keep the first four numbers then round up or down the last number. in this case i would keep 3499, then the last 9 would round up, making the number 35000. For only 4 sig numbers, you would have 3500.
there are 3 sig figs. 4, 0, and 5 are the sig figs
13.00 has 4 sig figs.
4
4... (:))
Depending on the number of significant figures, it could be the following:4.251x10^3 (4 sig figs)4.25x10^3 (3 sig figs)4.3x10^3 (2 sig figs)4x10^3 (1 sig fig)
0.0005482
143.4
there are 3 sig figs. 4, 0, and 5 are the sig figs
There are 4 sig figs in 20.13
13.00 has 4 sig figs.
4... (:))
4
4
Depending on how many significant figures you want, it can be2.421200x10^6 (7 sig figs)2.42120x10^6 (6 sig figs)2.4212x10^6 (5 sig figs)2.421x10^6 (4 sig figs)2.42x10^6 (3 sig figs)2.4x10^6 (2 sig figs)2x10^6 (1 sig fig)
4 sig figs the ending zero is not significant unless it was a measurement and the zero was an estimation that was part of the measurement.
there are 4 in the solution because the numbers are multiplied in the end, the solutions will have the same number of sig. figs. as the number in the problem with the lowest number of sig. figs. Ex: the lowest number of sig figs is 4 because 0.0107 has 4 sig figs.
4 of them.