3 sandwich choices 2 side item choices 4 beverage choices
Let's say the sandwich choices were PB&J, Ham, and Turkey Let's say the sides were fries and mandarin oranges Let's say the beverages were water, orange juice, milk, soda
PB&J Ham Turkey
Fries Mandarin Oranges
Water OJ Milk Soda
The combinations can be as follows:
PB&J, Fries, Water PB&J, Fries, OJ PB&J, Fries, Milk PB&J, Fries, Soda PB&J, Mandarin Oranges, Water PB&J, Mandarin Oranges, OJ PB&J, Mandarin Oranges, Milk PB&J, Mandarin Oranges, Soda Ham, Fries, Water Ham, Fries, OJ Ham, Fries, Milk Ham, Fries, Soda Ham, Mandarin Oranges, Water Ham, Mandarin Oranges, OJ Ham, Mandarin Oranges, Milk Ham, Mandarin Oranges, Soda Turkey, Fries, Water Turkey, Fries, OJ Turkey, Fries, Milk Turkey, Fries, Soda Turkey, Mandarin Oranges, Water Turkey, Mandarin Oranges, OJ Turkey, Mandarin Oranges, Milk Turkey, Mandarin Oranges, Soda
So your answer is correct: There's 24 lunch combinations that can be made.
The greatest common factor of 18 = 3·6 and 30 = 5·6 is 6.
The greatest common factor of x^2 and x^6 is x^2.
Factoring 6x^2 from both terms, we get ...
... 18x^2 +30x^6 = 6x^2(3 +5x^4)
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<em>Comment on the question</em>
Since this answer is not among the answer choices, I suggest you ask your teacher to demonstrate how this problem is worked.
It appears as though the answers go with the problem 18x^9 +30x^6. Maybe there's a typo somewhere. For that problem, the best choice is the 2nd answer.
