Answer:
an = -6 + n
a10 = -6+ 10 =4
Step-by-step explanation:
an should start at one behind the first number so it should be -6. Then we add whatever the sequence is increasing by, which is 1 in this case, multiplied by n.
an = -6 + n
a10 = -6+ 10 =4
Step-by-step explanation:
The question is not properly structured but we can as well find the sum of 5/7 and 10 as shown;

Hence the expression written in equivalent form of common denominator is 
The sum is 
Answer:
Step-by-step explanation:
<u>Use the Pythagorean theorem:</u>
<u>Verify this for each line:</u>
1.
- 28² + 45² = 53²
- 2809 = 2809
- Yes
2.
- 33² + 56² = 64²
- 4225 ≠ 4096
- No
3.
- 36² + 77² = 85²
- 7225 = 7225
- Yes
Answer:
![\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%265%5C%5C1%261%261%5C%5C4%266%265%5Cend%7Barray%7D%5Cright%5D%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx1%5C%5Cx2%5C%5Cx3%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C6%5C%5C7%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Let's find the answer.
Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.
Equations:
Eq. 1 : x1 + 2x2 + 5x3 = 5
Eq. 2 : x1 + x2 + x3 = 6
E1. 3 : 4x1 + 6x2 + 5x3 = 7
Coefficients for x1 ; x2 ; x3
From eq. 1 : 1 ; 2 ; 5
From eq. 2 : 1 ; 1 ; 1
From eq. 3 : 4 ; 6 ; 5
So matrix A is:
![\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%265%5C%5C1%261%261%5C%5C4%266%265%5Cend%7Barray%7D%5Cright%5D)
And the vector of vriables (X) is:
![\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx1%5C%5Cx2%5C%5Cx3%5Cend%7Barray%7D%5Cright%5D)
Now we can find the resulting vector (B) using the 'resulting values' from each equation:
![\left[\begin{array}{ccc}5\\6\\7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C6%5C%5C7%5Cend%7Barray%7D%5Cright%5D)
In conclusion, AX=B is:
![\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%265%5C%5C1%261%261%5C%5C4%266%265%5Cend%7Barray%7D%5Cright%5D%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx1%5C%5Cx2%5C%5Cx3%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C6%5C%5C7%5Cend%7Barray%7D%5Cright%5D)