Answer: 2sin^2x+sin2x+cos2x=0 ..... (1).
By using the trigonometric identities below :
sin2x=2sinxcosx
cos2x=cos^2x-sin^2x
We substitute the trigonometric identities into (1).
2sin^2x+2sinxcosx+cos^2x-sin^2x=0
By combining like terms .
sin^2x+2sinxcosx+cos^2x=0.....(2)
The equation (2) is equivalent to the following expression (3).
(sinx+cosx)(sinx+cosx)=0 .....(3).
sinx+cosx=0
cosx=-sinx
divide both sides by cosx
1=-sinx/cosx
-1=sinx/cosx
sinx/cosx=tanx
substitute
-1=tanx
tanx=-1
tangent is negative in 2nd and 4th quadrants
tan135º=-1 (one answer)
tan315º=-1 (second answer)
Step-by-step explanation:
Please refer to the trigonometric identities used and explained above .
<u>Answer:</u>
- The x coordinate is known as <u>abscissa</u><u> </u><u>(</u><u>B)</u>
<u>Step-by-step</u><u> Explanation</u><u>:</u>
In a Cartesian plane, we learn many such terms which represent a point, or a line, or cooridnates of some points etc
Such terms are:
- Origin (Intersection of x axis and y-axis.)
- x - axis (Horizontal line)
- y - axis (Vertical line)
- abscisaa (x -coordinate)
- ordinate (y - coordinate)
Answer:
slope = -4
Step-by-step explanation:
Given the following points on the graph: (-1, 4) (1, -4)
Let (x1, y1) = (-1, 4)
(x2, y2) = (1, -4)
Use the following slope formula:


Therefore, the slope of the line is -4
The distance between points A (3, -5) and A' (2, -3) is 2.4 units.
Given that,
The points A (3, -5) and A' (2, -3).
We have to determine,
The distance between point A and A'.
According to the question,
The distance between two points is determined by using the distance formula.

Then,
The distance between points A (3, -5) and A' (2, -3) is,

Hence, The distance between points A (3, -5) and A' (2, -3) is 2.4 units.
For more details refer to the link given below.
brainly.com/question/8069952