Answer:
The correct option is: Both -5 and 1 are true solutions
Step-by-step explanation:
The given expression is:
√c^2+2c-4=√1-2c
By squaring both sides we get:
(√c^2+2c-4)^2=(√1-2c)^2
c^2+2c-4=1-2c
Move all the values to the L.H.S
c^2+2c-4-1+2c=0
c^2+4c-5=0
Now we will perform factorization:
c^2+5c-c-5=0
c(c+5)-1
(c+5)=0
(c+5)(c-1)=0
c+5 = 0 , c-1=0
c= 0-5 , c=0+1
c= -5 , c=1
Thus the correct option is Both -5 and 1 are true solutions....
C I think but please mark me as Brainiest
Answer:
[2 , ∞)
Step-by-step explanation:
y = sqrt( x-2)
The domain is the values that x can take
Since we have a square root
Domain: [2 , ∞) since the square root must be greater than or equal to zero
The range is the values that y can take
The square root starts at 0 and increases
Range [0,∞)
The inverse swaps the domain and range
Range: [2 , ∞)
Domain [0,∞)
The required number will be 16
the expression will be like :
hope it helps you cap !!
Anti - Venom
Answer:
Step-by-step explanation:
