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(b) Using a scale of [tex]$2 \text{ cm}$[/tex] to 1 kilogram on the [tex]$x$[/tex]-axis and [tex]$2 \text{ cm}$[/tex] to [tex]$GHc 20$[/tex] on the [tex]$y$[/tex]-axis, copy and complete the table below for the relation [tex]$y = 20x + 20$[/tex], and draw the graph for the relation.

\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
[tex]$y$[/tex] & 40 & 60 & 80 & 100 & 120 & 140 & 160 & 180 & 200 \\
\hline
\end{tabular}

(i) From the graph, find the weight of meat that can be bought with [tex]$GHc 90$[/tex].

(ii) From the graph, find the cost of [tex]$3.5 \text{ kg}$[/tex] of meat.

(iii) Using the relation, how many kilograms of meat can be bought at a cost of [tex]$GHc 240$[/tex]?



Answer :

GinnyAnswer
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.

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