Talia is packing a moving box. She has a square-framed poster with an area of square feet. The cube-shaped box has a volume of cubic feet. Will the poster lie flat in the box? Explain?

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source

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.

just watch memes

step-by-step explanation:

it me < 3

here is me explaining it:

the pizza would be delivered around 5, give or take 15 minutes. this means it could be delivered anywhere from 4: 45 to 5: 15

let's say 5: 00 is equivalent to 0 on a number line and x represents time. 4: 45 would be equal to -15x because we are 15 minutes behind our starting point, 5: 00. 5: 15 would be equal to 15x because we are 15 minutes ahead of your starting point (which means you're late on the pizza delivery! )

absolute value is very confusing. the basic definition is how far you are from your starting point. if we deliver the pizza at 4: 45 (15 minutes before the starting point) we would be -15. this means we are 15 away from our starting point, 5: 00. that's the absolute value. |-15| (absolute value of negative fifteen) is equal to 15. if you did not understand this i suggest rereading this paragraph.

the normal way you would write this scenario would be 4.75 (4 and 3 quarters, or 4: 45. you can also write as a fraction.) is less than x and x is less than 5.25 (5 and 1 quarter or 5: 15). 4.75< x< 5.25 the reason why i am using numbers is because you cannot represent typical time as values on a graph, number line, or in an equation.

another way to write this scenario is with absolute value. we can write this as |x| being less than or equal to .25 (remember, this just means 15 minutes).           |x| ≤ .25.

sorry if the end was confusing. because we are talking about time this whole question is 10x harder. like i said earlier, you cannot represent time as, well, time, in a graph, equation, or number line. i set the values as follows: 1/4 or .25 for 15 minutes, and 1 for 1 hour. this is just a common way to set up graphs involving time.

|x| ≤ .25

i hope this ! : )

tell me if you need more assistance