Part A: Robert uses a wheelbarrow to haul wood from the back yard to his neighbor's house located 80 meters away. If he takes three breaks evenly when pushing the wheelbarrow to its destination, at which distance will he reach his first break? Part B: Robert has 36 pieces to take, but can only take 9 each time. What distance in meters will he travel to move all the wood?

Part A

He will reach his first break at 20 meters from the point he starts to travel to his neighbors house

Part B

640 meters

Step-by-step explanation:

Part A

The distance from which Robert hauls wood from his neighbors backyard = 80 meters

The number of evenly distributed breaks Robert takes = 3

The journey is presented graphically as follows;

Start 0  Break BreakBreak   Destination (80 meters)

Therefore, by taking three evenly distributed breaks, Robert shares the 80 meters distance into four (4) equal parts, which gives;

The length of each part = 80-m/4 = 20 m

Therefore, Robert will reach his first break, at 20 meters from the starting point

Part B

The number of pieces of wood Robert has to take = 36 pieces

The number of pieces of wood Robert can take each time = 9 pieces

Therefore;

The number of trips he has to travel to move all the wood = 36/9 = 4 trips

The distance of each trip to and from the destination = 80 meters × 2 = 160 meters

The total distance of the 4 trips Robert has to travel to move all the wood = 4 × The distance of each trip

∴ The total distance of the 4 trips = 4 × 160 = 640 meters

The total distance in meters of the 4 trips Robert has to travel to move all the wood = 640 meters.

1. Tall Dog

2. Small Dog

3. Regular Dog

4. Fat Dog

Step-by-step explanation:

It's simple. Look at the problem ;)