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### Problem Statement
If [tex]\(a\)[/tex] and [tex]\(b\)[/tex] vary inversely, the product [tex]\(a \cdot b = k\)[/tex], where [tex]\(k\)[/tex] is the constant of variation.
### Table (i)
Given values:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 16 & 20 & 8 & - & - \\ \hline y & 5 & - & - & 40 & 80 \\ \hline \end{array} \][/tex]
#### Step 1: Find the constant of variation [tex]\(k\)[/tex]
Using the given values [tex]\(x = 16\)[/tex] and [tex]\(y = 5\)[/tex]:
[tex]\[ k = 16 \cdot 5 = 80 \][/tex]
#### Step 2: Complete the table using [tex]\(k = 80\)[/tex]
1. For [tex]\(x = 20\)[/tex]:
[tex]\[ y = \frac{k}{x} = \frac{80}{20} = 4.0 \][/tex]
2. For [tex]\(x = 8\)[/tex]:
[tex]\[ y = \frac{k}{x} = \frac{80}{8} = 10.0 \][/tex]
3. For [tex]\(y = 40\)[/tex]:
[tex]\[ x = \frac{k}{y} = \frac{80}{40} = 2.0 \][/tex]
4. For [tex]\(y = 80\)[/tex]:
[tex]\[ x = \frac{k}{y} = \frac{80}{80} = 1.0 \][/tex]
Completing the table (i), we get:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 16 & 20 & 8 & 2.0 & 1.0 \\ \hline y & 5 & 4.0 & 10.0 & 40 & 80 \\ \hline \end{array} \][/tex]
---
### Table (ii)
Given values:
[tex]\[ \begin{array}{|l|l|l|l|l|l|} \hline x & 12 & 15 & - & - & 1.2 \\ \hline y & 10 & - & 20 & 5 & - \\ \hline \end{array} \][/tex]
#### Step 1: Find the constant of variation [tex]\(k\)[/tex]
Using the given values [tex]\(x = 12\)[/tex] and [tex]\(y = 10\)[/tex]:
[tex]\[ k = 12 \cdot 10 = 120 \][/tex]
#### Step 2: Complete the table using [tex]\(k = 120\)[/tex]
1. For [tex]\(x = 15\)[/tex]:
[tex]\[ y = \frac{k}{x} = \frac{120}{15} = 8.0 \][/tex]
2. For [tex]\(y = 20\)[/tex]:
[tex]\[ x = \frac{k}{y} = \frac{120}{20} = 6.0 \][/tex]
3. For [tex]\(y = 5\)[/tex]:
[tex]\[ x = \frac{k}{y} = \frac{120}{5} = 24.0 \][/tex]
4. For [tex]\(x = 1.2\)[/tex]:
[tex]\[ y = \frac{k}{x} = \frac{120}{1.2} = 100.0 \][/tex]
Completing the table (ii), we get:
[tex]\[ \begin{array}{|l|l|l|l|l|l|} \hline x & 12 & 15 & 6.0 & 24.0 & 1.2 \\ \hline y & 10 & 8.0 & 20 & 5 & 100.0 \\ \hline \end{array} \][/tex]
Thus, the completed tables are:
### Final Completed Tables:
(i)
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 16 & 20 & 8 & 2.0 & 1.0 \\ \hline y & 5 & 4.0 & 10.0 & 40 & 80 \\ \hline \end{array} \][/tex]
(ii)
[tex]\[ \begin{array}{|l|l|l|l|l|l|} \hline x & 12 & 15 & 6.0 & 24.0 & 1.2 \\ \hline y & 10 & 8.0 & 20 & 5 & 100.0 \\ \hline \end{array} \][/tex]
### Problem Statement
If [tex]\(a\)[/tex] and [tex]\(b\)[/tex] vary inversely, the product [tex]\(a \cdot b = k\)[/tex], where [tex]\(k\)[/tex] is the constant of variation.
### Table (i)
Given values:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 16 & 20 & 8 & - & - \\ \hline y & 5 & - & - & 40 & 80 \\ \hline \end{array} \][/tex]
#### Step 1: Find the constant of variation [tex]\(k\)[/tex]
Using the given values [tex]\(x = 16\)[/tex] and [tex]\(y = 5\)[/tex]:
[tex]\[ k = 16 \cdot 5 = 80 \][/tex]
#### Step 2: Complete the table using [tex]\(k = 80\)[/tex]
1. For [tex]\(x = 20\)[/tex]:
[tex]\[ y = \frac{k}{x} = \frac{80}{20} = 4.0 \][/tex]
2. For [tex]\(x = 8\)[/tex]:
[tex]\[ y = \frac{k}{x} = \frac{80}{8} = 10.0 \][/tex]
3. For [tex]\(y = 40\)[/tex]:
[tex]\[ x = \frac{k}{y} = \frac{80}{40} = 2.0 \][/tex]
4. For [tex]\(y = 80\)[/tex]:
[tex]\[ x = \frac{k}{y} = \frac{80}{80} = 1.0 \][/tex]
Completing the table (i), we get:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 16 & 20 & 8 & 2.0 & 1.0 \\ \hline y & 5 & 4.0 & 10.0 & 40 & 80 \\ \hline \end{array} \][/tex]
---
### Table (ii)
Given values:
[tex]\[ \begin{array}{|l|l|l|l|l|l|} \hline x & 12 & 15 & - & - & 1.2 \\ \hline y & 10 & - & 20 & 5 & - \\ \hline \end{array} \][/tex]
#### Step 1: Find the constant of variation [tex]\(k\)[/tex]
Using the given values [tex]\(x = 12\)[/tex] and [tex]\(y = 10\)[/tex]:
[tex]\[ k = 12 \cdot 10 = 120 \][/tex]
#### Step 2: Complete the table using [tex]\(k = 120\)[/tex]
1. For [tex]\(x = 15\)[/tex]:
[tex]\[ y = \frac{k}{x} = \frac{120}{15} = 8.0 \][/tex]
2. For [tex]\(y = 20\)[/tex]:
[tex]\[ x = \frac{k}{y} = \frac{120}{20} = 6.0 \][/tex]
3. For [tex]\(y = 5\)[/tex]:
[tex]\[ x = \frac{k}{y} = \frac{120}{5} = 24.0 \][/tex]
4. For [tex]\(x = 1.2\)[/tex]:
[tex]\[ y = \frac{k}{x} = \frac{120}{1.2} = 100.0 \][/tex]
Completing the table (ii), we get:
[tex]\[ \begin{array}{|l|l|l|l|l|l|} \hline x & 12 & 15 & 6.0 & 24.0 & 1.2 \\ \hline y & 10 & 8.0 & 20 & 5 & 100.0 \\ \hline \end{array} \][/tex]
Thus, the completed tables are:
### Final Completed Tables:
(i)
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 16 & 20 & 8 & 2.0 & 1.0 \\ \hline y & 5 & 4.0 & 10.0 & 40 & 80 \\ \hline \end{array} \][/tex]
(ii)
[tex]\[ \begin{array}{|l|l|l|l|l|l|} \hline x & 12 & 15 & 6.0 & 24.0 & 1.2 \\ \hline y & 10 & 8.0 & 20 & 5 & 100.0 \\ \hline \end{array} \][/tex]
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