Program p1;
var a,b,c,d : integer; {i presume you give integer numbers for the values of a, b, c }
x1, x2 : real;
begin
write('a='); readln(a);
write('b='); readln(b);
write('c=');readln(c);
d:=b*b - 4*a*c
if a=0 then x1=x2= - c/b
else
if d>0 then begin
x1:=(-b+sqrt(d)) / (2*a);
x2:=(-b - sqrt(d))/(2*a);
end;
else if d=0 then x1=x2= - b /(2*a)
else write ("no specific solution because d<0");
writeln('x1=', x1);
writeln('x2=',x2);
readln;
end.
Answer:
The most likely galaxy type to be identifiable regardless of orientation is: Irr
Explanation:
The Irr galaxies don't have a discernable or usual shape; that is why it is relatively easy to identify.
When we talk about E type galaxies, this statement proves itself by the way the cumulus of stars compounds the galaxy. The elliptical galaxies have the form of ellipses, with a reasonable distribution of stars. The degree of eccentricity is the number that complements the E letter; that's why E0 galaxies are almost spherical, while E7 is considerably elongated.
SBc, SBa galaxies are spiral; this means it can be flat in some angles difficulting their identification process; in this case, the last letter means the way the arms display their form, with "c" having a vague form and "a" well-defined arms. That's why in some angles can be mistreated as another type of galaxy.
The answer is True bc it saves you time and is efficient
I hope that this answer will help u to solve your problem. But u have tto take the importance clues only .