on For my final project in Creative Writing, I chose the first chapter of the novel option. My story, The Cabin, is a mystery novel that follows the disappearance of an NYU student who’s vacationing in southeast Connecticut for the weekend on a couple’s trip.
on on This semester in financial accounting 12 I used my critical thinking CC to get me through the course. For example, I used critical thinking to identify errors and inconsistencies. To elaborate, critical thinking helps in identifying errors or inconsistencies in financial records. It is important to question inconsistencies and seek explanations to ensure the accuracy of financial statements. For example, this skill is applied when working through the accounting workbook. The questions are very interactive and require critical thinking and logic to be able to analyze the questions.
on For my reflection I chose communication core competency. Having never come into an economics class before grade 12, I quickly learned the value that effective communication has within Economics 12. This class has shown me how to communicate my ideas through the coursework and tests that we have been assigned. For example, the quizes that we do in Economics 12 require through explanations of our point of view. Often times, there are no right or wrong answers, and we must rely on our communication skills to get our point across and explain our reasonings. Additionally, my communication skills have bettered as a result of the economics term map that we were tasked to create. I had to connect and explain how over 30 economics terms are all related, as well as generate examples for each term. This project really solidified my ability to communicate my ideas. I will use the communication skills I learned in economics 12 across many other boards in my education, and work life.
on This week in math we learned how to solve linear and quadratic rational equations, testing out different methods. For my blog post, I chose solving linear rational expressions because I find it satisfying to isolate x at the end. I also like that I can make a connection with other subjects I have learned in math this semester, and in years before. This undersyanding is important because it is a subject that will continue to advance, and having the foundational understandings of how to solve a linear rational expression is necessary in the success of your future math self.
Lets look at some examples…
Example a.
Step 1. My first step is to identify an LCM. Since one term has a denominator of x, and another of 9, the LCM which must include a 9 and an x can only be 9x.
Step 2. Identify non-pernissible values. Since our LCM is 9x, we need to identify a number that would make 9x = 0. Our non permissible value = 0; 9(0) = 0.
Step 3. Make the denominator common – while we know our LCM, we need to multiply each denominator by whatever term sets it to 9x and do the same to the numerator. In our first term, 9 is the denominator, so we multiply both the denominator and numerator by x. In our second term, we multiply the top and bottom by 9. Since everything is over the same denominator, there is no need to include the denominator in your work.
Step 4. Reorder the terms – now that we have distributed and gotten our new values for the numerator, we need to re order the terms so that the variable is on one side, and the constant on the other. I want to keep x positive, so I move x to the left with 4x making 3x.
Step 5. Now we can isolate for x – I divide everything by 3 to get rid of the coefficent infront of x, leaving me with x = 3. It is not our non permissible value so there is no need to reject the equation.
Example b.
Step 1. As always, we must determine our LCM. Looking at this equation I see there is a 4a, a 12, and 3a in our denominator’s place. I know that 12 includes 4 and 3, and since there is an a, this means our LCM is 12a.
Step 2. Now we must identify our non-permissible values, so we dont end up with a solution that is extraneous. a = 0, because if you multiplied each denomintaor by 0, you will end up with zero.
Step 3. We must now multiply each denominator by a value that will let it equal 12a. We must multiply the numerator by the same values as the denominator. I distribute the 3 into (a + 5), the a to the 11, and the 4 to the 2. Since each denominator is the same, we do not need to include it in our work.
Step 4. Reorder terms; Since both a values are on the left handside, i can keep them there and add them together. I move the 15 to the right hand side becoming 15 + 8 which equals -7.
Step 5. Isolate for X; I divide both sides by 14 to isolate x, leaving me with a negative fraction as my value for a. a =
Example C.
Step 1. I notice right away that one of my denominators contains a 
Step 2. Now I can find my non permissible values, look at the value being added to x and determine what would make that value equal to zero. Since I have a 4 and a 3, the non permissible values are -4 and -3.
Step 3. Now we identify an LCM; looking at this I know that our LCM is (x + 3)( x + 4)
Step 4. Make denominator equal to (x + 3)(x + 4) . I multiply my first term top and bottom by (x +4), my second term by (x+3) and my third term already contains both parts of the denominator. Now I can distribute and write out my values.
Step 5. Now I can add my like terms; 2x and 3x = 5x, and 8+9 = 17.
Step 6. Rearrange, I want to keep my x value positive, so I shift the constant to the right.
Step 7. All thats left to do is isolate; I divide everything by 5 to isolate x, giving me the answer x = -2.
on This week in math we covered solving one variable inequalities. This understanding is important because it ties in the aspects we learned earlier in the chapter and will continue to be put to use. I chose this topic because I understood how to solve this type of equation right away, as I was able to use the skills I have previously learned in this chapters.
Lets take a look at some examples…
Example a.
Step 1. My first step is to change the inequality into an equation with an ‘=’ sign so I can solve.
Step 2. I must now distribute the 2 to the terms inside the bracket.
Step 3. Now we rearrange so that the x values are to the left, and the constants to the left.
Step 4. Since I have a leading coefficent of 2 infront of my x value, I need to isolate the x by dividing both sides of the equation by 2. Finally it is important to substitute your inequality sign back in to get your final answer.
Example b.
Step 1. Our first step is to change the inequality to an equation.
Step 1. Since this example is written as a fraction, we want to eliminate the fraction by multiplying both sides of the equation by the denominator of 7. On the lefthand side, both 7s cancel out and on the right -3 is multilpied by 7 to get -21.
Step 3. Since the coefficent of x is one, we do not need to do any isolating; simply reearange the terms. X values to the left and constants to the right. Finally remember to change your ‘=’ sign back to an inequality.
Example C.
Step 1. As always, start by converting the inequality to an equation.
Step 2. Since in this example, both sides of the equation are written as fractions with different denominators we must multiply both sides of the equation by 21.
Step 3. Now that we have eliminated our fractions, we need to distribute the values outside our brackets to the inside terms.
Step 4. Finally we can rearrange. The x values to the left and constants to the right. Since the coefficent of x is 1, there is no need to isolate. Lastly, convert the equation back to an inequality.
Vocab :
Equation – contains an = sign.
Inequality – Compares two values, showing if one is less than, greater than, or simply not equal to another value.
on This week in math we learned how to convert equations in general form, where you can identify the y intercept, into standard form (or vertex) form. This is an important skill because in order to be able to graph a quadratic function you need the vertex, whether there is a stretch, and if the parabolla is facing upwards or down. I chose this topic because I like how similar it is to our last unit where we learned to complete the square.
Lets look at some examples…
Example A.
Step 1. In my first step I divide my B term AKA the middle term by 2 and square the product of that division. After dividing the term and squaring I am left with 9. To keep the equation in balance I add positive 9 and negative 9.
Step 2. My next step, the negative 9 can leave the brackets and there is no stretch that it needs to be multiplied with so its value remainss the same. Now we can add our two constants of negative 9 and one together leaving us with our terms in the brackets and negative 8 outside the brackets.
Step 3. My last step is to combine the inside terms. To do so; put x and b/2 into brackets and square it, with our constant of -8 remaining on the outside. To find the vertex simply take the value inside of the brackets and change its value which would give us -3 as our x value of the vertex. The Y value is our constant on the outside, -8. It remains negative.
Lets take a look at an uglier example…
Example B.
Step 1. My first step is to get rid of the leading coefficent infront of my A value. To do this out the A and B values in brackets and divide them by 2.
Step 2. My next step is to divide the B term by 2. Since my B term is a fraction over 2, multiply the denominator by 2 giving the new product after being divided. Square this term and add it into the brackets, as well as its negative value to balance out the equation. To get rid of the extra negative value we added, multiply it by the coefficent 2 we factored out so it can leave the brackets.
Step 3. Now we may simplify. Combine X plus the divided b term in brackets and square it. For our constants on the outside of the brackets, we can first divide -50/16 by 2 giving -25/8. Because we are combining -3 with -25/8; the denominators need to be the same so we must multiply -3/1 by 8. This leaves us with our final inside terms and our constant of -49/8. To find the vertex, take the value inside our bracket and change its sign to negative giving us our x value. This means there is a horizontal shift to the left. Our why value is the constant outside the brackets. This means there is a vertical shift downwards. The 2 outside the brackets tells us there is a stretch of 2.
Glossary :
Horizontal shift : inside changes that affect the input (x−) axis values and shift the function left or right.
Vertical shift : The outside changes that affect the output (y−) axis values and shift the function up or down.
Vertex/standard form : y = a ( x − h ) 2 + k
Stretch : (narrowing, widening) of the parabolla
on This week we learned how to predict the vertex of an equation without graphing it. We have not yet learned the equation to finding the vertex, but predicting the vertex lets us use what we know about quadratic equations without havinv to usw graphing technology. I chose this subjet because I find it to be straight forward and logical. This understanding is important because being able to indentify the vertex is a skill that will be used in math as we continue to develop our skillset.
Lets look at examples a and b that have been handwritten. 
We can verify these solutions using graphing techonology to ensure we are correct.
Example C
Verify with desmos
Glossary :
Vertex – the maximum or minimum point on the equation’s parabola.
Parent function –
Vertical translations – how far up or down along the y axis the vertex will be.
Horizontal translations – how far right or left along the x axis the equation will be.
on This week in math we learned the begging steps of graphing a quadratic equation, but before we learned how to do that we learned how to indentify if a function is linear, quadratic, or neither using a table of values. I chose this subject because I like having a visual, such as a table of values, to see what my function would look like. This understanding is important because knowing the difference between a linear equation, which we dealt with in earlier years, and a quadratic equation, which we are dealing with now and will continue to reappear in later math, is a fundemental skill and comprehension to have. For reference, lets look at a linear function on a graph and a quadratic function on a graph.
Linear :
Quadratic :
Example a.)
Step 1. My first step is to observe the pattern of x values in my table of values, notice how they are decreasing by 1. Therefore, the difference between the y values may determine the behaviour of the ordered pairs.
Step 2. Find the difference between the first two terms. -2 – (-3) = 1. Repeat this process to find the first difference. You should get 1, 2, 4, and 8. Notice how the first difference is not constant. This indicates we are NOT evaluating a linear function.
Step 3. We can now find the 2nd difference. Use the values of the first term to find the second difference. The first value is 2-(1) = 1. Repeat this process and get 1, 2 and 4. Notice the second difference is NOT constant. this means that it is NEITHER a linear function or quadratic.
Example b.)
Step 1. Notice that the x values are increasing by 2 units. The difference between the y values may detrmine the behaviours of the ordered pairs.
Step 2. Find the first difference. To get our first value, find the difference between 5 and 0; 0 -(5) = -5. If you continue this process for the rest of the values you get -5,-7,-9,-11. Notice how the first difference is NOT constant meaning that it is not a linear function.
Step 3. Find the second difference. Use the values of the first difference, and find the difference between each value. -7 – (-5) = -2. If you continue this process you get -2, -2, -2. The second difference IS constant meaning it is a quadratic function.
Glossary :
Linear function :a function whose graph is a straight line, that is, a polynomial function of degree zero or one.
on This week in math we learned the quadratic formula and how to solve quadratic equations with this formula. I chose this topic because I find using the quadratic formula to be much more consistent and straight forward than ‘complete the square’ method. This skill is important because solving quadratics, especially with the quadratic formula, is a topic that we will build of off in our next units, and the next math courses that I will take.
Lets look at some examples…
Example 1.
This example is a simple quadratic without a leading coefficent larger than one.
Step 1. My first step is to write out the quadratic formula, so I can refer to it in my next step.
Step 2. In my second step, I need to identify the a,b and c values so I know where to plug in my values. In this particular example, the equation is already in general form which makes things much more simple. The 
Step 3. Plug in your a,b,c values into the quadratic equation to solve for x1 and x2. Remeber, if you have two negatives it is equal to a positive so -6 = 6.
Step 4. Expand your values
Step 5. Simplify the inside terms
Step 6. Simplify 
Step 7. Since all terms are divisble by 2, we simplify and divide each term by 2. All terms must be divisble by the same value.
Example 2.
In this example we have a quadratic equation with a leading coefficent larger than 1.
Step 1. Convert the equation into general form. The -7x changes to a positive when it is moved to the other side of the equal sign.
Step 2. Identify values of a,b,c. Recall how we solved for these values in example 1.
Step 3. Plug each value into the quadratic formula.
Step 4. Expand the inside terms. BE CAREFUL. 4 X – 1 = -4; -4 X -4 = POSITIVE 16.
Step 5. Add like terms.
Step 6. see if you can simplify the radical, in this case we cannot which gives us our final answer. The +/- sign before the radical indicates there is two different answers. One positive and one negative. A quadratic equation always has 2 answers.
Glossary :
Constant :a number on its own, or sometimes a letter such as a, b or c to stand for a fixed number.
General Form : A formula that contains the degree of variables in decreasing order that = zero
Leading coefficent : the coefficient of the term of highest degree in a given polynomial.
Quadratic Equation : any equation containing one term in which the unknown is squared and no term in which it is raised to a higher power.
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