Answer:
The sum of the first five classroom numbers in a row is 5k + 20
Step-by-step explanation:
Since the smallest classroom number on the side of the building is numbered k and each consecutive odd integer is separated by a difference of
2.
Therefore:
k is the first class room.
k + 2 is the second class room.
k + 4 is the third class room.
k + 6 is the third class room.
k + 8 is the fifth class room.
The sum of the five consecutive class rooms are given as:
k + (k + 2) + (k + 4) + (k + 6) + (k + 8)
collecting alike terms we get
= k + k + k + k + k + 2 + 4 + 6 + 8
= 5k + 20
Therefore, The sum of the first five classroom numbers in a row is 5k + 20.
1. 1/a^8 (Sorry I can't find out what the others are)
Answer:
Step-by-step explanation:
divide by\[ 6 
\]
x=z/ 6 \pi y
Depending on the player, a spectator can figure out the chance of the player making the basket :D
Answer:
A maximum of 306
Step-by-step explanation:
You can estimate that since each bus had less than 52 students, that 51 were on each bus. 51 times 6=306
