ClubEnsayos.com - Ensayos de Calidad, Tareas y Monografias
Buscar

Progresion geometrica


Enviado por   •  19 de Abril de 2015  •  1.612 Palabras (7 Páginas)  •  268 Visitas

Página 1 de 7

EJERCICIO No. 1 Página 8

a^(2 )* a^(5 )= a^(2+5)= a^7

a^3* a^(8 )= a^(3+8)= a^11

a^2* a^4* a^5=a^(2+4+5)=a^11

b^ * b^3* b^2=b^(1+3+2)=b^6

〖(3b〗^ )* (5b^2 )* ( 〖6b〗^3 )=(3*b)* (5* b^2 )*( 6 * b^3 )=90b^(1+2+3)= 〖90b〗^6

c^3/c^8 = c^(3-8)= c^(-5)

h). (a^3* a^4)/a^5 = a^(3+4)/a^5 = a^7/a^5 = a^(7-5)= a^2

j) x^2* (x^5 ) 〖^3〗= x^2* (x^(5*3) )= x^2* (x^15 )= x^(2+15)= x^17

k) ((〖2y)〗^(2 )* (〖4y〗^3 ) 〖^4〗)/((2〖y)〗^7 )=(〖(2〗^2 x y^2) * (4^(4 ) x y^(3*4)))/(2^7 x 〖 y〗^7 )= ((〖4 x y〗^2 ) * (4^4 x y^12 ))/〖128 x y〗^7 = ((〖4 x y〗^2 ) * (〖256 x y〗^12 ))/〖128 x y〗^7 =〖1,024 * y〗^(2+12)/〖128 x y〗^7 =( 〖512 * y〗^14)/〖128 x y〗^7 =8 x y^(14-7)=8y^7

m) (a^2 b^3 ) 〖^4〗=(a^(2*4) b^(3*4) )= a^8 b^12

q) (1.05)^4 〖(1.05)〗^(10 )= 〖1.05〗^(4+10)= 〖1.05〗^14=1.979931599

r) (〖(1.30)〗^2 〖(1.30)〗^10 〖(1.30)〗^20)/1.30= 〖1.30〗^(2+10+20)/1.30= 〖1.30〗^(32-1)= 〖1.30〗^31=3,405.99

EJERCICIO No. 2 Página 8

x^0=1

a^0 b^3= 〖1b〗^3

a^(1/3) x a^(1/2)=a^(5/6)= √(6&a^5 )

e.) (a^(1/4) a^(3/5))/a^(1/2) = a^(17/20)/a^(1/2) = a^(17/20 - 1/2 )= a^(7/20)= √(20&a^7 )

h) (〖9x〗^(-2) ) 〖^(-5)〗=〖9^(-5)* x〗^(-2-5)= 9^(-5)*x^10= 1/9^5 *x^10=0.00001693508781*x^10

k) ((x^(2/3))〖^3〗)/x^(-2) = x^(2/3 x 3/1)/x^(-2) = x^2/x^(-2) = x^(2-(-2))=x^4

n) (a^(-3)/b^(-6) ) 〖^(-1/4)〗= (〖ab〗^(-3-6) ) 〖^(-1/4)〗=(ab^(-9) ) 〖^(-1/4)〗= 〖ab〗^(-9 *1/4)= 〖ab〗^(9/4)= ∜(〖ab〗^9 )

EJERCICIO No. 3 Página 9

√(x^3 )= x^(3/2)

(∛(x^2 ))(√(x^3 ))= x^(2/3) .x^(3/2)= x^(2 1/6)

(b^(2 ) x √b)/√(b^(3 ) )= (b^(2 ) x b^(1/2))/b^(3/2) = b^(2+ 1/2)/b^(3/2) = b^(3/2)/b^(3/2) = b^(3/2-3/2)=b

EJERCICIO No. 4 Página 9

√32=5.6568542

∛(〖25〗^5 )= ∛9,765,625=213.75

∜0.485 ∛0.36= 0.834517473*0.71137866=0.593657921

e.) ((1+0.18) 〖^4〗- 1)/0.18= (〖1.18〗^4- 1)/0.18=(1.93877776-1)/0.18= 0.93877776/0.18=5.215432

EJERCICIO No. 5 Página 9

100 (1+i)² = 200

(1+i)² = 200 / 100

(1+i)² = 2

1+i = √(2 )=2^(1/2)=1.414213562

i = 1.414213562 – 1

i = 0.414213562

5,000 (1+i)³ = 1,500

(1+i)³ = 1,500 / 5,000

(1+i)³ = 0.30

1+i = ∛0.30=〖0.30〗^(1/3)=0.66943295

i = 0.66943295 – 1

i = - 0.330567050

1,250 (1+i)〖^60〗 = 25,000

(1+i)〖^60〗 = 25,000 / 1,250

(1+i)〖^60〗 = 20

1+i = √(60&20)= 〖20〗^(1/60)=1.051196323

i = 1.051196323 – 1

i = 0.051196323

50,000 (1+i)〖^20〗=3,000

(1+i)〖^(-20)〗=3,000/50,000

(1+i)〖^(-20)〗 = 0.06

1+i = √(-20&0.06)= 〖0.06〗^((-1)/20)=1.151045357

i = 1.151045357 – 1

i = 0.151045357

10,000 (1+i)〖^(-4)〗=6,000

(1+i)〖^(-4)〗=6,000/10,000

(1+i)〖^(-4)〗 = 0.60

1+i = √(-4&0.60)= 〖0.60〗^((-1)/4)=1.136219366

i = 1.136219366 – 1

i = 0.136219366

EJERCICIO No. 16 Página 24

Determinar el último término y la suma de las progresiones siguientes:

11, 23, 35 . . . . . 12 Términos

U= ti + ( n- 1 ) d

U= 11 + (12-1)12

U= 11 + (11) 12

U= 11+132

U= 143

S= n/2 [ 2 ti + (n-1) d ]

S= 12/2 [ 2 (11) + (12-1) 12]

S= 6 [ 22 + (11) 12 ]

S= 6 ( 154)

S= 924

5, -3, -11. . . . . . . . 10 Términos

U= ti + ( n- 1 ) d

U= 5 + (10-1) -8

U= 5 + (9) -8

U= 5 + -72

U= -67

S= n/2 [ 2 ti + (n-1) d ]

S= 10/2 [ 2 (5) + (10-1) -8]

S= 5 [ 10 + (9) -8 ]

S= 6 (-62)

S= - 310

½, 5/8, 3/4, . . . . . . . . . 7 Términos

U= ti + ( n- 1 ) d

U= ½ + (7-1) -1/8

U= ½ + (6) 1/8

U= ½ + ¾

U= 1 ¼

S= n/2 [ 2 ti + (n-1) d ]

S= 7/2 [ 2 (1/2) + (7-1) 1/8]

S= 3.5 [ 1 + (6) 1/8 ]

S= 3.5 (1 3/4)

S= 6 1/8

¼, 1/12, -1/12, . . . . . . . . . 20 Términos

U= ti + ( n- 1 ) d

U= ¼ + (20-1) -1/6

U= ¼ + (19) -1/6

U= ¼ - 3 1/6

U= -2 11/12

S= n/2 [ 2 ti + (n-1) d ]

S= 20/2 [ 2 (1/4) + (20-1) -1/6]

S= 10 [ ½ + (19) -1/6 ]

S= 10 (-3 1/6)

S= - 31 2/3

1.00, 1.05, 1.10, . . . . . . . . . 12 Términos

U= ti + ( n- 1 ) d

U= 1 + (12-1) 0.05

U= 1 + (11) 0.05

U= 1 + 0.55

U= 1.55

S= n/2 [ 2 ti + (n-1) d ]

S= 12/2 [ 2 (1) + (12-1) 0.05]

S= 6 [ 2 + (11) 0.05]

S= 6 (2.55)

S= 15.30

EJERCICIO No. 17 Página 24

Determine la suma de:

Los números pares de 1 a 100

S= n/2 [ 2 ti + (n-1) d ]

S= 50/2 [ 2 (2) + (50-1) 2]

S= 25 [ 4 + (49) 2]

S= 25 (102)

S= 2,550

Los números nones de 9 a 100

S= n/2 [ 2 ti + (n-1) d ]

S= 46/2 [ 2 (9) + (46-1) 2]

S= 23 [ 18 + (45) 2]

S= 23 (108)

S= 2,484

Los números múltiplos

...

Descargar como (para miembros actualizados) txt (9 Kb)
Leer 6 páginas más »
Disponible sólo en Clubensayos.com