I thibk its 5 tell if im right
<span>It is false since the rational function is discontinuous when the denominator is zero. But the denominator is a polynomial and a polynomial has only finitely many zeros. So the discontinuity points of a rational function is finite. </span>
Step-by-step explanation:
(a) A histogram is like a bar graph, but with ranges instead of values.
(b) The distribution is approximately uniform, with slightly more people at the younger and older ends than in the middle.
(c) The ages do not appear to come from a normal distribution. Normal distributions are bell-shaped, tall in the middle and short on the sides.
The question is missing parts. Here is the complete question.
Let M =
. Find
and
such that
, where
is the identity 2x2 matrix and 0 is the zero matrix of appropriate dimension.
Answer: 

Step-by-step explanation: Identity matrix is a sqaure matrix that has 1's along the main diagonal and 0 everywhere else. So, a 2x2 identity matrix is:
![\left[\begin{array}{cc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
![M^{2} = \left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]](https://tex.z-dn.net/?f=M%5E%7B2%7D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D)
![M^{2}=\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]](https://tex.z-dn.net/?f=M%5E%7B2%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D31%2610%5C%5C-2%2615%5Cend%7Barray%7D%5Cright%5D)
Solving equation:
![\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+c_{1}\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right] +c_{2}\left[\begin{array}{cc}1&0\\0&1\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D31%2610%5C%5C-2%2615%5Cend%7Barray%7D%5Cright%5D%2Bc_%7B1%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D%20%2Bc_%7B2%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%260%5C%5C0%260%5Cend%7Barray%7D%5Cright%5D)
Multiplying a matrix and a scalar results in all the terms of the matrix multiplied by the scalar. You can only add matrices of the same dimensions.
So, the equation is:
![\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+\left[\begin{array}{cc}6c_{1}&5c_{1}\\-1c_{1}&-4c_{1}\end{array}\right] +\left[\begin{array}{cc}c_{2}&0\\0&c_{2}\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D31%2610%5C%5C-2%2615%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6c_%7B1%7D%265c_%7B1%7D%5C%5C-1c_%7B1%7D%26-4c_%7B1%7D%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dc_%7B2%7D%260%5C%5C0%26c_%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%260%5C%5C0%260%5Cend%7Barray%7D%5Cright%5D)
And the system of equations is:

There are several methods to solve this system. One of them is to multiply the second equation to -1 and add both equations:




With
, substitute in one of the equations and find
:





<u>For the equation, </u>
<u> and </u>
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