The first one is independent, and the second one is dependent.
:
.
-- :
Bike > ________300km_________< Van
20km/h 80km/h
In one hour, the bike will have moved 20km and the van would have moved 80km.
The distance between them was 300km. But the bike moved 20km and the van moved 80km, since its been one hour. Which means the distance between them is less.
300km-(20km+80km) = 200km
So now the distance between them is 200km.
Now another hour passes. The distance between them was 200km. But the bike moved 20km and the van moved 80km, since its been one hour. Which means the distance between them is less.
200km-(20km+80km) = 100km
So now the distance between them is 100km
Now another hour passes. The distance between them was 100km. But the bike moved 20km and the van moved 80km, since its been one hour. Which means the distance between them is less.
100km-(20km+80km) = 0km
So now the distance between them is 0km, which means they have met. If we count, it took them 3 hours.
:
Step 1)
Add both speeds. 20km+80km=100km
Step 2)
Total distance between them = 300km
So do 300km/100km = 3hours.
______
The solution is: .
————
Hope this helped!
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 !
Answer:
-2
Step-by-step explanation:
you will group the like terms and add and subtract the remaining digits
1. 12
2. 1
3. 18
4. 9
5. 2
6. 20
7. 1
8. 6
^^ I am so sorry if any are wrong, I tried as best as I could
