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muminat
3 years ago
10

Use the numbers 1-12 one time each to creat four true statements with addition, multiplication, subtraction and division

Mathematics
1 answer:
laila [671]3 years ago
7 0

If you mean that you can only use each number along the four expressions, it looks impossible to me. Here's the proof.

Multiplications and divisions are the most restrictive operations, because only some triplets will work. In particular, you can choose

(2,3,6),\ (2,4,8),\ (2,5,10),\ (2,6,12),\ (3,4,12)

I'm listing them in triplets because you can use them in many ways. For example, the first triple can be used to generate

2\cdot 3 = 6,\quad 3\cdot 2 = 6,\quad 6\div 3 = 2,\quad 6\div 2 = 3

But it doesn't really matter, in all cases you used numbers 2,3, and 6.

If you look closely, all triplets but the last one involve 2. This means that we must use the last one, because otherwise we would use two triplets with two, and we would have repetitions.

So, we surely have to use (3,4,12), either to write 3\cdot 4=12 or 12\div 4=3.

This means that we can't use 3, 4 and 12 anymore. The only triplet remaining is (2,5,10).

For example, let's say that our multiplication is 2\cdot 5 = 10 and our division is 12\div 4 = 3.

We're left with the following numbers:

1,\ 6,\ 7,\ 8,\ 9,\ 11

From here, we're left with a few choices for the addition: if we choose 1+6=7 we're left with 8,9 and 11. We can't write any subtraction with these numbers.

If we choose 1+7=8, we're left with 6, 9 and 11. We can't write any subtraction with these numbers.

If we choose 1+8=9, we're left with 6, 7 and 11. We can't write any subtraction with these numbers.

To recap, we're forced to use 2,3,4,5,10 and 12 for the multiplication and the division, and the remaining numbers don't allow to write an addiction and a subtraction.

Thus, you can't use all the operations involving numbers 1-12 only once.

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The domain of all polynomials is all real numbers.  To find the range, let's solve that quadratic for its vertex.  We will do this by completing the square.  To begin, set the quadratic equal to 0 and then move the -10 over by addition. The first rule is that the leading coefficient has to be a 1; ours is a 2 so we factor it out.  That gives us:

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