Answer :
To fill in the table, we will evaluate the function \( h(t) = 16 - t^2 \) for each given \( t \) and write down the corresponding \((t, h(t))\) values.
1. When \( t = -5 \):
[tex]\[ h(-5) = 16 - (-5)^2 = 16 - 25 = -9 \][/tex]
So, the pair is \((-5, -9)\), which is already filled.
2. When \( t = -4 \):
[tex]\[ h(-4) = 16 - (-4)^2 = 16 - 16 = 0 \][/tex]
So, the pair is \((-4, 0)\).
3. When \( t = -3 \):
[tex]\[ h(-3) = 16 - (-3)^2 = 16 - 9 = 7 \][/tex]
So, the pair is \((-3, 7)\).
4. When \( t = -2 \):
[tex]\[ h(-2) = 16 - (-2)^2 = 16 - 4 = 12 \][/tex]
So, the pair is \((-2, 12)\).
5. When \( t = -1 \):
[tex]\[ h(-1) = 16 - (-1)^2 = 16 - 1 = 15 \][/tex]
So, the pair is \((-1, 15)\).
6. When \( t = 0 \):
[tex]\[ h(0) = 16 - 0^2 = 16 \][/tex]
So, the pair is \((0, 16)\).
7. When \( t = 1 \):
[tex]\[ h(1) = 16 - 1^2 = 16 - 1 = 15 \][/tex]
So, the pair is \((1, 15)\).
8. When \( t = 2 \):
[tex]\[ h(2) = 16 - 2^2 = 16 - 4 = 12 \][/tex]
So, the pair is \((2, 12)\).
9. When \( t = 3 \):
[tex]\[ h(3) = 16 - 3^2 = 16 - 9 = 7 \][/tex]
So, the pair is \((3, 7)\).
10. When \( t = 4 \):
[tex]\[ h(4) = 16 - 4^2 = 16 - 16 = 0 \][/tex]
So, the pair is \((4, 0)\).
11. When \( t = 5 \):
[tex]\[ h(5) = 16 - 5^2 = 16 - 25 = -9 \][/tex]
So, the pair is \((5, -9)\).
Now, let's fill in the table with the calculated values:
[tex]\[ \begin{array}{|r|l|l|} \hline x & h(t)=16-t^2 & (x, y) \\ \hline -5 & -9 & (-5, -9) \\ \hline -4 & 0 & (-4, 0) \\ \hline -3 & 7 & (-3, 7) \\ \hline -2 & 12 & (-2, 12) \\ \hline -1 & 15 & (-1, 15) \\ \hline 0 & 16 & (0, 16) \\ \hline 1 & 15 & (1, 15) \\ \hline 2 & 12 & (2, 12) \\ \hline 3 & 7 & (3, 7) \\ \hline 4 & 0 & (4, 0) \\ \hline 5 & -9 & (5, -9) \\ \hline \end{array} \][/tex]
1. When \( t = -5 \):
[tex]\[ h(-5) = 16 - (-5)^2 = 16 - 25 = -9 \][/tex]
So, the pair is \((-5, -9)\), which is already filled.
2. When \( t = -4 \):
[tex]\[ h(-4) = 16 - (-4)^2 = 16 - 16 = 0 \][/tex]
So, the pair is \((-4, 0)\).
3. When \( t = -3 \):
[tex]\[ h(-3) = 16 - (-3)^2 = 16 - 9 = 7 \][/tex]
So, the pair is \((-3, 7)\).
4. When \( t = -2 \):
[tex]\[ h(-2) = 16 - (-2)^2 = 16 - 4 = 12 \][/tex]
So, the pair is \((-2, 12)\).
5. When \( t = -1 \):
[tex]\[ h(-1) = 16 - (-1)^2 = 16 - 1 = 15 \][/tex]
So, the pair is \((-1, 15)\).
6. When \( t = 0 \):
[tex]\[ h(0) = 16 - 0^2 = 16 \][/tex]
So, the pair is \((0, 16)\).
7. When \( t = 1 \):
[tex]\[ h(1) = 16 - 1^2 = 16 - 1 = 15 \][/tex]
So, the pair is \((1, 15)\).
8. When \( t = 2 \):
[tex]\[ h(2) = 16 - 2^2 = 16 - 4 = 12 \][/tex]
So, the pair is \((2, 12)\).
9. When \( t = 3 \):
[tex]\[ h(3) = 16 - 3^2 = 16 - 9 = 7 \][/tex]
So, the pair is \((3, 7)\).
10. When \( t = 4 \):
[tex]\[ h(4) = 16 - 4^2 = 16 - 16 = 0 \][/tex]
So, the pair is \((4, 0)\).
11. When \( t = 5 \):
[tex]\[ h(5) = 16 - 5^2 = 16 - 25 = -9 \][/tex]
So, the pair is \((5, -9)\).
Now, let's fill in the table with the calculated values:
[tex]\[ \begin{array}{|r|l|l|} \hline x & h(t)=16-t^2 & (x, y) \\ \hline -5 & -9 & (-5, -9) \\ \hline -4 & 0 & (-4, 0) \\ \hline -3 & 7 & (-3, 7) \\ \hline -2 & 12 & (-2, 12) \\ \hline -1 & 15 & (-1, 15) \\ \hline 0 & 16 & (0, 16) \\ \hline 1 & 15 & (1, 15) \\ \hline 2 & 12 & (2, 12) \\ \hline 3 & 7 & (3, 7) \\ \hline 4 & 0 & (4, 0) \\ \hline 5 & -9 & (5, -9) \\ \hline \end{array} \][/tex]
