Tag: burtonmath10C

A system: looking at two lines at the same time to see where they intercept.

Even though I wrote down that definition on Monday, I didn’t remember it until Friday.

This week, one question that stumped me was this one:

I first started with rearranging the equation to isolate a variable. For this one, I chose to isolate y in the second equation since it has no coefficient (“work smarter, not harder” says a Ms.Burton voice in my head).This is where my I made my mistake the first time because I forgot to make 3 a negative when I moved it.

 

From there, I substituted this equation into the second one wherever I saw y. This is what I got.

Then I expanded and collected all like variables. After that, I was able to do some algebra and find x.

Finally, I substituted my x value into an equation so I could find the value for y. When I got my answer, I plugged in the two values into both the equations to check my answer.

Ta-da! I probably could have saved a lot of time of my first attempt if I was more careful with my numbers.

Wonky Initials graphing project

A question that I had some trouble on this week was this one:

Write the equation, in general form, of a line perpendicular to 3x-2y+5=0 and with the same y-intercept as 3x–y+18=0.

From the work below, we know that the y-intercept is 18.

In class, we learned that the slope of a line that is perpendicular will be the reciprocal and the opposite symbol- negative or positive. From the given formula, we can change it to slope-intercept form and find the slope. Then, we will reciprocate it and change it to a negative or a positive.

Now that we know the y-intercept and the slope, we can write a formula in slope-intercept form with this information.

From the slope-intercept form, all we need to do is to rearrange to make it equal to 0 and that will be the final answer.

On Friday, we started slopes. I remember doing a little bit of this in grade 8 and 9 but all I remember were slopes being positive (or only the positive ones) so I assumed that all slopes were positive. This week, I learned that slopes can be negative.

The concept of negative slopes didn’t wrap around my head easily and I was quite confused about how they could be negative. For me to understand how a slope could be negative, I first figured out that the ones go towards the 2nd quadrant were negative. They usually look like \.

I always count the run first so the run of the two points I chose in the picture below is 3. Then I count the rise and the rise of this is also 3. Since the formula for slope is rise/run it will be written as 3/3 and it will be simplified to 1.

However, you can see that the number for the run seems odd. That’s because it is, how I like to say it, “going to the dark side” because it is going towards the negatives. Since I always count the rise upwards, it will always be positive for me but the run can go left or right. If it goes towards the left, or “to the dark side,” the slope will be negative. So the slope of this graph is not 1 but it is -1.

 

From grade 9, we learn that x is our input number and y is our output number. In grade 10, we build on that. This week, I learned two new terms that are very useful for this unit. The first one is domain and the second is range.

The domain of a graph are all the numbers that can be used for x in an equation. In a graph like this,

You would write your domain like this: D= {0 < x < 10} because the domain is the dots along the x-axis. The curly brackets are used to show that these numbers all have something in common. The less than symbols are to show that x is more than 0 and less and 10 since the dots at the end are open dots meaning the values in between will be more or less than where the dot is. There is also that part here: x ε R, this symbol, ε, means that x is an element of R.

The range in the graph would be the highest and lowest point along the y-axis. So, the range of the graph above would be written like this R= {-1 ≤ y ≤ 1}. For this one, we used the less than symbol because the dots are closed dots.

If the graph has arrows, like this one:

You would have to include xεR in the domain and yεR in the range because it is continuous. This symbol, ε, means that x or y is an element of R.

Can divers pee easily underwater? Answer: no, but the acronym in that silly question, CDPEU, can be easily used with trinomials. Instead of ‘can divers pee easily underwater’ we’ll replace them with mathematical terms that can be used with trinomials.

When you are unsure about a question, ask yourself “is it common? Is it a difference of squares,” and so on. If you answer ‘no’ to all of them, the given question is not factorable. This is one thing that I learned this week that I found extremely helpful when I’m not sure what to do or when I am stuck on a question.

 

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 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.

 

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

((64a

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 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).