We can model this situation with an arithmetic series.
we have to find the number of all the seats, so we need to sum up the number of seats in all of the 22 rows.
1st row: 23
2nd row: 27
3rd row: 31
Notice how we are adding 4 each time.
So we have an arithmetic series with a first term of 23 and a common difference of 4.
We need to find the total number of seats. To do this, we use the formula for the sum of an arithmetic series (first n terms):
Sₙ = (n/2)(t₁ + tₙ)
where n is the term numbers, t₁ is the first term, tₙ is the nth term
We want to sum up to 22 terms, so we need to find the 22nd term
Formula for general term of an arithmetic sequence:
tₙ = t₁ + (n-1)d,
where t1 is the first term, n is the term number, d is the common difference. Since first term is 23 and common difference is 4, the general term for this situation is
tₙ = 23 + (n-1)(4)
The 22nd term, which is the 22nd row, is
t₂₂ = 23 + (22-1)(4) = 107
There are 107 seats in the 22nd row.
So we use the sum formula to find the total number of seats:
S₂₂ = (22/2)(23 + 107) = 1430 seats
Total of 1430 seats.
If all the seats are taken, then the total sale profit is
1430 * $29.99 = $42885.70
Bru are u serious ? it’s C 7 x the parable equates with the equinox of traversal is 36/5
the answer does be 7875810503.17
Answer:
he would need 31 tanks
30 tanks would hold 240 fish with 6 left over
Step-by-step explanation:
246/8= 30.75
30*8 = 240
