V1=(-2,4)=(x1,y1)→x1=-2, y1=4
V2=(4,0)=(x2,y2)→x2=4, y2=0
V3=(2,-3)=(x3,y3)→x3=2, y3=-3
V4=(x4,y4)→x4=?, y4=?
V1-V2
dx=x2-x1=4-(-2)=4+2→dx=6
dy=y2-y1=0-4→dy=-4
V4-V3
dx=x3-x4→6=2-x4
Solving for x4:
6=2-x4→6-2=2-x4-2→4=-x4→(-1)(4=-x4)→-4=x4→x4=-4
dy=y3-y4→-4=-3-y4
Solving for y4:
-4=-3-y4→-4+3=-3-y4+3→-1=-y4→(-1)(-1=-y4)→1=y4→y4=1
V4=(x4, y4)→V4=(-4, 1)
Answer: The coordinates of the fourth vertex are (-4,1)
Answer:
46
Step-by-step explanation:
14-60 equals 46 and just go to your answer and put that in and there u go buddy
Answer:
tenth place= 5
hundredth= 8
Step-by-step explanation:
tenth is 5 since 5÷10=0.5
Strictly speaking, x^2 + 2x + 4 doesn't have solutions; if you want solutions, you must equate <span>x^2 + 2x + 4 to zero:
</span>x^2 + 2x + 4= 0. "Completing the square" seems to be the easiest way to go here:
rewrite x^2 + 2x + 4 as x^2 + 2x + 1^2 - 1^2 = -4, or
(x+1)^2 = -3
or x+1 =i*(plus or minus sqrt(3))
or x = -1 plus or minus i*sqrt(3)
This problem, like any other quadratic equation, has two roots. Note that the fourth possible answer constitutes one part of the two part solution found above.
Answer:
9
Step-by-step explanation:
The equation says y = x + 9
Therefore, y = 9 since x = 0 on the y axis.
Picture of graph also attached :)