How do you do inverse operation to solve the equation 7p=28
2 answers:
You might be interested in
I hope this helps you
A) 1/6^2=6^-2
B) 6^-3
C) 1/216=1/6^3=6^-3
D) 1/6^3=6^-3
A) 1/6^2=6^-2
B) 6^-3
C) 1/216=1/6^3=6^-3
D) 1/6^3=6^-3
Answer:
There are no real solutions. The complex solutions are ...
(x, y) = (2.5-i√1.75, -2.5+i√1.75) or (2.5+i1.75, -2.5-i1.75)
Step-by-step explanation:
Subtract the first equation from the second.
y -y = (x^2 -6x +8) -(-x)
0 = x^2 -5x +8
Rewrite to facilitate completing the square:
x^2 -5x = -8
x^2 -5x +6.25 = -1.75 . . . . . add 6.25 to complete the square
(x -2.5)^2 = -1.75 . . . . . write as a square
x -2.5 = ±i√1.75 . . . . . . take the square root
x = 2.5 ± i√1.75 . . . . . . add 2.5; y is the opposite of this
Solutions are ...
- x = 2.5 +i√1.75, y = -2.5 -i√1.75
- x = 2.5 -i√1.75, y = -2.5 +i√1.75
_____
The graph shows the equations have no point of intersection, meaning there are no real solutions. The complex solutions are shown above.
