It is rounded to 20 because 23 is more closer to 23 than 30
Answer:
sorry i dont know...
Step-by-step explanation:
Let x and y be the dimensions of the rectangle. If the perimeter is 40, we have
![2(x+y)=40 \iff x+y=20](https://tex.z-dn.net/?f=2%28x%2By%29%3D40%20%5Ciff%20x%2By%3D20)
We can expression one variable in terms of the others as
![x+y=20 \iff x=20-y](https://tex.z-dn.net/?f=x%2By%3D20%20%5Ciff%20x%3D20-y)
Since the area is the product of the dimensions, we have
![xy=(20-y)y=-y^2+20y](https://tex.z-dn.net/?f=xy%3D%2820-y%29y%3D-y%5E2%2B20y)
This is a parabola facing down, so it's vertex is the maximum:
![f(y)=-y^2+20y \implies f'(y)=-2y+20](https://tex.z-dn.net/?f=f%28y%29%3D-y%5E2%2B20y%20%5Cimplies%20f%27%28y%29%3D-2y%2B20)
So, the maximum is
![f'(y)=0 \iff -2y+20=0 \iff 2y=20 \iff y=10](https://tex.z-dn.net/?f=f%27%28y%29%3D0%20%5Ciff%20-2y%2B20%3D0%20%5Ciff%202y%3D20%20%5Ciff%20y%3D10)
And since we know that
, we have
as well.
This is actually a well known theorem: out of all the rectangles with given perimeter, the one with the greatest area is the square.
Answer:
4 cm, 7 cm, 11 cm
Step-by-step explanation:
The sum of two shorter sides of a triangle is always greater than the larger side.
1 cm + 5 cm < 7 cm ( can't be the third side )
5 cm + 4 cm > 7 cm ( can be )
5 cm + 7 cm > 7 cm ( can be )
5 cm + 7 cm > 11 cm ( can be )
5 cm + 7 cm < 13 cm ( can't be )
5 cm + 7 cm < 15 cm ( can't be )
Answer:
Help me on my question please
Step-by-step explanation: