guia de-estudios-para-la-evaluacion-diagnostica-2016-2017
TRANSCRIPT
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𝑃 = 1 + (𝑛 − 1) (2) 𝑃
(𝑛, 𝑃)
𝑛 = 1
𝑛 = 2
𝑛 = 3
𝑛 = 4
𝑛 = 5
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5
3
35
3
35
3
35 X 350 mililitros =
1050
35= 30 mililitros de alcohol
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6
1
41
3
24
6
24
7
24
8
24
10
5 1 6
7 7 7
5 1 4
7 7 7
4
5
6
1
12
215
12
17
4
5
6
1
12
215
12
13
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7
1 −7
12=
12 − 7
12=
5
12
12
12−
7
12=
5
12
5
12(
2
2) =
10
24
3/2
3
8
(1
2) (
3
8) (
8
1) =
24
16
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8
24
16=
12
8=
6
4=
3
2
2500ml1
351
=2500 ml
35=
500
7= 71.429 ml
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𝑟 = 3/2
(3 + 2 = 5)
80/5 = 16
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(𝐴 = 𝑙𝑎) 𝐴
𝐴 𝑙 𝐴
𝐵 𝐴
Mosaicos tipo 𝐵
Mosaico tipo 𝐴
5 m
3 m
3 m
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𝐴
𝐴 𝐵
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𝐴 𝐵
𝐴 𝐵
𝐴 𝐵
𝐴 𝐵
𝐴
5𝐴 + 1 = 𝐵
𝐴
𝐵 = 5(64) + 1; 𝐵 = 320 + 1; 𝐵 = 321
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𝒎
𝒎 𝒙
𝒎𝒙
𝒃
𝒎
𝒚 = 𝒎𝒙 + 𝒃
𝑚 = 3000
𝑏 = 4 000 − 3 000 = 1 000
𝑦 = 3 000𝑥 + 1 000
𝒚
𝒙
𝒎
𝒃
𝑦 = 3000(1) + 1000 𝑦 = 3000 + 1000 𝑦 = 4000
𝑦 = 3000(2) + 1000 𝑦 = 6000 + 1000 𝑦 = 7000
𝑦 = 3000(3) + 1000 𝑦 = 9000 + 1000 𝑦 = 10,000
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26, 56, 86, 116, ….
𝒅
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2
3
2
31
2(𝑥 −
2
3𝑥) =
1
2(
1
3𝑥) =
1
6𝑥
2
3𝑥 +
1
6𝑥 = 20 litros
2(2𝑥)+ 1𝑥
6𝑥 4𝑥 + 𝑥 = 6(20)
5𝑥 = 120 𝑥 =120
5𝑥 = 24 𝑙𝑖𝑡𝑟𝑜𝑠
𝑥
𝐴1𝑥 + 𝐵1𝑦 = 𝐶1
𝐴2𝑥 + 𝐵2𝑦 = 𝐶2
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𝑥
𝑦
𝑥
𝑦
3𝑥 + 𝑦 = 36
𝑥 + 4𝑦 = 56
(−3)
(−3)(𝑥 + 4𝑦) = (56)(−3)
−3𝑥 − 12𝑦 = −168
−3𝑥 − 12𝑦 = −168
3𝑥 + 𝑦 = 36
−11𝑦 = −132
𝑦 =132
11
𝒚 = 𝟏𝟐
𝑥 = (36 − 𝑦
3) = (
36 − 11
3) =
24
3= 8
𝑥 𝑦
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𝑥
𝑥2 = 902 + 902
𝑥 = √8100 + 8100
𝑥 = √16200
𝑥 = 127.27 pies
𝑥 = 127.27 pies
𝐴 =(𝑃 𝑎)
2
𝑃 = 5𝑙 = (5)(6) = 30 cm
𝐴 =(30)(4)
2= 60 cm2
𝑥
90
90
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√10
√10
𝑐 𝑎
𝑐2 = 𝑎2 + 𝑏2 𝑎2 = 𝑏2 + 𝑐2
𝑐 = √𝑎2 + 𝑏2
𝑎 = √𝑐2 − 𝑏2
𝑏 = √𝑐2 − 𝑎2
𝑎
𝑏
𝑐 𝑎
𝑏
𝑐
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𝑐
𝑐 = √1 + 9
𝑐 = √10
√10
3
1
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𝑥
2=
𝑥 + 8
4
16
2
𝜋 ∙(4)2 ∙(16)
3−
𝜋 ∙(2)2 ∙(8)
3
1000
234.572
𝑉1
3𝜋ℎ(𝑟2 + 𝑟′2 + 𝑟𝑟′) 𝑟′
4 cm
8 cm
X cm
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24
5
12
15
8
15 años
14 años
13 años
12 años
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25
30°
45°
72°
288°
40
5=
360
𝑥𝑥 =
(360)(5)
40 =
1800
40 = 45
𝐷𝑚 =|11−12|+|11−11|+|11−11|+|11−10|+|11−11|
5=
2
5= 0.4
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𝐶𝐵𝐸 𝐴𝐵𝐷 𝐴𝐵𝐶 𝐴𝐵𝐷
𝐵𝐴𝐶 𝐴𝐶𝐵
𝐵𝐴𝐶 = 30° 𝑦 𝐴𝐶𝐵 = 30°
𝐵𝐴𝐶 = 60° 𝑦 𝐴𝐶𝐵 = 30°
𝐵𝐴𝐶 = 60° 𝑦 𝐴𝐶𝐵 = 60°
𝐵𝐴𝐶 = 30° 𝑦 𝐴𝐶𝐵 = 60°
𝑥 + 2𝑥 + 3𝑥 = 180 𝑥 = 30 𝐵𝐴𝐶 = 𝑥 𝐴𝐶𝐵 = 2𝑥
𝐵𝐴𝐶 = 30 𝐴𝐶𝐵 = 60
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1.5
𝑥=
0.6
20𝑥 =
(20)(1.5)
0.6=
30
0.6 = 50 𝑚
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𝑑𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑐𝑖ó𝑛 𝑓𝑖𝑛𝑎𝑙−𝑑𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑐𝑖ó𝑛 𝑖𝑛𝑖𝑐𝑖𝑎𝑙
𝑡𝑖𝑒𝑚𝑝𝑜 𝑓𝑖𝑛𝑎𝑙−𝑡𝑖𝑒𝑚𝑝𝑜 𝑖𝑛𝑖𝑐𝑖𝑎𝑙=
750−950
5−1=
−200
4= −50
𝑦 = −50𝑥 + 1000
𝑦 = −50(8) + 1000 = 600
(𝑥)
(𝑦)
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𝑥
𝑦 = 5𝑥 + 15
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Dulces vendid
Mes
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𝑚 𝑣𝑒𝑛𝑡𝑎 𝑓𝑖𝑛𝑎𝑙−𝑣𝑒𝑛𝑡𝑎 𝑖𝑛𝑖𝑐𝑖𝑎𝑙
𝑡𝑖𝑒𝑚𝑝𝑜 𝑓𝑖𝑛𝑎𝑙−𝑡𝑖𝑒𝑚𝑝𝑜 𝑖𝑛𝑖𝑐𝑖𝑎𝑙
𝑚
𝑚 =𝑣𝑒𝑛𝑡𝑎 𝑓𝑖𝑛𝑎𝑙−𝑣𝑒𝑛𝑡𝑎 𝑖𝑛𝑖𝑐𝑖𝑎𝑙
𝑡𝑖𝑒𝑚𝑝𝑜 𝑓𝑖𝑛𝑎𝑙−𝑡𝑖𝑒𝑚𝑝𝑜 𝑖𝑛𝑖𝑐𝑖𝑎𝑙
700−200
3−1=
500
2=
250
1
𝑚
𝑥
𝑥2 + 3𝑥 + 8
𝑥2 + 4𝑥 + 7
𝑥2 + 5𝑥 + 3
𝑥2 + 6𝑥 + 2
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𝑥2 + 6𝑥 + 2
o
o
o
𝑐
𝑐
𝑐
𝑐
𝑦 = 15𝑥
𝑦 𝑥
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𝑇 = 3𝑥 + 8
𝑥
25
40+
15
40= 1
𝑇 = 3𝑥 + 8
𝑇 = 3(1) + 8 = 3 + 8 = 11
𝑇 = 3(2) + 8 = 6 + 8 = 14
𝑇 = 3(3) + 8 = 9 + 8 = 17
𝑇 = 3(4) + 8 = 12 + 8 = 20
𝑇 = 3(5) + 8 = 15 + 8 = 23
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LAS TIC
SON
INCLUYEN Tecnologías
desarrolladas para INFLUYEN
PERMITEN
las tecnologías para almacenar
información y recuperarla después
la adquisición el almacenamiento la producción
el tratamiento la comunicación el registro
en el avance educativo
y social
DE LA INFORMACIÓN A
TRAVÉS DE IMÁGENES,
SONIDO Y VIDEOS
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24
5
35
7
5
1
60
10
2/3
1/4
1/3
2/7
7/7
4/12
1/12
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𝑦 𝑥
𝑦 = 150𝑥 − 1500
𝑦 = 150𝑥 + 1500
𝑦 = 1500𝑥 − 20 000
𝑦 = 1500𝑥 + 150
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𝐴 2𝑥2 +𝜋𝑥2
2
𝐴 3𝑥 +𝜋𝑥2
4
𝐴 3𝑥 +𝜋𝑥2
2
𝐴 2𝑥2 +𝜋𝑥2
4
𝑥 2𝑥
1/𝑥 1/2𝑥
2𝑥
𝑥
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𝑥 + 𝑦 = $5000
( 𝑥 + 0.1𝑥) + ( 𝑦 + 0.15𝑦) = $ 5560
(𝜋 = 3.14).
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𝐴 𝐵
𝐶
𝐴 𝐵
𝐶 = 5𝑥 − 4
2𝑥 + 11 4𝑥 + 8
𝐵 𝐴
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𝐵𝐶
𝐵𝐶
A)
B)
C)
D)
𝐴
𝐸
𝐶
𝐷
𝐵
50
40
20
𝑥
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0
10
20
30
40
50
60
Grande Mediana Chica
Tallas de uniformes
15
55
30
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𝑌 = 2700 + 38𝑥
𝑌 = 2700 + 8𝑥
𝑌 = 2700 + 8𝑥
𝑌 = 2700 + 38𝑥
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A) 3/10
1/2
C) 2/5
D) 7/10
Degustación
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𝑥
A) 𝑥2 + 3𝑥 + 8 B) 𝑥2 + 4𝑥 + 7 C) 𝑥2 + 6𝑥 + 2 D) 𝑥2 + 5𝑥 + 3
𝑡
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A) 𝑡2
B) 2𝑡
C) 2𝑡
D) 𝑡 + 2
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𝜋 = 3.14
9
1512
8
4
0
5
10
15
20
Núm Alumnos
Núm Alumnos
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3
1
39
11
3
2
37
13
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1/5
3/8 1/4
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1. Análisis gráfico de tendencia de obesidad en niños de 1er año de primaria. Mexicali, B.C.
2. Análisis gráfico de tendencia de obesidad en niños de 5to año de primaria. Mexicali, B.C.
3. Análisis gráfico de tendencia de obesidad en niñas de 1er año de primaria. Mexicali, B.C.
4. Análisis gráfico de tendencia de obesidad en niñas de 5to año de primaria. Mexicali, B.C.
DESNUTRICIÓN NORMAL SOBREPESO
CATEGORÍA POR PESO
DESNUTRICIÓN NORMAL SOBREPESO
CATEGORÍA POR PESO
DESNUTRICIÓN NORMAL SOBREPESO
CATEGORÍA POR PESO
DESNUTRICIÓN NORMAL SOBREPESO
CATEGORÍA POR PESO
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