This is an infographic I created about the short story The Lottery, that we read in my English Honours class:
You can read The Lottery here.
This is an infographic I created about the short story The Lottery, that we read in my English Honours class:
You can read The Lottery here.
As an English student, I would describe myself as someone who is hard working, creative, and often an over-achiever. I love to write both creatively and persuasively, always aiming to receive the highest mark I can. I strive to improve on my mistakes and I can act on feedback as something to work on in the future. My strengths in English include creative writing, as well as proper grammar and punctuation. I can analyze online or written sources to check if they are accurate or not. I also enjoy working in a group and can respectfully share my ideas while still listening to others, or helping them if they need. If I am passionate about something, I can pursue that interest and carry it into my writing as something of importance to me. This semester I would like to work on improving my skills of essay writing, specifically planning and organizing the points before writing the essay itself. I would also like to practice editing my writing so it is more concise and less wordy. Overall, I am excited to experience this semester of English and I look forward to every new skill I will learn.
These are the top 5 things I learned this year in my Precalculus 11 class. I chose some of the most important topics from this semester that I didn’t know how to do in previous years of math. Below are the 5 topics with a definition of each, a few examples, and the reasons why I chose them as part of this blog post.
1: Mixed and Entire Radicals
Mixed and entire radicals are both forms in which to write a radical (root of a number). Entire radicals are simply the root of any numbers, written normally. Mixed radicals are a simplified form of these, where a number is multiplied by a root. For example:
√3 √17 ∛44
2√7 3√5 2∛4
As you can see, the mixed radicals are usually smaller numbers, as they were originally large numbers that were distributed outside the root sign. To write an entire radical as a mixed radical, follow these steps. √45 – find multiples of the number, with at least one square (or cube) root. Since 9 x 5 is 45, and 9 can be square rooted, these numbers will work. (If no multiples are squares, then creating a mixed radical is not possible). Find the square root of 9, which is 3. Put the 3 outside the root symbol, leaving the remaining number inside, which was 5. It now looks like this: 3√5 To go back to the original entire radical, multiply the outside number by itself, since the symbol is a square root (and not a cube or fourth root). This would be 9. Put it back inside the symbol and multiply by the number there – √9×5 = √45
Here is one more example, using a cube root. ∛16 – find multiples that include a cube root. You could also make a factor tree for 16. Since 8×2 is 16, you can make this into a mixed radical. The cube root of 8 is 2, so put that outside the symbol: 2∛2 To go backwards, multiply the outside number by itself two times, because it is a cube root. This makes 8. Then put the 8 inside and solve: ∛8×2 = ∛16
I chose this topic as one of my top 5 because it was something unique I hadn’t learned about until grade 11. I found this interesting because if there is a question or answer with a root, you can simplify it to make it easier to read. Personally, I find math more straightforward if it is written in lowest or most simple terms, and this is one way to do that.
2: Rationalizing the Denominator
Rationalizing the denominator is helpful when there is a fraction with a root on the bottom. This technique puts the fraction into a simpler form, with an integer on the bottom instead. Here is an example of a non-rationalized fraction:
To rationalize the denominator, just multiply the top and bottom of the fraction by √3. Because you are multiplying the top and bottom by the same value, it is as if you are multiplying by 1. This doesn’t change the ratio of the fraction, it just makes it easier to read.
Here is a final example, where I have continued simplifying the fraction:
This skill was in my top 5 because it is a quick, easy technique that I find super useful. I still remember it today, even after learning it a few months ago. I also think it makes any answer look cleaner and easier to read.
3: Graphing a Parabola
This is something new that I only learned in precalc 11. A parabola is a u-shaped line that curves on a graph. I learned that every quadratic function or equation creates a graph of a parabola, with certain characteristics. Equations can come in standard, vertex, or factored form. Vertex form is the easiest to graph. It looks like this: y = (x – p)². This is an example of a simple equation in vertex form, and it’s resulting parabola. y = x² – 3
The middle point, at the highest or lowest spot, is called the vertex. This is either at the top or bottom of the parabola. A parabola is symmetrical, with the axis of symmetry being in the center. Based on the sign (positive or negative) of the x², the parabola will open upwards or downwards. Here are a few other examples:
For the first graph, the equation would be y = -x². The negative means it opens downward. The vertex, or highest point, is at the coordinates (0,0), and the axis of symmetry is x = 0. The second graph below shows the equation y = -(x – 3)² + 2. The vertex is (3,2), and the axis of symmetry is x = 2. As you can see the -3 turned into +3 when finding the vertex. Make sure you switch to the opposite sign when finding the number for the x-coordinate in vertex form.
Parabolas are a new and interesting way of graphing, making them one of my top 5 things I learned. When I could see the relation between the equation and a graph, it made it much easier for me to read and understand a question because there was a visual.
4: Multiplying & Dividing Rational Expressions
This was one skill that I found really fun once I learned to properly solve the expressions. When multiplying and dividing these expressions, you can cancel out like terms, or terms that are the exact same. These terms must be either in both the numerator and denominator of the same fraction, or in opposite fractions, with one on the top and the other on the bottom. Here is an example showing the usual way to multiply fractions, which can take a while. Below it is a more efficient way, where the 4 is cancelled out to find the same answer.
Sometimes you have to factor an expression first, to check for any terms that can be cancelled out. Remember that terms in brackets are connected, and cannot be cancelled out unless another term in brackets is the exact same. For example, (x + 3) and x are not like terms, but (x – 4) and (x – 4) are. When you have cancelled out all like terms, always check to make sure a fraction cannot be simplified any further. Don’t forget to include restrictions on your answer (values that would be impossible because they would make the denominator 0). With dividing, just flip the second fraction to it’s reciprocal, and continue. These examples show more complicated questions:
This technique turns a complicated equation into a simple fraction, without having to do lots of difficult calculations. That is why I chose it for one of the top things I learned, because it is so simple and fun!
5: The Cosine Law
The cosine law is a way to find the angles and side lengths or a non-right triangle. This law is a formula which works for any non-right triangle where there are 3 sides and no angles, or finding a third side when 2 sides and the angle between them are given. The cosine law looks like this when you need to find a side:
To find an angle instead, just switch the same formula to this:
Here are 2 examples using the cosine law to find a missing side/angle:
This was one of my top 5 because I didn’t realize trigonometry was possible with non-right triangles before math 11. The cosine law makes it easy to find missing values on a triangle by putting them into a simple formula.
This is a reflection I wrote about practicing my Creative and Critical Thinking skills in precalc 11:
In this week of precalc 11, we continued trigonometry by learning about the sine law. This is a way to determine unknown angles or sides in a non-right triangle. The sine law can look different depending on if you are looking for an angle or side. If you are looking for a missing side, the sine law looks like this:
If you need to find an angle, just flip each fraction to look like this:
Notice the letters are capital and lowercase. This is very important, as they mean different things. A capital letter usually refers to an angle, while the lowercase letter refers to the corresponding side. In this diagram, you can see that the sides opposite each angle have the same letter:
The sine law can be used to find a missing side, as shown in this example:
Just put the known values into the sine law equation, then select the two fractions with the least variables. There should only be 1 variable to make the equation possible to solve. Next, just complete the equation like algebra, using a calculator.
To find a missing angle, follow a similar procedure, like shown below. Don’t forget to use inverse sine at the end to find the correct angle.
In precalc 11 this week, we learned about trigonometry, and more specifically, special triangles in which you don’t need a calculator. There are two types of triangles we learned to memorize, which look like this:
As you can see, these triangles involve the angles 45°, 30°, and 60°. They also have the three primary trigonometric ratios: sine, cosine, and tangent. For the second triangle, the ratios can be based off both the 30° or 60°. Here are the ratios:
Below are some questions where the angle is given, and you need to find the ratio (or fraction) for either sine, cosine, or tangent. You can start by drawing a diagram of the four quadrants, labelling each with sine, cosine, tangent, or all. This tells you in what quadrant each of the ratios will be positive. Then figure out where the angle is from the question, and use it to draw the triangle. By doing this, you can find the reference angle, and therefore choose which of the special triangles to use.
Next, refer back to the original ratios stated above, and find the one to fit the reference angle. If the sign (positive or negative) of the angle is in a quadrant that matches the sign, keep it positive. If not, change the answer to negative.
Remember, these questions can all be done without a calculator if you remember the two special triangles. As long as either 45°, 30°, or 60°, is present, they are easy to solve.
In precalc 11 this week, we continued learning about rational equations, and how to apply them into real life word problems. Most word problems can be solve creating a rational equation and solving it. This blog post will explain one type of word problem I learned, which involves work and time. Here is an example:
It takes John 5 hours to clean his basement. When John and his friend Barry clean it together, it only takes 3 hours. How long would it take Barry to clean the basement by himself?
To solve this, you can create a table with John, Barry and Both as the categories on top. On the left side, write the time. Choose values of time that represent the information given in the question. Because it takes John 5 hours alone and 3 hours together, I chose to use 1 hour, 3 hours, and 5 hours. Just make sure the value of time for which they can finish the job together is one of the options (in this case, 3).
Next, add in the values you know. It takes John 5 hours to finish the job, so in 1 hour John will have cleaned 1/5 of the basement. In 3 hours, he will have done 3/5 of the job. In 5 hours, he will have finished 5/5 (which equals 1), which means all of the basement.
Because we don’t know the values for Barry, write similar fractions in the table, but use a variable. In one hour, he will have done 1/x of the work. Repeat this for 3 and 5 hours.
Finally, complete the section with both people working as a team. Add the values from each row, as shown below. In the section with 3 hours, we know they finished the job. Therefore, the two values added together equal one.
Now, just solve the equation you created:
This means that it would take Barry 7.5 hours to clean the basement on his own.
This is a different example using the same strategy:
Mary takes three times as long as her daughter to set up their Christmas decorations. When they do it together, it takes 9 hours to finish the job. How long would it take each person on their own to set up the decorations?
In math 11 this week, we learned about adding rational expressions. This is different than multiplying expressions, because you usually cannot cancel out numbers. To find out what the variable is for this kind of expression, you can usually find a common denominator, to multiply with each fraction. If you multiply the numerator and denominator, the denominator will become the same, so the numerators can be added. However, this is the same as multiplying by 1, so it does not change the ratio of the equation. Here are a few rational expressions using addition, and the steps on how to solve them:
Remember, always include restrictions, or non-permissible values with the answers. Here is one more example. Don’t forget to distribute the negative into all terms after it (in the numerator only):
In this week of precalc 11, we learned how to solve rational expressions, which are fraction questions with variables. One thing in specific we learned about was non-permissible values. These are values that x cannot be in each equation. First, here are some examples of what rational expressions can look like:
These expressions cannot have square roots, or exponents that are variables.
A non-permissible value is also known as a restriction, the domain, or values that the expression cannot be defined by. This means there are certain numbers, that when put in place of x, make the denominator zero. When the numerator is zero the equation can still be simplified. However, when the bottom of the fraction (the denominator) is zero, this is not possible. Therefore, these are non-permissible. When you write these restrictions, use the equal symbol with a line through it, meaning “x cannot be equal to…”
Here are the non-permissible values for the expressions shown above:
For example, in the second expression, x cannot be -12. If it was, it would look like this:
Since the denominator is now zero, which is impossible, it means -12 is the restriction on this expression. There can also be multiple restrictions, depending on the fraction and number of variables. The last example had a fraction as a restriction, which is not as easy to solve in your head. This is how I solved to find -3/4:
Here are a couple more examples:
In this week of precalc 11, we learned about solving inequalities. This was partly a review from grade 9, with new skills added on. Linear equations make a line on a graph, while quadratic equations create a parabola (a curved symetrical shape). Both of these equations can be inequalities as well.
Here are some examples of linear inequalities:
6x > 18
4x + 1 < 9
3x – 3 ≥ 4x + 1
These are all quadratic inequalities:
x² + 3x + 2 ≤ 0
2x² > -3x – 7
x² – 7x + 9 ≥ 9
To solve an inequality, you must get the variable on its own. For example:
For this inequality: 6x > 18, you divide both sides by 6, so the x is by itself. Therefore, x would be greater than 3 (x > 3)
To solve this inequality: 3x – 3 ≥ 4x + 1, start by subtracting the 4x from both sides. Then add the 3 from the left side to the right. This would create the equation: – 1x ≥ 4. To make x positive, divide both sides by negative 1. This would make x less than or equal to 4 (x ≤ -4), because when you multiply or divide by a negative, the sign changes.
Here is one final example using the quadratic inequality: x² – 4x > 5
This is done a bit differently than the linear inequalities because you need to factor. First, move the 5 to the left side so the inequality has the zero on the right: x² – 4x – 5 > 0
Then, factor the right side: (x – 5) (x + 1) > 0
Next, find the roots. This is the two numbers, 5 and 1, but with opposite signs: +5, -1.
Finally, write out the solution: x < -1, and x > 5. This is the answer because the original question asked where the inequality was greater than 0. Because this equation would open up and have a minimum vertex (the x² was positive), it is greater than zero to the left (less than -1) and to the right (greater than 5) of the parabola. In between the parabola, x is less than 0, so it wouldn’t satisfy the inequality.
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