9514 1404 393
Answer:
D.) a+2b
Step-by-step explanation:
The integers 'a' and 'b' can be any, so you can choose a couple and evaluate these expressions to see what you get. For example, we can let a=1 and b=0. For these values, the offered expressions evaluate to ...
A) 3(0) = 0 . . . even
B) 1 +3 = 4 . . . even
C) 2(1+0) = 2 . . . even
D) 1 +2(0) = 1 . . . odd
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<em>Additional comment</em>
These rules apply to even/odd:
- odd × odd = odd
- odd × even = even
- even × even = even
- odd + odd = even
- odd + even = odd
- even + even = even
Then A is (odd)(even) = even; B is (odd)+(odd) = even; C is (even)(whatever) = even; D = (odd)+(even) = odd.
Answer:
10
Step-by-step explanation:
Each box is worth 2 units, because the x and y axis are counting by 2s
Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).
Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2
6 is d, 7 is b, 8 is c, 9 is c
8x^3 -5x^2 + 8x + 9+5x^3 + 3x^2 - 5x + 4 =
8x^3+5x^3-5x^2 + 3x^2+ 8x - 5x+ 9<span>+ 4 = </span>
13x^3-2x^2+3x+13
