Answer:
1 hour and 35 minutes!! :)
Step-by-step explanation:
1
Step-by-step explanation:
simply we equate the equation with 3 then sove it
The answer would be D if it’s not including the number six, but if it’s 6 or less then it would be E
Let
be the dimensions of the rectangle. We know the equations for both area and perimeter:
![A=xy=36](https://tex.z-dn.net/?f=A%3Dxy%3D36)
![P=2(x+y)=36 \iff x+y=18](https://tex.z-dn.net/?f=P%3D2%28x%2By%29%3D36%20%5Ciff%20x%2By%3D18)
So, we have the following system:
![\begin{cases}xy=36\\x+y=18\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dxy%3D36%5C%5Cx%2By%3D18%5Cend%7Bcases%7D)
From the second equation, we can deduce
![y=18-x](https://tex.z-dn.net/?f=y%3D18-x)
Plug this in the first equation to get
![xy=x(18-x)=-x^2+18=36](https://tex.z-dn.net/?f=xy%3Dx%2818-x%29%3D-x%5E2%2B18%3D36)
Refactor as
![x^2-18x+36=0](https://tex.z-dn.net/?f=x%5E2-18x%2B36%3D0)
And solve with the usual quadratic formula to get
![x=9\pm3\sqrt{5}](https://tex.z-dn.net/?f=x%3D9%5Cpm3%5Csqrt%7B5%7D)
Both solutions are feasible, because they're both positive.
If we chose the positive solution, we have
![x=9+3\sqrt{5} \implies y=18-x=18-9-3\sqrt{5}=9-3\sqrt{5}](https://tex.z-dn.net/?f=x%3D9%2B3%5Csqrt%7B5%7D%20%5Cimplies%20y%3D18-x%3D18-9-3%5Csqrt%7B5%7D%3D9-3%5Csqrt%7B5%7D)
If we choose the negative solution, we have
![x=9-3\sqrt{5} \implies y=18-x=18-9+3\sqrt{5}=9+3\sqrt{5}](https://tex.z-dn.net/?f=x%3D9-3%5Csqrt%7B5%7D%20%5Cimplies%20y%3D18-x%3D18-9%2B3%5Csqrt%7B5%7D%3D9%2B3%5Csqrt%7B5%7D)
So, we're just swapping the role of
and
. The two dimensions of the rectangle are
and ![9-3\sqrt{5}](https://tex.z-dn.net/?f=9-3%5Csqrt%7B5%7D)