Hilbert axioms changed Euclid's theorem by identifying and explaining the concept of undefined terms
<h3>What was Hilbert's Axiom?</h3>
These were the sets of axioms that were proposed by the man David Hilbert in the 1899. They are a set of 20 assumptions that he made. He made these assumptions as a treatment to the geometry of Euclid.
These helped to create a form of formalistic foundation in the field of mathematics. They are regarded as his axiom of completeness.
Hilbert’s axioms are divided into 5 distinct groups. He named the first two of his axioms to be the axioms of incidence and the axioms of completeness. His third axiom is what he called the axiom of congruence for line segments. The forth and the fifth are the axioms of congruence for angles respectively.
Hence we can conclude by saying that Hilbert axioms changed Euclid's theorem by identifying and explaining the concept of undefined terms.
Read more on Euclid's geometry here: brainly.com/question/1833716
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complete question
Hilbert’s axiom’s changed Euclid’s geometry by _____.
1 disproving Euclid’s postulates
2 utilizing 3-dimensional geometry instead of 2-dimensional geometry
3 describing the relationships of shapes
4 identifying and explaining the concept of undefined terms
Answer:
Corresponding angles converse theorem
Step-by-step explanation:
If two lines (AB and CD) are intersected by a transversal (BD) and corresponding angles are congruent (∠B ≅ ∠D), then the lines are parallel.
The Answer is D. 16.
XZ = 8/sin 30
Answer:
The radius of the circle 'r' = 5
Equation of the circle
(x - 2 )² + (y+1)² = (5)²
center of the circle ( h,k) = (2 , -1) , 'r' = 5
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given circle is x² - 4 x + y² + 2 y - 20 =0
x² - 2(2) x + (2)²-(2)²+ y² + 2 y(1) +(1)²-(1)² - 20 = 0
We know that
(a+b)² = a²+2 a b+b²
(a-b)² = a² -2 a b+b²
⇒(x - 2 )² - 4 + (y+1)² -21 =0
⇒ (x - 2 )² + (y+1)² = 25
<u><em>Step(ii):</em></u>-
Equation of the circle form
( x- h)² + (y -k)² = r²
Given circle (x - 2 )² + (y+1)² = (5)²
center of the circle ( h,k) = (2 , -1)
radius of the circle 'r' = 5