Answer :
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!
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!
