CLAVE-101-2-M-1-00-2018_sK UNIVERSIDAD DE SAN CARLOS DE GUATEMALA FACULT FACULTAD DE INGENIE IN GENIERÍA RÍA DEPARTAMENTO DE MATEMÁTICA
CURSO:
Matemática Básica 1
SEMESTRE:
Primero
CÓDIGO DEL CURSO:
101
TIPO DE EXAMEN:
Segundo Examen Parcial
FECHA DE EXAMEN:
15 de marzo de 2018
HORA DE EXAMEN:
7:00 a.m.
RESOLVIÓ EL EXAMEN:
Eddy Brandon de León
REVISÓ EL EXAMEN:
Ing. Mario Rivera
COORDINADOR:
Ing. Arturo Samayoa
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TEMA 1 Solución:
Primero se rotulan los demás ángulos en la figura,
Se utilizan los teoremas: Teorema 1:
1 AB + CD φ = 2
Teorema 2:
(1)
1 AB 2
ψ =
(2)
Ángulo B AE No.
Ex Explicación
Operatoria
1
Ángulos suplementarios,
B AE
2
Poner valores numéricos,
B AE = 180
3
Simplificar,
+ B AF = 180 ◦
◦ − B AF = 180 ◦ − 75◦
B AE = 105
◦
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Ángulo φ No.
Ex Explicación
Operatoria
1
Util Utiliizar zar el teor teorem emaa (2), donde ψ = B AF = 70 ◦
AB = 2 ψ
2
Colocar el valor numérico
AB = 2 (75◦ ) = 150 ◦
3
Util Utiliizar zar el teor teorem emaa (1), donde CD = 70 ◦
φ = ( 70◦ + 150◦ ) = 220 ◦
1 2
Respuesta: φ = 110 ◦ Ángulo β No.
Ex Explicación
Operatoria
1
Por ángulo ánguloss suplem suplement entaarios,
2
Simplificar,
β + φ = 180 ◦
β = 180 ◦
− φ = 180◦ − 110◦
Respuesta: β = 70 ◦ Ángulo θ No.
Ex Explicación
Operatoria 1 AB θ= 2
1
Utilizar el teorema (2),
2
U i li
l
éi
θ
1
( 150◦ )
75 ◦
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TEMA 2 Encuentre la ecuación de la circunferencia que pasa por el punto (0, 3) y cuyo centro es el punto de intersección de las rectas − x + y = 3 y − x − y = 1.
Solución:
No.
Ex Explicación
1
Se resuelve el sistema de ecuaciones para encontrar el centro.
2
Sumar las dos ecuaciones,
3
Operatoria
−
− −
−2x = 4 ⇒ x = −2
Sustituir x en alguna de las ecuaciones,
4
El centro ( h, k ) es
5
Hallar Hallar el radio radio con la fór fór-mula de la distancia,
6
Poner valores numéricos,
7
Utilizar Utilizar la ecuación ecuación general de la circunferencia,
7
Sustituir Sustituir los valores valores hallahallados, Respuesta: La Respuesta: La ecuación de la circunferencia es:
x + y = 3 x y = 1
y = 3 + x = 3
− 2 = 1
( h, k ) = ( 2, 1)
−
r 2 = ( x1
r2 = (0
(x
(x
− h)2 + ( y1 − k )2
− (−2))2 + (3 − 1)2 = 8
− h)2 + ( y − k )2 = r 2
− (−2))2 + ( y − 1)2 = 8.
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TEMA 3 Un tanque tiene 10 metros de largo y su sección transversal es un semicírculo de 4 metros de diámetro, como se muestra en la figura. Calcule:
a. El volumen total del tanque. Solución:
Datos: L = 10 m r = 2 m V = A L
·
No.
Ex Explicación
1
Calcular el área,
2
Colocar los valores,
3
Calcular el volumen,
Operatoria A =
A =
1 2 π r 2
1 m2 π (2m)2 = 2 π m 2
V = (2π m m2 )( 10 m) = 20 π m m3
Respuesta: El Respuesta: El volumen total es 20 m 3, aproximadamente 62.8 m 3 .
b. Volum olumeen de agu agua cuand uandoo la altu altura ra de la supe superfi rfici ciee del del agu agua es h = 1. Solución:
El área sombreada es la diferencia entre el área del sector circular y el triángulo isósceles.
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A = ASC
− A
Notar que el área del triángulo isósceles es el mismo que el del equilátero que se muesmues tra a continuación.
No.
Ex Explicación
1
Área del sector circular, circular,
2
El ángulo es 120◦ = 2π /3 /3 radianes,
3
Área del sector, sector,
4
Área del triángulo triángulo equiláequilátero,
Operatoria 1 2
ASC = θ r2
ASC
1 2π = (2 m ) 2 = 2 3
ASC =
A =
4π 2 m 3
√ 3 4
s2
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Respuesta: El Respuesta: El volumen total cuando h = 1 es 4π √ 10 3 m3 − 3
c. El área del espejo de agua cuando h = 1. Solución:
A = b L
·
No. 1
Ex Explicación Calcular b. Se utiliza un triángu triángulo lo rectá rectángul nguloo con catetos 1 m y b /2 e hipotehipotenusa 2 m.
2
Despejar b
3
Calcular el área,
Operatoria (2)2 = (1)2 + ( b/2)2
√
b = 2 3 m
√
√
A = (2 3 m)(10 m) = 20 3 m2
Respuesta: El Respuesta: El área del espejo de agua cuando h = 1 es
√
20 3 m2
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TEMA 4 Calcule el área sombreada de la figura que se muestra, si los lados del cuadrado miden 6 metros de longitud. Solución:
Se hacen las relaciones geométricas que se muestran en la siguiente figura:
No. 1
2
Ex Explicación
Calcular x por relaciones de triángulos semejantes, Área sombreada,
Operatoria 6−x x
=
A = ASC
6 ⇒ x = 2 3
− 2 · A − A
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No.
Ex Explicación
Operatoria
5
Área del triángulo,
6
Colocar los valores de x ,
7
Área del cuadrado,
8
Área sombreada,
m2 A = 9 π m
9
Área sombreada,
A = ( = ( 9π
A =
1 x (6 − x ) 2
1 2
− 2) m2 = 4 m2
A = (2)(6
A = x 2 = 4 m 2
Respuesta: El Respuesta: El área sombreada es: = ( 9π A = (
− 12) m2
− 2(4 m 2 ) − 4 m 2
− 12) m2 ≈ 16.3 m2
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TEMA 5 Dada la gráfica de la función de f ( x) que se muestra.
a. Determine el dominio y el rango de la función. Solución:
En base a la gráfica, el dominio es:
−3 ≤ x ≤ 7 Y el rango es:
0 ≤ y ≤ 3
b. Grafique f ( x − 2) , f (2x ) y 2 f ( x ). Solución:
La función f ( x − 2) es la traslación de f ( x) dos unidades a la derecha.
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La función 2 f ( x) es el estiramiento vertical de f ( x) dos unidades a la derecha.