Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Our Q&A platform provides quick and trustworthy answers to your questions from experienced professionals in different areas of expertise. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
Certainly! Let's solve the problem step-by-step, ensuring clarity and thorough understanding.
### 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]
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.