Here is a photo of how to solve it
Answer:
The product is:
![\left[\begin{array}{cc}15 & 14\\-1 & 9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D15%20%26%2014%5C%5C-1%20%26%209%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
For this problem you need to multiply the first row only for the two first column of the others matrix and get the desired result:
![\left[\begin{array}{ccc}1&3&1\\-2&1&0\end{array}\right] \times \left[\begin{array}{cc}2&-2\\3&5\\4&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%263%261%5C%5C-2%261%260%5Cend%7Barray%7D%5Cright%5D%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%26-2%5C%5C3%265%5C%5C4%261%5Cend%7Barray%7D%5Cright%5D)

So the value of the element in the position
is 15

So the value of the element in the position
is 14
Then with these two values you can determinate the result matrix.
![\left[\begin{array}{cc}15 & 14\\-1 & 9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D15%20%26%2014%5C%5C-1%20%26%209%5Cend%7Barray%7D%5Cright%5D)
Answer:
6
10
3
57
10
Step-by-step explanation:
just divide and find x:1
Answer:
3.4625
Step-by-step explanation:
Given the equation:
Q(t)=0.05t^2 + 0.1t + 3.4 parts per million
Where t = time in years
By approximately how much will the carbon monoxide lvel change during the coming 6 months?
t = 6 months = 6/12 = 0.5 year
Q(0.5) = 0.05(0.5)^2 + 0.1(0.5) + 3.4 parts per million
Q(0.5) = 0.05(0.25) + 0.05 + 3.4
Q(0.5) = 0.0125 + 0.05 + 3.4
Q = 3.4625
Carbon monoxide level will change by approximately 3.4625 parts per million in 6 months