Certainly! Let's solve this problem step by step.
We are given the following information:
1. The charge for 163 units of electricity is \$453.80.
2. The charge for 42 units of electricity is \$139.20.
We are also told that the charge for electricity consumption, \( E \), is partly constant and partly varies with the number of units used, \( N \).
This means we can represent the total charge \( E \) as:
[tex]\[ E = C + kN \][/tex]
where \( C \) is the constant charge and \( k \) is the charge per unit of electricity.
### Part (a): Calculate the charge per unit of electricity
Let's set up two equations based on the given information:
1. For 163 units:
[tex]\[ 453.80 = C + k \times 163 \][/tex]
2. For 42 units:
[tex]\[ 139.20 = C + k \times 42 \][/tex]
To find \( k \) (the charge per unit), we'll subtract the second equation from the first:
[tex]\[ 453.80 - 139.20 = (C + k \times 163) - (C + k \times 42) \][/tex]
[tex]\[ 314.60 = k \times (163 - 42) \][/tex]
[tex]\[ 314.60 = k \times 121 \][/tex]
[tex]\[ k = \frac{314.60}{121} \][/tex]
From simplifying this, we find:
[tex]\[ k = 2.60 \][/tex]
So, the charge per unit of electricity is \$2.60.
### Part (b): Obtain a formula for \( E \) in terms of \( N \)
Next, we need to determine the constant charge \( C \). We can use either of the original equations along with the value of \( k \) to find \( C \). Let's use the equation for 163 units:
[tex]\[ 453.80 = C + 2.60 \times 163 \][/tex]
[tex]\[ 453.80 = C + 423.80 \][/tex]
Solving for \( C \):
[tex]\[ C = 453.80 - 423.80 \][/tex]
[tex]\[ C = 30.00 \][/tex]
Now, we have both the charge per unit \( k = 2.60 \) and the constant charge \( C = 30.00 \).
The formula for \( E \) in terms of \( N \) is:
[tex]\[ E = 30.00 + 2.60N \][/tex]
To summarize:
- The charge per unit of electricity, \( k \), is \$2.60.
- The formula for the total charge [tex]\( E \)[/tex] in terms of the number of units [tex]\( N \)[/tex] is [tex]\( E = 30.00 + 2.60N \)[/tex].
