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Question

Lightbulbs act as resistors. Janine is building a circuit that contains two lightbulbs in parallel. One of the lightbulbs has a resistance of 120 ohms, but the resistance of the second lightbulb is unknown. She models the total resistance in the circuit, [tex]t[/tex], with this equation, in which [tex]r[/tex] represents the resistance of the second lightbulb.
[tex]\[ t = \frac{120r}{r + 120} \][/tex]

Part A
Find the inverse of Janine's equation.
[tex]\[ r = \][/tex]

Part B



Answer :

GinnyAnswer
To solve for [tex]\( r \)[/tex] in terms of [tex]\( t \)[/tex] from the given equation:

[tex]\[ t = \frac{120r}{r + 120} \][/tex]

we need to isolate [tex]\( r \)[/tex]. Here’s the step-by-step process to find the inverse:

1. Start with the given equation:
[tex]\[ t = \frac{120r}{r + 120} \][/tex]

2. Multiply both sides by [tex]\((r + 120)\)[/tex] to eliminate the fraction:
[tex]\[ t(r + 120) = 120r \][/tex]

3. Distribute [tex]\( t \)[/tex] on the left side:
[tex]\[ tr + 120t = 120r \][/tex]

4. Move all terms involving [tex]\( r \)[/tex] to one side of the equation:
[tex]\[ tr - 120r = -120t \][/tex]

5. Factor out [tex]\( r \)[/tex] on the left side:
[tex]\[ r(t - 120) = -120t \][/tex]

6. Divide both sides by [tex]\((t - 120)\)[/tex] to solve for [tex]\( r \)[/tex]:
[tex]\[ r = \frac{-120t}{t - 120} \][/tex]

Thus, the inverse of Janine’s equation, solved for [tex]\( r \)[/tex] in terms of [tex]\( t \)[/tex], is:

[tex]\[ r = \frac{-120t}{t - 120} \][/tex]

This expression allows you to find the resistance of the second lightbulb when given the total resistance [tex]\( t \)[/tex] in the parallel circuit.