Let {xn} be a sequence of random variables. p (a) if xn ~ 0, then for any p > 0 ixnip ~ o 1 + ixnip and p (b) if for some p > 0 (6.21) holds, then xn ~ 0. (c) suppose p > 0. show xn ~ 0 iff (6.22).

answer: standard form is ax^2 + bx + c = 0

step-by-step explanation:

standard form is ax^2 + bx + c = 0

(2x + 1)(x - 2) = 0

distribute to get into standard form

2x^2 - 4x + x - 2 = 0

2x^2 - 3x - 2 = 0

a = 2, b= -3, c = -2

a is your answer

51

step-by-step explanation:

73-22=51 the   mass number minus the amount of protons gives you the amount of neutrons

ok then what happened.

b

step-by-step explanation:

it have 5 sides and 5 identicle angle.the rest does not have the same property as b.

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