Answer:
Therefore, the inverse of given matrix is

Step-by-step explanation:
The inverse of a square matrix
is
such that
where I is the identity matrix.
Consider, ![A = \left[\begin{array}{ccc}4&3\\3&6\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%263%5C%5C3%266%5Cend%7Barray%7D%5Cright%5D)








Therefore, the inverse of given matrix is

6/x= -2/18
Cross multiply. ( Multiply the numerators together ) . ( Multiply the denominators together (.
6*18= - 2*x
108= -2x
Divide by -2 for 108 and -2x
108/-2= -2x/2
x= -54
Answer: x= -54
Answer:
The diameter will increase at a rate of 1/30π cm/min
Step-by-step explanation:
Here we want to calculate the rate at which the diameter will increase
Mathematically, the area of a sphere is given as;
A = 4πr^2
But r = d/2
so A = 4 * π * d/2 * d/2 = πd^2
dA/d(d) = 2πd
Thus dd/dA = 1/2πd = 1/2 * π * 15 = 1/30π
Given dA/dt = 10
Mathematically;
d(d)/dt = d(d)/dA * dA/dt
dd/dt = 1/30π * 10 = 10/30π = 1/3π cm/min