A theater sold 4 adult tickets and 3 child tickets for 156 dollars they also sold 3 adult tickets and 4 child tickets for 145 dollars. How much are adult and child tickets

Adult tickets cost 27 dollars while Child tickets cost 16 dollars

Step-by-step explanation: The question clearly states that adult tickets and child tickets are sold for a combined total sales amount. However the combination of adult and child tickets always yield a different sales total depending on the mix of tickets. This implies that both tickets have different amounts attached to them.

For a start, we shall assign an alphabet to each of the unknown ticket values. Let’s assume that adult tickets are represented by letter A, and child tickets are represented by B. If 4 adult and 3 child tickets are sold for 156 dollars, then what we have is

4A + 3B = 156 —————(1)

Also if 3 adult tickets and 4 child tickets are sold for 145 dollars, then what we have here is

3A + 4B = 145 ————-(2)

Now we have a pair of simultaneous equations which are;

4A + 3B = 156 —————(1)

3A + 4B = 145 —————(2)

We shall apply the elimination method to solve for A and B. First of all we shall eliminate the coefficients of A by multiplying equation (1) with 3, and multiplying equation (2) with 4. This now gives us

12A + 9B = 468 ———-(3)

12A + 16B = 580 ———(4)

Subtract equation 3 from equation 4 and we arrive at

7B = 112

(Note that 12A - 12A = 0)

Divide both sides of the equation by 7

B = 16

Having that in mind, we shall now substitute for the value of B = 16 in equation 1.

4A + 3B = 156

4A + 3(16) = 156

4A + 48 = 156

Subtract 48 from both sides of the equation

4A = 108

Divide both sides of the equation by 4

A = 27

Hence, adult tickets are sold for 27 dollars while child tickets are sold for 16 dollars.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

the answer is 2/54

step-by-step explanation: