Answer:
would that be 31 if if im wrong im just nine Step-by-step explanation:
Yes, (1,7) is a solution.
(x,y)
y < x + 8
y ≤ 6x + 1
7<1+8 = 7<9 Yes, true.
7≤6(1)+1 = 7≤ 7 Yes, true.
Yes, (1,7) is a solution to the inequalities.
Let student tickets be s and adult tickets be a. The number of tickets sold of both adult and student then is s + a = 396. If each student ticket costs $3, then we represent the money equation by tacking the dollar amount onto the ticket. 3s is the cost of one student ticket. 4a is the cost of an adult ticket. The total money from the sales of both is 4a + 3s = 1385. We now have a system of equations we can solve for a and s. If s+a=396, then s = 396-a. We will sub that into the second equation to get 4a + 3(396-a) = 1385. Distributing we have 4a+1188-3a=1385. a = 197. That means there were 197 adult tickets sold. If s + a = 396, then s + 197 = 396 and s = 199. 197 adult tickets and 199 student tickets. There you go!
Answer:
They have no minimum y -value
