Answer:
The domain of the inverse of a relation is the same as the range of the original relation. In other words, the y-values of the relation are the x-values of the inverse.
Thus, domain of f(x): x∈R = range of f¯¹(x)
and range of f(x): x∈R =domain of f¯¹(x)
The slope and intercept form is the form of the straight line equation that includes the value of the slope of the line
Reason:
The slope and intercept form is the form y = m·x + c
Where;
m = The slope
Two equations are parallel if their slopes are equal
Two equations are perpendicular if the relationship between their slopes, m₁, and m₂ are;
1. The given equations are in the slope and intercept form
The slope, m₁ = 3
The slope, m₂ =
Therefore, the equations are <u>neither</u> parallel or perpendicular
2. y = 5·x - 3
10·x - 2·y = 7
The second equation can be rewritten in the slope and intercept form as follows;
Therefore, the two equations are <u>parallel</u>
3. The given equations are;
-2·x - 4·y = -8
-2·x + 4·y = -8
The given equations in slope and intercept form are;
Slope, m₁ =
Slope, m₂ =
The slopes
Therefore, m₁ ≠ m₂
The lines are <u>Neither</u> parallel nor perpendicular
4. The given equations are;
2·y - x = 2
m₁ =
y = -2·x + 4
m₂ = -2
Therefore;
Therefore, the lines are <u>perpendicular</u>
5. The given equations are;
4·y = 3·x + 12
-3·x + 4·y = 2
Which gives;
First equation,
Second equation,
Therefore, m₁ = m₂, the lines are <u>parallel</u>
6. The given equations are;
8·x - 4·y = 16
Which gives; y = 2·x - 4
5·y - 10 = 3, therefore, y =
Therefore, the two equations are <u>neither</u> parallel nor perpendicular
7. The equations are;
2·x + 6·y = -3
Which gives
12·y = 4·x + 20
Which gives
m₁ ≠ m₂
8. 2·x - 5·y = -3
Which gives;
5·x + 27 = 6
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