F(x) = 4^x and g(x) = 4^1/2x use the values given in the table and the functions to determine the missing values. a = 1/16 1/8 1/4 1/2 b = 0 1/2 1 2 c = 1 2 4 8

Functions can be represented on graphs and tables.

The values of A, B and C are: 1/4, 1 and 4

The functions are given as:

The value of A

x = -2, when g(x) = A.

So, we have:

Express as a fraction

So, we have:

The value of B

x = 0, when g(x) = B.

So, we have:

So, we have:

The value of C

x = 2, when g(x) = C.

So, we have:

So, we have:

Hence, the values of A, B and C are: 1/4, 1 and 4

Read more about functions at:

link

There are a few different ways to do this one. We can just note

so

That's one way. We might also note

so we can just take the square root of the corresponding value of f,

Finally nothing prevents us from substituting directly into the given g(x):

That's three different ways and they all give the same answer, which they have to, otherwise math doesn't work.

A=1/4, B=1, C=4

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.

hmu if ur willing to do a algebra a online class i’ll pay 600 hmu on the gram profitpreme