Use synthetic division to solve (4x2 – 3x2 +5x+6) + (x+8). What is the quotient? 4x2 - 27x+167-
996
X-6
792
4x2+21X+ 131+
X+ 6
792
4x2 + 21X+131+
X-6
4x2 - 27x+167-
996
X+6

answer d

4x² -27x +167 -996 / (x+6)

Step-by-step explanation:

you wrote down the equation wrong. Looking at the (unclear) picture, I see this:

(4x³ -3x² +5x +6)

/ (x+6)

x+6 = 0 gives x = -6

So synthetic division using -6

Term 1 = 4

-6 (4) = -24 and -3 + -24 = - 27 so Term 2 = -27

-6 (-27) = 162 and 5 + 162 = 167 so Term 3 = 167

-6 (167) = -1002 and 6 + -1002 = -996 so the last term is the remainder and it is -996 / (x+6)

Now you have found your answer, which is answer d.

4x² -27x +167 -996 / (x+6)

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐

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.