# How do you convert 2.3 x10^-2 into expanded notation?

$6 \cdot {10}^{-} 2 x$

#### Explanation:

By definition, expanded notation produces an expression which represents each place holder's value.

NB: Standard notation is always represented by the sums of all the values multiplied by ${10}^{\text{to the power of any number}}$ which should equal the original expression.

for e.g.
428 = $4 \cdot {10}^{2} + 2 \cdot {10}^{1} + 8 \cdot {10}^{0}$

However, this question takes this idea one step further and introduces decimal places.

By using the above example, if we were given 428.39,
this would be expressed as:

428.39 = $4 \cdot {10}^{2} + 2 \cdot {10}^{1} + 8 \cdot {10}^{0} + 3 \cdot {10}^{-} 1 + 9 \cdot {10}^{-} 2$

Beautiful isn't it.

As one can see, the degree (the exponents) of the $x ' s$ are consecutively decreasing by one as one moves down every unit placeholder.

With this in mind, by looking at your question, $2.3 x {10}^{-} 2$, we can apply the same trend.

Step one: (turn the expression into decimal format)

$2.3 x {10}^{-} 2$,

$= \frac{6}{{10}^{2}} x$

$= \frac{6}{100} x$

$= 0.06 x$

Step Two: (convert to Standard notation)

Since this expression only has one numerical value of 6 which is occupying the hundredth's position, we multiply 6 by ${10}^{-} 2$ as this is equal to 0.06.

Now with regards to the algebraic terms, such as $x$ in this case, since these variables are unknown and can have any value, they are left unaltered.

For term clarity,
"Numerical Form" of 42 has a "Standard Form" of 40 + 2 which can be expressed in the "Standard Notation" of $4 \cdot 10 + 2 \cdot {10}^{0}$