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Turn both equations into slope-intercept form [ y = mx + b ].
x + 3y = 3
~Subtract x to both sides
3y = 3 - x
~Divide 3 to everything
y = 1 - x/3
~Reorder
y = -1/3x + 1
4x + 3y = -6
~Subtract 4x to both sides
3y = -6 - 4x
~Divide 3 to everything
y = -2 - 4x/3
~Reorder
y = -4/3x - 2
Graph of the equations will be shown below. Note that the solution of graphing two equations will be where both equations intersect. Both lines intersect at (-3, 2), hence making that the solution.
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Answer:
Step-by-step explanation:
The first term is 2, so part of the recursive definition is ...
f(1) = 2
The common ratio is 6/2 = 3, so each term is 3 times the previous one. That part of the recursive definition is ...
f(n) = 3·f(n -1)
These two parts of the definition match choice C.
Part (a)
<h3>Answer: y1 and y3 are perpendicular</h3>This is because the two slopes 2 and -1/2 multiply to -1. Perpendicular slopes multiply to -1 assuming neither line is vertical or horizontal.
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Part (b)
Graph each line to see where they cross. The three points of intersection are
(0,4)
(2,-2)
(4,2)
The order of the points doesn't matter.
You could also form three systems of equations pairing up the equations, and solving each system. That way you can find the points of intersection. Graphing may be a better and faster route in my opinion. See the diagram below.
