I’m pretty sure it’s
6x-10y+x^2-21
Unless I read it wrong then sorry
R^4-2(r-14)>0
r^4-2r+28>0
r^4-2r+28=0
There is no real solution but there is all real numbers
:)
Answer:
Step-by-step explanation:
In ordered pairs (a,b) a is the x value and b is a y value.
if we have 3x+y=6
if x=0, y=6 --> (0,6)
if y=0, x=2 -->(2,0)
if x=3, y=-3--> (3, -3)
if x=6, y= -12 --->(6, -12)
if x=6, y= -9 ---> (5, -9)
