Answer:
the top right box
Step-by-step explanation:
Answer:
6 min
Step-by-step explanation:
5 kilometers = 5,000 meters
500/5,000 = 1/10 meaning it would take him 1/10 of an hour or 6 minutes to cover this distance.
89,170,326
millions is the place value of 9
hope that helps :)
Answer:
(a)
(b)
Step-by-step explanation:
(a) For using Cramer's rule you need to find matrix
and the matrix
for each variable. The matrix
is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get
more easily.

![\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]](https://tex.z-dn.net/?f=%5Ctherefore%20A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%265%5C%5C1%264%5Cend%7Barray%7D%5Cright%5D)
To get
, replace in the matrix A the 1st column with the results of the equations:
![B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]](https://tex.z-dn.net/?f=B_1%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%265%5C%5C2%264%5Cend%7Barray%7D%5Cright%5D)
To get
, replace in the matrix A the 2nd column with the results of the equations:
![B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]](https://tex.z-dn.net/?f=B_2%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%261%5C%5C1%262%5Cend%7Barray%7D%5Cright%5D)
Apply the rule to solve
:

In the case of B2, the determinant is going to be zero. Instead of using the rule, substitute the values of the variable
in one of the equations and solve for
:

(b) In this system, follow the same steps,ust remember
is formed by replacing the 3rd column of A with the results of the equations:

![\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]](https://tex.z-dn.net/?f=%5Ctherefore%20A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%260%5C%5C1%262%261%5C%5C0%261%262%5Cend%7Barray%7D%5Cright%5D)
![B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]](https://tex.z-dn.net/?f=B_1%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%260%5C%5C0%262%261%5C%5C0%261%262%5Cend%7Barray%7D%5Cright%5D)
![B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]](https://tex.z-dn.net/?f=B_2%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%260%5C%5C1%260%261%5C%5C0%260%262%5Cend%7Barray%7D%5Cright%5D)
![B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]](https://tex.z-dn.net/?f=B_3%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%261%5C%5C1%262%260%5C%5C0%261%260%5Cend%7Barray%7D%5Cright%5D)



You might be able to do direct substitution.