Ahigh school drama club is selling tickets to their show. there is a maximum of 400 tickets. the tickets cost $10 (if bought before day of show) and $15 (if bought day of show). let x represent the number of tickets sold before the day of the show and y represents the number of tickets sold the day of the show. to meet the expenses of the show, the club must sell at least $3500 worth of tickets. the club sells 300 tickets before the day of the show. is it possible to sell enough additional tickets on the day of the show to meet the expenses of the show? justify your answer.

  yes

Step-by-step explanation:

The club can sell 100 more tickets before they reach their maximum. If they sell those at the "day-of" price, they will have additional revenue of ...

  100 × $15 = $1500

Then their total revenue would be ...

  300 × $10 + $1500 = $4500 . . . . more than enough to meet expenses.

To make up the additional $500 they need, the drama club only needs to sell ...

  ceiling($500/$15) = 34 . . . tickets at the "day-of" price.

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

i think it is c, not positive though

step-by-step explanation: