So u already know that the radius is half of the circle which in this case is 28 mm. And both circles at the top touch both sides of the rectangle.
Process:
28x4=112.
This is the top side, but because you could fit 2 circles at the top and you know the measurement, it is common sense to think that it would be the same on the other side. So the area of the square/ rectangle:
112x112= 12,534 mm
Answer:
The set of solutions is ![\{\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}12\\-7-r\\r\end{array}\right]: \text{r is a real number} \}](https://tex.z-dn.net/?f=%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D12%5C%5C-7-r%5C%5Cr%5Cend%7Barray%7D%5Cright%5D%3A%20%5Ctext%7Br%20is%20a%20real%20number%7D%20%20%5C%7D)
Step-by-step explanation:
The augmented matrix of the system is
.
We will use rows operations for find the echelon form of the matrix.
- In row 2 we subtract
from row 1. (R2- 2/3R1) and we obtain the matrix ![\left[\begin{array}{cccc}3&6&6&-9\\0&1&1&-7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D3%266%266%26-9%5C%5C0%261%261%26-7%5Cend%7Barray%7D%5Cright%5D)
- We multiply the row 1 by
.
Now we solve for the unknown variables:
The system has a free variable, the the system has infinite solutions and the set of solutions is ![\{\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}12\\-7-r\\r\end{array}\right]: \text{r is a real number} \}](https://tex.z-dn.net/?f=%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D12%5C%5C-7-r%5C%5Cr%5Cend%7Barray%7D%5Cright%5D%3A%20%5Ctext%7Br%20is%20a%20real%20number%7D%20%20%5C%7D)
The correct answer is 4 and 1/8
Answer:
Step-by-step explanation:
a) r = √(1² + (-5²)) = √26 = 5.09901...
θ = tan⁻¹(-5/1) = 4.9097... radians
(5.1, 4.9)
b) r = - 5.09901...
θ = 4.9097... - π = 1.76819...
(-5.1, 1.8)