Let's fill in the missing values in the table step-by-step.
We start with the given information:
1. [tex]\( 1 \square J = 10^{-3} J \)[/tex]
Next, we interpret the notation for [tex]\( kJ \)[/tex] and [tex]\( dJ \)[/tex] and fill in the blanks accordingly:
### Row 2: [tex]\( 1 kJ = 10^x J \)[/tex]
The prefix [tex]\( k \)[/tex] stands for "kilo", meaning [tex]\( 10^3 \)[/tex].
So, [tex]\( 1 kJ = 10^3 J \)[/tex].
To fill in the value for [tex]\( x \)[/tex], we see that [tex]\( x = 3 \)[/tex].
Thus, [tex]\( \boxed{3} \)[/tex]
### Row 3: [tex]\( 1 dJ = 10^y J \)[/tex]
The prefix [tex]\( d \)[/tex] stands for "deci", meaning [tex]\( 10^{-1} \)[/tex].
So, [tex]\( 1 dJ = 10^{-1} J \)[/tex].
To fill in the value for [tex]\( y \)[/tex], we see that [tex]\( y = -1 \)[/tex].
Thus, [tex]\( \boxed{-1} \)[/tex]
Finally, we interpret the given relation for the square Joule [tex]\( \square J \)[/tex]:
### Row 4: [tex]\( 1 \square J = 10^6 J \)[/tex]
This is already filled out, and it tells us that [tex]\( 1 \square J \)[/tex] is a large quantity, specifically [tex]\( 1 \square J = 10^6 J \)[/tex].
Thus, the completed table is:
[tex]\[
\begin{array}{l}
1 \square J = 10^{-3} J \\
1 kJ = 10^3 J \\
1 dJ = 10^{-1} J \\
1 \square J = 10^6 J \\
\end{array}
\][/tex]
The important points are:
- [tex]\( x = 3 \)[/tex] for kilojoules (kJ).
- [tex]\( y = -1 \)[/tex] for decijoules (dJ).
