Answer:
m∠AOC= 120°
, m∠BOD = 130°
m∠COE = 110°
m∠COD.= 60°
Step-by-step explanation:
Let's note that
AOF = COD= 60°
BOC = FOE= 70°
AOB = DOE= 50°
Given: m∠AOB=50°, m∠FOE=70°. m∠AOC
, m∠BOD,
m∠COE
m∠COD. = AOF = (360-(2(70)+2(50)))/2
AOF = (360-240)/2
AOF = 120/2
AOF = 60°= COD
COE = COD+DOE= 60+50= 110°
BOD = BOC + COD = 70+60= 130°
AOC = AOB + BOC = 50+70 = 120°
1500 volume x 1 per minute
_________________________
24 hours or 1440 minutes
1500/1440 = 1.04
Flow Rate is 1.04
Answer:
268
Step-by-step explanation:
7*35=245
245+23=268
#5 -.66
#6 -.425
#9 -.888
#10 -.1666
Temporarily subdivide the given area into two parts: a large rectangle and a parallelogram. Find the areas of these two shapes separately and then combine them for the total area of the figure.
By counting squares on the graph, we see that the longest side of the rectangle is the hypotenuse of a triangle whose legs are 8 and 2. Applying the Pyth. Thm., we find that this length is √(8^2+2^2), or √68. Similarly, we find the the width of this rectangle is √(17). Thus, the area of the rectangle is √(17*68), or 34 square units.
This leaves the area of the parallelogram to be found. The length of one of the longer sides of the parallelogram is 6 and the width of the parallelogram is 1. Thus, the area of the parallelogram is A = 6(1) = 6 square units.
The total area of the given figure is then 34+6, or 40, square units.
