5(2x+3) by adding them you get this equation.
I believe D, but I am unfamiliar with this particular terminology
For independent events,
P(X ∩ Y) = P(X)* P(Y)
=>
1/3 = P(X)*(5/6)
solve for P(X) =>
P(X) = (1/3)*(6/5) = 2/5 = 0.4
(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
-6 + x = -7
+6 +6 add 6 to both sides, 6's cancel out on the left.
________
x = -1, is the answer.
Hope this helps! Let me know if you have any other questions related to this problem. :)
