• Asignatura: Física
  • Autor: Dayhi
  • hace 9 años

Expresar en notacion escrita, notacion cientifica, notacion de calculador y notacion de informatica las siguientes operaciones

(4,48*103)+(2,4848*103)= (7,6548*10-6)+(4,6*10-6)+(9,32*10-6)= (4,8*103)-(2,8*103)= (3,8*103)+(4,8*104)+(5,78*104)= (7,2*105)+(8,3*10-2)= (3*105)*(3,9*1013)= (2,5*10-3)(3,2*10-5)= (6*10-5)(4,8*106)= (4,5*106)/(0,9*10-3)= (6*106)/(0,15*103)=

Respuestas

Respuesta dada por: Anónimo
1

(4,48*10³)+(2,4848*10³)=

4480 + 2484,8 = 6964,8 = 6,9648 *10^3

 

(7,6548*10^-6)+(4,6*10^-6)+(9,32*10^-6)=

0,0000076548 + 0,0000046 + 0,00000932

0,0000215748 = 2,15748 *10^-5

 

(4,8*10³)-(2,8*10³)=

4800 - 2800 = 2000 = 2*10^3

 

(3,8*10³)+(4,8*10^4)+(5,78*10^4)=

3800 + 48000 + 57800

109600 = 1,096 * 10^5

 

(7,2*10^5)+(8,3*10^-2)=

720000 + 0,083 =

720000,083 = 7,20000083*10^5

 

(3*10^5)*(3,9*10^13)=

11,7 * 10^18

1170000000000000000

 

(2,5*10^-3)(3,2*10^-5)=

8 * 10^-8 = 0,00000008

 

(6*10^-5)(4,8*10^6)=

28,8 * 10^1 = 28,8

 

(4,5*10^6)/(0,9*10^-3)=

5 * 10^9 = 5000000000

 

(6*10^6)/(0,15*10³)=

40 *10^9 = 40000000000

 

espero que te sirva, salu2!!!

 

 
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