What is the graph of the solution to the following compound inequality? 2 <6x-4 < 14
A.
-10-9-8-7-6-5-4-3 -2 -1
0 1 2 3 4
5 6
7 8
9 10
B.
O
-10-9-8-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
C.
-10-9-8-7-8-5-4-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
D.
0
-10-9-8-7-6-5-4-3-2-1 1 2 3 4 5 6 7 8 9 10

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!

8

step-by-step explanation:

just put √(2) * √(32) into a calculator. that will make the answer be 8.

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92