Identify ONE way that the images show how the role of political leaders in Western Europe changed during the period from 1200 to 1450 CE.Use these images to answer all parts of the question that follows. Please answer in a non-plagiarized form preferably
Answers
The images show the importance of political leaders and the influence of the Catholic Church in this process.
What is seen in the images?In the first image you can see a King and his subjects and on the right side the warriors of the kingdom fighting a group of people. This image demonstrates the absolute power that the kings had during the 12th century. However, no images or symbols that refer to the Catholic Church are shown.
In the second image, a king and a priest (Pope) are seen discussing on equal terms, which reflects the power that the Church and the priests had achieved to equate themselves with the Kings of the time.
How did the roles of European leaders change?The roles of European leaders changed from the twelfth century to the fourteenth century because the Church gained more influence in the government, so the Kings allowed them to interfere in their decisions and mandates because they considered that this was the way to go to paradise after his death.
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Related Questions
Find the limit of the function (if it exists). (If an answer does not exist, enter DNE.) x3 - 8 lim X-2 X-2 Write a simpler function that agrees with the given function at all but one point. Use a graphing utility to confirm your result. g(x)
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Answer:
The answer to the limit would be [tex]\lim_{x \to 2} \frac{x^3-8}{x-2}=12[/tex]. The simpler equation would be: [tex]\frac{x^2+2x+4}{1} \\[/tex]
Explanation:
The equation has a hole at at x=2. A hole is a type of removable discontinuity which allows us to solve the limit problem without receiving a DNE result. When examining the equation [tex]\frac{x^3-8}{x-2}[/tex], it is noted that the numerator can be expanded using the formula for the difference of cubes. The formula for the difference of cubes is: [tex]a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]. We identify [tex]x[/tex] as [tex]a[/tex] and [tex]2[/tex] as [tex]b[/tex]. Plugging these numbers into the difference of cubes expansion formula we get: [tex]\frac{(x-2)(x^2+2x+4)}{(x-2)}[/tex]. Notice that [tex](x-2)[/tex] is present in both the numerator and denominator. Because of this, we can cancel both of them out, resulting in the simplified equation: [tex]x^2+2x+4[/tex]. We then plug [tex]2[/tex] in for [tex]x[/tex] and get [tex]12[/tex] as the final answer to the limit.