(-1,1) and (0,0) are probably it
(x-8) ^ 2 = 121
(x-8) = + / - root (121)
x = 8 +/- root (121)
The solutions are:
x1 = 8 + root (121)
x2 = 8 - root (121)
2a ^ 2 = 8a-6
2a ^ 2-8a + 6 = 0
a ^ 2-4a + 3 = 0
(a-1) (a-3) = 0
The solutions are:
a1 = 1
a2 = 3
x ^ 2 + 12x + 36 = 4
x ^ 2 + 12x + 36-4 = 0
x ^ 2 + 12x + 32 = 0
(x + 4) (x + 8) = 0
The solutions are:
x1 = -8
x2 = -4
x ^ 2-x + 30 = 0
x = (- b +/- root (b ^ 2 - 4 * a * c)) / 2 * a
x = (- (- 1) +/- root ((- 1) ^ 2 - 4 * (1) * (30))) / 2 * (1)
x = (1 +/- root (1 - 120))) / 2
x = (1 +/- root (-119))) / 2
x = (1 +/- root (119) * i)) / 2
The solutions are:
x1 = (1 + root (119) * i)) / 2
x2 = (1 - root (119) * i)) / 2
Well i think that it would be 8×8-40 and that would come out to a solution of 28
16n+8=4
16n=4-8
16n=-4
n=-4/16
n=-1/4.
Answer:
La afirmación es falsa, no todos los divisores de 100 son divisores de 50, ya que solo se toman en cuenta sus divisores comunes, los cuales son todos los divisores de 50. Expresamos a 100 en sus factores primos: 100 = 2 · 2 · 5 · 5 = 2² · 5² Divisores de 100: {1, 2, 4, 5, 10, 20, 25, 50, 100}
Step-by-step explanation:
