2(a+bc)=2a+2bc
3.(ab-cb-ba+bc)= 3ab-3cb-3ba+3bc
Since the brackets and multiplication have a higher order than the rest.
Cb=Bc
Ab=ba
Ac=Ca
Because multiplication is commutative.
2a+2bc + cb +ab + ac -(3ab-3cb-3ba+3bc)
2a + 2bc + cb + ab + ac(+ -3ab - - 3cb - -3ba +-3bc)
2a + 2bc + cb + ab + ac -3ab + 3cb + 3ba -3bc
2a + (2bc-3bc +cb + 3cb ) + (ab -3ab + 3ba) + ac
=2a + 3cb + ba+ ac OR 2a +3bc + ab + ac
The Or is just a different way to write it.
Hope this helps
The are 40320 ways in which the 5 indistinguishable rooks be can be placed on an 8-by-8 chess- board so that no rook can attack another and neither the first row nor the first column is empty
<h3 /><h3>What involves the
rook polynomial? </h3>
The rook polynomial as a generalization of the rooks problem
Indeed, its result is that 8 non-attacking rooks can be arranged on an 8 × 8 chessboard in r8.
Hence, 8! = 40320 ways.
Therefore, there are 40320 ways in which the 5 indistinguishable rooks be can be placed on an 8-by-8 chess- board so that no rook can attack another and neither the first row nor the first column is empty.
Read more about rook polynomial
brainly.com/question/2833285
#SPJ4
<span>The rate for one unit of a given quantity is called a u</span>nit rate.
<span>
1.) Will yield consecutive odd integers.
k+10, k+12, k+14
or
k+2, k+3, k+4
The answer is k + 10, k +12 , k + 14
Because consecutive integers have difference of 2.
If k is odd, then k+10, k+12, and k+14 are consecutive odd integers.
For example, assume k = 1, then,
k+10=11
k+12=13
k+14=15
And 11, 13 and 15 are consecutive odd integers.
2.) Will yield consecutive integers.
k+1, k+2, k+3
or
k+6, k+8, k+10
The answer is k+1, k+2, k+3
Let k be any integer number, k+1, k+2, k+3 are consecutive integers.
For example, let k = 23
k+1=24
k+2=25
k+3=26
24,25, and 26 are consecutive integers.
</span>
