Problem situation:
Janelle is at the movie theater and has $20 to spend.
She spends $8 on a ticket and wants to buy some snacks. Each snack costs $4.99.
How many snacks, , can Janelle buy?

Inequality that represents this situation:
20≥8+4.99

Drag each number to show if it is a solution to both the inequality and the problem situation, to the inequality only, or if it is not a solution.

FOR BOTH :  1  AND  2

FOR ONLY INEQUALITY : -4  / 2.3  /  3/4

FOR NOT A SOLUTION : 2.5

Step-by-step explanation:

I SUBSTITUTED THE 6 NUMBERS AND SOLVED THE EQUASION.

FIRST, FOR FRACTIONS AND DECIMALS, WE CANT BUY A THING WITH PART OF IN SO THEY CAN BE EITHER IN ONLY INEQUALITY OR NOT A SOLUTION.  

2.3  AND 3/4 CAN BE A SOLUTION ONLY WITH INEQUALITY.

-4 CAN BU USED IN THE INEQUALITY BUT ITS A NEGATIVE NUMBER NOT A WHOLE NUMBER (WHOLE NUMBERS ARE FROM 1 TO INFENITY)

BUT 2.5 CANT BE BECAUSE IT GIVES US A NUMBER GREATER THAN 20

IN ANOTHER SIDE, 1 AND 2 CAN BE USED BOTH IN INEQUALITY AND IN REAL LIFE.

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!