A stadium has 50,000 seats. Seats sell for $28 in Section A, $24 in Section B, and $20 in Section C. The number of seats in Section A equals the total number of seats in Sections B and C. Suppose the stadium takes in $1,258,000 from each sold-out event. How many seats does each section hold

Section A holds 25,000 seats, section B holds 14,500 seats and section C holds 10,500 seats.

Step-by-step explanation:

From the information given you can write the following equations:

A+B+C=50,000 (1)

28A+24B+20C=1,258,000 (2)

A=B+C (3)

A= number of seats in section A

B= number of seats in section B

C= number of seats in section C

You can replace (3) in (1) and (2) to get two equations:

B+C+B+C=50,000

2B+2C=50000

28(B+C)+24B+20C=1,258,000

28B+24B+28C+20C=1,258,000

52B+48C=1,258,000

The two equations are:

2B+2C=50000 (4)

52B+48C=1,258,000 (5)

You can isolate B in (4):

2B=50,000-2C

B=(50,000/2)-(2C/2)

B=25,000-C

Now, you can replace B in (5):

52(25,000-C)+48C=1,258,000

1,300,000-52C+48C=1,258,000

1,300,000-1,258,000=4C

42,000=4C

C=42,000/4

C=10,500

Now, you can replace the value of C in B=25,000-C:

B=25,000-10,500

B=14,500

Finally, you can replace the values of B and C in A=B+C to find A:

A=14,500+10,500

A=25,000

According to this, the answer is that section A holds 25,000 seats, section B holds 14,500 and section C holds 10,500.

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2