This is obviously a square that cannot exist in our universe, where all sides of a square ere equal....
Let ABCD be the border of square garden of side 30m and the shaded region be the path Then AD = 30m , XP=1m, SY = 1m and XY = AD ⇒ XP + PS + SY = AD ⇒ PS = AD - XP - SY = 30m - 1m - 1m = 28m ∴ PQRS in a square of side 28m ∴ The area of path = Area of ABCD - Area of PQRS
2m4(m2-15m+18) i hope this is right! srry if its not i just did it for hw
The formula for area of a circle is Pi x r (squared), A = Pi(rxr). So since the diameter is 30 you divide it by two and get 15 and then square it and multiply it by pi and get 706.86 meters.
4m because V=BxHso H =120/30=4
As two dimensions of the park are given, it can be assumed to be a rectangle in shape. → perimeter of fence = 40m + 30m + 40m + 30m = 140m For the area of the path, it runs along all the perimeter of the fence so its length is 140m and its width is 1.5m → the area of the footpath next to the fence is 140m x 1.5m = 210 m2. However, this area only includes the footpath that has one side next to the fence but excludes those parts of the footpath that go round the corners of the fence. As the path is given as 1.5m wide, it could be assumed that the path at each of the corners will be a quarter circle of radius 1.5m. As there are four of these quarter circles, their total area is the area of a whole circle: → area of the path = 210m2 + π x (1.5m)2 = 140m2 + 2.25π m2 ≈ 217.07 m2 The corners could also be squared off, so that each is a square of side 1.5m: → area of the path = 210m2 + 4 x (1.5m)2 = 219 m2.
15m and 30 m
The GCF of 12m, 15m and 30m is 3m.
The dimensions of the rectangular lot are: 30m by 15m
The GCF is: 3m
15m+3/2 = -6 30m+3 = -12 30m = -15 m = -1/2
4*7.5 = 30m
30m-50m double action 15m-25m reapeter
As the rectangular garden is 10m wide, then two sides will require 10m of fencing. Thus 2*10 = 20m of fencing. It is also 15m long, so two sides will require 15m of fencing. Thus 2*15 = 30m of fencing. Adding up all 4 sides => 20+30 = 50m of fencing.
30 meters (7.5m x 4 sides = 30m)
Day traders will usually use 1m, 5m, 15m or 30m time frames. Which time frame to choose is depends on the trader's character, an aggressive trader will probably use the 1m and 5m, a more serene trader will probably use the 15m or 30m.
30*1 = 30 square metres.
Let ABCD be the border of square garden of side 30m and the shaded region be the path Then AD = 30m , XP=1m, SY = 1m and XY = AD ⇒ XP + PS + SY = AD ⇒ PS = AD - XP - SY = 30m - 1m - 1m = 28m ∴ PQRS in a square of side 28m ∴ The area of path = Area of ABCD - Area of PQRS