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just olya [345]
3 years ago
13

Simplify -2.9f + .9f -12 - 4

Mathematics
1 answer:
Natalka [10]3 years ago
6 0

\huge\text{Hey there!}

\large\text{-2.9f + 0.9f -12 - 4}\\\\\large\text{COMBINE the LIKE TERMS}\\\\\large\text{(-2.9f + 0.9f) + (-12 - 4)}\\\\\large\text{(-2.9f + 0.9f) = \bf{-2f}}\\\\\large\text{(-12 - 4) = \bf{-16}}\\\\\large\text{= \bf{-2f - 16}}\\\\\\\boxed{\boxed{\large\text{Answer: \bf{-2f - 16}}}}\huge\checkmark

\text{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

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What do you have to do to this fraction before we can at the numerators ? 1/4 + 2/7​
motikmotik

Answer:

you have to convert them to equivalent units (the same denominators)

Step-by-step explanation:

first, multiply  2 and 7 by 4.

now you have 8/21.

then, make the other fraction the same value.

multiply 1 and 4 by 7.

7/21.

now your equation is 7/21 + 8/21= 15/21.

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3 years ago
Suppose you have a complete, weighted graph with 8 vertices. How many Hamilton Circuits are there in this graph?
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The number of Hamilton Circuits with 8 vertices are 5040.

Given that, a complete, weighted graph with 8 vertices.

<h3>What are Hamilton Circuits?</h3>

A Hamiltonian circuit is a circuit that visits every vertex once with no repeats. Being a circuit, it must start and end at the same vertex. A Hamiltonian path also visits every vertex once with no repeats, but does not have to start and end at the same vertex.

For N vertices in a complete graph, there will be (n−1)!=(n−1)(n−2)(n−3)…3⋅2⋅1 routes. Half of these are duplicates in reverse order, so there are (n−1)!/2 unique circuits.

A complete graph with 8 vertices would have = 5040 possible Hamiltonian circuits. Half of the circuits are duplicates of other circuits but in reverse order, leaving 2520 unique routes.

Therefore, the number of Hamilton Circuits with 8 vertices are 5040.

Learn more about the Hamilton Circuits here:

brainly.com/question/24725745.

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1 year ago
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