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Drag each tile to the table to multiply each row heading by each column heading.

\begin{tabular}{|l|l|l|}
\hline & [tex]$5t$[/tex] & [tex]$-4$[/tex] \\
\hline [tex]$4t$[/tex] & & \\
\hline [tex]$5$[/tex] & & \\
\hline
\end{tabular}

Sagot :

GinnyAnswer
Sure! Let's fill out the given table by multiplying each row heading by the column headings step-by-step.

First, we will set up a table to organize our products:

[tex]\[ \begin{array}{|c|c|c|} \hline & 5t & -4 \\ \hline -16t & & \\ \hline 4t & & \\ \hline 5 & & \\ \hline \end{array} \][/tex]

Now we will go row by row to calculate the products:

### First row:
- For the first cell: multiply \(-16t\) by \(5t\) to get \(\left(-16t \cdot 5t\right) = -80t^2\).
- For the second cell: multiply \(-16t\) by \(-4\) to get \(\left(-16t \cdot (-4)\right) = 64t\).

So the table now looks like this:
[tex]\[ \begin{array}{|c|c|c|} \hline & 5t & -4 \\ \hline -16t & -80t^2 & 64t \\ \hline 4t & & \\ \hline 5 & & \\ \hline \end{array} \][/tex]

### Second row:
- For the first cell: multiply \(4t\) by \(5t\) to get \(\left(4t \cdot 5t\right) = 20t^2\).
- For the second cell: multiply \(4t\) by \(-4\) to get \(\left(4t \cdot (-4)\right) = -16t\).

So the table now looks like this:
[tex]\[ \begin{array}{|c|c|c|} \hline & 5t & -4 \\ \hline -16t & -80t^2 & 64t \\ \hline 4t & 20t^2 & -16t \\ \hline 5 & & \\ \hline \end{array} \][/tex]

### Third row:
- For the first cell: multiply \(5\) by \(5t\) to get \(\left(5 \cdot 5t\right) = 25t\).
- For the second cell: multiply \(5\) by \(-4\) to get \(\left(5 \cdot (-4)\right) = -20\).

So the final table looks like this:
[tex]\[ \begin{array}{|c|c|c|} \hline & 5t & -4 \\ \hline -16t & -80t^2 & 64t \\ \hline 4t & 20t^2 & -16t \\ \hline 5 & 25t & -20 \\ \hline \end{array} \][/tex]

This is our completed table with all the products calculated!
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