Round each number to the nearest whole number. then
use the appropiate symbol to compare the rounded numbers.
368.35 and 368.32

On rounding first number 368.35 we get 368  

On rounding second number 368.32 we get 368.

368=368 so appropriate comparison operator is “=”.

Solution:

Need to round two given numbers.

First number is 368.35

Given number resides between two whole numbers 368 and 369.

If three numbers are visualize on number line,it can be said that 368.35 is more closer to 368 than 369. And as we need to round the number to nearest whole number. On rounding 368.35 we get 368.

Second number is 368.32

Given number resides between two whole numbers 368 and 369.

If three numbers are visualize on number line, it can be said that 368.32 is more closer to 368 than 369. And as we need to round the number to nearest whole number. On rounding 368.32 we get 368.

On rounding first number 368.35 we get 368  

On rounding second number 368.32 we get 368.

That is both the output is same after rounding.

On comparison we get 368 = 368. So required appropriate comparison operator is “=”.

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!