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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]

Find the inverse of Janine's equation.

[tex]\[ r = \][/tex]

Sagot :

GinnyAnsweridn
To find the inverse of Janine's equation, we need to solve for [tex]\( r \)[/tex] in terms of [tex]\( t \)[/tex] from the given equation.

First, we start with the original equation:
[tex]\[ t = \frac{120r}{r + 120} \][/tex]

We want to isolate [tex]\( r \)[/tex]. To do that, let's follow these steps:

1. Multiply through by [tex]\( r + 120 \)[/tex] to clear the fraction:
[tex]\[ t(r + 120) = 120r \][/tex]

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

3. Rearrange the equation to group terms involving [tex]\( r \)[/tex]:
[tex]\[ tr - 120r = -120t \][/tex]

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

5. Solve for [tex]\( r \)[/tex]:
[tex]\[ r = \frac{-120t}{t - 120} \][/tex]

Now we have expressed [tex]\( r \)[/tex] in terms of [tex]\( t \)[/tex]. Thus, the inverse of Janine's equation is:
[tex]\[ r = \frac{-120t}{t - 120} \][/tex]

This is the required inverse equation.
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