History, 07.07.2019 04:30, savyblue1724707

In100-150 words compare and contrast the four sections of the codex justinian

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The Justinian Code was important as it simplified and streamlined centuries of existing Roman laws and Justinian's own laws and into one system. Four sections made up the code, and these were the Codex Constitutionum, Digesta, Institutiones and the Novellae Constitutiones Post Codicem.

Justinian I was Emperor of the Roman Byzantine Empire from 527 to 565. The first code, which included only the Codex was completed in 529. The Digesta and Institutiones were added in 534.

In the effort to simplify the code of laws, Justinian appointed 10 men to research all known laws and ordinances in the history of the empire. The commission siphoned out any outdated laws or laws that contradicted each other, and the result was known as the Codex Constiutionum.

The Digesta was a collection of 50 books of jurists rulings over the centuries. Any rulings that didn't appear in these books were then unable to be cited by lawyers in any court tribunal. This collection was built upon the research of 16 lawyers. The Institutiones amounted to a textbook for law students.

Finally, the Novellae Constiutiones Post Codicem acted as a supplement to the code and included Justinian's own ordinances that he announced between 534 and 565. This collection was added to the code after the first revision of the code.The Justinian Code was important as it simplified and streamlined centuries of existing Roman laws and Justinian's own laws and into one system. Four sections made up the code, and these were the Codex Constitutionum, Digesta, Institutiones and the Novellae Constitutiones Post Codicem.

answered: Guest

answer: this problem has only trivial solutions, with all points on the same line. x=5 or x=7/5.

step-by-step explanation:

there are three triangles, and three unknowns: x, y=length of middle segment, and z=cos(angles marked with red arc).

we need

$\cos(2\gamma)=(2\cos(\gamma)^2-1)$

and law of cosines,

$c^2=a^2+b^2-2ab\cos(\gamma)$

the three equations are

$(2x-1)^2=9^2+y^2-2(9yz)$

$(3x)^2=15^2+y^2-2(15yz)$

$(3x+2x-1)^2=9^2+15^2-2(9\times15(2z^2-1))$

see attached image for solution.

What is the value of x? enter your answer in the box. x =

answered: Guest

answer: they arrived in new york.

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In100-150 words compare and contrast the four sections of the codex justinian...