Let c be any constant number. Which of the following will always be perpendicular to -3x + y = -1?

Question

lgbasallote · Answer

where's c?

anonymous · Answer

c= -1

parthkohli · Answer

First take it to the y = mx + b form.

$ \color{Black}{\Rightarrow y = 3x - 1}$

Now the slope of a perpendicular line would be the negative reciprocal. I'm not sure what you mean by the c here.

anonymous · Answer

y=3x+c
y=-3x+c
y=1/3x+c
y=-1/3x+c

anonymous · Answer

A perpendicular line will always have a slope that is the negative reciprocal of the original line.
y = 3x - 1
y = -1/3x - 1
That is a line perpendicular to the line. The "b" in y = mx + b can be changed interchangeably.

anonymous · Answer

Since in this problem, c is the constant, y = -1/3x + c

anonymous · Answer

Can you also do the simplified form of sqrt(72x^5y^4)

anonymous · Answer

or What is the value of x in the solution to the following system of equations?

x - 2y = 2
3x + y = 6

anonymous · Answer

7x = 14
x = 2


√(72x^5y^4) = √(36*2x^5y^4) = 6x^2y^2√2x