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 !
<h2>Vertical distance gained from takeoff is 493.72 ft</h2>
Step-by-step explanation:
Refer the figure.
Horizontal distance = AB = 2800 ft
Angle = ∠B = 10°
We need to find AC that is h.
We have
Vertical distance gained from takeoff = 493.72 ft
Answer:
x=6
y=3 i think
Step-by-step explanation:
becuase it looks like they were cutting the mesuments from the triangle so the y would equal 3, andx would be 6.
Question:
Taylor and his Dad plan to build a deck on the back of their house. First, they make a scale drawing of the deck. The scale is 1.25 inches = 2 feet.
This is the drawing: the scale drawing deck is 17 by 11
Using this drawing, enter the length, in feet, Taylor must use for the deck?Answer:
The length, Taylor must use for the deck is 27 .2 feet
Step-by-step explanation:
Given:
The dimensions of the deck = 17 by 11
Scale = 1.25 inches = 2 feet.
To Find:
The length, in feet, Taylor must use for the deck
Solution:
According to the scale
1.25 inches = 2 feet
Then
1 inch = 
The length of the deck = 17 inches
So 17 inches = 
17 inches = 27.2 feet
Answer:
The answer to the expression (5x^4-9x^3+7x-1) = -98
Step-by-step explanation:
