Q: How do you remove 2 from 5 and leave 4?

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It is LCM(3, 4, 5) + 2 = 62

2-4, 3-2, 4-5, 4-4, 5-5

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All of them. Only 4 does not leave a remainder of 1.

2 ÷ 4/5 = 5/2 or 21/2

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There are 64 subsets, and they are:{}, {A}, {1}, {2}, {3}, {4}, {5}, {A,1}, {A,2}, {A,3}, {A,4}, {A,5}, {1,2}, {1,3}, {1,4}, {1,5}, {2,3}, {2,4}, {2,5}, {3,4}, {3, 5}, {4,5}, {A, 1, 2}, {A, 1, 3}, {A, 1, 4}, {A, 1, 5}, {A, 2, 3}, {A, 2, 4}, {A, 2, 5}, {A, 3, 4}, {A, 3, 5}, {A, 4, 5}, {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}, {A, 1, 2, 3}, {A, 1, 2, 4}, {A, 1, 2, 5}, {A, 1, 3, 4}, {A, 1, 3, 5}, {A, 1, 4, 5}, {A, 2, 3, 4}, {A, 2, 3, 5}, {A, 2, 4, 5}, {A, 3, 4, 5}, {1, 2, 3, 4}, {1, 2, 3, 5}, {1, 2, 4, 5}, {1, 3, 4, 5}, {2, 3, 4, 5}, {A, 1, 2, 3, 4}, {A, 1, 2, 3, 5}, {A, 1, 2, 4, 5}, {A, 1, 3, 4, 5}, {A, 2, 3, 4, 5}, {1, 2, 3, 4, 5} {A, 1, 2, 3,,4, 5} .

It is {1, 2, 2, 4, 4, 4, 4, 5, 5, 5, 5, 5, 9, 12}

I discovered the solution by accident about 2 weeks ago, but didn't realize it. There is one 6 cone that fits on to a 5 space and one 5 cone that fits on to a 6 space. I will leave it up to you to find the corresponding colors that work. Good luck and have fun.... {| |- | 5 3 2 1 4 6 4 1 5 2 5 3 6 5 3 4 1 2 1 2 6 3 6 4 2 4 1 6 3 5 3 6 4 5 2 1 |}

The greatest of these fractions is 3/4 and the last is 2/5. The fraction 2/3 would fall in the middle of the other two. The least of these fractions would be 2/5 and the greatest would be 3/4. That would leave the middle number being 2/3.

2x+5=9 2x+5-5=9-5 you leave only the x on one side. 2x=4 x=2

The first five numbers which when divided by 5 leave a remainder of 4 are: 4 = 4/5 = 0 remainder 4 9 = 9/5 = 1 remainder 4 14 = 14/5 = 2 remainder 4 19 = 19/5 = 3 remainder 4 24 = 24/5 = 4 remainder 4 The pattern continues in this way.