It's D. the median for town A, 30, is less than the median for town B, 40.
Answer:
a = 6, b = 3, and c = 2
1. ab1 = 6 × 3 × 1 = 18
2. c + 42 = 2 + 42 = 44
3. 18 = 18
4. a − b4 = 6 - 3(4) = 6 - 12 = - 6
5. 2c3 = 2 × 2 × 3 = 12
6. b ÷ 35 = 3÷35 = 0.086
7. a − 1 6 = 6 - 16 = -10
8. 6 + c8 = 6 + 2 (8) = 6 + 16 = 22
Step-by-step explanation:
Answer:
A
Step-by-step explanation:
we should first find the value of x.
<em><u>Option A</u></em>
<em><u>The solution is:</u></em>

<em><u>Solution:</u></em>

We have to solve the equation f(x) = 0
Let f(x) = 0

Solve the above equation


Take square root on both sides

Thus the solution is found
Answer:
Option C
Step-by-step explanation:
We are given a coefficient matrix along and not the solution matrix
Since solution matrix is not given we cannot check for infinity solutions.
But we can check whether coefficient matrix is 0 or not
If coefficient matrix is zero, the system is inconsistent and hence no solution.
Option A)
|A|=![\left[\begin{array}{ccc}4&2&6\\2&1&3\\-2&3&-4\end{array}\right] =0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%262%266%5C%5C2%261%263%5C%5C-2%263%26-4%5Cend%7Barray%7D%5Cright%5D%20%3D0)
since II row is a multiple of I row
Hence no solution or infinite
OPtion B
|B|=![\left[\begin{array}{ccc}2&0&-2\\-7&1&5\\4&-2&0\end{array}\right] \\=2(10)-2(10)=0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%260%26-2%5C%5C-7%261%265%5C%5C4%26-2%260%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%3D2%2810%29-2%2810%29%3D0)
Hence no solution or infinite
Option C
![\left[\begin{array}{ccc}6&0&-2\\-2&0&6\\1&-2&0\end{array}\right] \\=2(36-2)=68](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%260%26-2%5C%5C-2%260%266%5C%5C1%26-2%260%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%3D2%2836-2%29%3D68)
Hence there will be a unique solution
Option D
=0
(since I row is -5 times III row)
Hence there will be no or infinite solution
Option C is the correct answer