If the length is three times the width and the area of the rectangle is
300 
then the length of rectangle is 30m and width be 10m.
Given that the length is three times the width and the area of the rectangle is 300 .
We are required to find the length and breadth of the rectangle.
Area of the rectangle is basically the product of length and breadth.
let the breadth of rectangle be w.
The length of rectangle will be 3w.
Area=Length*Breadth
300=3w*w
300=3
100=
w=+10,-10 but ignore -10 because width cannot be negative.
Length=3w=3*10=30m
Width=w=10m
The area can be written as 3 because the area of rectangle is product of length and breadth.
Hence if the length is three times the width and the area of the rectangle is 300 then the length of rectangle is 30m and width be 10m.
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(a+h)^2+3(a+h)+5(a+h)
a^2+2ah+h^2+3a+3h+5
(x^2+3ax+5)(a+h)
<span>H^2+3a+3h+5</span>
Answer:
your answer will be Michelle
Step-by-step explanation:
Actually there are three types of construction that were never accomplished by Greeks using compass and straightedge these are squaring a circle, doubling a cube and trisecting any angle.
The problem of squaring a circle takes on unlike meanings reliant on how one approaches the solution. Beginning with Greeks Many geometric approaches were devised, however none of these methods accomplished the task at hand by means of the plane methods requiring only straightedge and a compass.
The origin of the problem of doubling a cube also referred as duplicating a cube is not certain. Two stories have come down from the Greeks regarding the roots of this problem. The first is that the oracle at Delos ordered that the altar in the temple be doubled over in order to save the Delians from a plague the other one relates that king Minos ordered that a tomb be erected for his son Glaucus.
The structure of regular polygons and the structure of regular solids was a traditional problem in Greek geometry. Cutting an angle into identical thirds or trisection was another matter overall. This was necessary to concept other regular polygons. Hence, trisection of an angle became an significant problem in Greek geometry.
Roosters to chickens is 2 to 8
That means where there are 2 roosters, there are 8 chickens.
Simplify the ratio 2:8
The ratio can be simplified to 1:4
So, for 1 rooster, there are 4 chickens.
Now, the question says that if there are 5 roosters, how many chickens will there be?
1 rooster to 4 chickens
5 roosters to ? chickens
We went from 1 rooster to 5 rooster. We multiplied by 5.
To find out how many chickens there are for 5 roosters, multiply 4 chickens by 5.
4 * 5 = 20
Where there are 5 roosters, there will be
20 chickens.
Another way to solve is as follows:
Your final answer is
20 chickens.<span />
