Guia N. 1 Mercadeo
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GUIA N. 1 TALLER DE REPASO DOCENTE MEDIADOR: CARLOS OBANDO
1. Realizar las siguientes operaciones A. 1/2 + 1/2 = 1 B. 1/2 + 1 = 3/2 C. 1/2 - 1/4 = 1/4 D. 10/4 ÷ 2 = 5/4 E. (1/2 + 4/2) + 1/2 = 3 F. 1/10 ÷ 4/3 = 3/40 G. [(4/4 - 1/2) * 3/10) / 6/5] = 1/8 H. 4 1 3 I. 2 − * 2 2500 2 4 2 4 + = 10 1 = 3 9 6 6 5
2.
1 1 6 3 + 3 2 3 5 Calcular el valor de la incógnita X= + = = 2 3 6 6
12 +
3 2 =1
27 24 + 3 2 = X =1 2
X
2X +3X −16 = 20 2X +3X = 20 +16 5X = 36 36 X= 5 X = 7.2
18 X −10 =
20 2
20 +10 2 20 + 20 40 20 18 X = = = 2 2 1 20 X = 18 10 X = 9 18 X =
X = 1.
27 2
X=
27 2
3 X + 4 = 29
3 X = 29 − 4 3 X = 25 25 X = 3 X = 8.33
X + 4 = 16 X = 16 − 4 X = 12 3. Para Xcada una de las funciones determinar = 12
2
2
2
A.
3. 4 − X + 6 f ( XX) == X f ( 0) f (−2) 3
2
f (a )
f ( 0) = x 3 − x 2 + 6
f (−2) = −2 − ( −2) + 6 f (−2) = −8 − 4 + 6 f (−2) = −12 + 6 f (−2) = −6 3
f ( 0) = 0 3 − 0 2 + 6 f ( 0) = 6
2
f (a ) = a 3 − (a) 2 + 6 f (a ) = a − a + 6 f (a ) = a + 6
3 x −8 f ( x) = x− 1 f (3) f (−1) 2
B.
f ( x − 2)
3. 32 − 8 f (3) = 3− 1 92 − 8 f (3) = 2 81 − 8 f (3) = 2 73 f (3) = 2 f (3) = 36.5
F ( x −2 ) =
(3( x −2)
2
−8
)
( x −2 ) −1 ( 3( x −2 )( x −2 ) −8) F ( x −2 ) = ( x −2 ) −1 F ( x −2 ) = F ( x −2 ) = F ( x −2 ) =
(3(x
2
(3(x
2
(3 x
2
)
2 x −2 x +4 −8 ( x −2) −1
)
−4 x +4 −8
( x −2 ) −1
)
−12 x +12 −8
( x −2 ) −1
3 x 2 −12 x +4 F ( x −2 ) = ( x −2) −1 3 x 2 −12 x +4 F ( x −2 ) = x −3
)
)
3.( − 1) 2 − 8 f (− 1) = (− 1) − 1 − 32 − 8 f (− 1) = − 2 9− 8 f (− 1) = − 2 1 f (− 1) = − 2 f (− 1) = − 0.5
C.
f (1)
f (−2)
f ( x) = 3x − x 2 f (1) = 3(1) −12 f (1) = 3 −1 f (1) = 2
f ( x) = 3 x − x 2 1 1 1 f ( ) = 3( ) − ( ) 2 h h h 2 1 3 1 f( )= − h h h 1 3h 2 − 1h f( )= h h3
F ( a ) = 3( a ) − a 2 F ( a ) = 3a − a 2
f ( x) = 3x − x 2
f (a) f ( x) = 3x − x 2 f (−2) = 3(−2) − (−2) 2 f (−2) = −6 − 4 f (−2) = −10
1. Realizar las siguientes operaciones A. 1/2 + 1/2 = 1 B. 1/2 + 1 = 3/2 C. 1/2 - 1/4 = 1/4 D. 10/4 ÷ 2 = 5/4 E. (1/2 + 4/2) + 1/2 = 3 F. 1/10 ÷ 4/3 = 3/40 G. [(4/4 - 1/2) * 3/10) / 6/5] = 1/8 H. 4 1 3 I. 2 − * 2 2500 2 4 2 4 + = 10 1 = 3 9 6 6 5
2.
1 1 6 3 + 3 2 3 5 Calcular el valor de la incógnita X= + = = 2 3 6 6
12 +
3 2 =1
27 24 + 3 2 = X =1 2
X
2X +3X −16 = 20 2X +3X = 20 +16 5X = 36 36 X= 5 X = 7.2
18 X −10 =
20 2
20 +10 2 20 + 20 40 20 18 X = = = 2 2 1 20 X = 18 10 X = 9 18 X =
X = 1.
27 2
X=
27 2
3 X + 4 = 29
3 X = 29 − 4 3 X = 25 25 X = 3 X = 8.33
X + 4 = 16 X = 16 − 4 X = 12 3. Para Xcada una de las funciones determinar = 12
2
2
2
A.
3. 4 − X + 6 f ( XX) == X f ( 0) f (−2) 3
2
f (a )
f ( 0) = x 3 − x 2 + 6
f (−2) = −2 − ( −2) + 6 f (−2) = −8 − 4 + 6 f (−2) = −12 + 6 f (−2) = −6 3
f ( 0) = 0 3 − 0 2 + 6 f ( 0) = 6
2
f (a ) = a 3 − (a) 2 + 6 f (a ) = a − a + 6 f (a ) = a + 6
3 x −8 f ( x) = x− 1 f (3) f (−1) 2
B.
f ( x − 2)
3. 32 − 8 f (3) = 3− 1 92 − 8 f (3) = 2 81 − 8 f (3) = 2 73 f (3) = 2 f (3) = 36.5
F ( x −2 ) =
(3( x −2)
2
−8
)
( x −2 ) −1 ( 3( x −2 )( x −2 ) −8) F ( x −2 ) = ( x −2 ) −1 F ( x −2 ) = F ( x −2 ) = F ( x −2 ) =
(3(x
2
(3(x
2
(3 x
2
)
2 x −2 x +4 −8 ( x −2) −1
)
−4 x +4 −8
( x −2 ) −1
)
−12 x +12 −8
( x −2 ) −1
3 x 2 −12 x +4 F ( x −2 ) = ( x −2) −1 3 x 2 −12 x +4 F ( x −2 ) = x −3
)
)
3.( − 1) 2 − 8 f (− 1) = (− 1) − 1 − 32 − 8 f (− 1) = − 2 9− 8 f (− 1) = − 2 1 f (− 1) = − 2 f (− 1) = − 0.5
C.
f (1)
f (−2)
f ( x) = 3x − x 2 f (1) = 3(1) −12 f (1) = 3 −1 f (1) = 2
f ( x) = 3 x − x 2 1 1 1 f ( ) = 3( ) − ( ) 2 h h h 2 1 3 1 f( )= − h h h 1 3h 2 − 1h f( )= h h3
F ( a ) = 3( a ) − a 2 F ( a ) = 3a − a 2
f ( x) = 3x − x 2
f (a) f ( x) = 3x − x 2 f (−2) = 3(−2) − (−2) 2 f (−2) = −6 − 4 f (−2) = −10
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