Category: Grade 10

Week 10- Math 10

This week was the beginning of polynomials. They’re quite simple to understand and get the hang of, but the “ugly” ones are more difficult to figure out. A trick that I learned, and that I found helpful, was using the area models.

If you are given a question like this 2x^2+7x+6, there is an easier way to factorize it without doing everything in your head. First, we start by multiplying your first and last term, as well as starting your area model with what you know.

Then we list the multiples of 12 and see which ones add up to 7, which is our middle term. If the question has a negative term in it, you will need at least one negative term. From the list here, we see that 3 and 4 add up to 7.

Now we know what to put into the two blank squares. You can put whichever in each blank, the order of those 2 numbers doesn’t matter.

On the side, put what is common from each row horizontally on one side and the common numbers or variables, vertically, on another side. Don’t forget your signs like I did.

Now, put what you have on each side into brackets and you have your answer! For me, when I have a brain fart, this will be very useful.

 

Week 9- Math 10

This week was about polynomials. I was like a review of what we did in grade 9 plus an extension on it. We started with distributing binomials like (x-1)(x+3) and then to trinomials such as (x+1)(x-4+y). There was this one trick we learned that I find very helpful and useful. Here’s how it works:

Say you have a question like this (3m+7)(m^2 -3m +6). Drawing out the algebra tiles for this would be a hassle so I will do the area diagram trick! First, I would draw this

It looks like a weird hashtag. Next, I will place each term in along the sides like so and it should look like the picture below:

Then, I will multiply the numbers and simplify to get my final answer.

Just like that and you’re done!

 

 

Week 8- Math 10

Ahhh solving triangles. You get one angle and one side, figure out the rest.

Once you get the hang of it, it’s quite fun. You’re given a triangle like this

To start, label the sides and list what you need to find.

Find everything that is missing (aka the stuff you just listed). We’ll start with angle F here. Since we know angle E is 90°, angle D is 50° and that the sum of the angles in a triangle must be 180°, we can just do 180-(90+50) to find angle F.

We can solve side EF and DE with the information we now have and SOH CAH TOA like so:

Week 7- Math 10

The most fun and interesting part of this week was learning how to find missing angles using sin, cos and tan on my calculator.

If I had a question like this (I labeled each side already and written SOH CAH TOA to figure out which one to use):

Just having sin, cos or tan would not be helpful because I am not given the degree of the angle inside. Instead, I would use each of those to the power of -1. Since I now know which sides I’m given and which formula to use, I can start to write my equation and solve it, like so:

Week 6- Math 10

Dear Math Blog Post,                                                                                                                          Week 6

My brain has gotten stuffed with more information. Every day, it’s something new. Do I always remember what I learned after the lesson? Not always but then they force me to use it the next day so I am forced to remember, and the information gets drilled into my head. This week, I remember learning how to find the volume and surface area of a hemisphere.

To find the volume of a hemisphere, you will use this formula:

For example, if you wanted to find the volume of a hemisphere with a radius of 2, this is what you would do:

When you do the surface area, it gets slightly more math-y. There are 2 exposed parts in the surface area of a hemisphere; the bottom half and the circle on top. When you calculate the surface area, the two must be added together.

The formula and an example question for finding the surface area of a hemisphere are in the picture below.

Week 5- Math 10

In math 10 this week, I learned how to convert units such as feet to inches or imperial to metric.

For example, I had a question where the I got was in meters cubed but the question asked me to have it in liters. The answer was 689.64^2m. To convert it into meters I did this:

Because the m^2 cancels each other out and the L is left which is the only unit so the answer would be, in this case, 78.37L.

However, it is not simply that. In the picture above, the units were already in m^2. If it were not, I would have to multiply the unit twice (or three times if it is cubed) to match up to the squared (or cubed).

Like this,

Math 10- Week 4

This week, my ah-ha moment was while I was working on this question:

((64a)^\frac{1}{3})^\frac{1}{2}

At first, I got the answer to the left. I then checked the answer only to realize that it was wrong but I could not figure out why. A few moments later, I realized that 64 was a coefficient and that \frac{1}{6} also had to be distributed to that as well. So then I redid my work and got the right answer (the work on the right).

Math 10- Week 3

Week 3 was almost like a review of exponents from grade 9 but with a bit more “pizazz” because we learned about negative exponents, which was entirely new to me.

I understood it quickly and got the hang of it after a few examples, however, I did still have some trouble with the assignments. Mainly this one: Simplify. Write the final answer with positive exponents. (5 a^3 b^2)(-2a^-2b)^-3 ÷ (-5a^8b^-9)^-2

At first, I got confused while I was simplifying and ended up with this:

I realized where I went wrong- the exponents inside do not change when you move it “upstairs” or “downstairs.” So then I changed how I did my work and got this:

which was correct. I learned to be extremely careful with negatives because I can be forgetful with the rules.

Week 2- Math 10

This week, we covered more prime factorization and a few square and cube root related questions.

One question I had a bit of trouble with was this one:

The mixed radical \frac{1}{12} \sqrt[3]{128} can be converted to a mixed radical in simplest form a \sqrt[3]{b} . The value of a+b to the nearest tenth is ______. 

At first, I did this:

Then I checked the answer and it wasn’t right. I realized what I did wrong so I redid my math which resulted in this:

And it was the right answer.