Help me with this and give the correct answer and get 25 points

x =6.0

The base is 6x  = 35.8

The side is 3x =  17.9

Step-by-step explanation:

We can use the pythagorean theorem

a^2 + b^2 = c^2

(6x)^2 +(3x)^2 = 40^2

36x^2 + 9x^2 = 1600

Combine like terms

45x^2 = 1600

Divide each side by 45

45x^2 /45 =1600/45

x^2 =35.555555555

Take the square root of each side

sqrt(x^2) = sqrt(35.555555555)

x =5.96284794 = 6.0

The base is 6x = 6*5.9628479 =35.77708 = 35.8

The side is 3x = 3*5.9628479 =17.88854382= 17.9

x = 6.0

3x = 17.9

6x = 35.8

Step-by-step explanation:

Using pythagoras theorem

40² = (3x)² + (6x)²

1600 = 9x² + 36x²

1600 = 45x²

x² = 320/9

x = 5.96284794

3x = 17.88854382

6x = 35.77708764

answer: do we write it

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.