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skelet666 [1.2K]
3 years ago
7

Robert uses a wheelbarrow to haul wood from the back yard to his neighbor's house located 9 meters away. If he takes two breaks

evenly when pushing the wheelbarrow to its destination, at which distance will he reach his first break?
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?
Mathematics
1 answer:
Naily [24]3 years ago
6 0

Answer:

3 meters

63 meters

Step-by-step explanation:

In order for his two breaks to occur after even distances, simply divide his total path in three:

B = \frac{9}{3}\\ B=3\ meters

He will reach his first break after 3 meters.

If he can only take 9 pieces at a time the number of trips required is:

n=\frac{36}{9}\\ n=4

Note that he will have to back to the back yard for the first three trips (traveling back the 9 meters path). The total distance traveled is:

d= 3*(9+9)+9\\d=63\ meters

He will travel 63 meters.

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Answer:

    x=12 groups

This will give 5 Students each in 12 different equally sized groups

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This also shows that students can grouped in equal groups of 12 in 5 places

Step-by-step explanation:

From the question we are told that

60 children in equal sized groups

Generally interpreting 60\div 5 can be mathematically represented as

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    x=\frac{60}{5}

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This will give 5 Students each in 12 different equally sized groups

b)

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This also shows that students can grouped in equal groups of 12 in 5 places

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\bf ~~~~~~~~~~~~\textit{internal division of a line segment}&#10;\\\\\\&#10;A(-4,-8)\qquad B(11,7)\qquad&#10;\qquad \stackrel{\textit{ratio from A to B}}{3:7}&#10;\\\\\\&#10;\cfrac{A\underline{C}}{\underline{C} B} = \cfrac{3}{7}\implies \cfrac{A}{B} = \cfrac{3}{7}\implies 7A=3B\implies 7(-4,-8)=3(11,7)\\\\[-0.35em]&#10;~\dotfill\\\\&#10;C=\left(\frac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \frac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\[-0.35em]&#10;~\dotfill


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