November 2018 1 16 Report
Proszę o rozwiązanie zadania z Góry dziękuje ;)
Z potęg o wykładniku całkowitym ....
Oblicz :
13^2
(-1)^6
0^10
10^6
(-25)^1
-2^6
(1/3)^3
(-8/11)^2
(2 i 1/2)^2
-(1 i 3/4)^2
(0,1)^4
(-0,05)^3
(-1,2)^2
( (\sqrt{5} ) ^{2}
( \sqrt[4]{2,5} ) ^{4}
( \sqrt[3]{4} ) ^{9}
More Questions From This User See All

Recommend Questions



Life Enjoy

" Life is not a problem to be solved but a reality to be experienced! "

Get in touch

Social

© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.