<span>v = 4/3*pi*r^3
derivating both sides with respect to t
dv/dt = 4*pi*r^2*dr/dt
when d = 1.7, r = 0.85, and dv/dt = 2:
2 = 4*pi*(0.85)^2*dr/dt
thus
dr/dt = 1/(2pi*(0.85)^2)
=1/(2*3.14*0.85^2)
=0.22</span><span />
-5 because the denominator can’t equal zero
Consider the operation is
.
Given:
The augmented matrix below represents a system of equations.
![\left[\left.\begin{matrix}1&0&1\\1&3&-1\\3&2&0\end{matrix}\right|\begin{matrix}-1\\-9\\-2\end{matrix}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cleft.%5Cbegin%7Bmatrix%7D1%260%261%5C%5C1%263%26-1%5C%5C3%262%260%5Cend%7Bmatrix%7D%5Cright%7C%5Cbegin%7Bmatrix%7D-1%5C%5C-9%5C%5C-2%5Cend%7Bmatrix%7D%5Cright%5D)
To find:
Matrix results from the operation
.
Step-by-step explanation:
We have,
![\left[\left.\begin{matrix}1&0&1\\1&3&-1\\3&2&0\end{matrix}\right|\begin{matrix}-1\\-9\\-2\end{matrix}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cleft.%5Cbegin%7Bmatrix%7D1%260%261%5C%5C1%263%26-1%5C%5C3%262%260%5Cend%7Bmatrix%7D%5Cright%7C%5Cbegin%7Bmatrix%7D-1%5C%5C-9%5C%5C-2%5Cend%7Bmatrix%7D%5Cright%5D)
After applying
, we get
![\left[\left.\begin{matrix}1&0&1\\-3(1)&-3(3)&-3(-1)\\3&2&0\end{matrix}\right|\begin{matrix}-1\\-3(-9)\\-2\end{matrix}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cleft.%5Cbegin%7Bmatrix%7D1%260%261%5C%5C-3%281%29%26-3%283%29%26-3%28-1%29%5C%5C3%262%260%5Cend%7Bmatrix%7D%5Cright%7C%5Cbegin%7Bmatrix%7D-1%5C%5C-3%28-9%29%5C%5C-2%5Cend%7Bmatrix%7D%5Cright%5D)
![\left[\left.\begin{matrix}1&0&1\\-3&-9&3\\3&2&0\end{matrix}\right|\begin{matrix}-1\\27\\-2\end{matrix}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cleft.%5Cbegin%7Bmatrix%7D1%260%261%5C%5C-3%26-9%263%5C%5C3%262%260%5Cend%7Bmatrix%7D%5Cright%7C%5Cbegin%7Bmatrix%7D-1%5C%5C27%5C%5C-2%5Cend%7Bmatrix%7D%5Cright%5D)
Therefore, the correct option is A.