surface area of the outside of a cube shaped closed-top water tank whose edge length is 7.2 ft, can be solved by solving the area of each face of a cube. since the cube has 6 faces and all the faces are congruent because it all has the same edge.
so the formulaA = 6s^2where s is the edge lengthA = 6 ( 7.2^2)A = 311.04 sq ft.
Answer:
Ans is 59
Step-by-step explanation:
46 +51+y=180 (sum of triangle)
y=180-46-51
y=83
now,
83+x+38=180
x=180-83-38
x=59
value of x is 59
The associative property of addition is pretty much just any numbers in place of these: a + (b + c) = (a + b) + c. Therefore, your answer would be d) (6 + 1) + 9 = 6 + (1 + 9)
The answer is Hundredths place
A)
It looks like the [irregular] hexagon has 3 rectangles and 2 triangles within it.
So let's exclude the triangular corners on bottom left and top right for now.
First we have a large rectangle covering most of the upper left of the polygon. 20 ft × 7 ft = 140 sq.ft.
Now we have a rectangle on the bottom right. The width is 11 ft, so take the 7 away from that, 4 ft. × 14 ft. on bottom. 4 ft × 14 ft = 56 sq.ft.
The last small rectangle fits on the right between the 2 other rectangles. It is 24-20 on top/bottom × 7-6 right/left. 4 ft × 1 ft = 4 sq.ft.
Now for the triangles: bottom left is 11-7 × 24-14 = 4 ft × 10 ft. 1/2bh = 1/2×10×4 = 20 sq.ft.
Top right is 24-20 × 11-5 = 4 ft × 6 ft. 1/2bh = 1/2×4×6 = 12 sq.ft.
B)
Add them all together for the total area (A):
A = 140 + 56 + 4 + 20 + 12 = 140+60+32
= 232 sq.ft.
Hope that explains it well enough! ;)
