Numerical solution of a system of linear equations by
1) LU FACTORIZATION METHOD.
2) GAUSS ELIMINATION METHOD.
3) MATRIX INVERSION BY GAUSS ELIMINATION METHOD.
Gauss jordan and Guass elimination methodMeet Nayak
This ppt is based on engineering maths.
the topis is Gauss jordan and gauss elimination method.
This ppt having one example of both method and having algorithm.
Numerical solution of a system of linear equations by
1) LU FACTORIZATION METHOD.
2) GAUSS ELIMINATION METHOD.
3) MATRIX INVERSION BY GAUSS ELIMINATION METHOD.
Gauss jordan and Guass elimination methodMeet Nayak
This ppt is based on engineering maths.
the topis is Gauss jordan and gauss elimination method.
This ppt having one example of both method and having algorithm.
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Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
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Honest Reviews of Tim Han LMA Course Program.pptxtimhan337
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
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The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
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1. DE LA GRÁFICA A LA
FUNCIÓN CUADRÁTICA
INTRODUCCIÓN
En los módulos anteriores aprendimos que la representación algebraica de la
función cuadrática puede tener tres formas distintas y que en cada una de
éstas se puede obtener algún punto clave para su gráfica. Por ejemplo en la
forma general 𝑓(𝑥) = 𝒂𝑥2
+ 𝑏𝑥 + 𝒄 se puede obtener fácilmente el corte con
el eje "𝑦" al evaluar la función en cero se obtiene 𝑓(0) = 𝒄. En la forma
factorizada 𝑓(𝑥) = 𝒂(𝑥 − 𝑥1)(𝑥 − 𝑥2) se pueden conocer fácilmente los
cortes con el eje "𝑥", en 𝑥 = 𝑥1 y 𝑥 = 𝑥2. En la forma estándar 𝑓(𝑥) =
𝒂(𝑥 − 𝒉)𝟐
+ 𝒌 se tiene directamente la coordenada del vértice en 𝑉(𝒉, 𝒌).
En éste apartado revisaremos como pasar de la gráfica de una función
cuadrática, a su representación algebraica, y esto dependerá de los puntos
que se muestren en su gráfica.
2. 2 Mtra. Ana R. Faraco Pérez
Ejemplo 1: En la gráfica se presenta el corte con el eje "𝒚" y otros dos
puntos.
Al tener en la gráfica el corte con el
eje "𝑦", es conveniente trabajar con
la forma general de la función
cuadrática:
𝑓(𝑥) = 𝒂𝑥2
+ 𝑏𝑥 + 𝒄.
Dado que el punto 𝐴(0, 𝟏𝟐) en la
función representa que 𝑓(0) = 𝟏𝟐
entonces se tiene que el valor del
parámetro 𝒄 = 𝟏𝟐.
𝑓(𝑥) = 𝒂𝑥2
+ 𝑏𝑥 + 𝟏𝟐
De igual forma, los puntos 𝐵(1,7) y 𝐶(3,9), que se pueden obtener de la
gráfica representan en la función que 𝑓(1) = 7 y que 𝑓(3) = 9. Si sustituimos
esto en la forma general de la función cuadrática se forma un sistema de
ecuaciones.
𝑓(1) = 7 ⟹ 𝑓(1) = 𝒂(1)2
+ 𝑏(1) + 𝟏𝟐 = 𝟕 ⟹ 𝒂 + 𝑏 = −𝟓 (Ecuación 1)
𝑓(3) = 9 ⟹ 𝑓(1) = 𝒂(3)2
+ 𝑏(3) + 𝟏𝟐 = 𝟗 ⟹ 9𝒂 + 3𝑏 = −𝟑 (Ecuación 2)
Despejamos 𝒂 en la ecuación 1⟹ 𝒂 = −𝟓 − 𝑏, y sustituimos en la ecuación 2:
⟹ 9(−𝟓 − 𝑏) + 3𝑏 = −𝟑 resolvemos esta ecuación:
⟹ −45 − 9𝑏 + 3𝑏 = −𝟑
⟹ −6𝑏 = 𝟒𝟐
Obtenemos el valor del parámetro ⟹ 𝑏 = −
𝟒𝟐
𝟔
= −𝟕
Lo sustituimos en la ecuación 1⟹ 𝒂 = −𝟓 − 𝑏 obtenemos el valor del
parámetro 𝒂
⟹ 𝒂 = −𝟓 (
− −𝟕) = −𝟓 + 𝟕 = 𝟐
3. 3 Mtra. Ana R. Faraco Pérez
Sustituyendo los valores de los parámetros 𝒂 = 𝟐, 𝑏 = −𝟕 y 𝒄 = 𝟏𝟐 en la
forma general de la función cuadrática se obtiene:
𝑓(𝑥) = 𝟐𝑥2
− 𝟕𝑥 + 𝟏𝟐
Ejemplo 2: En la gráfica se presentan los puntos de corte con el eje "𝒙" y
otro punto.
Al tener en la gráfica los puntos
de corte con el eje "𝑥", es
conveniente trabajar con la
forma factorizada de la función
cuadrática:
𝑓(𝑥) = 𝒂(𝑥 − 𝑥1)(𝑥 − 𝑥2)
Dado que la gráfica corta al eje el
"𝑥" en los puntos 𝐴(−𝟐, 0) y
𝐵(𝟓, 0), entonces los factores de
la función serán:
𝑓(𝑥) = 𝒂(𝑥 − (−2))(𝑥 − 5)
𝑓(𝑥) = 𝒂(𝑥 + 2)(𝑥 − 5)
Para poder obtener el valor del parámetro 𝒂 podemos sustituir el otro punto
𝐶(2,4) en la función anterior y establecemos una ecuación, puesto que éste
punto representa que si evaluamos 𝑓(2) obtenemos 4
𝑓(2) = 4 ⟹ 𝑓(2) = 𝒂(2 + 2)(2 − 5) = 𝟒 ⟹ 𝒂(4)(−3) = 𝟒 (Ecuación 1)
Resolviendo ésta ecuación se obtiene 𝒂 = −
𝟏
𝟑
.
De ahí que la forma factorizada de la función es:
𝑓(𝑥) = −
𝟏
𝟑
(𝑥 + 2)(𝑥 − 5)
Al resolver el producto de los binomios con término común y multiplicar por
el −
𝟏
𝟑
se llega a la forma general.
4. 4 Mtra. Ana R. Faraco Pérez
Aplicando propiedad distributiva 𝑓(𝑥) = −
𝟏
𝟑
[𝑥(𝑥 − 5) + 2(𝑥 − 5)]
Propiedad distributiva 𝑓(𝑥) = −
𝟏
𝟑
[𝑥2
− 5𝑥 + 2𝑥 − 2(5)]
Reduciendo términos semejantes 𝑓(𝑥) = −
𝟏
𝟑
[𝑥2
+ (−5 + 2)𝑥 − 2(5)]
Propiedad distributiva 𝑓(𝑥) = −
𝟏
𝟑
[𝑥2
− 3𝑥 − 10]
Forma general 𝑓(𝑥) = −
𝟏
𝟑
𝑥2
+ 𝑥 +
10
3
Ejemplo 3: En la gráfica se presentan el vértice y otro punto.
Al tener las coordenadas del
vértice en el punto 𝐴(𝟑, −𝟐)
se elige la forma estándar
𝑓(𝑥) = 𝒂(𝑥 − 𝒉)𝟐
+ 𝒌 para
sustituir 𝒉 = 𝟑 y 𝒌 = −𝟐:
𝑓(𝑥) = 𝒂(𝑥 − )
𝟑 𝟐
− 𝟐
Para poder obtener el valor del parámetro 𝒂 podemos sustituir el otro punto
𝐵(2,2) en la función anterior y establecemos una ecuación, puesto que éste
punto representa que si evaluamos 𝑓(2) obtenemos 2
𝑓(2) = 2 ⟹ 𝑓(2) = 𝒂(2 − )
𝟑 𝟐
− 𝟐 = 𝟐 ⟹ 𝒂(−1)2
= 𝟒 (Ecuación 1)
Resolviendo ésta ecuación se obtiene 𝒂 = 𝟒 .
De ahí que la forma estándar de la función es:
𝑓(𝑥) = 𝟒(𝑥 − )
𝟑 𝟐
− 𝟐
5. 5 Mtra. Ana R. Faraco Pérez
Desarrollando el binomio al cuadrado: 𝑓(𝑥) = 𝟒(𝑥2
− 6𝑥 + 9) − 𝟐
Aplicando propiedad distributiva: 𝑓(𝑥) = 4𝑥2
− 24𝑥 + 36 − 𝟐
y reduciendo términos semejantes se llega a la forma general de la función:
𝑓(𝑥) = 4𝑥2
− 24𝑥 + 34