August 2021 0 14 Report
касательная к графику функции f(x) = x^3 + 6x^2 + 14x - 3 параллельна прямой y - 2x = 1 найдите абсциссу точки касания. Срочно надо!!
Please enter comments
Please enter your name.
Please enter the correct email address.
You must agree before submitting.
More Questions From This User See All
napisat uravnenie okislitelno vosstanovitelnoj reakciikmno4hclsncl2sroch
Answer
zakonchit i urovnyat reakcii1 ch3cooh baoh22mncl2 nh4oh3niso4 nh42s
Answer
v chetverg akcii kompanii podorozhali na nekotoroe chislo procentov a v pyatnicu po39e0b5b1d9d1bdca9700356c6a300a12 85494
Answer
sosud yomkostyu 5l soderzhit 7g azota pri temperature 273 k opredelit davlenie g
Answer

Helpful Links

Helpful Social

Copyright © 2024 SCHOLAR.TIPS - All rights reserved.