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Talia took the bus from her home to the bank and then walked back to her home along the same route. The bus traveled at an average speed of 40 km/h and she walked at an average speed of 5 km/h. To determine the time, x, that it took Talia to walk home, she used the equation

40(0.9 – x) = 5x.

The time of Trip 2 is
hours.

The distance of Trip 1 is
km.

Image to Text
Rate (km/h) | Time (h) | Distance (km)
Trip 1 | 40 | 0.9 - x | 40(0.9 - x)
Trip 2 | 5 | x | 5x

Summary
The image presents a table showing the rate, time, and distance for two trips. For Trip 1, the rate is 40 km/h, time is 0.9 - x hours, and distance is 40(0.9 - x) km. For Trip 2, the rate is 5 km/h, time is x hours, and distance is 5x km.

3 answers

To solve for the time it took Talia to walk home (Trip 2) and the distance of Trip 1, we start with the given equation:

\[ 40(0.9 - x) = 5x \]

First, let's simplify this equation:

  1. Distribute the 40 on the left side:

    \[ 40 \times 0.9 - 40x = 5x \]

    \[ 36 - 40x = 5x \]

  2. Combine like terms by adding \(40x\) to both sides:

    \[ 36 = 45x \]

  3. Solve for \(x\):

    \[ x = \frac{36}{45} \]

    \[ x = \frac{4}{5} \]

    \[ x = 0.8 \text{ hours} \]

So, the time it took Talia to walk home (Trip 2) is 0.8 hours.

Next, we need to find the distance of Trip 1. We already know that the distance for Trip 1 is given by:

\[ \text{Distance} = 40(0.9 - x) \]

Substituting \(x = 0.8\) into the equation:

\[ \text{Distance} = 40(0.9 - 0.8) \]

\[ = 40(0.1) \]

\[ = 4 \text{ km} \]

Therefore, the distance of Trip 1 is 4 km.

Summary of Results:

  • The time of Trip 2 is 0.8 hours.
  • The distance of Trip 1 is 4 km.
Which equation can be used to solve for the rate, r, for Sam to change all the light bulbs in the office building by himself?

3r = 3 r equals StartFraction 3 Over 8 EndFraction.
3r + 3 r plus StartFraction 3 Over 8 EndFraction equals 1. = 1
r + StartFraction 1 Over 3 EndFraction r plus StartFraction 3 Over 8 EndFraction r equals 1.r = 1
r + StartFraction 1 Over 3 EndFraction r plus StartFraction 1 Over 8 EndFraction r equals 1.r = 1

Working together, Sam and Charlie can change all the light bulbs in the office building in 3 hours. It would have taken Charlie 8 hours to do the job alone.

Rate (part/h) | Time (h) | Part of Light Bulbs Changed
Sam | r | 3 | r(3)
Charlie | 1/8 | 3 | 1/8(3)

Summary
The table compares the rate at which Sam and Charlie change light bulbs over a period of 3 hours. Sam’s rate is denoted by ‘r’ and after 3 hours, Sam changes ‘r(3)’ parts of the light bulbs. Charlie’s rate is 1/8 part per hour, and after 3 hours, Charlie changes 1/8(3) parts of the light bulbs.

To find the correct equation to solve for Sam's rate, \( r \), we need to analyze the situation provided.

We know the following:

  • Together, Sam and Charlie can change all the light bulbs in 3 hours.
  • Charlie’s rate of changing light bulbs is \( \frac{1}{8} \) parts per hour, meaning he can complete the job alone in 8 hours.
  • Let Sam's rate be \( r \) parts per hour.

When they work together for 3 hours, they complete the entire job (which is considered 1 complete part or all the light bulbs).

Let's express the total amount of work done (part of light bulbs changed) by both Sam and Charlie in 3 hours:

  1. Sam's Contribution:

    • Sam's rate is \( r \) parts/hour.
    • In 3 hours, Sam will change \( r \times 3 = 3r \) parts of the light bulbs.

  2. Charlie's Contribution:

    • Charlie's rate is \( \frac{1}{8} \) parts/hour.
    • In 3 hours, Charlie will change \( \frac{1}{8} \times 3 = \frac{3}{8} \) parts of the light bulbs.

Now, combining their contributions gives us the total work done:

\[ 3r + \frac{3}{8} = 1 \]

This equation states that the total amount of work done by Sam and Charlie together in 3 hours is equal to the full job (1 part).

Therefore, the correct equation to use to solve for Sam's rate, \( r \), is:

\[ 3r + \frac{3}{8} = 1 \]

Summary

  • Sam's rate equation is correctly given by \( 3r + \frac{3}{8} = 1 \).