The document discusses solving polynomial inequalities, including:
- Linear inequalities have the general form ax + b ≷ 0 and their solution sets are intervals.
- Quadratic inequalities can be solved using critical points or factoring depending on the discriminant.
- Examples are provided of solving linear, quadratic, and polynomial inequalities using critical points or factoring.
Properties of parallelogram applies to rectangles, rhombi and squares.
In a parallelogram,
Opposite sides of a parallelogram are parallel.
A diagonal of a parallelogram divides it into two congruent triangles.
Opposite sides of a parallelogram are congruent.
Opposite angles of a parallelogram are congruent.
If one angle of a parallelogram is right, then all the angles are right.
Consecutive angles of a parallelogram are supplementary.
Diagonals of a parallelogram bisect each other.
https://www.youtube.com/channel/UCOuMfD4sggCh7XeiAHlus6Q
Properties of parallelogram applies to rectangles, rhombi and squares.
In a parallelogram,
Opposite sides of a parallelogram are parallel.
A diagonal of a parallelogram divides it into two congruent triangles.
Opposite sides of a parallelogram are congruent.
Opposite angles of a parallelogram are congruent.
If one angle of a parallelogram is right, then all the angles are right.
Consecutive angles of a parallelogram are supplementary.
Diagonals of a parallelogram bisect each other.
https://www.youtube.com/channel/UCOuMfD4sggCh7XeiAHlus6Q
How to Make a Field invisible in Odoo 17Celine George
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
Introduction to AI for Nonprofits with Tapp NetworkTechSoup
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
A Strategic Approach: GenAI in EducationPeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
Embracing GenAI - A Strategic ImperativePeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
3. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
RESOLVER
INECUACIONES
LINEALES
APLICAR EL
MÉTODO DE
LOS PUNTOS
CRÍTICOS
RESOLVER
INECUACIONES
CUADRÁTICAS
4. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
𝑰𝑵𝑬𝑪𝑼𝑨𝑪𝑰𝑶𝑵𝑬𝑺
𝑷𝑶𝑳𝑰𝑵𝑶𝑴𝑰𝑨𝑳𝑬𝑺
En esta ocasión, desarrollamos el tema
de inecuaciones que permiten encontrar
un conjunto de valores que permiten
explicar una necesidad.
en el mercado libre y nos indica el punto E
de equilibrio y las zonas de excedente del
consumidor y del productor, generadas por
Inecuaciones.
En el gráfico se muestra los excedentes
𝒂𝒙 + 𝒃 > 𝟎
5. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Inecuaciones polinomiales
𝐃𝐞𝐟𝐢𝐧𝐢𝐜𝐢ó𝐧: Una inecuación es una desigualdad entre
dos expresiones donde aparezca al menos una variable.
𝐄𝐣𝐞𝐦𝐩𝐥𝐨𝐬
a) 2𝑥 − 3 > 𝑥 + 2
Para resolver una inecuación se utilizan los teoremas
de las desigualdades
b) 𝑥2 − 3𝑥 + 4 ≥ 2𝑥 − 1
2𝑥 − 3 > 𝑥 + 2
→
(+3) 2𝑥 > 𝑥 + 5
→
(−𝑥) 𝑥 > 5
Luego: C. S = 5; +∞
𝐈𝐍𝐄𝐂𝐔𝐀𝐂𝐈Ó𝐍 𝐋𝐈𝐍𝐄𝐀𝐋
Tiene como forma general 𝑎𝑥 + 𝑏 ≷ 0 ; 𝑎 ≠ 0
𝐄𝐣𝐞𝐦𝐩𝐥𝐨𝐬
a) 4𝑥 − 7 ≥ 0 b) −5𝑥 + 8 < 0
Para resolver una inecuación lineal, se despeja la
variable por los teoremas de las desigualdades
a) 4𝑥 − 7 ≥ 0
→
(+7) 4𝑥 ≥ 7
→
(÷ 4) 𝑥 ≥
7
4
Luego:
C. S = ቈ
7
4
; ۧ
+∞
𝐄𝐣𝐞𝐦𝐩𝐥𝐨
6. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
b) − 5𝑥 + 8 ≥ 0
→
(−8) −5𝑥 ≥ −8
→
÷ (−5) 𝑥 ≤
8
5
Luego: C. S = ۦ−∞; ൨
8
5
¡NOTA!
El C.S de una inecuación lineal es
un intervalo no acotado
𝐄𝐧 𝐠𝐞𝐧𝐞𝐫𝐚𝐥:
𝑎𝑥 + 𝑏 > 0
Si tenemos
→
(−𝑏) 𝑎𝑥 > −𝑏
Si 𝑎 > 0: Si 𝑎 < 0:
𝑎𝑥 > −𝑏 𝑎𝑥 > −𝑏
→
÷ (𝑎) 𝑥 > −
𝑏
𝑎
→
÷ (𝑎) 𝑥 < −
𝑏
𝑎
Luego:
Luego:
C. S = ൽ−
𝑏
𝑎
; + ۧ
∞ C. S = ۦ−∞; − ඁ
𝑏
𝑎
Tenemos 2 posibles casos, estos son:
𝐄𝐣𝐞𝐦𝐩𝐥𝐨
7. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Criterio de los puntos críticos
Es un método que nos permite observar como
𝐄𝐣𝐞𝐦𝐩𝐥𝐨
Analizar 𝑃 𝑥 = 𝑥 + 3 𝑥 − 4
Sus raíces son: −3; 4 los ubicamos en la recta real
−3 4
𝐼
𝐼𝐼
𝐼𝐼𝐼
−3 < 𝑥 < 4 4 < 𝑥
𝑥 < −3
Analizando el signo en cada zona
ZONA
4 < 𝑥
−3 < 𝑥 < 4
𝑥 < −3
𝑥 + 3 𝑥 − 4 𝑥 + 3 𝑥 − 4
(+) (+) (+)
(−)
(−)
(+)
(−) (−) (+)
𝑃 𝑥
cambia de signo los factores lineales (con
coeficientes reales), en una multiplicación indicada
Se deduce:
−3 4
(−) (+)
(+)
+∞
−∞
−∞ +∞
𝑆𝑖
𝑆𝑖
𝑃 𝑥 > 0
𝑃 𝑥 < 0
↔
↔
𝑥 ∈
𝑥 ∈
−∞; −3 ∪ 4; +∞
−3; 4
8. C R E E M O S E N L A E X I G E N C I A
ቐ
C U R S O D E Á L G E B R A
𝐄𝐣𝐞𝐦𝐩𝐥𝐨 𝟏
Resolver: (𝑥 + 7) (𝑥 − 5) ≤ 0
𝐑𝐞𝐬𝐨𝐥𝐮𝐜𝐢ó𝐧
Encontramos sus puntos críticos (P. C):
𝑥 + 7 = 0 → 𝑥 = −7
𝑥 − 5 = 0 → 𝑥 = 5
P. C
→ P. C = −7; 5
−∞ +∞
5
−7
→ 𝑥 ∈ −7; 5
𝐄𝐣𝐞𝐦𝐩𝐥𝐨 𝟐
Resolver: (𝑥 + 1) (𝑥 − 4)(𝑥 + 8) > 0
𝐑𝐞𝐬𝐨𝐥𝐮𝐜𝐢ó𝐧
Encontramos sus puntos críticos (P. C):
𝑥 + 1 = 0
𝑥 = −1
𝑥 − 4 = 0
𝑥 = 4
𝑥 + 8 = 0
𝑥 = −8
→ P. C = −1; 4;−8
−∞ +∞
−1
−8 4
→ 𝑥 ∈ −8; −1 ∪ 4; +∞
9. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
𝐀𝐩𝐥𝐢𝐜𝐚𝐜𝐢ó𝐧 𝟏
Resolver
𝑥 + 5 𝑥 − 2 ≥ 0
Aplicando el método de puntos críticos
Los P. C son: −5; 2
−∞ +∞
2
−5
→ 𝑥 ∈ ۦ−∞; ሿ
−5 ∪ ሾ2; ۧ
+∞
𝑥2 + 3𝑥 − 10 ≥ 0
Factorizando
𝑥2 + 3𝑥 − 10 ≥ 0
𝑥
𝑥
+5
−2
𝐀𝐩𝐥𝐢𝐜𝐚𝐜𝐢ó𝐧 𝟐
Resolver
2𝑥 + 1 𝑥 − 2 < 0
Aplicando el método de puntos críticos
Los P. C son: −1/2; 2
−∞ +∞
2
−1/2
→ 𝑥 ∈ ۦ−1/2; ۧ
2
2𝑥2 − 3𝑥 − 2 < 0
Factorizando
2𝑥2 − 3𝑥 − 2 < 0
2𝑥
𝑥
+1
−2
10. C R E E M O S E N L A E X I G E N C I A
Inecuación cuadrática
C U R S O D E Á L G E B R A
Tiene como forma general
𝑎𝑥2 + 𝑏𝑥 + 𝑐 ≷ 0 ; 𝑎 ≠ 0
𝐑𝐞𝐬𝐨𝐥𝐮𝐜𝐢ó𝐧
Considerando 𝑎 > 0 se tiene 3 casos según el
análisis de su discriminante (∆= 𝑏2 − 4𝑎𝑐)
𝑃 𝑥 =
CASO ∆> 𝟎 ∆= 𝟎
FORMA Es
factorizable
Trinomio cuadrado
perfecto
MÉTODO Puntos
críticos
𝑥2 ≥ 0
𝐄𝐣𝐞𝐦𝐩𝐥𝐨 𝟏
Resolver: 3𝑥2 − 5𝑥 − 2 ≥ 0
Analizando ∆= (−5)2−4(3)(−2) = 49 → ∆ > 0
→ 𝐄𝐥 𝐩𝐨𝐥𝐢𝐧𝐨𝐦𝐢𝐨 𝐞𝐬 𝐟𝐚𝐜𝐭𝐨𝐫𝐢𝐳𝐚𝐛𝐥𝐞
3𝑥2 − 5𝑥 − 2 ≥ 0
3𝑥
𝑥
+1
−2
→ (3𝑥 + 1)(𝑥 − 2) ≥ 0
→ Los P. C son: −
1
3
; 2
−∞ +∞
2
−1/3
→ 𝑥 ∈ ۦ−∞; ሿ
−1/3 ∪ ሾ2; ۧ
+∞
11. C R E E M O S E N L A E X I G E N C I A
ቐ
C U R S O D E Á L G E B R A
𝐄𝐣𝐞𝐦𝐩𝐥𝐨 𝟐
𝑦 𝑐𝑜𝑖𝑛𝑐𝑖𝑑𝑒𝑛 𝑐𝑜𝑛 𝑙𝑎𝑠 𝑟𝑎í𝑐𝑒𝑠 𝑑𝑒𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜
𝐴𝑙 𝑟𝑒𝑠𝑜𝑙𝑣𝑒𝑟 𝑢𝑛𝑎 𝑖𝑛𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑐𝑢𝑎𝑑𝑟á𝑡𝑖𝑐𝑎,
𝑙𝑜𝑠 𝑒𝑥𝑡𝑟𝑒𝑚𝑜𝑠 𝑓𝑖𝑛𝑖𝑡𝑜𝑠 𝑑𝑒 𝑠𝑢 𝐶. 𝑆 𝑠𝑜𝑛 𝑃. 𝐶
𝑆𝑖 𝑎𝑥2 + 𝑏𝑥 + 𝑐 < 0 ↔ 𝐶. 𝑆 = 𝑚; 𝑛
→ ∧
𝑚 + 𝑛 = −
𝑏
𝑎
𝑚𝑛 =
𝑐
𝑎
Si al resolver la inecuación
2𝑥2 + 𝑚𝑥 − 𝑛 < 0
Se obtiene C. S = 3;7
Calcule 𝑚. 𝑛
𝐑𝐞𝐬𝐨𝐥𝐮𝐜𝐢ó𝐧
Como C. S = 3; 7 Tenemos que 3 y 7 son P. C y raíces. Por teorema de Cardano
→ 3 + 7 = −
𝑚
2
→ 𝑚 = −20
→ 3.7 =
−𝑛
2
→ 𝑛 = −42
𝑚. 𝑛 = 840
12. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
¡ 𝐑𝐞𝐜𝐮𝐞𝐫𝐝𝐚! ∀𝑥 ∈ ℝ ; 𝑥2 ≥ 0
Entonces
(𝑥 − 3)2 ≥ 0 → 𝑥 ∈ ℝ
(𝑥 − 3)2 > 0 → 𝑥 ∈ ℝ − 3
(𝑥 − 3)2 < 0 → 𝑥 ∈ ∅
(𝑥 − 3)2 ≤ 0 → 𝑥 ∈ 3
Esto es:
𝑥2 − 6𝑥 + 9 ≥ 0 → C. S = ℝ
𝑥2 − 6𝑥 + 9 > 0 → C. S = ℝ − 3
𝑥2 − 6𝑥 + 9 < 0 → C. S = ∅
𝑥2 − 6𝑥 + 9 ≤ 0 → C. S = 3
𝐓𝐞𝐨𝐫𝐞𝐦𝐚: Si 𝑎𝑥2 + 𝑏𝑥 + 𝑐 ≤ 0 tiene 𝐶. 𝑆 = 𝛼
→ 𝑎 > 0 ∧ ∆= 0
𝐄𝐣𝐞𝐦𝐩𝐥𝐨
Si 𝑥2 − 4𝑥 + 𝑛 ≤ 0 tiene 𝐶. 𝑆 = 𝛼 . Calcule el valor de 𝑛
𝐑𝐞𝐬𝐨𝐥𝐮𝐜𝐢ó𝐧
Como tiene 𝐶. 𝑆 = 𝛼 → ∆= 0
→ ∆= (−4)2−4(1)(𝑛) = 0
→ 16 − 4𝑛 = 0
→ 𝑛 = 4
13. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
𝐓𝐞𝐨𝐫𝐞𝐦𝐚: Si 𝑎𝑥2 + 𝑏𝑥 + 𝑐 > 0 tiene 𝐶. 𝑆 = ℝ − 𝛼
→ 𝑎 > 0 ∧ ∆= 0
𝐄𝐣𝐞𝐦𝐩𝐥𝐨
Si 2𝑥2 − 6𝑥 + (𝑛 − 2) > 0 tiene 𝐶. 𝑆 = ℝ − 𝛼 . Calcule el valor de 𝑛
𝐑𝐞𝐬𝐨𝐥𝐮𝐜𝐢ó𝐧
Como tiene 𝐶. 𝑆 = ℝ − 𝛼 → ∆= 0
→ ∆= (−6)2−4(2)(𝑛 − 2) = 0
→ 52 − 8𝑛 = 0
→ 𝑛 =
13
2
¡ 𝐈𝐦𝐩𝐨𝐫𝐭𝐚𝐧𝐭𝐞!
𝐴𝑙 𝑟𝑒𝑠𝑜𝑙𝑣𝑒𝑟 𝑢𝑛𝑎 𝑖𝑛𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛
𝑐𝑢𝑎𝑑𝑟á𝑡𝑖𝑐𝑎 𝑝𝑟𝑖𝑚𝑒𝑟𝑜 𝑣𝑒𝑟𝑖𝑓𝑖𝑐𝑎 𝑠𝑖 𝑠𝑒
𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎, 𝑑𝑒 𝑙𝑜 𝑐𝑜𝑛𝑡𝑟𝑎𝑟𝑖𝑜 𝑎𝑛𝑎𝑙𝑖𝑧𝑎
𝑠𝑢 𝑑𝑖𝑠𝑐𝑟𝑖𝑚𝑖𝑛𝑎𝑛𝑡𝑒
14. w w w . a c a d e m i a c e s a r v a l l e j o . e d u . p e