The 8th grade is selling tickets to the 8th grade formal the cost per ticket is 15 the decorations and food cost 600 each ticket cost 0.50 to print. How many tickets need to be sold in order for the 8th grade to make profit

600 > 620.  This is not true, so 40 tickets isn't enough.

Explanation:

Each ticket is $15.  The number of tickets is what we are trying to solve for.  The class spends a certain amount of money to prepare for the formal.  They hope that the money they make in ticket sales is MORE than what they spend.  The expression that represents the number of tickets at $15 each is 15x, where x is the number of tickets.  They hope that the sales are greater than what they spend, so what we have so far is

15x >

Greater than what, though?  What do they spend?  They spend 600 for the food, so

15x > 600...

but they also have to print a certain, unknown number of tickets at .50 each.  The expression that represents the printing of each ticket is .5x (we can drop the 0; it doesn't change the answer or make it wrong if we drop it off).  So the cost for this affair is the food + the printing.

15x > 600 + .5x

Solve this inequality for x.  Begin by subtracting .5 from both sides to get

14.5x > 600 so

x > 41.3

Because we are not selling (or printing) .3 of a ticket, it's safe to say (and also correct!) that they need to sell (and print) 41 tickets.  If they sell 41 tickets, the profit is found by

15(41) > 600 + .5(41)

615 > 600

This means that at 41 tickets, they make a profit.  At 40 tickets, the inequality looks like this:

15(40) > 600 + .5(40) and

600 > 620.  This is not true, so 40 tickets isn't enough.

the ileocecal valve (also called bauhin valve), which limits the rate of food passage into the cecum and may prevent material from returning to the small intestine.

elevated levels of blood glucose (hyperglycemia) lead to spillage of glucose into the urine, hence the term sweet urine.