Answer:
3
Step-by-step explanation:
because 3 is the only number that goes into both 3 and 27
Given the graph of the function

and the graph of the function


when f(x) = g(x).
This occurs at the point(s) of intersection of the graphs of the function f(x) and g(x).
From the graph, we can approximate the points of intersection of the graphs of the function f(x) and g(x) to pe points
(-1.9, 13.7) and (2.7, 0).
Answer:
The rectangular coordinates of the point are (3/2 , √3/2)
Step-by-step explanation:
* Lets study how to change from polar form to rectangular coordinates
- To convert from polar form (r , Ф) to rectangular coordinates (x , y)
use these rules
# x = r cos Ф
# y = r sin Ф
* Now lets solve the problem
∵ The point in the rectangular coordinates is (√3 , π/6)
∴ r = √3 and Ф = π/6
- Lets find the x-coordinates
∵ x = r cos Ф
∵ r = √3
∵ Ф = π/6
∴ x = √3 cos π/6
∵ cos π/6 = √3/2
∴ x = √3 (√3/2) = 3/2
* The x-coordinate of the point is 3/2
- Lets find the y-coordinates
∵ y = r sin Ф
∵ r = √3
∵ Ф = π/6
∴ y = √3 sin π/6
∵ sin π/6 = 1/2
∴ y = √3 (1/2) = √3/2
* The y-coordinate of the point is √3/2
∴ The rectangular coordinates of the point are (3/2 , √3/2)
2/5 I believe. Correct me if I am wrong
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