What you have to do in such a case, is to muliply each term of each factor by each term of other factors. As ax+bx = (a+b)x, (a+b)(x+y) = (a+b)x+(a+b)y = ax+bx+ay+by, so (a+b)(x+y) = ax+ay+bx+by. Here, one way is to first distribute...

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Hi, my name is Rassoul, I'm from Senegal, and I'm a native French speaker. I finished my high-school years in June 2012 and I graduated from the department of mathematics/physics. I also obtained a baccalaureate diploma with a math/physics option.I'm now pursuing an associate's degree in Electrical Engineering at CCRI. For math, I studied advanced levels of algebra, calculus, geometry (both 2D and 3D) and arithmetic in secondary school, and just finished Calculus III in college.

So far, I have never been paid for tutoring, but I have occasionally given a few hours of tutoring to people I knew. I have helped some of my classmates raise their grades. I've developed some good tutoring qualities in the process and am looking to continue improving. What I have learned from my experience is that patience is a significant factor in tutoring and I think that each student has to develop his own ways of thinking and solving problems he/she will face.

Science:

Physics
Language:

French
Corporate Training:

French
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Having received a solid education in mathematics, algebra has been a step I mastered quite early. I also used to help my classmates and other students, which gives me a little bit of experience.

Having received a solid education in mathematics, algebra has been a step I mastered quite early. I also used to help my classmates and other students, which gives me a little bit of experience.

Besides the fact that I am a native french speaker, not only have I attended french classes but I studied history, geography, philosophy, math and other subjects in french too... That gives me a large variety of assimilated concepts.

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What you have to do in such a case, is to muliply each term of each factor by each term of other factors. As ax+bx = (a+b)x, (a+b)(x+y) = (a+b)x+(a+b)y = ax+bx+ay+by, so (a+b)(x+y) = ax+ay+bx+by. Here, one way is to first distribute...

From the fact that for m and n,integers, and a, real number, (am)n = amn, (ambn)p = ampbnp, am * an = am+n, am / an = am-n, (a2b2c-3)-4 * a4c-3 = a-8b-8c12a4c-3 = a-4b-8c9 So, [(a2b2c-3)-4 * a4c-3]/2a-2b3 = a-4b-8c9/2a-2b3 = a-2b-11c9/2 =...

Hi Rebekah. If I read correctly, the equation is 216-r-1 * 36-3r-2 = 362r-1. In that case, first note that 216 = 63 and that 36 = 62 So, knowing that for m and n,integers, and a, real number, (am)n = amn, 216-r-1 * 36-3r-2 = 362r-1 <=> 6-3r-3 * 6-6r-4 = 64r-2 ....

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