If (a-2)^2+(b+1)^2+(c-3)2 =0 then the value of a^2/4+b^2+c^2/3

5

Step-by-step explanation:

(a-2)^2+(b+1)^2+(c-3)^2 =0

Since all 3 terms are square numbers, they must be = 0 to have a sum of 0.

so a-2= 0, b+1=0 and c-3= 0

a^2/4+b^2+c^2/3= 4/4+1+9/3= 1+1+3= 5

what's the question?

step-by-step explanation:

there is no question.

it would open into a square depending on the type of envelope i’m not sure about the second part