# Question #a89be

##### 1 Answer

#### Explanation:

For starters, you know that **all non-zero digits** are significant, so you can say that your number has *at least* **significant figures** because it contains

#color(black)(color(red)(1)0color(red)(9)0.00color(red)(1)0) -> " 3 non-zero digits: " { color(red)(1), color(red)(9), color(red)(1) }#

Now, you should also know that all zeros that are *sandwiched* between two **non-zero digits** are significant. In other words, if a zero follows a non-zero digit and is followed by a non-zero digit, regardless if other zeros are adjacent, it is significant.

In your case, you have **sandwiched zeros**. The first sandwiched zero follows

#color(black)(color(red)(1)color(blue)(0)color(red)(9)0.00color(red)(1)0)#

The second, third, and fourth sandwiched zeros follow

#color(black)(color(red)(1)color(blue)(0)color(red)(9)color(blue)(0). color(blue)(00)color(red)(1)0) -> " 4 sandwiched zeros: " {color(blue)(0), color(blue)(0), color(blue)(0), color(blue)(0) }#

Finally, you should know that *trailing zeros*, i.e. zeros that follow a non-zero digit and **are not** followed by a non-zero digit, that **follow a decimal point** are significant.

In this case, you have one trailing zero that follows the decimal point and the

#color(black)(color(red)(1)color(blue)(0)color(red)(9)color(blue)(0). color(blue)(00)color(red)(1)color(green)(0)) -> " 1 significant trailing zero: " {color(green)(0)}#

Therefore, you can say that your number has a total of

#color(white)(aaaaaaa)color(red)("3 non-zero digits") " "+#

#color(white)(aaaaa)color(blue)("4 sandwiched zeros")#

#color(green)("1 significant trailing zero")#

#color(white)(aaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#

#color(white)(aaaaa)"8 significant figures"#