding up the distances, we get:

(7.299x10^1 km - 3.58x10^0 km) + 2.64x10^1 km.

Simplifying the expression:

= 7.299x10^1 km - 3.58x10^0 km + 2.64x10^1 km

= 7.299x10^1 km + 2.64x10^1 km - 3.58x10^0 km

Now we can perform the addition:

= (7.299 + 2.64) x 10^1 km - 3.58x10^0 km

= 9.939x10^1 km - 3.58x10^0 km

= 9.939x10^1 km - 3.58 km

= 9.939x10^1 km - 3.58 km

= 96.19 km

Therefore, the displacement of the whale, referenced to the East, is 96.19 km (to 2 significant figures).