Answer:
equation 1
y = -2x
equation 2
y = x-3
Solution to both
(1,-2)
Step-by-step explanation:
We need to get the equations of both lines
General form is;
y = mx + c
where m is slope and c is the y-intercept
Table 1
since we have a point 0,0; the y-intercept here is zero
Let us get the slope. We can do this by selecting any two points
m = (y2-y1)/(x2-x1)
m = (2-10)/(-1+5) = -8/4 = -2
So the equation of the first line is;
y = -2x
Table 2
we get the slope
m = (4+2)/(7-1) = 6/6 = 1
The partial equation is;
y = x + c
To get c, we select any two point and substitute
4 = 7 + c
c = 4-7
c = -3
So the equation is;
y = x-3
To get the solution to both systems, we equate the y
-2x = x - 3
-2x-x = -3
-3x = -3
x = -3/-3
x = 1
To get y, we substitute;
recall; y = -2x
y = -2(1)
y = -2
Solution to the system is;
(1,-2)
Answer:
17 is the answers to the question
Answer:
pqr²+pr-rp²+rq²
Step-by-step explanation:
pq(r²+1)-r(p²+q²)
pqr²+pr-rp²+rq²
Answer:
A. They're not simalier because two pairs of corresponding angles in the two trapezoids are congurant
