Find solutions to your problems with the expert advice available on IDNLearn.com. Explore thousands of verified answers from experts and find the solutions you need, no matter the topic.
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]
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for choosing IDNLearn.com for your queries. We’re here to provide accurate answers, so visit us again soon.
