Week 11 of Math 10 has us moving on from factoring trinomials to relations in mathematics. The second week of quarantine learning has been great so far, as I’ve been absorbing and understanding the topics well despite the limitations of online teaching. This blog post will be covering relations and how to express them in various forms.

A relation is a set of data that – as the name suggests – are related to each other. Some examples of relations include:

• The value of a vintage wine and its age
• The battery charge of a phone compared to how long its been plugged in
• Time to drive to a supermarket and distance

Relations are are comprised of two variables which are the independent variable and dependent variable. Although these can be renamed, these two values are commonly referred to as ordered pairs which is an X and Y value inside a parenthesis like (x , y). The independent variable which is X does not rely upon other values, whereas the dependent variable relies upon the independent variable to find the answer.

Relations can be represented in a variety of ways like with diagrams, words, sets of ordered pairs, but for this blog post I will be covering how to solve for x through equations and map them with a table of values

Our relation will be y=6x-2 and we’ll first start off by making a table of values to add them into.

To keep it simple, we will be using numbers 1-8 as our input values, in order to find the value of Y we have to solve the right side. So replace x with 1 and solve accordingly.

Now simply input the numbers into the X to solve for the output of Y.

Now that there’s a full table, you can actually turn these into ordered pairs, which can then be used for other diagrams like graphs.

And that’s how you solve relations.

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Week 10 marks the first blog post while during the COVID-19 Pandemic quarantine. Most, if not all of our lessons are going to be in online classes within the confines of our homes. Personally, I find it hard adjusting to this new way of schooling, but I’ll try my best to both learn and display the knowledge gained from these classes. This blog will be covering what I learned while inside my bedroom as opposed to the classroom.

We reviewed the last topic that we were learning before spring break, which was factoring trinomials. Generally there are three steps to go through when assessing a trinomial question, and it goes like this:

Step 1: Check if there is something common within all the terms, or look for the Greatest Common Factor. If there is a common factor, divide all terms and place the GCF behind a parenthesis.

Note that ALL terms need to have a common factor in order for this step to apply, if even one term is out of line, disregard the step and go straight to the next step.

Step 2: Check if there are two terms. If there are two terms, it’s likely that it’s a Difference of Squares question. A difference of squares question is when the constants are perfect squares and the variables are even, and so we can factor by square rooting.

Two key things to note is that Difference of Squares questions –like the name suggests– need to have subtraction as its operation, if it also has a variable power that is not even or a constant/coefficient that isn’t a perfect square, then the whole step is to be disregarded and we move on to the next step.

Step 3: Check if it follows the formula: . This pattern means that its possible to turn the trinomial into two simple binomials by finding two numbers that can add into the second term while multiplying into the third term.

Sometimes however, you would go through all the steps and find that the trinomial simply can’t be factored and say its not factorable. But although some trinomials may look VERY unappealing, there is a fourth step you can do to try and factor “ugly” trinomials. One such example of an “ugly” trinomial is  which I will show how to factor.

Start off by drawing a 2×2 grid, then place the first term on the upper-left box and the third term on the lower-left box

Now, multiply both and find a factor that adds into the second term and place them into the remaining boxes.

With the grid complete you can now find the final answer. One way to find the factored answer is by looking for numbers that are in common with each other, here and 15x both have 5x in common, so you should align them together like so

In this case, in order to get  you need another x to multiply into it, and in order to get 15x you need to multiply 5x by 3, so align them by their respective boxes like so

Now there’s only one term left, and 2 is the number that can multiply into 2x and 6 from the x and 3 from earlier.Getting the one term usually helps you find the remaining terms. Our final answer is (5x+2)(x+3).

And that’s how you factor trinomials.

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Week 7, the final week before Spring Break and the inevitable COVID-19 quarantine. One of the new topics we learned is how to factor trinomials, previously we were expanding upon polynomial questions but now we’re working backwards. In this blog post I’ll be covering a few different ways to factor trinomials.

The first way to solve trinomials  can only apply to simple trinomials. Simple trinomials are characterized by having a degree of two, no leading coefficient, and have three terms. One example of a simple trinomial is . To find the factored version of this trinomial, you first have to look for the constant which in this case is 15. Next, you have to factor 15 in all the ways you can, such as here.

Then you have to find the factorizations that add up to the middle term’s – or x’s- coefficient which is 7. Here, only 4+3 can add up to 7 while multiplying into 12. Finally, turn the factored numbers into a simple binomial. Our final answer would be (x+3)(x+4)

That’s one way to factor trinomials, but the downside to this is that it can only apply to simple trinomials, so the next method is a more all-encompassing method to factoring any kind of trinomial.

The next trinomial example will be . The first step to solving this is to find the GCF or Greatest Common Factor of all the coefficients. In this case, 6 is the GCF as it divides into 12 and 18. However, it doesn’t just stop here; not only is 6 a common factor, x is also common among all the terms, making 6x our GCF. The next step is to divide the GCF by all the terms, like this:

If possible you can simplify even further, the factored trinomial just so happens to also be a simple trinomial, so we can keep going. The only numbers that can multiply into 2 while adding up to 3 is 2 x 1, so our final factored trinomial would be 6x (x+2)(x+1).

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Week 6 was arguably the best week for me because I had a great understanding of what the topic was and I managed to learn some new things too. We’re still covering polynomials but we expanded a bit more and learned multiple ways to distribute when multiplying two polynomials. This blog will be covering those methods and the benefits and downsides to using them.

The first method is using the distributive property of multiplication. This method is straightforward and simple, as you are placing all terms of the first polynomial behind the second polynomial and multiplying. If our question was (3x-4)(2x-2) it would be formatted like this:

After finding the simplified versions to the terms add them together, the answer will be . The benefits to this method is that it is simple and easy to do, and can apply to most polynomials, the downsides are that it is quite time consuming to multiply each individual term, especially if there are three or more terms that need distributing.

The second method is using Algebra tiles. This method has you using a grid-like pattern using rectangles and squares, with x being long rectangles and squares being constant 1’s. If our question were (2x+3)(x-2) it would be formatted like this:

Here, you stack all the shapes in a column facing vertically or horizontally, then you draw a straight line until all lines intersect and make a grid-shaped square. After that you label which numbers are positive and negative, if two positive rectangles intersect then its positive, if one positive and one negative rectangle intersect then its negative, if two negative rectangles intersect then its positive again. Once that’s finished sort all like terms together and simplify, the answer to this equation is .

The benefits to this method is that it uses visuals like rectangles and squares that can aid those who are visual learners. The downsides to this are that this can take a while to draw and shade the grid in, it isn’t very helpful when the terms are large, and it can only solve questions that have a degree of 1.

The third method is what’s called the FOIL method. FOIL is an acronym that stands for First, Outside, Inside, Last and is the order of which you multiply terms together. If our question were (6x+9)(4x-3) it would be formatted like this:

The first term of the first polynomial is multiplied with the first term of the second polynomial, the outer terms are multiplied with each other, the terms furthest inside the parenthesis are multiplied, then the last terms are multiplied. The answer would be

The benefit of this method is that it is very fast if you can multiply numbers quickly, but the downside is that it you may be liable to miscalculate and get the wrong answer as there aren’t any visuals to aid you in solving.

The final method is the multiplication grid method. This method is essentially a better, improved version of Algebra tiles. You start off by making a grid with each term next to a square, you then multiply across with all terms. If our question were (9x+7)(8x-5) it would be formatted like this:

After adding all like terms together the answer would be @latex 72x^2+11x-35$. The benefits to this method is that it is easy to do, has visuals, and can even apply to polynomials that have more than 3 terms, as you can add more tiles to the grid if need be. The only downside is that you might not have the space to draw out the grid, especially on test paper.

And those are the four methods of multiply binomials I learned.

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For Week 5 of Math 10 we’ve been re-introduced to Polynomials. The lessons have been a refresher and review for those of us who have already learned this, but there have been a few interesting things I managed to learn that never really crossed my mind. For this blog post I will be covering how to identify and label parts of a polynomial question and solve one.

The polynomials we are covering are essentially long, complicated ways to read math problems. However, they are not equations as they do not have the equals (=) sign, so all you have to do is simplify and remove all the unnecessary parts of the problem.

Our example of a polynomial question is . The first thing to do is label what kind of polynomial it is.

Polynomial is a blanket term for any type of math expression that has variables and coefficients, but there are specific names for polynomials that have a certain amount of terms. Terms are the number of operations that occur in a polynomial, so 2x+2 would have two terms while would have one term.

Here is a small chart that can help you identify the types of polynomials.

The next thing that can be labeled is the degree. The degree is another classification that can be applied to polynomials, and is found by looking for the largest exponent in the expression. So in the highest exponent is 5 which means the polynomial has a degree of 5. There are also names for degrees as well, which this chart can also help you identify.

There are two final labels that I’ve learned from Week 5: the leading coefficient and the constant. The constant is always the number in the polynomial that does not have a variable, so in , three would be the constant. The leading coefficient on the other hand is the coefficient of the variable with the largest exponent. Usually finding the degree will help you find the leading coefficient.

Now that all the labeling is out of the way its time to simplify our example polynomial. The most ideal way to simplify polynomials is to align all like terms and solve. This is how it would turn out:

Now with the polynomial simplified we can label it. Our example polynomial is a Cubic Polynomial with a leading coefficient of 16 and constant of 8. And there you have it, how to label and solve a polynomial.

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Week 4 of Math 10 has been relatively uneventful for me, lots of stress and pressure from other subjects alongside some difficulty with certain topics in Math made things more difficult. Lots of  difficulty mostly stemmed from attempting to interpret word questions, so I’ll try to make and solve my own. I have a deep interest in creative writing so I’ll attempt to make a story.

My word question is this:

“An elite spy is tasked with eliminating a high-ranking general during a secret arms deal. After finding the location of the meeting the spy heads to a 643 ft high cliff overlooking the general’s position. The spy planned on using a sniper rifle, so he determined that his angle of depression from the enemy general is 43°. Find the distance from the spy to the enemy general.”

So the first step to solving a word question is to find all of the crucial information that can make up an equation. A good tip would be to try to visualize what the scenario looks like so it can be easier to draw out the equation. The question says a few key things: the angle of depression and the height of the cliff.

It is also important to make sure that your information is also accurate. There are two different angles that act as the reference point, the angle of depression and the angle of elevation. The angle of elevation is simple, it’s how much the angle elevates or looks up, while the angle of depression is how much the angle depresses or looks down.

Now that the information you know is accurate, now you can draw a visual to help solve the question.

We’re trying to solve for the distance between the spy and the commander, so the side we are trying to solve for is the adjacent.

Now we can create an equation of what we need, since we have the adjacent and opposite, this question requires using the tangent. So simply evaluate, for this question we’ll round to the nearest foot.

After evaluating, the distance from the spy to the general is 690 ft. Now he knows how much he has to compensate for muzzle velocity, wind speeds, bullet yaw and pitch, and all the other factors needed to calculate firing a gun, but that’s for the spy to figure out.

Mission complete, they’re gonna tell stories of this one boss!

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This week’s new math topic is Trigonometry, I’ve always heard about it and Pre-Calculus from my older sister and was curious to see what it was. Now I’ve found out that its solving for triangle angles using the weird buttons on my scientific calculator. It was real exciting to finally find out what “sin” “cos” and “tan” meant.

The equation begins with a right-angled triangle, one like this:

The first step to solving this would be to label the variables in the equation. There are three labels you need to use: Hypotenuse, Adjacent, and Opposite. But before you can find the labels, you need to determine what the Reference Point is.

The reference point is usually a number inside the triangle that is measured in degrees (°) In this case the reference point is 42. For the labeling, the Hypotenuse is the longest side of the triangle and can be found by looking at the side across from the 90° angle. The Opposite is the side across and furthest away from the Reference Point, which is 69. Finally, the Adjacent is the side that is right next to the Reference Point.

Note that the Adjacent cannot be the Hypotenuse and vice versa, so always look for the Hypotenuse first before everything else so that you don’t get confused. After labeling the triangle should now look like this.

The next step is to find out which ratio is needed to solve for this question. Sine, Cosine or Tangent. The ratios for each are as follows:

A good way to remember what each ratio has is “Opposite is always on top” “Hypotenuse is always on bottom” and “Tangent has both Opposite and Adjacent”. Next is to write down an acronym called SOH CAH TOA. This acronym can help you determine which ratio is going to be used for the equation.

What you should do with SOH CAH TOA is you must look for the side that has is blank or empty, then cross out all the letters corresponding to your blank side. For this equation the side that is empty is the Hypotenuse or H so cross out all H’s in the acronym. The end result should look like this:

TOA or Tangent is the only ratio that isn’t crossed out, so Tangent is the ratio we will be using for this equation. The ratio’s number is the reference point, so the final equation is Tangent 42 =

Now we can solve the equation. First thing to do is to isolate X, but in this case X is the denominator which makes things more complicated. The best thing to do is to cross multiply X with the Tangent which results in this:

A scientific calculator is needed to solve this. After inputting the required variables the result is 76.63226352 which isn’t pretty to look at, so depending on the question or teacher the final result would be limited to a certain decimal point. For this question we’ll just be limiting it to the hundredths or 2 decimal points. So the final answer would be X = 76.63

Voila, your work is done, and so is mine.

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Things are finally getting interesting with Week 2 of Math 10. New topics like scientific notation and negative exponents are things that I have never learned before, and I’m quite excited to learn something new. In this blog post I will be explaining what Scientific Notation is and how to solve and write them down.

Scientific notation in essence is the simplified writing of very large numbers like 23000000, or very small numbers like 0.000000000030 into shorter numbers such as and

As you can tell, Scientific Notation is much cleaner and nicer than their previous counterparts, and everyone likes nice things right?

To demonstrate solving for large positive numbers I’ll be using 4000000

Starting off, find the decimal point of the number that you’re solving for

After finding the decimal point, move it backwards until you’ve reached the first number of the whole, while also keeping in mind how many times the point moves.

In this case, the decimal point moved six times before stopping at 4. The format for scientific notation usually goes as (first number) 10 raised to the power of (# of times the decimal point moved).
So our final answer would be

Numbers like millions, billions and trillions are actually scientific notation with a million at which slowly increases by three for billions and trillions.

While the method to get the answer may seem straightforward, add the first number 10 and just count the number of zeroes,  some things don’t come so easily. 6320000000000 for example IS NOT
To find the right answer, move the decimal point back until you’ve reached the first number again. Then add any numbers in front of the first that are higher than one as a decimal.

So in this case, the correct answer is actually . The best thing to do is the second method, as it makes sure that the answer you get is correct, as its not as simple as counting the zeroes. For large decimal numbers I will be using 0.0000000000000576

Start off by finding the decimal point and moving it forwards until you’ve reached the first number. Instead of a positive, use a negative exponent.

This follows the same rules as the second method for large positive numbers, but instead of going backwards until you reach the first number, you go forwards until you reach the first number

And that’s how you solve for scientific notation.

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Week 1 has been par for the course for Math 10 with a review to get the minds of students back to numbers. So far, everything has been a familiar sight  with topics such as square roots, perfect squares, BEDMAS and the like. However, one new strategy to approaching topics like finding out the GCF of numbers has caught my eye, and I am going to demonstrate what I learned.

So to begin, GCF stands for Greatest Common Factor, and it is the largest factor that is shared in two or more different numbers. So to demonstrate let’s have 12 and 16 as our comparison.

Here, we try to find the different factors that can make up 12 and 16. Afterwards, we look for the largest factor that is present in both numbers, and in this case it is 4.

This is a simple process for smaller numbers as it is easier to find the factors and thus find the GCF, but it is more difficult the larger the numbers get. So in order to solve for bigger numbers we’ll have to approach the question from a different direction. For demonstration we’ll be using 90 and 135

Start by creating a factor tree or division table to find the prime factorization of 90 and 135

Then put them side by side. From there, look for the like terms and highlight them.

Then simply multiply all of them together to get the GCF

To double check, you can either divide the GCF by the numbers you’re solving for, or in this case multiply until you’ve ended up with the numbers you’re comparing.

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