Let's address each part of the question step-by-step, using the relation \( y = 20x + 20 \):
### Completing the Table
We need to use the equation \( y = 20x + 20 \) to find the missing values for \( y \) corresponding to given \( x \)-values.
Here are the calculations:
- For \( x = 1 \): \( y = 20(1) + 20 = 40 \)
- For \( x = 2 \): \( y = 20(2) + 20 = 60 \)
- For \( x = 4 \): \( y = 20(4) + 20 = 100 \)
- For \( x = 6 \): \( y = 20(6) + 20 = 140 \)
- For \( x = 8 \): \( y = 20(8) + 20 = 180 \)
- For \( x = 9 \): \( y = 20(9) + 20 = 200 \)
Now we can complete the table:
[tex]\[
\begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline
x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
y & 40 & 60 & 80 & 100 & 120 & 140 & 160 & 180 & 200 \\
\hline
\end{array}
\][/tex]
### Drawing the Graph
To draw the graph for the relation \( y = 20x + 20 \), use the following steps:
1. Draw a set of Cartesian coordinates using the given scale.
- On the x-axis: Use a scale of \( 2 \text{ cm} \) to \( 1 \text{ kg} \).
- On the y-axis: Use a scale of \( 2 \text{ cm} \) to \( \text{GH}c 20 \).
2. Plot the points from the table onto the graph:
- \((1, 40)\)
- \((2, 60)\)
- \((3, 80)\)
- \((4, 100)\)
- \((5, 120)\)
- \((6, 140)\)
- \((7, 160)\)
- \((8, 180)\)
- \((9, 200)\)
3. Connect these points with a straight line since the relation \( y = 20x + 20 \) is linear.
### Part (i): Finding the Weight of Meat That Can Be Bought with GHc 90
From the given relation, we need to find \( x \) when \( y = 90 \):
Solving \( 90 = 20x + 20 \):
[tex]\[ 90 = 20x + 20 \][/tex]
[tex]\[ 90 - 20 = 20x \][/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 GHc 90 is \( 3.5 \text{ kg} \).
### Part (ii): Finding the Cost of 3.5 kg of Meat
For \( x = 3.5 \), we need to find \( y \):
[tex]\[ y = 20(3.5) + 20 \][/tex]
[tex]\[ y = 70 + 20 \][/tex]
[tex]\[ y = 90 \][/tex]
So, the cost of \( 3.5 \text{ kg} \) of meat is GHc 90.
### Part (iii): Finding the Kilograms of Meat That Can Be Bought for GHc 240
From the given relation, we need to find \( x \) when \( y = 240 \):
Solving \( 240 = 20x + 20 \):
[tex]\[ 240 = 20x + 20 \][/tex]
[tex]\[ 240 - 20 = 20x \][/tex]
[tex]\[ 220 = 20x \][/tex]
[tex]\[ x = \frac{220}{20} \][/tex]
[tex]\[ x = 11 \][/tex]
So, the kilograms of meat that can be bought at a cost of GHc 240 is [tex]\( 11 \text{ kg} \)[/tex].
