<span>Use a proportion:
x/(5 1/2) = 50/4
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x = (11/2)(50/4)
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x = 550/8
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x = $68.75</span>
Answer:
The choice A ;

Step-by-step explanation:

Hello!
To solve this problem we first need to find the value of each given item.
12^4 is equal to 12·12·12·12, or 20,736
12·12·12·12·12 is equal to 12^5, or 248,832
With this information being provided, it can be said that 12^4 < 12x12x12x12x12
Hope this helps.
Answer:
1)6.32
2)12.80
3)13
Step-by-step explanation:
1)

2)

3)

We will investigate how to determine Hamilton paths and circuits
Hamilton path: A path that connect each vertex/point once without repetition of a point/vertex. However, the starting and ending point/vertex can be different.
Hamilton circuit: A path that connect each vertex/point once without repetition of a point/vertex. However, the starting and ending point/vertex must be the same!
As the starting point we can choose any of the points. We will choose point ( F ) and trace a path as follows:

The above path covers all the vertices/points with the starting and ending point/vertex to be ( F ). Such a path is called a Hamilton circuit per definition.
We will choose a different point now. Lets choose ( E ) as our starting point and trace the path as follows:

The above path covers all the vertices/points with the starting and ending point/vertex are different with be ( E ) and ( C ), respectively. Such a path is called a Hamilton path per definition.
One more thing to note is that all Hamilton circuits can be converted into a Hamilton path like follows:

The above path is a hamilton path that can be formed from the Hamilton circuit example.
But its not necessary for all Hamilton paths to form a Hamilton circuit! Unfortunately, this is not the case in the network given. Every point is in a closed loop i.e there is no loose end/vertex that is not connected by any other vertex.