The square piece of wood measures 32 in per side,
Then the circle that can be cut off it has a radius of 16 in (half of the diameter 32 in)
Therefore, the area of the circle is going to be given by the formula:

And the area of the initial square was:
Area = 32^2 in^2 = 1024 in^2
Therefore the left-overs of the piece of wood would be the difference:
1024 in^2 - 803.84 in^2
Define
![{x} = \left[\begin{array}{ccc}x_{1}\\x_{2}\end{array}\right]](https://tex.z-dn.net/?f=%7Bx%7D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_%7B1%7D%5C%5Cx_%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20)
Then
x₁ = cos(t) x₁(0) + sin(t) x₂(0)
x₂ = -sin(t) x₁(0) + cos(t) x₂(0)
Differentiate to obtain
x₁' = -sin(t) x₁(0) + cos(t) x₂(0)
x₂' = -cos(t) x₁(0) - sin(t) x₂(0)
That is,
![\dot{x} = \left[\begin{array}{ccc}-sin(t)&cos(t)\\-cos(t)&-sin(t)\end{array}\right] x(0)](https://tex.z-dn.net/?f=%5Cdot%7Bx%7D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-sin%28t%29%26cos%28t%29%5C%5C-cos%28t%29%26-sin%28t%29%5Cend%7Barray%7D%5Cright%5D%20x%280%29)
Note that
![\left[\begin{array}{ccc}0&1\\-1&09\end{array}\right] \left[\begin{array}{ccc}cos(t)&sin(t)\\-sin(t)&cos(t)\end{array}\right] = \left[\begin{array}{ccc}-sin(t)&cos(t)\\-cos(t)&-sin(t)\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%261%5C%5C-1%2609%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcos%28t%29%26sin%28t%29%5C%5C-sin%28t%29%26cos%28t%29%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-sin%28t%29%26cos%28t%29%5C%5C-cos%28t%29%26-sin%28t%29%5Cend%7Barray%7D%5Cright%5D%20)
Therefore
Answer:
1 - Variable
2 - Corfficient
3 - Term
4 - Constant
Step-by-step explanation:
So, I am pretty srue the way the boxes are now is the correct way already. (I should be wrong?)
Answer:
3/40 :D
Step-by-step explanation:
The width is half the length, so is
width = (1/2)*length
width = (1/2)*(<span>3.2a + 0.18b) cm
width = (1.6a +0.09b) cm
The perimeter of the rectangle is twice the sum of length and width.
perimeter = 2*(length + width)
perimeter = 2*((3.2a +0.18b) cm + (1.6a +0.09b) cm)
perimeter = 2*(4.8a +0.27b) cm)
perimeter = (9.6a +0.54b) cm
Sasha did not get this answer, so apparently ...
her reasoning was not correct.</span>