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Sagot :
Let's go through each part of the question step by step and determine the values required to complete the flow diagram and the table.
### 1. Copy and complete this flow diagram:
The flow diagram is not provided, so this part is ambiguous. However, if it’s about connecting the steps logically, let’s proceed to the given tasks.
### 2. Determine the value of [tex]\( t \)[/tex] using the equation [tex]\( t = p \times 2 + 2 \)[/tex] for [tex]\( p = 0, -3, -5, -7, \)[/tex] and [tex]\( -9 \)[/tex]
We will substitute each value of [tex]\( p \)[/tex] into the equation [tex]\( t = p \times 2 + 2 \)[/tex].
- For [tex]\( p = 0 \)[/tex]:
[tex]\[ t = 0 \times 2 + 2 = 2 \][/tex]
- For [tex]\( p = -3 \)[/tex]:
[tex]\[ t = (-3) \times 2 + 2 = -6 + 2 = -4 \][/tex]
- For [tex]\( p = -5 \)[/tex]:
[tex]\[ t = (-5) \times 2 + 2 = -10 + 2 = -8 \][/tex]
- For [tex]\( p = -7 \)[/tex]:
[tex]\[ t = (-7) \times 2 + 2 = -14 + 2 = -12 \][/tex]
- For [tex]\( p = -9 \)[/tex]:
[tex]\[ t = (-9) \times 2 + 2 = -18 + 2 = -16 \][/tex]
So, the values of [tex]\( t \)[/tex] are:
[tex]\[ t = [2, -4, -8, -12, -16] \][/tex]
### 3. Use the formula [tex]\( A = s^2 \)[/tex] where [tex]\( A \)[/tex] is the area of a square and [tex]\( s \)[/tex] is its side, to find outputs for the following inputs: [tex]\( s = 1, 3, 6, 8, 10 \)[/tex]
We'll use the formula [tex]\( A = s^2 \)[/tex] to calculate the area for each side length [tex]\( s \)[/tex]:
- For [tex]\( s = 1 \)[/tex]:
[tex]\[ A = 1^2 = 1 \][/tex]
- For [tex]\( s = 3 \)[/tex]:
[tex]\[ A = 3^2 = 9 \][/tex]
- For [tex]\( s = 6 \)[/tex]:
[tex]\[ A = 6^2 = 36 \][/tex]
- For [tex]\( s = 8 \)[/tex]:
[tex]\[ A = 8^2 = 64 \][/tex]
- For [tex]\( s = 10 \)[/tex]:
[tex]\[ A = 10^2 = 100 \][/tex]
So, the values of [tex]\( A \)[/tex] are:
[tex]\[ A = [1, 9, 36, 64, 100] \][/tex]
### 4. Complete the table below using the formula for the side of a square: [tex]\( s = \sqrt{A} \)[/tex] and list the domain
Using the formula [tex]\( s = \sqrt{A} \)[/tex], calculate [tex]\( s \)[/tex] for each given [tex]\( A \)[/tex]:
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Side (s) & 3 & 6 & 8 & 10 & 12 \\
\hline
Area (A) & 9 & 36 & 64 & 100 & 144 \\
\hline
\end{tabular}
So, the completed table is:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Side (s)} & 3 & 6 & 8 & 10 & 12 \\ \hline \text{Area (A)} & 9 & 36 & 64 & 100 & 144 \\ \hline \end{array} \][/tex]
And the domain for [tex]\( s = \sqrt{A} \)[/tex] is:
[tex]\[ s = [3.0, 6.0, 8.0, 10.0, 12.0] \][/tex]
### 1. Copy and complete this flow diagram:
The flow diagram is not provided, so this part is ambiguous. However, if it’s about connecting the steps logically, let’s proceed to the given tasks.
### 2. Determine the value of [tex]\( t \)[/tex] using the equation [tex]\( t = p \times 2 + 2 \)[/tex] for [tex]\( p = 0, -3, -5, -7, \)[/tex] and [tex]\( -9 \)[/tex]
We will substitute each value of [tex]\( p \)[/tex] into the equation [tex]\( t = p \times 2 + 2 \)[/tex].
- For [tex]\( p = 0 \)[/tex]:
[tex]\[ t = 0 \times 2 + 2 = 2 \][/tex]
- For [tex]\( p = -3 \)[/tex]:
[tex]\[ t = (-3) \times 2 + 2 = -6 + 2 = -4 \][/tex]
- For [tex]\( p = -5 \)[/tex]:
[tex]\[ t = (-5) \times 2 + 2 = -10 + 2 = -8 \][/tex]
- For [tex]\( p = -7 \)[/tex]:
[tex]\[ t = (-7) \times 2 + 2 = -14 + 2 = -12 \][/tex]
- For [tex]\( p = -9 \)[/tex]:
[tex]\[ t = (-9) \times 2 + 2 = -18 + 2 = -16 \][/tex]
So, the values of [tex]\( t \)[/tex] are:
[tex]\[ t = [2, -4, -8, -12, -16] \][/tex]
### 3. Use the formula [tex]\( A = s^2 \)[/tex] where [tex]\( A \)[/tex] is the area of a square and [tex]\( s \)[/tex] is its side, to find outputs for the following inputs: [tex]\( s = 1, 3, 6, 8, 10 \)[/tex]
We'll use the formula [tex]\( A = s^2 \)[/tex] to calculate the area for each side length [tex]\( s \)[/tex]:
- For [tex]\( s = 1 \)[/tex]:
[tex]\[ A = 1^2 = 1 \][/tex]
- For [tex]\( s = 3 \)[/tex]:
[tex]\[ A = 3^2 = 9 \][/tex]
- For [tex]\( s = 6 \)[/tex]:
[tex]\[ A = 6^2 = 36 \][/tex]
- For [tex]\( s = 8 \)[/tex]:
[tex]\[ A = 8^2 = 64 \][/tex]
- For [tex]\( s = 10 \)[/tex]:
[tex]\[ A = 10^2 = 100 \][/tex]
So, the values of [tex]\( A \)[/tex] are:
[tex]\[ A = [1, 9, 36, 64, 100] \][/tex]
### 4. Complete the table below using the formula for the side of a square: [tex]\( s = \sqrt{A} \)[/tex] and list the domain
Using the formula [tex]\( s = \sqrt{A} \)[/tex], calculate [tex]\( s \)[/tex] for each given [tex]\( A \)[/tex]:
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Side (s) & 3 & 6 & 8 & 10 & 12 \\
\hline
Area (A) & 9 & 36 & 64 & 100 & 144 \\
\hline
\end{tabular}
So, the completed table is:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Side (s)} & 3 & 6 & 8 & 10 & 12 \\ \hline \text{Area (A)} & 9 & 36 & 64 & 100 & 144 \\ \hline \end{array} \][/tex]
And the domain for [tex]\( s = \sqrt{A} \)[/tex] is:
[tex]\[ s = [3.0, 6.0, 8.0, 10.0, 12.0] \][/tex]
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