Answer:
1,3,4,5,6,10,11,14,16,17,18,20
Step-by-step explanation:
I've been there so many times. Hope you have a great day!
Answer: 45r
Step-by-step explanation:
Answer:
x = -122/13 OR 9.3846
Step-by-step explanation:
First, take a look at the second equation. Add 8x to the other side.
-7y= 8x -18
Then, divide by -7 to get a regular "y=" equation.
y= 8/7x -18/7
Move on the the first equation. Let's get "y" by itself. Add 3x to the other side.
y= 3x +15
Considering both equations are equal to Y, set them equal ( except the "y" part )
8/7x - 18/7 = 3x + 15
Multiply all by seven so that there are no fractions.
8x - 18 = 21x + 105
Subtract 8x from both sides.
-17 = 13x + 105
Subtract 105 from both sides.
-122 = 13x
Divide by 13 on both sides.
x = 9.3846
Or if you don't want decimals, just say x = -122/13
Quick Note: I made a mistake. I had looked at the second equation at -18. The answer is incorrect but the method of solving is correct. Also, make sure to plug in the value of x to get Y.
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 !
