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)
Question I read your answer to another student about how to solve a question similar to this one:
"The hypotenuse of a triangle is 15 feet long. The longer of the two legs is three feet longer than the other. Find the length of the shorter leg."
Now here is what I have done. I have already given values to a,b and c. They are "x", "x+2" and 15^2 in that order. So I have in front of me:
15^2=x^2+(x+2)^2
But what I don't know is how to go from there... how do I use the quadratic formula to solve for "a" and "b".
Thankyou,
Tim
Answer Hi Tim,
Start by expanding your squares.
15²=x²+(x+2)²
(15)(15)=x²+(x+2)(x+2)
225=x²+x²+2x+2x+4
Collect like terms.
225=2x²+4x+4
Bring all the terms to one side of the equation, so that the other equals 0.
0=2x²+4x-221
At this point, you have 2, 4, and -221 as your a, b, and c values respectively.
