Ejercicios en Geogebra Colaborativo Yuselli Ardila

7/21/2019 Ejercicios en Geogebra Colaborativo Yuselli Ardila

http://slidepdf.com/reader/full/ejercicios-en-geogebra-colaborativo-yuselli-ardila 1/15

UNIVERSIDAD NACIONAL ABIERTA Y A

DISTANCIA

ALGEBRA, TRIGONOMETRIA Y GEOMETRIA ANALITICA

APORTE INDIVIDUAL

PRESENTADO A:

DIBER ALBEIRO VAQUIRO PLAZAS

ELABORADO POR:

YUSELLI ARDILA TRIANA

CÓD. 1100.954.970

GRUPO: 301301_ 44

UNIVERSIDAD NACIONAL ABIERTA Y A DISTANCIA ! UNAD !CEAD!BUCARA"ANGA

30 DE "ARZO DEL 015




DISTANCIA

1. D#$#%&'(# #) *+&'('+ *# ), -(/'(:

f ( x )=√ 4−3

x2−4

x2−4=0 4 x−3≥0

x2=4 4 x ≥3

x=±√ 4 x ≥ 3

4

x=±2

E) *+&'('+ # R− {2,−2 } E) *+&'('+ #: [34 ,∞) ,−{2 }

2. D#$#%&'(# #) %,(2+ *# ), -(/'(

f ( x )= x+6√ x−5

y= x+6

√ x−5

( y )2=( x+6

√ x−5 )2

y2= ( x+6 )2

(√ x−5 )2

y2=

( x+6 )2

x−5




DISTANCIA

( x−5 ) y 2= x2+12 x+36

x y 2−5 y2= x2+12 x+36

− x2+ x y

2−12 x−36−5 y2=0 (−1 )

se multiplic por−1 todalaecuacion

x2− x y

2+12 x+36+5 y

2=0

x2+ x (12− y

2 )+36+5 y2

x=−b ±√ b2−4ac

2a

a=1; b=12− y2

; c=36+5 y2

x=−(12− y

2 )±√ (12− y2 )2−4 (1 ) (36+5 y

2 )2(1)

x= y

2−12±√ 144−24 y2+ y

4−144−20 y2

2

x= y

2−12±√ y4−44 y 2

2

y4−44 y

2≥0

y2 ( y2−44 )≥0

y2=0∨ y

2−44≥0

y2

≥44




DISTANCIA

y ≥ ±√ 44

Rango f(x)= ±√ 44

3. D,*, ), -(/'+(# f ( x )=

2 x−1

2 ; g ( x )= x2+2

a) (f + g)(2)

2 x−1

2+

x2+2

1

2 x−1+2 x2+4

2

f (2)=2 (2)2+2(2 )+3

2

8+4+32

f (2 )=15

2

b) (f-g)(2)

2 x−1

2−( x

2+21 )

2 x−1− (2 x2+4 )

2

f (2 )=2 (2 )−1−(2 (2)2+4)

2




DISTANCIA

4−1−8−4

2

f (2)=−9

2

c) (f*g)(3)

2 x−1

2∗ x

2+2

1

2 x3− x

2+4 x−2

2

f (3 )=2 (3 )3−(3 )2+4 (3 )−2

2

54−9+12−2

2

f (3 )=55

2

d) (f/g)(-3)

( 2 x−1

2 ) x

2+21

2 x−1

2 x2+4

f (−3 )= 2 (−3 )−1

2 (−3 )2+4




DISTANCIA

−6−1

18+4

f (−3 )=−7

22

4. D,*, ), -(/'+(# f ( x )=√ x+2 g ( x )= x2−1 D#$#%&'(#

, 6- + 26

(f ∘g )=√ x2−1+2

¿√ x2+1

(f ∘g )= x+1

8 62 + -6

( g∘ f )=(√ x+2 )2−1

¿ x+2−1

(g∘ f )= x+1

/ 6- 26

(f +g )=√ x+2+ x2−1

(f +g )= x2+√ x+2−1

d) 6 - 26

(f −g )=√ x+2−( x2−1 )

(f −g)=√ x+2− x2+1

5. V#%'-';# ), '2'#($# '*#($'*,*<




DISTANCIA

2 senxcosx−cosx

1−senx+se n2 x−co s

2 x=cotx

cosx (2 senx−1 )

se n2 x−senx+se n

2 x=cotx

cosx (2 senx−1)

2 se n2 x−senx

=cotx

cosx (2 senx−1)senx (2 senx−1)

=cotx

cosx

senx=cotx

cotx=cotx

6. D#&#$%# ), '2'#($# '*#($'*,* ,(*+ ), *#-'('/'+(# *# ), *#%'=,*,'*#($'*,*# >'?#%8)'/, -(*,&#($,)#:

tanhx1−tanh

2 x=senh

2 x

tanhx

sech2 x=senh

2 x

tanhx

1

1

cosh2 x

=senh2 x

tanhx∗cosh2 x=sen h

2 x

senhx

coshx∗cosh

2 x=senh

2 x




DISTANCIA

senhx∗coshx=senh2 x

N+ >,@ '2,)*,* #( #$, '*#($'*,*#.

7. U( ,='( ;# ?,, 0 &#$%+ +8%# ), ,+$#, *# ( #*'-'/'+ *# 40 &#$%+ *#,)$%, *#/'#(*# 00 &#$%+ >,$, $+/,% $'#%%, #( ( )2,% A. C+( ;# (2)+*#/#(*' QF *'$,(/', >,@ #($%# ), 8,# *#) #*'-'/'+ @ #) )2,% A

,

?

60

200

60

40

A

x




DISTANCIA

cosθ=100m

200m

cosθ=0,5

θ=cos−1❑(0,5 )

θ=60°

∠=90°−60°

∠=30° el angulo decon el descendio fue de30°

8

a2+b

2=c2

x=√ (200m)2−(100m )

2

x=√ 40000m2−10000m

2

x=√ 30000m2

x=173,205mes ladistancia entre labase del edificio y ellugar A

. D#*# )+ ,)$+ *# ( 2)+8+ # +8#%=, (, /'*,* A /+( ( (2)+ *# 50H @ +$%,/'*,* B '$,*, ,) +$%+ ),*+ @ #( )(#, %#/$, /+( ( (2)+ *# 0H. S,8'#(*+;# #) 2)+8+ # #(/#($%, , (, *'$,(/', *# J')&#$%+ *# ), /'*,* A @ , 4J')&#$%+ *# ), /'*,* B. D#$#%&'(# ), *'$,(/', #($%# ), /'*,*# A @ B.

6050




DISTANCIA

180°=50°+60°− x

x=180°−50°−60°

x=70° esel valor del angulo

a2=b

2+c2−2bc∗cosθ teoremadel coseno

x2=(6km )2+ (4 km)2−2 (6km ) (4 km )∗cos (70° )

x2=36k m

2+16k m2−16,41k m

2

x2=35,59k m

2

x=√ 35,59k m2

x=5,96 km esla distanciaentre la ciudad A y

9. E(/#($%# #) =,)+% *# ;# ,$'-,/# ), '2'#($# #/,/'( ?,%, (2)+ #($%#0HK K 30H

70

6

X

4 km

BA




DISTANCIA

2cos2 x+√ 3 senx=−1

2cos2 x+√ 3 senx+1=0

2(1−sen2 x )+√ 3 senx+1=0

2−2 se n2 x+√ 3 senx+1=0(−1)

−2+2 se n2 x−√ 3senx−1=0

2se n2 x−√ 3 senx−3=0

a=2; b=−√ 3 ; c=−3

senx=−b ±√ b2−4ac

2a

senx=−(√ 3) ±√ (√ 3)

2

−4 (2) (−3 )2(2 )

senx=√ 3±√ 3+24

4

senx=√ 3±√ 27

4

senx=√ 3+√ 27

4senx=√ 3−√ 27

4

senx=√ 3 senx=−0,866

x=sin−1 (√ 3 ) x=sin

−1 (0,866 )

x=no tiene x=−60

x=360−60




DISTANCIA

x=300es elangulo !ue satisface laecuaci"n

G%,-'/, *# ##%/'/+ 1

G%,-'/, *# ##%/'/'+




DISTANCIA

G%,-'/, *# ##%/'/'+ 3

G%,-'/, *# ##%/'/'+ 4

G%,-'/, *# ##%/'/'+ 5




DISTANCIA

G%,-'/, *# ##%/'/'+

G%,-'/, *# ##%/'/'+ 9




DISTANCIA

Ejercicios en Geogebra Colaborativo Yuselli Ardila

Documents

Transcript of Ejercicios en Geogebra Colaborativo Yuselli Ardila