Answer:
True, because whenever you just put x= any number then it would be a vertical line because it is crossing the x-axis.
Step-by-step explanation:
Answer:

Step-by-step explanation:
as said in the question that the triangle has three equal sides. i.e. this is an equilateral triangle. So,
let us consider one side of the triangle to be x . and 3x( x+x+x) should be equal to 1/5

Answer:
466 and 286
Step-by-step explanation:
Sum means add
the number is x
13 divided by a number is 13/x
the number divided by the number is x/13
so

also can be simplified to

or

translated it is

where x is the number
Given:

x lies in the III quadrant.
To find:
The values of
.
Solution:
It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.
We know that,




x lies in the III quadrant. So,


Now,



And,





We know that,



Therefore, the required values are
.