Home
О нас
Products
Services
Регистрация
Войти
Поиск
Alexander270
@Alexander270
July 2022
1
3
Report
Помогите решить все номера
Please enter comments
Please enter your name.
Please enter the correct email address.
Agree to
terms of service
You must agree before submitting.
Send
Answers & Comments
petyaGavrikov
Verified answer
1.
100(10) = 1100100(2), 350(10) = 101011110(2), 70(10) = 1000110(2)
Решение - в прилагаемом файле.
2.
111(2) = 1*2^2+1*2^1+1*2^0 = 4+2+1 = 7(10)
203(4) = 2*4^2+0*4^1+3*4^0 = 32+3 = 35(10)
150(7) = 1*7^2+5*7^1+0*7^0 = 49+5*7 = 84(10)
3.
101010(2) = 1*2^5+1*2^3+1*2^1 = 32+8+2 = 42(10)
42(10) = 132(5)
Решение - в прилагаемом файле.
0 votes
Thanks 1
More Questions From This User
See All
Alexander270
August 2022 | 0 Ответы
napishite pozhalujsta nebolshoe neformalnoe pismo
Answer
Alexander270
July 2022 | 0 Ответы
kakoe imya samoe topovoe
Answer
рекомендуемые вопросы
rarrrrrrrr
August 2022 | 0 Ответы
o chem dolzhny pozabotitsya v pervuyu ochered vzroslye pri organizacionnom vyvoze n
danilarsentev
August 2022 | 0 Ответы
est dva stanka na kotoryh vypuskayut odinakovye zapchasti odin proizvodit a zapcha
myachina8
August 2022 | 0 Ответы
najti po grafiku otnoshenie v3v1 v otvetah napisano 9 no nuzhno reshenie
ydpmn7cn6w
August 2022 | 0 Ответы
Choose the correct preposition: 1.I am fond (out,of,from) literature. 2.where ar...
millermilena658
August 2022 | 0 Ответы
opredelite kak sozdavalas i kto sozdaval arabskoe gosudarstvo v kracii
MrZooM222
August 2022 | 0 Ответы
ch ajtmanov v rasskaze krasnoe yabloko ispolzuet metod rasskaz v rasskaze opi
timobila47
August 2022 | 0 Ответы
kakovo bylo naznachenie kazhdoj iz chastej vizantijskogo hrama pomogite pozhalujsta
ivanyyaremkiv
August 2022 | 0 Ответы
moment. 6....
pozhalujsta8b98a56c0152a07b8f4cbcd89aa2f01e 97513
sarvinozwakirjanova
August 2022 | 0 Ответы
pomogite pozhalusto pzha519d7eb8246a08ab0df06cc59e9dedb 6631
×
Report "Помогите решить все номера..."
Your name
Email
Reason
-Select Reason-
Pornographic
Defamatory
Illegal/Unlawful
Spam
Other Terms Of Service Violation
File a copyright complaint
Description
Helpful Links
О нас
Политика конфиденциальности
Правила и условия
Copyright
Контакты
Helpful Social
Get monthly updates
Submit
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
Verified answer
1.100(10) = 1100100(2), 350(10) = 101011110(2), 70(10) = 1000110(2)
Решение - в прилагаемом файле.
2.
111(2) = 1*2^2+1*2^1+1*2^0 = 4+2+1 = 7(10)
203(4) = 2*4^2+0*4^1+3*4^0 = 32+3 = 35(10)
150(7) = 1*7^2+5*7^1+0*7^0 = 49+5*7 = 84(10)
3.
101010(2) = 1*2^5+1*2^3+1*2^1 = 32+8+2 = 42(10)
42(10) = 132(5)
Решение - в прилагаемом файле.