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)



Answer:
10.626
Step-by-step explanation:
There's multiple ways to go about it. See how
5.06 * 2.1 = (5 + 0.06)(2 + 0.1) = 5*2 + 0.06*2 + 5*0.1 + 0.06*0.1 =
= 10 + 0.12 + 0.5 + 0.006 = 10.626
Please notice how breaking up the product into smaller parts has let us make all the computations without too much hassle.
X | Y
-------
-5| 10
-4| 8
-3| 6
-2| 4
-1| 2
0 | 0
1 | -2
2 | -4
3 | -6
4 | -8
5 | -10
Answer:
- using the rule given: 2.5
- using an exponential rule: 7
Step-by-step explanation:
Evaluating the linear rule given, for n = 1, we have ...
a1 = 7(1/2)(1) -1 = 7/2 -1 = 5/2 = 2.5
_____
We suspect you may intend the exponential function ...
an = 7(1/2)^(n-1)
Then, for n = 1, we have ...
a1 = 7(1/2)^(1 -1) = 7(1) = 7 . . . . the first term is 7
_____
When writing an exponential expression in plain text, it requires the exponential operator, a caret (^). If the exponent contains any arithmetic, as this one does, it must be enclosed in parentheses.
Answer: I'm thinking it's reflectional symmetry.
Step-by-step explanation: