Answer:
The domain is all real numbers.
The initial value is 3
The simplified base is 
Step-by-step explanation:
The given function is 
.
To find the initial value, we put x=0 into the function:
.
.
The initial value is actually 3.
The given function is an exponential function, therefore the domain is all real numbers.
The range of this function refers to all values of y for which the function is defined.
The line y=0, is the horizontal asymptote.
The range is 
The simplified base is 
.
Answer:
The correct answer is 175.4379
Step-by-step explanation:
(9.9)^2 • 1.79
=98.01 • 1.79
=175.4379
Answer:
y = 2/3(x - 3)
Step-by-step explanation:
slope of the line is 2/3
we can use point (3,0)
formula: y - y = m(x - x)
y - 0 = 2/3 (x - 3)
y = 2/3(x - 3)
Answer:
mat hhway u gives all the answers
Step-by-step explanation:
Answer: The required characteristic polynomial of the given matrix A is 
Step-by-step explanation: We are given to find the characteristic polynomial of the following 3 × 3 matrix A with unknown variable x :
![A=\left[\begin{array}{ccc}0&0&1\\4&-3&4\\-2&0&-3\end{array}\right].](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%261%5C%5C4%26-3%264%5C%5C-2%260%26-3%5Cend%7Barray%7D%5Cright%5D.)
We know that
for any square matrix M, the characteristic polynomial is given by
where I is an identity matrix of same order as M.
Therefore, the characteristic polynomial of matrix A is
![|A-xI|=0\\\\\\\Rightarrow \left|\left[\begin{array}{ccc}0&0&1\\4&-3&4\\-2&0&-3\end{array}\right]-x\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\right|=0\\\\\\\Rightarrow \left|\left[\begin{array}{ccc}-x&0&1\\4&-3-x&4\\-2&0&-3-x\end{array}\right] \right|=0\\\\\\\Rightarrow -x(3+x)^2+1(0-6-2x)=0\\\\\Rightarrow (x+3)(-3x-x^2-2)=0\\\\\Rightarrow (x+3)(x^2+3x+2)=0\\\\\Rightarrow x^3+6x+11x+6=0.](https://tex.z-dn.net/?f=%7CA-xI%7C%3D0%5C%5C%5C%5C%5C%5C%5CRightarrow%20%5Cleft%7C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%261%5C%5C4%26-3%264%5C%5C-2%260%26-3%5Cend%7Barray%7D%5Cright%5D-x%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%261%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%5Cright%7C%3D0%5C%5C%5C%5C%5C%5C%5CRightarrow%20%5Cleft%7C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-x%260%261%5C%5C4%26-3-x%264%5C%5C-2%260%26-3-x%5Cend%7Barray%7D%5Cright%5D%20%5Cright%7C%3D0%5C%5C%5C%5C%5C%5C%5CRightarrow%20-x%283%2Bx%29%5E2%2B1%280-6-2x%29%3D0%5C%5C%5C%5C%5CRightarrow%20%20%28x%2B3%29%28-3x-x%5E2-2%29%3D0%5C%5C%5C%5C%5CRightarrow%20%28x%2B3%29%28x%5E2%2B3x%2B2%29%3D0%5C%5C%5C%5C%5CRightarrow%20x%5E3%2B6x%2B11x%2B6%3D0.)
Thus, the required characteristic polynomial of the given matrix A is
