Category Archives: Grade 11

Grade 11

La Recette Spéciale

November 23, 2018 jessicap2015 Leave a comment

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Grade 11, Math 11

Precalculus 11 : Math Equations

June 26, 2018 jessicap2015 Leave a comment

Grade 11, Math 11

Top 5 Things I learned During Pre-Calculus 11

June 19, 2018 jessicap2015 Leave a comment

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This week will be my very last blog post for Pre-Calculus 11. I have learned so many things and so many skills from taking this course and I’m really happy I chose it. Some of my favorite parts of this class was getting to sit with multiple people and learn from them. I liked learning new techniques and hearing ideas and helping others, which helped me become a better teacher and helped me understand things better. To be honest I learned that I like writing blog posts, for me it is very time consuming but it is very useful and I can always go back and look at my work and review. It is a skill that I need to have because I need have to be able to transfer what I’m thinking to others and be able to interpret how to do things. I think blog posts will help me on my numeracy assessment because I will have to be able to write down what I’m thinking and why I am thinking in that way. Anyway, This whole post is to share what the top 5 things I learned in this class, so I should get on that.

Number 1: My allowance

In our first chapter we learned about geometric series. I liked this the most because I learned how to trick my parents with my allowance. It didn’t really work for that long because they realized that they would be giving me more money, and I mean a lot more money than I would normally get. But I think that it was cool that I was at least one step ahead of the game for once. I could have been rich in a month. How exciting 😀

Number 2: Improving My Blogging Capabilities

During this semester, I think I really improved my blog posting skills. I learned some new ways to make my blog posts look pretty dang good. I tried really hard when working on my blog posts because if I got stuck on something I could always go back and look at my previous work. It has also helped me so much with reviewing for test and finals. This is also a way for me to be able to teach others and has helped me learn how to teach others, which is a very valuable skill to have.

Number 3: Learned something that connected with socials

I always wondered what pre-calculus could help with outside of class. It too, connects with sequences and series unit. In Socials, we learned a lot about population, and the downfalls with it. Our most common debating topic was if the world was over populated or not. We came to learn that Thomas Malthus wrote an essay on how food production is arithmetic, and population growth is geometric. I learned that he believed that we would one day become overpopulated because we would run out of land to produce food and at the same time run out of room for everyone to live. I thought this was a pretty cool connection.

Number 4: Better Studying Habits

Personally, I learned what was best for me when I Study. Math is a subject where you actually have to do work to learn and understand. For me I really like to break down what I was doing to truly understand. That’s is why I worked so hard in my blog posts. I like to see every step so I know exactly what to do and then I can take shortcuts after. I also learned that it’s not a subject where you can just wait for the day before a test to cram everything in. Working a little bit after school everyday helped me dramatically.

Number 5: How to Work with Others

I think this was a very important thing for me to get used to because sometimes I don’t really want to ask for help. But with working with my peers I learned that its okay to not know how to do something, because when you ask you give the chance for two people to get better. I really enjoyed working with multiple people in my class especially when switching up the table groups because I got to know almost everyone and how to do things in a new way. I think that I got better when working with others because I wasn’t as afraid of my peers when I got to know someone new.

Overall I loved Pre-Calculus 11. It was challenging at times, but I feel like I pushed through and did very well. I am really proud of my effort and positivity this year, as well as my overall performance. I learned many things, not just about math but things outside of class. I had lots of fun and I am really looking forward to Pre-calculus 12 🙂

Grade 11, Socials 11, Uncategorized

Do periods of economic hardship bring forth periods of heightened racism?

June 18, 2018 jessicap2015 Leave a comment

The link to our presentation

https://docs.google.com/presentation/d/10jZAiXxxsHVf4P7hWHAok0SdLcw5T8zmGvt-VCw1ckQ/edit?usp=sharing

Grade 11, Socials 11

Poverty Cycle: Discussion Questions

June 17, 2018 jessicap2015 1 Comment

Explain how the person/people in each story from the documentary is trapped in the poverty cycle.

Danse: trapped in the poverty cycle because he needs a source of income to get housing, but he doesn’t want to go on welfare because he is afraid that if he signs government papers he is somehow giving up his aboriginal rights
Amanda: was stuck in the poverty cycle because she was in a toxic relationship, that involved drugs, she was also didn’t have any education which didn’t help her while trying to find a job. It was also hard for her because she didn’t have a source of income for a really long time because the process of getting welfare was long and difficult but she eventually got it.
Michael: was stuck in the poverty cycle because he had such an addiction to drugs and he didn’t have much money accept for his little flower business. He really needed others to step in to help him get some housing and a source of income.

Refer to the poverty cycle on page 404 of your textbook. Pretend that you are the government and provide an intervention at 4 different points of the cycle and explain how and why this intervention would break the poverty cycle at that point.

number 1: Give an allowance to families with a low source of income or no income. This would be able to help them because the family is not so focused on having to feed themselves but now they can also look for a home and or clothing. This would also help the family have a better health situation and would help the child to develop.
number 2: I would also give free medical care. This would help because not everyone can afford it so when it is provided they can get sufficient help and their money can go towards more important things.
Number 3: Free education and when it comes to university, it should cost less. I think this would go hand in hand with my previous statements of their money can go towards other important things. This would also help with the child’s literacy level which would help with jobs and economic success.
Number 4: Provide family plan: This could go hand in hand with medical care because it could teach you about contraceptives or it could help with how to find help from different charities or organizations.

What should be our general rules/attitude we should take when approaching the issue of solving poverty in our own communities?

I think we should be helpful when trying to solve poverty because it is easy to get stuck in the poverty cycle and it is hard to get out of the poverty cycle. I think we should be more sympathetic and not jump to conclusions when we see a homeless person on the street because they may have or may have not gotten themself into the situation they are in. I think we should pitch in and have more charities and organizations who could help, or we should at least spread word about these current organizations so we can all help people and break the poverty cycle, or lower the rate of people getting stuck in the poverty cycle.

Grade 11, Math 11

Week 17 Pre Calculus 11: Sine Law

June 11, 2018 jessicap2015 Leave a comment

Image result for sine law cartoon jokes

Hey there, thanks for coming back to my blog!! This week in Pre Calculus 11 we learned two very important formulas that will make solving triangles so much easier. I am so thankful that I learned how to use this. If you stay tuned you too, will learn the secrets of trigonometry. Super helpful, keep reading.

Mathematicians are very organized and you will see them use CAPITAL LETTERS, the Theta symbol ( $\theta$ ), small case letters, and x’s now and then

The capital letters in a triangle will represent each angle of a triangle.
( $\theta$ ) which will represent the unknown angle.
small case letters will represent the sides of the triangle. Neat enough the small letter represent the opposing side of the big letter.
X represents the unknown angle or side length.

This is the Sine law

$\frac{a}{ Sin A} = \frac{b}{Sin B} = \frac{c}{Sin C}$

If you are looking for a side length it is easier to put the side lengths (small letters on the top. You can reciprocate the whole Sin law if you are looking for an angle.

EXAMPLE: Triangle ABC

Find the side length of AB (also known as c)

Step 1: Find the given angles and side lengths

AB (c) = x
BC (a) = unknown
AC (b) = 5.0 cm
$A = 180 - 80 - 30 = 70^o$
$B = 80^o$
$C = 30^o$

Step 2: Fill in the Sine Law

$\frac{a}{ Sin A} = \frac{b}{Sin B} = \frac{c}{Sin C}$
$\frac{a}{ Sin 70^o} = \frac{5.0cm}{Sin 80^o} = \frac{x}{Sin 30^o}$

Step 3: Find out which equation to use

$\star$ We need side c, and we have the numerator and denominator of $\frac{b}{Sin B}$ . So we can use…

$\frac{b}{Sin B} = \frac{c}{Sin C}$
$\frac{5.0cm}{Sin 80^o} = \frac{x}{Sin 30^o}$

Step 4: Solve for c

$\frac{5.0cm}{Sin 80^o} = \frac{x}{Sin 30^o}$
$Sin 30^o\cdot \frac{5.0cm}{Sin 80^o} = x$
x = 2.5 cm

Final Answer: Side AB (side c) = 2.5cm

Trying to find an Angle using Sine Law.

Steps:

Use the reciprocated version of the Sine Law $\frac {Sin A}{a} = \frac {Sin B}{b} = \frac {Sin C}{c}$
Find the given angles and side lengths
Fill in the Sine Law
Find which equation to use
Solve
To get theta ( $\theta$ ) by itself you have to use inverse sine (2nd function sine)
Check which quadrants they are in to make sure that you get the correct angle.

Example: Find Angle B

Step 1: Use the reciprocated version of the Sine Law $\frac {Sin A}{a} = \frac {Sin B}{b} = \frac {Sin C}{c}$

Step 2: Find the given angles and side lengths

AB (c) = 10 cm
BC (a) = unknown
AC (b) = 7.0 cm
$A = unknown$
$B = \theta$
$C = 150^o$

Step 3: Fill in the Sine Law

$\frac {Sin A}{a} = \frac {Sin B}{b} = \frac {Sin C}{c}$
$\frac {Sin A}{a} = \frac {Sin B}{7.0 cm} = \frac {Sin 115^o}{10 cm}$

Step 4: Find which equation to use..

$\star$ I have enough information to use .

$\frac {Sin B}{b} = \frac {Sin C}{c}$
$\frac {Sin B}{7.0 cm} = \frac {Sin 115^o}{10 cm}$

Step 5: Solve

$\frac {Sin B}{7.0 cm} = \frac {Sin 115^o}{10 cm}$
$Sin B= \frac {Sin 115^o}{10 cm}\cdot 7.0 cm$
B= $Sin^{-1}( \frac {Sin 115^o}{10 cm} \cdot 7.0 cm)$
$\angle B \doteq 39^o$

Step 6: Check which quadrants they are in to make sure that you get the correct angle.

$\star$ Using the cast rule

This means that $Sin 39^o$ can be in either Quadrant 1 or 2 Because it will be positive where All and Sine is.

$\star$ The reference angle in both Q1 and Q2 would be $39^o$

$\star$ The rotation angle in Q1 would be…

0 + 39 = $39^o$

$\star$ The rotation angle in Q2 would be…

180 – 39 = $141^o$

$\angle B$ can either be $39^o$ or $141^o$

But since $141^o + 115^o = 256^o$ and exceeds the maximum amount a triangle can be with all three angles added together to make $180^o$ $\angle B \neq 141^o$

$\therefore \angle B = 39^o$ and not $141^o$

This is how you would use the Sine Law in Trigonometry

Grade 11, Math 11

Pre Calculus 11 Week 16: Multiplying and Dividing Rational Expressions

June 4, 2018 jessicap2015 Leave a comment

During the Unit of Rational Expressions I have learned a few things like multiplying and dividing rational expressions. Compared to adding and subtracting these expressions it is a tad bit easier. I have learned to Just Do IT!

Steps When Multiplying Rational Expressions

Simplify: Can do so by factoring to see if anything from the numerator and the denominator will cancel out.
Multiply across: Just Do It!
Simplify: By taking out a common factor if possible
Variables: You can simplify variables too! If there is a $x^2$ on the denominator and a single x in the numerator we know that at least one pair can cancel out. The leftover variable will be left on the same side that it is on.
State any Non Permissible values

Example 1: $\frac{10x^2}{32x^3}\cdot\frac{4x}{22x^2}$

Step 1: Simplify: Get the simplest form of each fraction by taking out a common factor.

In this case it is two

$\star$ could be a different number for the two fractions, it’s not always the same.

$\frac{10x^2}{32x^3}\cdot\frac{4x}{22x^2}$
$\frac{5x^2}{16x^3}\cdot\frac{2x}{11x^2}$

You can also simplify the the variables

$\frac{5x^2}{16x^3}\cdot\frac{2x^3}{11x^2}$
$\frac{5}{16x}\cdot\frac{2x}{11}$

Step 2: Multiply across (Just Do It!)

$\frac{5}{16x}\cdot\frac{2x}{11}$
$\frac{10x}{176x}$

Step 3: Simplify by taking out common factors if possible

$\frac{10x}{176x}$
$\frac{5x}{88x}$

Step 4: Variables

$\frac{5x}{88x}$
$\frac{5}{88}$

Step 5: State any Non Permissible Values

$x\neq 0$

This is how you would multiply rational expressions

How to Divide Rational Expressions

Steps When Dividing Rational Expressions

Dividing rational expressions is very similar to that of multiplying. There is just one extra step. It is to reciprocate the fraction that is to right after the divide sign.

Simplify: Can do so by factoring
Reciprocate
Cancel any pair terms
Multiply across: Just Do It!
Simplify: By taking out a common factor if possible
Variables: You can simplify variables too! If there is a $x^2$ on the denominator and a single x in the numerator we know that at least one pair can cancel out. The left over variable will be left on the same side that it is on.
State any Non Permissible values

Example 2: $\frac{3x-27}{x^2 - 1}\cdot\frac{x^2 -7x - 8}{x^2 - 81}\div\frac{5x}{2x - 2}$

Step 1: Simplify (factor)

$\frac{3x-27}{x^2 - 1}\cdot\frac{x^2 -7x - 8}{x^2 - 81}\div\frac{5x}{2x - 2}$
$\frac{3(x-9)}{(x -1)(x +1)}\cdot\frac{(x-8)(x+1)}{(x-9)(x+9)}\div\frac{5x}{2(x - 1)}$

Step 2: Reciprocate (this gets rid of the dividing sign and makes it a multiplying sign)

$\frac{3(x-9)}{(x -1)(x +1)}\cdot\frac{(x-8)(x+1)}{(x-9)(x+9)}\div\frac{5x}{2(x-1)}$
$\frac{3(x-9)}{(x -1)(x +1)}\cdot\frac{(x-8)(x+1)}{(x-9)(x+9)}\cdot\frac{2(x - 1)}{5x}$

Step 3: Cancel out any pair terms

$\frac{3(x-9)}{(x -1)(x +1)}\cdot\frac{(x-8)(x+1)}{(x-9)(x+9)}\cdot\frac{2(x - 1)}{5x}$
$\frac{3}{1}\cdot\frac{(x-8)}{(x+9)}\cdot\frac{2}{5x}$

Step 4: Multiply Across (Just Do It!)

$\frac{3}{1}\cdot\frac{(x-8)}{(x+9)}\cdot\frac{2}{5x}$
$\frac {2\cdot 3\cdot (x-8)}{5x\cdot (x+9)}$
$\frac {6(x-8)}{5x(x+9)}$

State the Non Permissible Values

$x\neq { -9, -1, 0, 1, 9}$

FINAL ANSWER: $\frac {6(x-8)}{5x(x+9)}$

This is how you would divide rational expressions.

Grade 11, Socials 11

Socials 11: Defining Poverty

May 30, 2018 jessicap2015 Leave a comment

In your own words, explain the problems with measuring poverty (standard of living, absolute vs relative, HDI)

There are many problems when trying to define poverty. Many countries live in different situations, for example a less developed country may seem poor to a more developed country. This is called standard of living. Some countries like Canada have a higher standard of living than others. When comparing absolute poverty vs. relative poverty on one country compared to another it can be different. The needs of a person in a developed country is higher for example the low standard of living for a higher developed country could very well be the higher standard of living for a less developed country. Absolute poverty is when you do not have enough support or money to afford the essentials in life, and relative poverty is being able to afford your basic needs. Canada’s absolute poverty could be another countries relative poverty. This is why it is difficult to determine if a country is poor while being compared to another country.

What do you think is the best way to measure poverty in Canada and the world.

I think the best way to measure poverty in Canada and the world is to use the Human Development Index. I think gathering number relative to that countries standard of living adding it together and then comparing it might help even out and help determine if a country is in actual poverty or not. Basic Human Needs, which includes medical care, sanitation, and shelter foundations of wellbeing, which covers education, access to technology, and life expectancy and opportunity, which looks at personal rights, freedom of choice, and general tolerance is a way that could help determine if a country is poor when all these factors are added up and compared.
I think that this would be better than just comparing the economic growth of a country because it adds more factors into effect and will be more accurate and comparable.

Grade 11, Math 11

Pre Calculus 11 Week 15: Adding and Subtracting Rational Expressions With Binomial and Trinomial Denominators

May 27, 2018 jessicap2015 Leave a comment

Image result for adding and subtracting rational expressions with binomial and trinomial denominators jokes

This week in Pre calculus 11 we learned how to add and subtract rational expressions that contained a binomial or trinomial denominator. These are quite ugly and difficult and require focus. I will show an example of each, and hopefully help you understand how to simplify these kinds of expressions.

Steps: Binomial Denominator

Simplify the Denominator if possible
Find the Common Denominator
Make Equivalent Fractions
Add and or Subtract
Simplify/Reduce
State the Non-Permissible Values

Example: $\frac{8}{6x+9} + \frac{3}{4x-4}$

Step 1: Simplify the Denominator

$\frac{8}{6x+9} + \frac{3}{4x-4}$
$\frac{8}{3(2x+3)} + \frac{3}{4(x-1)}$

Step 2: Find the Common Denominator

$3(2x+3)\cdot 4(x-1)$
12(2x+3)(x-1) = Common Denominator

Step 3: Make Equivalent Fractions

Multiply the denominator of one side to the other side (top and bottom) and then use the other denominator and multiply it to the other expression. Only multiply by what is need to get to the common denominator.

$\frac{8}{3(2x+3}\cdot\frac{4(x-1)}{4(x-1)} + \frac{3}{4(x-1)}\cdot\frac{3(2x+3)}{3(2x+3)}$
$\frac{32(x-1)}{12(2x+3)(x-1)} + \frac{9(2x+3)}{12(2x+3)(x-1)}$
$\frac{32x-32}{12(2x+3)(x-1)} + \frac{18x+27}{12(2x+3)(x-1)}$

Step 4: Add or Subtract $\star$ Make into one big fraction

$\frac{32x-32 + 18x+27}{12(2x+3)(x-1)}$

Step 5: Simplify/Reduce

$\frac{32x - 32 + 18x + 27}{12( 2x + 3 )( x - 1)}$
$\frac{32x + 18x - 32 + 27}{12( 2x + 3 )( x - 1)}$
$\frac{50x - 5 }{12( 2x + 3 )( x - 1)}$

$\star$ Simplify the Numerator

$\frac{50x - 5 }{12( 2x + 3 )( x - 1)}$
$\frac{5(10x - 1) }{12( 2x + 3 )( x - 1)}$

Step 6: Non-Permissible Values

$\frac{5(10x - 1) }{12( 2x + 3 )( x - 1)}$
12(2x+3)(x-1)
2x + 3 = 0
2x = -3
$x = -\frac {3}{2}$
$x\neq -\frac {3}{2} , 1$

FINAL ANSWER: $\frac{5(10x - 1) }{12( 2x + 3 )( x - 1)}$

This is how you would simplify a rational expression with a binomial denominator

Steps: Trinomial Denominator

Simplify the Denominator if possible
Find the Common Denominator
Make Equivalent Fractions
Add and or Subtract
Simplify/Reduce
State the Non-Permissible Values

Example: $\frac{b}{b^2 + 10b + 24} + \frac{2b}{b^2 + 12b + 32}$

Step 1: Simplify the Denominator

$\frac{b}{b^2 + 10b + 24} + \frac{2b}{b^2 + 12b + 32}$
$\frac{b}{(b + 6)(b + 4)} + \frac{2b}{(b + 8)(b+4)}$

Step 2: Find the Common Denominator

$\star$ We know that we need a (b+8), a (b+4) and a (b+6) we don’t use the extra (b+4) because there is already one being used, we don’t want have any duplicates.

(b+8)(b+6)(b+4) = Common Denominator

Step 3: Make Equivalent Fractions

Only multiply by what is needed to get to the common denominator

$\frac{b}{(b + 6)(b + 4)}\cdot\frac {b+8}{b+8} + \frac{2b}{(b + 8)(b + 4)}\cdot\frac {b+6}{b+6}$
$\frac{b(b+8)}{(b + 8)(b + 6)(b+4)} + \frac{2b(b+6)}{(b + 8)(b+6)(b+4)}$
$\frac{b^2+8b}{(b + 8)(b + 6)(b+4)} + \frac{2b^2+12b}{(b + 8)(b+6)(b+4)}$

Step 4: Add or Subtract $\star$ Make into one big fraction

$\frac{b^2+8b}{(b + 8)(b + 6)(b+4)} + \frac{2b^2+12b}{(b + 8)(b+6)(b+4)}$
$\frac{b^2+8b + 2b^2+12b}{(b + 8)(b+6)(b+4)}$

Step 5: Simplify/Reduce

$\frac{b^2+8b + 2b^2+12b}{(b + 8)(b+6)(b+4)}$
$\frac{b^2 + 2b^2+ 8b+ 12b}{(b + 8)(b+6)(b+4)}$
$\frac{3b^2+ 20b}{(b + 8)(b+6)(b+4)}$

$\star$ Simplify the Numerator

$\frac{3b^2+ 20b}{(b + 8)(b+6)(b+4)}$
$\frac{b(3+ 20)}{(b + 8)(b+6)(b+4)}$

Step 6: Non-Permissible Values

$\frac{b(3+ 20)}{(b + 8)(b+6)(b+4)}$
$x\neq -8, -6, -4$

This is how you would simplify Binomial and Trinomial Rational expressions when adding and or subtracting. I hope this blog post really helped you understand how to simplify these expressions. Feel free to look back at some of my other blog posts. They may help you with anything you are struggling with. Thanks 🙂

Below are two videos to help further understanding

here is a video that helped me understand how to add and subtract (binomial denominator) kinds of expressions.

And here is another video that helped me understand how to add and subtract rational expressions with a trinomial denominator (in the video the trinomial has already been factored).

Grade 11, Math 11

Pre-Calc 11: Week 14 Multiplying and Dividing Rational Expressions

May 23, 2018 jessicap2015 Leave a comment

Last week in Pre-Calculus 11 we learned about multiplying and dividing rational expressions. A thing to remember about expressions is that it does not have an equal sign and you do not need to solve, you only need to simplify. I will be teaching you what non-permissible values are and how to determine what they are.

Non-permissible value : Values that cause the fraction to have a denominator with a value of zero. (In math, we cannot divide by zero).

Here is an example of multiplying rational expressions

Steps:

Step 1: Simplify $\frac{3x}{2(x-3)}\cdot\frac{8(x-3)}{9x^2}$

Non permissible values: $x\neq -3$

$\frac{3x}{2(x-3)}\cdot\frac{8(x-3)}{9x^2}$
remove (x-3) from top and bottom because they cancel eachother out
then it becomes… $\frac {3x}{2}\cdot\frac{8}{9x^2}$

Step 2: Simplify by multiplying across (Just do it)

$\frac {3x}{2}\cdot\frac{8}{9x^2}$
$\frac{24x}{18x^2}$

Step 3: Take the highest common factor from both the numerator and the denominator.

In this case 6 is the highest number that goes into both.

$\frac{24x}{18x^2}$
$\frac{4x}{18x^2}$

Step 4: Notice that x is on the bottom and the top, if it has a pair it can cancel out. two x’s on the bottom one on the top. when they cancel out each other you will be left with only 1 on the bottom.

$\frac{4x}{3x^2}$
$\frac{4}{3x}$

Final Answer: $\frac{4}{3x}$

Dividing Rational Expressions

There are many steps when dividing rational expressions

Simplify the fraction: can factor or take out the common denominator.
State the non permissible values.
Reciprocate the second fraction and it will because a multiplication expression.
State the restrictions again because there are new values in the denominator and could be non-permissible.
Simplify (cancel out like terms that have a pair on the numerator and denominator.
Multiply across (Just do it)
Simplify again if possible.

Step 1: Simplify $\frac{x+5}{x-4}\div\frac{x^2 - 25}{3x-12}$