Question

Proszę was o pomoc w zadania z matematyki z góry wam dziękuje.

Answer 1

Zad.1

a)

7^-2 = (1/7)^2 = 1/49

(-7)^-2 = [1/(-7)]^2 = 1/49

-7^-2 = -(1/7)^2 = -1/49

b)

(-2)^5 = -32

(-2)^-5 = [1/(-2)]^5 = -1/32

2^-5 = (1/2)65 = 1/32

 

Zad.2

a)   1,(2)

x = 1,(2)

10x = 12,2

 

9x= 12,2 -1,2

9x = 11

x = 11/9 = 1 2/9

Zatem  1,(2) = 1 2/9

b)   0,(06)

x = 0,(06)

100x = 6,06

 

99x = 6,06 - 0,06

99x = 6

x = 6/99 = 2/33

Zatem  0,(06) = 2/33

 

Zad.3

c)

(4^3 * 2^-8) : (8^3 * 4^-2)^-1 =

= [(2^2)^3 * 2^-8) : [(2^3)^3 * (2^2)^-2]^-1 =

= (2^6 * 2^-8) : (2^9 *2^-4)^-1 =

= 2^-2 : (2^5)^-1 =

= 2^-2 : 2^-5 =

= 2^3 = 8