Use the distributive properly to remove the parentheses. -(4w-y-3)

Mathematics

1 answer:

Brrunno [24]3 years ago

7 0

-4w+y+3 is the correct answer

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Suppose you have the following recursion formula a1 = 1, a2 = 2, and an = a(n - 1)+ a(n - 2) for integers n ≥ 3. How would you d

Firlakuza [10]

$\left\{\begin{array}{ccc}a_1=1\\a_2=2\\a_n=a_{n-1}+a_{n-2}&for\ n\geq3\end{array}\right\\\\a_3=a_{3-1}+a_{3-2}=a_2+a_1\to a_3=2+1=3\\\\a_4=a_{4-1}+a_{4-2}=a_3+a_2\to a_4=3+2=5\\\\a_5=a_{5-1}+a_{5-2}=a_4+a_3\to a_5=5+3=8\\\vdots$

6 0

3 years ago

Find the number of ways of arranging the numbers

Doss [256]

First of all, note that all integers are either 0,1, or 2 modulo 3 (if you're not familiar with this terminology, it means that every integer is either a multiple of 3, or it is 1 or 2 away from a multiple of 3).

So, we can think of our numbers as

$\begin{array}{c|c}x&x\mod 3\\0&0\\1&1\\2&2\\3&0\\4&1\\5&2\\6&0\\7&1\\8&2\\9&0\end{array}$

In order to make sure that the sum of any three adjacent numbers is divisible by 3, we have to make sure that any group of 3 three adjacent numbers contains a 0, a 1 and a 2. This is possible only if we arrange our 9 numbers in 3 groups of 3 numbers containing 0,1 and 2 exactly once, repeating always the same pattern.

For example, we could arrange our numbers following the pattern

$0,1,2,0,1,2,0,1,2$

$2,0,1,2,0,1,2,0,1$

We have $3!=6$ possible patterns. Suppose for example that we choose the pattern

$0,1,2,0,1,2,0,1,2$

One possible way of following this pattern would be the arrangement

$3,1,2,6,4,5,9,7,8$

In fact, we substituted every '0' with a multiple of 3 (3, 6 or 9), every '1' with a number 1 away from a multiple of 3 (1, 4 or 7) and every '2' with a number 2 away from a multiple of 3 (2, 5 or 8).

This means that, once we fix a patter, we have 3 choices for the first 3 slots, 2 choices for the next 3 slots, and the final slot will be fixed. So, we have

$3\cdot 3\cdot 3\cdot 2 \cdot 2 \cdot 2 = 216$