A coach needs both small and large cones to set up an obstacle course. The coach knows that there are fewer than 30 cones in the storage room. The coach needs at least a dozen large cones. Let x represent the number of small cones and y represent the number of large cones.

Which inequalities model the situation?

x + y < 30

y ≥ 12

Step-by-step explanation:

Here, we want to write inequalities

From the question, we are told that there are fewer than 30 cones

Mathematically, we have this as;

x + y < 30

Secondly, at least 12 ( a dozen) large cones are needed

When we talk of at least, we mean it can be greater than but not less than

So the inequality in this case since y represents the number of large cones will be;

y ≥ 12

hi

step-by-step explanation:

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!