The drama club is selling tickets to their play to raise money for the show's expenses.

Each student ticket sells for $8 and each adult ticket sells for $13. The auditorium can

hold at most 70 people. The drama club must make at least $730 from ticket sales to

cover the show's costs. If 41 adult tickets were sold, determine the minimum number

of student tickets that the drama club must sell in order to meet the show's expenses.

If there are no possible solutions, submit an empty answer.

25

Step-by-step explanation:

Ticket sale (student) = $8

Ticket sale (adult) = $13

The hall can contain 70 people.

Number of adult tickets sold = 41

Cost of show = $730

Let the number of student ticket sold = x

Setting an algebraic equation to show represent the cost of the show and set minimum target of ticket to be sold.

730 = (41 * 13) + (x * 8)

730 = 533 + 8x

8x = 730 - 533

8x = 197

X = 197 / 8

x = 24.6 = 25

The minimum number of student tickets to be sold is 25

answe

its a

step-by-step explanation:

true

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2