60
2 and 30
30
2 and 15
15
3 and 5
96
8 and 12
circle for 8 is 2 and the square is 4
circle for 12 is 2 and the last circle is also 2
Yes it is, 6.52 and 6.52 are the same but one has an extra 0.
Take the root of both sides, and then solve.
c = sqrt a^2 + b^2, - sqrt a^2 + b^2.
When discriminant >0, then it contrains 2 distinct real zeros.
Discriminant=6 and 6>0 so, you have 2 solutions.
Answer:
(1, 2 )
Step-by-step explanation:
x + y = 3 (subtract x from both sides )
y = 3 - x → (1)
2x - y = 0 → (2)
Substitute y = 3 - x into (2)
2x - (3 - x) = 0
2x - 3 + x = 0
3x - 3 = 0 ( add 3 to both sides )
3x = 3 ( divide both sides by 3 )
x = 1
Substitute x = 1 into (1)
y = 3 - 1 = 2
solution is (1, 2 )
