We can reject the last one: subtracting a non-zero value will result in a smaller value.
So now we have:
<span>2(A + B)
(A + B)2
A2 + B2
If all of them are mulptiplications, then they are all equivalent!
I mean by this, if what you meant is this:
</span>
<span>2*(A + B)
(A + B)*2
A*2 + B*2
If there is no sign, then the multiplication sign is implicit,
and all of these expressions say exactly the same: two of A and two of B.
</span>
Hey!
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Solution:
4(9x + 7) = x - 4
36x + 28 = x - 4
36x + 28 + 28 = x - 4 + 28
36x = x + (-32)
36x - x = x - (-24) - x
35x = -32
35x/35 = -32/35
x = -32/35
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Answer:
x = -32/35
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Hope This Helped! Good Luck!
Answer:
네, 죄송합니다. 알았어 알았어. 알았어. 그러면 너는 내 대답을 위해 이것을하고있다.
Wow !
OK. The line-up on the bench has two "zones" ...
-- One zone, consisting of exactly two people, the teacher and the difficult student.
Their identities don't change, and their arrangement doesn't change.
-- The other zone, consisting of the other 9 students.
They can line up in any possible way.
How many ways can you line up 9 students ?
The first one can be any one of 9. For each of these . . .
The second one can be any one of the remaining 8. For each of these . . .
The third one can be any one of the remaining 7. For each of these . . .
The fourth one can be any one of the remaining 6. For each of these . . .
The fifth one can be any one of the remaining 5. For each of these . . .
The sixth one can be any one of the remaining 4. For each of these . . .
The seventh one can be any one of the remaining 3. For each of these . . .
The eighth one can be either of the remaining 2. For each of these . . .
The ninth one must be the only one remaining student.
The total number of possible line-ups is
(9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) = 9! = 362,880 .
But wait ! We're not done yet !
For each possible line-up, the teacher and the difficult student can sit
-- On the left end,
-- Between the 1st and 2nd students in the lineup,
-- Between the 2nd and 3rd students in the lineup,
-- Between the 3rd and 4th students in the lineup,
-- Between the 4th and 5th students in the lineup,
-- Between the 5th and 6th students in the lineup,
-- Between the 6th and 7th students in the lineup,
-- Between the 7th and 8th students in the lineup,
-- Between the 8th and 9th students in the lineup,
-- On the right end.
That's 10 different places to put the teacher and the difficult student,
in EACH possible line-up of the other 9 .
So the total total number of ways to do this is
(362,880) x (10) = 3,628,800 ways.
If they sit a different way at every game, the class can see a bunch of games
without duplicating their seating arrangement !
