2(8-4)^2-5
16-32-5
-21
you can change the -10/2 to a -5 and it will be easier.
...hehehe beh ènehbnsjajwuwijwuwiiwhwniqkwiqkwjwmjwujw
Answer:25
Step-by-step explanation:Count all the wholes then put all the halves and there is 25
X+y+5=0
y=-x-5
If a solution exists y=y so we can say
x^2-9x+10=-x-5 add x+5 to both sides
x^2-8x+15=0 now factor
x^2-3x-5x+15=0
x(x-3)-5(x-3)
(x-5)(x-3) so x=3 and 5, using y=-x-5
y(3)=-8 and y(5)=-10
So the two solutions are:
(3,-8) and (5,-10)
