Answer:
Linear algebra is the study of lines and planes, vector spaces and mappings that are required for linear transforms. It is a relatively young field of study, having initially been formalized in the 1800s in order to find unknowns in systems of linear equations.
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Given data:
The first set of equations are x+y=4, and x=6.
The second set of equations are 3x-y=12 and y=-6.
The point of intersection of first set of te equations is,
6+y=4
y=-2
The first point is (6, -2).
The point of intersection of second set of te equations is,
3x-(-6)=12
3x+6=12
3x=6
x=2
The second point is (2, -6).
The equation of the line passing through (6, -2) and (2, -6) is,
Thus, the required equation of the line is y=x-8.
Answer:
C. 4
Step-by-step explanation:
2x³+2x¹+2x+5+4/x-1
It is a i think sorry if wrong
Is RS perpendicular to DF? Select Yes or No for each statement. R (6, −2), S (−1, 8), D (−1, 11), and F (11 ,4) R (1, 3), S (4,7
guajiro [1.7K]
I'll do the first one to get you started.
Find the slope of the line between R (6,-2) and S (-1,8) to get
m = (y2-y1)/(x2-x1)
m = (8-(-2))/(-1-6)
m = (8+2)/(-1-6)
m = 10/(-7)
m = -10/7
The slope of line RS is -10/7
Next, we find the slope of line DF
m = (y2 - y1)/(x2 - x1)
m = (4-11)/(11-(-1))
m = (4-11)/(11+1)
m = -7/12
From here, we multiply the two slope values
(slope of RS)*(slope of DF) = (-10/7)*(-7/12)
(slope of RS)*(slope of DF) = (-10*(-7))/(7*12)
(slope of RS)*(slope of DF) = 10/12
(slope of RS)*(slope of DF) = 5/6
Because the result is not -1, this means we do not have perpendicular lines here. Any pair of perpendicular lines always has their slopes multiply to -1. This is assuming neither line is vertical.
I'll let you do the two other ones. Let me know what you get so I can check your work.
