Solucion Primer Taller Mercadeo
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SOLUCION PRIMER TALLER 1.
1 1 2+2 4 + = = =1 2 2 4 4
2.
1 1 1+ 2 3 + = = 2 1 2 2
3.
1 1 4−2 2 1 − = = = 2 4 8 8 4
4.
1 1 1 + = 8 9 72
5.
10 4 2 = 10 = 5 1 8 4
6. 1 + 4 + 1 = 2 + 8 + 1 = 10 + 1 = 20 + 4 = 24 = 3 = 3 8 8 1 2 2 2 4 2 4 2 7.
1 10 4 = 3 3 40
8. 4 − 1 + 3 4 2 10 = 6 5 2
8 − 4 3 8 + 10 = 6 5 2
2
64 4 3 40 + 24 80 8 + 10 80 = = 6 = 320 = 80 = 40 = 20 = 16 = 2 6 6 5 480 120 60 30 15 3 5 5 2
9. 4 2 + 2 = 16 + 2 = 48 + 2 = 50 = 2500 3 3 3 3 9
X=
1 16 + 2 3
CALCULA EL VALOR DE LA INCOGNITA (X) 3 + 32 X= 6 X=
35 6
1.
2.
3 12 + 2 =1 X 3 12 + 2 = =X 1 24 + 3 2 = =X 1 27 27 20 = 2 = 18 x1 − 10 2= 2
3. 2 x + 3 x − 16 = 20 20 + 10 ==218 x x+ =3x 2= 20 + 16 = 5 x = 36 36 == x18=x = 20 + 20 5 2 4. 3 x + 4 = 29 40 = 18 x = = 3 x = 292− 4 = 3 x = 25 25 40 x= = x =3 2 18
5.
=x=
40 36
=x=
20 18
=x=
10 9
6.
( x ) 2 + 4 = 16 = ( x ) = 16 − 4 2
= ( x ) = 12 2
= x = 12 = x = 3,48 DETERMINAR FUNCIONES 1.
f ( 0) = ( x ) − ( x ) + 6 3
2
• = f ( 0 ) = 03 − 0 2 + 6 3 2 f ( − 2 ) = ( − 2) − ( − 2 ) + 6 •
= f ( 0) = 6 = f ( − 2 ) = ( − 8) − ( − 4 ) + 6 = f ( − 2) = −8 − 4 + 6 = f ( − 2) = −12 + 6 = f ( − 2) = −6
•
f ( a) = ( a) − ( a) + 6 3
2
= f ( a) = a + 6 2. •
f ( 3) =
(3( 3)
−8 3 −1
= f ( 3) = f ( 3) =
2
)
( 3( 9) − 8) 3 −1
27 − 8 2
19 2 2 3( − 1) − 8 f ( − 1) = −1−1 2 3( x − 2 ) − 8 f ( x − 2) = ( 3(1( )x−−82) ) − 1 = f ( − 1) = −1−1 ( 3( x − 2)( x − 2) − 8) = f ( x − 2) = 3 − 8 ( x − 2) − 1 = f ( − 1) = −2 3 x 2 − 2x − 2x + 4 − 8 = f ( x − 2) = − 5 5 ( x − 2) − 1 = f ( − 1) = = −2 2 = f ( 3) =
•
(
)
(
)
((
= f ( x − 2) = •
(3( x
) )
) )
− 4x + 4 − 8 ( x − 2) − 1
2
= f ( x − 2) =
3 x 2 − 12 x + 12 − 8 ( x − 2) − 1
= f ( x − 2) =
3 x 2 − 12 x + 4 x − 2 −1
= f ( x − 2) =
3 x 2 − 12 x + 4 x−3
3. •
f (1) = 3(1) − (1)
2
= f (1) = 3 − 1 = f (1) = 2 •
f ( − 2 ) = 3( − 2 ) − ( − 2 ) = f ( − 2) = −6 − ( 4) = f ( − 2) = −6 − 4 = f ( − 2) = −10 1 1 1 f = 3 − h h h 1 3 1 = f = − 2 h h h
2
2
1 1 2+2 4 + = = =1 2 2 4 4
2.
1 1 1+ 2 3 + = = 2 1 2 2
3.
1 1 4−2 2 1 − = = = 2 4 8 8 4
4.
1 1 1 + = 8 9 72
5.
10 4 2 = 10 = 5 1 8 4
6. 1 + 4 + 1 = 2 + 8 + 1 = 10 + 1 = 20 + 4 = 24 = 3 = 3 8 8 1 2 2 2 4 2 4 2 7.
1 10 4 = 3 3 40
8. 4 − 1 + 3 4 2 10 = 6 5 2
8 − 4 3 8 + 10 = 6 5 2
2
64 4 3 40 + 24 80 8 + 10 80 = = 6 = 320 = 80 = 40 = 20 = 16 = 2 6 6 5 480 120 60 30 15 3 5 5 2
9. 4 2 + 2 = 16 + 2 = 48 + 2 = 50 = 2500 3 3 3 3 9
X=
1 16 + 2 3
CALCULA EL VALOR DE LA INCOGNITA (X) 3 + 32 X= 6 X=
35 6
1.
2.
3 12 + 2 =1 X 3 12 + 2 = =X 1 24 + 3 2 = =X 1 27 27 20 = 2 = 18 x1 − 10 2= 2
3. 2 x + 3 x − 16 = 20 20 + 10 ==218 x x+ =3x 2= 20 + 16 = 5 x = 36 36 == x18=x = 20 + 20 5 2 4. 3 x + 4 = 29 40 = 18 x = = 3 x = 292− 4 = 3 x = 25 25 40 x= = x =3 2 18
5.
=x=
40 36
=x=
20 18
=x=
10 9
6.
( x ) 2 + 4 = 16 = ( x ) = 16 − 4 2
= ( x ) = 12 2
= x = 12 = x = 3,48 DETERMINAR FUNCIONES 1.
f ( 0) = ( x ) − ( x ) + 6 3
2
• = f ( 0 ) = 03 − 0 2 + 6 3 2 f ( − 2 ) = ( − 2) − ( − 2 ) + 6 •
= f ( 0) = 6 = f ( − 2 ) = ( − 8) − ( − 4 ) + 6 = f ( − 2) = −8 − 4 + 6 = f ( − 2) = −12 + 6 = f ( − 2) = −6
•
f ( a) = ( a) − ( a) + 6 3
2
= f ( a) = a + 6 2. •
f ( 3) =
(3( 3)
−8 3 −1
= f ( 3) = f ( 3) =
2
)
( 3( 9) − 8) 3 −1
27 − 8 2
19 2 2 3( − 1) − 8 f ( − 1) = −1−1 2 3( x − 2 ) − 8 f ( x − 2) = ( 3(1( )x−−82) ) − 1 = f ( − 1) = −1−1 ( 3( x − 2)( x − 2) − 8) = f ( x − 2) = 3 − 8 ( x − 2) − 1 = f ( − 1) = −2 3 x 2 − 2x − 2x + 4 − 8 = f ( x − 2) = − 5 5 ( x − 2) − 1 = f ( − 1) = = −2 2 = f ( 3) =
•
(
)
(
)
((
= f ( x − 2) = •
(3( x
) )
) )
− 4x + 4 − 8 ( x − 2) − 1
2
= f ( x − 2) =
3 x 2 − 12 x + 12 − 8 ( x − 2) − 1
= f ( x − 2) =
3 x 2 − 12 x + 4 x − 2 −1
= f ( x − 2) =
3 x 2 − 12 x + 4 x−3
3. •
f (1) = 3(1) − (1)
2
= f (1) = 3 − 1 = f (1) = 2 •
f ( − 2 ) = 3( − 2 ) − ( − 2 ) = f ( − 2) = −6 − ( 4) = f ( − 2) = −6 − 4 = f ( − 2) = −10 1 1 1 f = 3 − h h h 1 3 1 = f = − 2 h h h
2
2
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