Stadium has 48,000 seats seats sell for $30 in section 8 $24 in section B and $18 in section C the number of seats in section a equals the total number of seats in section BNC supposed to stadium takes in 1,239,600 From each sold out that how many seats does each section whole

there are  24,000 sears in section A,  14,600 in section B  and  19,400 in section C.

Step-by-step explanation:

Step 1

let  the number of seats in section B = x

and the number of seats in section C.=y

Given that the  number of seats in section A is the sum of seats in sections B and C, we can then say A =x + y

From the given question,  We have that

seatsA + seatsB + seatsC = 48,000 which also equals

(x+y)  + x      + y      = 48,000

2x + 2y = $48,000

2(x+y) = 48,000

x + y = 48,000/2 =24,000  which also equals

B+C =24,000

Now the number of seats in section A =24,000 and the sum of seats in sections B and C is 24000.

We now have

B+C = 24,000 eqn  (1)

Also

  since we already have the number of seats for Section A, We will find Section  B and C  

A +B+C =1,239,600,

$30 x 24,000 + 24B + 18C= 1,239,600

720,000 + 24B + 18C= 1,239,600

 24B + 18C= 1,239,600-720,000

24B + 18C= 519,600 eqn 2

Step 2 We will now solve with the following equations

B+C = 24,000 eqn  (1)

24B + 18C= 519,600 eqn 2

Multiply eq(1) by 24 (both sides). Then subtract eq(2) from it:

24B + 24C = 576,000eqn    (3)

24B+ 18C =   519,600-- ---eqn(4)

6C = 576,000- 519,600

6C=56,400

C= 9,400

TO get B.

B+C = 24,000

B= 24,000 - 9,400

B=14,600

Therefore, there are 24,000 sears in section A,  14,600 in section B  and  19,400 in section C.

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.