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read this excerpt from "a modest proposal" by jonathan swift:
infant's flesh will be in season throughout the year, but more plentiful in march, and a little before and after, for we are told by a grave author, an
eminent french physician, that fish being a prolifick dyet, there are more children born in roman catholick countries about nine months after
lent, the markets will be more glutted than usual, because the number of popish infants, is at least three to one in this kingdom, and therefore it
will have one other collateral advantage, by lessening the number of papists among us
i have already computed the charge of nursing a beggar's child (in which list i reckon all cottagers, labourers, and four. fifths of the farmers) to
be about two shillings per annum, rags included, and i believe no gentleman would repine to give ten shillings for the carcass of a good fat child,
which, as i have said, will make four dishes of excellent nutritive meat, when he hath only some particular friend, or his own family to dine with
him. thus the squire will learn to be a good landlord, and grow popular among his tenants, the mother will have eight shillings neat profit, and be
fit for work till she produces another child
those who are more thrifty (as i must confess the times require) may flea the carcass, the skin of which, artificially dressed, will make admirable
gloves for ladies, and summer boots for fine gentlemen
in which point of view is this excerpt written?
a.
first person
b. second person
c.
third-person limited
od. third-person omniscient
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x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

sorry i just need to anwers this to get points

step-by-step explanation:

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.