3y+(-y+3)=5
2y+3=5
2y=2
y=1
x=-(1)+3
x=2
Your answer is (2,1)
Hope this helps
I’m not sure but u need more information on your question
Answer:
C. 0
Step-by-step explanation:
–4j^2 + 3j – 28 = 0
The discriminant is b^2-4ac if >0 we have 2 real solutions
=0 we have 1 real solutions
<0 we have 2 imaginary solutions
a = -4, b =3 c = -28
b^2 -4ac
(3)^2 - 4(-4)*(-28)
9 - 16(28)
9 -448
This will be negative so we have two imaginary solutions.
Therefore we have 0 real solutions
We find the area of the total square which is 8in • 8in=64in^2 . Now we find the area of the small quadrilateral which is 4in•4in=16 in^2.
Now we subtract the area of the small quadrilateral from the big quadrilateral’s area 64in^2 -16in^2 leaving us with the are of 48in^2 of the figure
