The answer is yes.
A line and a point outside the line define a unique plane. In other words, there is a single plane that contains a line and a point outside the line. If you are given line XY and point C outside the line, there is a single plane containing both line XY and point C. Therefore, line XY and point C must lie in the same plane.
you don't list any expressions to choose from but
if width = x
and length = 2x
Area = 2x * x
Area = 2x^2
Answer:
8/5 < x< 12/5
Step-by-step explanation:
(2 - x)^2 < 4/25
Take the square root of each side
sqrt((2 - x)^2) <±sqrt( 4/25)
Make two equations
2-x < 2/5 2-x > -2/5
Subtract 2 from each side
2-x-2 < 2/5 -2 2-x-2 > -2/5-2
-x < 2/5 - 10/5 -x > -2/5 - 10/5
-x < -8/5 -x > -12/5
Multiply by -1, remembering to flip the inequality
x> 8/5 x < 12/5
8/5 < x< 12/5
Answer:
(1+√7,0),(1−√7,0)
Step-by-step explanation:
You can't factor the expression evenly, so use the quadratic formula.
a = 2
b= -4
c= -12






End result: (1+√7,0),(1−√7,0)