AboutAzeem Hussain Expertise I can answer virtually any kind of question dealing with of Math 536 and below, my forte being in parabolic functions and analytical geometry.
I'm currently learning Linear Algebra, and cannot answer questions that deal with subject matter more advanced than that.
Experience I am neither a professor nor a teacher of this subject. I am merely a student who is gifted at mathematics and enjoys being of service to his community. I frequently tutor people in math and the results are usually great.
Publications Reflections, Riverside School Board (2005, 2006)
Education/Credentials Diploma of Secondary Studies from Chambly Academy High School, and IBO-MYP certificate as well. My lowest mark on a high school math final was 97%, peaking at 99% in 2006 and 2007 (second-highest Math 436 mark in the province). Being a Quebecer, I am fluent in English and French and can respond to questions easily in both languages.
Awards and Honors Pascal Math Competition, School Champion(2007)
please show me how to do this questions step by steo I understand I am asking alot but please help me out
ANSWER: Hello Kira,
What you need to understand in order to solve the question is written immediately before the question.
For question 1, use v_t=g/k and plug in the values of k to solve for the terminal velocities.
For question 2, the terminal velocity is inversely proportional to the k value. As k increases, v_t decreases, and vice-versa.
For question 3, plug t=12 into the equation and solve for v.
These questions are straightforward. Be sure to read them over a few times along with the theory that comes before them before solving them.
Thanks for asking,
Azeem
---------- FOLLOW-UP ----------
QUESTION: Oh ok thank you
I did not read the equation right
but how would you do part B question 4
Answer Hi Kira,
"98% of its terminal velocity" means v=(0.98)45=44.1. Plug this into the equation and solve for t.
v=45[1-(0.804)^t]
44.1=45[1-(0.804)^t]
44.1/45=1-(0.804)^t
0.98=1-(0.804)^t
0.804^t=0.02
log0.02/log0.804=t
t=17.93
