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Sagot :
Alright, let's tackle this question step-by-step:
### Step 1: Completing the Table
Given the relation \( y = 20x + 20 \), we can fill in the missing \( y \)-values for \( x = 1, 2, 4, 6, 8 \).
For each \( x \):
1. \( x = 1 \):
[tex]\[ y = 20(1) + 20 = 40 \][/tex]
2. \( x = 2 \):
[tex]\[ y = 20(2) + 20 = 60 \][/tex]
3. \( x = 4 \):
[tex]\[ y = 20(4) + 20 = 100 \][/tex]
4. \( x = 6 \):
[tex]\[ y = 20(6) + 20 = 140 \][/tex]
5. \( x = 8 \):
[tex]\[ y = 20(8) + 20 = 180 \][/tex]
So, the completed table is:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
\( x \) & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
\( y \) & 40 & 60 & 80 & 100 & 120 & 140 & 160 & 180 & 200 \\
\hline
\end{tabular}
### Step 2: Drawing the Graph
Use the table with \( x \)-values from 1 to 9 and their corresponding \( y \)-values to plot points on a graph.
1. Use a scale of 2 cm for each unit on the \( x \)-axis (representing 1 kg).
2. Use a scale of 2 cm for each GH¢ 20 units on the \( y \)-axis.
For example:
- Plot the point \( (1, 40) \) at 2 cm on the \( x \)-axis and 4 cm on the \( y \)-axis.
- Plot \( (2, 60) \) at 4 cm on the \( x \)-axis and 6 cm on the \( y \)-axis, and so on.
Connect these points to form a straight line, which represents the relation \( y = 20x + 20 \).
### Step 3: Solving the Problems
#### (i) Weight of Meat that can be Bought with GH¢ 90
From the relation \( y = 20x + 20 \), solve for \( x \) when \( y = 90 \):
[tex]\[ 90 = 20x + 20 \][/tex]
[tex]\[ 70 = 20x \][/tex]
[tex]\[ x = \frac{70}{20} \][/tex]
[tex]\[ x = 3.5 \][/tex]
So, the weight of meat that can be bought with GH¢ 90 is 3.5 kg.
#### (ii) Cost of 3.5 kg of Meat
To find the cost of 3.5 kg of meat, use the relation \( y = 20x + 20 \):
[tex]\[ y = 20(3.5) + 20 \][/tex]
[tex]\[ y = 70 + 20 \][/tex]
[tex]\[ y = 90 \][/tex]
So, the cost of 3.5 kg of meat is GH¢ 90.
#### (iii) Kilograms of Meat that can be Bought at a Cost of GH¢ 240
From the relation \( y = 20x + 20 \), solve for \( x \) when \( y = 240 \):
[tex]\[ 240 = 20x + 20 \][/tex]
[tex]\[ 220 = 20x \][/tex]
[tex]\[ x = \frac{220}{20} \][/tex]
[tex]\[ x = 11 \][/tex]
So, 11 kg of meat can be bought at a cost of GH¢ 240.
### Step 1: Completing the Table
Given the relation \( y = 20x + 20 \), we can fill in the missing \( y \)-values for \( x = 1, 2, 4, 6, 8 \).
For each \( x \):
1. \( x = 1 \):
[tex]\[ y = 20(1) + 20 = 40 \][/tex]
2. \( x = 2 \):
[tex]\[ y = 20(2) + 20 = 60 \][/tex]
3. \( x = 4 \):
[tex]\[ y = 20(4) + 20 = 100 \][/tex]
4. \( x = 6 \):
[tex]\[ y = 20(6) + 20 = 140 \][/tex]
5. \( x = 8 \):
[tex]\[ y = 20(8) + 20 = 180 \][/tex]
So, the completed table is:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
\( x \) & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
\( y \) & 40 & 60 & 80 & 100 & 120 & 140 & 160 & 180 & 200 \\
\hline
\end{tabular}
### Step 2: Drawing the Graph
Use the table with \( x \)-values from 1 to 9 and their corresponding \( y \)-values to plot points on a graph.
1. Use a scale of 2 cm for each unit on the \( x \)-axis (representing 1 kg).
2. Use a scale of 2 cm for each GH¢ 20 units on the \( y \)-axis.
For example:
- Plot the point \( (1, 40) \) at 2 cm on the \( x \)-axis and 4 cm on the \( y \)-axis.
- Plot \( (2, 60) \) at 4 cm on the \( x \)-axis and 6 cm on the \( y \)-axis, and so on.
Connect these points to form a straight line, which represents the relation \( y = 20x + 20 \).
### Step 3: Solving the Problems
#### (i) Weight of Meat that can be Bought with GH¢ 90
From the relation \( y = 20x + 20 \), solve for \( x \) when \( y = 90 \):
[tex]\[ 90 = 20x + 20 \][/tex]
[tex]\[ 70 = 20x \][/tex]
[tex]\[ x = \frac{70}{20} \][/tex]
[tex]\[ x = 3.5 \][/tex]
So, the weight of meat that can be bought with GH¢ 90 is 3.5 kg.
#### (ii) Cost of 3.5 kg of Meat
To find the cost of 3.5 kg of meat, use the relation \( y = 20x + 20 \):
[tex]\[ y = 20(3.5) + 20 \][/tex]
[tex]\[ y = 70 + 20 \][/tex]
[tex]\[ y = 90 \][/tex]
So, the cost of 3.5 kg of meat is GH¢ 90.
#### (iii) Kilograms of Meat that can be Bought at a Cost of GH¢ 240
From the relation \( y = 20x + 20 \), solve for \( x \) when \( y = 240 \):
[tex]\[ 240 = 20x + 20 \][/tex]
[tex]\[ 220 = 20x \][/tex]
[tex]\[ x = \frac{220}{20} \][/tex]
[tex]\[ x = 11 \][/tex]
So, 11 kg of meat can be bought at a cost of GH¢ 240.
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