This week in Math we finished off working on polynomial operations and started off on factoring polynomials.

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For Polynomial Operations, we mainly focused expanding out binomials and eventually trinomials, from there we would use distributive property and solve the equation to how we see fit.

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For Factoring Polynomials, we just started remembering how to find the GCF and learnt that the opposite of expanding is factoring, we then applied those concepts to solve questions in the workbook.

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My “a-ha” moment of this week wasn’t necessarily from the new unit, as I already grasp the concept of Polynomials fairly well but moreso on reviewing my Trig test. Before I would normally skim through questions and ignore all logical thinking to just do my calculations, but as it turns out, that shouldn’t be the case. I made an error in which I thought all angles could be used for solving a triangle, I never took into account the 90° angle since it is always there. Needless to say, I failed that question. So, only TWO angles can be used to solve for triangles due to the 90° angle.

This week in Math we began our unit on Polynomials, starting off on our review of the Grade 9 Polynomials unit and applying something new towards the end of the week.

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I can grasp the concept of Polynomials easily as it requires only substitution and logical thinking. However my only experience in it was doing things algebraically, this week I learnt (or reviewed, depending) on how to do algebra tiles again as well as area diagrams. Those would be my “a-ha” moments as before, solving it alone as an equation would be my only method; now that I’ve learnt two other methods, I can apply it to different questions in the workbook.

This week in Math I focused on finding the measurement of a missing angle for Trigonometry.

Generally, this is in the same line on how to deal with SOHCAHTOA, but rather than multiplying by its’ first function button, I pressed the secondary function button to get (sin-1) or however other ones there are. Afterwards, I divided the two measurements given and from then on I found the answer to my question.

My “a-ha” moment of this week featured two things:

1) To find the ratios of a triangle, the longest side represents the hypotenuse and is directly across from a right angle. Secondly, to find the opposite, it is simply opposite directly from the reference angle. Thirdly, by default the last missing side would be the adjacent side.

2) I learnt two vocabulary words: Angle of depression, angle of elevation. The latter I assumed would just be a regular triangle whereas the former I had little to no idea what I could do. Then the idea stemed from the workbook itself as I looked at both picture examples of angle of depression and angle of elevation, as I looked at it carefully I realized the angle of depression only represented an upside down triangle. Afterwards, I just input the measurements and right angle to form my triangle diagram and solved questions relating to that.

This week in Math I learnt how to effectively apply the formula SOHCAHTOA with triangle equations.

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It is rather more or less the same rules to apply when doing BEDMAS, however you should pay attention that the calculator is set in degrees rather than another function.

My “a-ha” moment of this week was where to place the fraction when it came to sin, tan, or cos: sin and tan both have their numbers on the denominator whereas the cos has it on the top.

Overall, this week was easy considering trigonometry uses calculators to solve for the product.

This week in Math I learnt the formulas for the different types of shapes there are (ex. spheres, cones, cylinders).

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This is the formula sheet we were handed to us, however I struggled with cylinders for the longest time.

After seeking help, I learnt a better method to do the formula: 2(pi)r² + (pi)d×h. As it is more or less the same formula that was handed to us on the pink sheet. – This would be considered my “a-ha” moment.

A question that stumped me was on the practice test on page 55, clearly, I did not know how to use prime factorization let alone try to do it on a test. The question stated: The sum of all prime factors of 160 797, I didn’t even know how to find the GCF in the first place. Then I looked back on my workbook to really understand what a prime number was as well as a GCF, completing examples in the process. Afterwards, one obstacle remained which was to use prime factorization, to overcome this I was deep in thought on how to use a factor table. Then I realized, I can use a factor tree instead. After a tedious process I found the solution to the problem, with the answer being 73.

There weren’t necessairly any a-ha moments this week besides figuring out how to use prime factorization to my full advantage as well as figuring out the difference between the GCF and the LCM. Which was that the GCF represented the number that could factor into all products such as 8, 16 and 20 (4), whereas the LCM represents the lowest product number you can find between quotients (ex. 5 and 7 have an LCM of 35).

A question that stumped me this week was on the assignment of radicals (1.5), question #13 on page 38. Here the question asked me to find the fifth root of -7/8 and add it to the fourth root of 7/8 with a coefficient of 2. While the answer in the Math 10 workbook stated that it was 0.96 (rounding it to nearest hundrenth), I did not understand as to how to arrive at that answer. The confusing part of it all really was that it was dealing with fractions and roots and I was incapable of solving it, despite asking for help from my friend – I really did not understand how to solve it so I left it behind, promising to come back to it.

This week I was really unsure about how to convert mixed radicals into whole and reversing the process back, that is until a friend at lunch helped describe the steps to take in order to do them. He explained to me how to go upon doing it, in which we tried with a series of different questions; the results were, to say the least, successful.