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Problem 7)
The answer is choice B. Only graph 2 contains an Euler circuit.
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To have a Euler circuit, each vertex must have an even number of paths connecting to it. This does not happen with graph 1 since vertex A and vertex D have an odd number of vertices (3 each). The odd vertex count makes it impossible to travel back to the starting point, while making sure to only use each edge one time only.
With graph 2, each vertex has exactly two edges attached to it. So an Euler circuit is possible here.
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Problem 8)
The answer is choice B) 5
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Work Shown:
abc base 2 = (a*2^2 + b*2^1 + c*2^0) base 10
101 base 2 = (1*2^2 + 0*2^1 + 1*2^0) base 10
101 base 2 = (1*4 + 0*2 + 1*1) base 10
101 base 2 = (4 + 0 + 1) base 10
101 base 2 = 5 base 10
It's a trick question. There are an infinite number of mixed numbers between 3 and 4 that can multiply to equal 12 (for example, 3 and 3/7 times 3 and 1/2), but there are no mixed numbers between 3 and 4 that can multiply to equal 9. 3 times 3 is not between them but is 3, but that quantity is excluded because 3<x<4. Anything even a small bit above the number 3 would have to be multiplied by 2 and some fraction, which would not be between 3 and 4.
Answer: lol wut
Step-by-step explanation:
