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4. Carbon Migration Calculation

\begin{tabular}{|l|l|l|l|l|}
\hline 4 & [tex]$x$[/tex] & 5 & [tex]$=$[/tex] & 3 \\
\hline 4 & & 4 & & \\
\hline 6 & 1 & 6 & [tex]$=$[/tex] & 36 \\
\hline 10 & & [tex]$=$[/tex] & & \\
\hline 10 & & 19 & [tex]$=$[/tex] & 190 \\
\hline
\end{tabular}

\begin{tabular}{|l|l|l|l|l|}
\hline & 1 & 50 & [tex]$=$[/tex] & 80 \\
\hline 4 & & & & 4 \\
\hline & 1 & 5 & [tex]$=$[/tex] & 30 \\
\hline [tex]$=$[/tex] & & & & [tex]$=$[/tex] \\
\hline & 1 & 36 & [tex]$=$[/tex] \\
\hline
\end{tabular}

12


Sagot :

Sure, let's work through this question step-by-step in a detailed manner.

### First Table
1. Row 1: [tex]\(4 \cdot x \cdot 5 = 3\)[/tex]
- From this, we need to solve for [tex]\( x \)[/tex].
[tex]\[ 4 \cdot x \cdot 5 = 3 \][/tex]
[tex]\[ 20x = 3 \][/tex]
[tex]\[ x = \frac{3}{20} = 0.15 \][/tex]

2. Row 3: [tex]\( 6 \cdot 1 \cdot 6 = 36 \)[/tex]
- This equation is straightforward.
[tex]\[ 6 \cdot 1 \cdot 6 = 36 \][/tex]
- It is already correct.

3. Row 5: [tex]\( 10 \cdot 19 = 190 \)[/tex]
- This equation is straightforward.
[tex]\[ 10 \cdot 19 = 190 \][/tex]
- It is already correct.

### Second Table
4. Row 1: [tex]\( 1 \cdot 50 = 80 \)[/tex]
- This should be solved as:
[tex]\[ 1 \cdot 50 \neq 80 \][/tex]
- So there's a possibility of missing operation or incorrect data. If nothing is to be changed, we'll move forward.

5. Row 3: [tex]\( 1 \cdot 5 = 30 \)[/tex]
- The operation isn't accurate:
[tex]\[ 1 \cdot 5 = 5 \][/tex]
- Hence, assuming the multiplication should match the desired output context.

Now assembling up,

### Solutions in Simple Form
1. From the first Table:
- [tex]\( x = 0.15 \)[/tex]
- Direct continuity confined with mistakes on direct terms, signaling it's actual interpretation in each individual mannerism.

### Conclusion
Primarily though data can be:
[tex]\[ 4 \cdot 0.15 \cdot 5 = 3 \][/tex]
Which is balanced when data interpreted correctly corresponding tables.
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