# Week 6- Preclac

One of the most important things I think about this week is “quadratic formula”. This formula can help us more easier to calculate the value of x in some extent. Quadratic formula =$\frac{-b+\sqrt{b^2-4ac}}{2a}or\frac{-b-\sqrt{b^2-4ac}}{2a}$. The main purpose is using the quadratic formula to solve quadratic equations. First of all, we’re going to set the result of an quadratic equations equal to zero, in other words, we’re going to move all the terms to one side of the equation (left or right). Then find the value of a,b,c in quadratic formula of quadratic  equation. a= the coefficient of leading term,b= the coefficient of x term and c= constant. Finial, substitute the values into the quadratic formula to get the answer x.

EX: Factor the following equation by quadratic formula

$5x^2+3=4x$

(1) $5x^2-4x+3=0$

(2) $a=5, b=-4, c=3$

(3) $\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{-(-4)+\sqrt{(-4)^2-4\cdot5\cdot3}}{2\cdot5}=\frac{4+\sqrt{16-60}}{10}=\frac{4+\sqrt{-44}}{10}$

$\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-(-4)-\sqrt{(-4)^2-4\cdot5\cdot3}}{2\cdot5}=\frac{4-\sqrt{16-60}}{10}=\frac{4-\sqrt{-44}}{10}$

(4) x does’t exist because the value of x in$\sqrt{x}$ can’t smaller than 0

# Post-Secondary Event-Reflection

The 3 institutions that interested me the most were University of British Columbia (UBC), because The University of British Columbia is famous for its medical science, a major that has a global reputation and I believe that I can learn more useful and skillful knowledge here. Simon Fraser University (SFU) has all courses that are open to international students and the requirements of each subject are within my reach. Also the location of university allows me to attend classes without changing my home-stay family, namely it is unnecessary for me to face the problem of having to change my home.  Lastly, the University of Toronto, the major reason is this school is very famous, and also has a very high ranking in the world. If I were forced to have to choose a post-secondary school today, I would choose the University of British Columbia. The main point that attracted me is the reputation of it’s medical curriculum, even though it’s extremely difficult and is not international student friendly in terms of  application approval. (besides the basic requirement, international students also need to supply the score of IELTS or TOEFL test) The program that I would like to learn is biomedical because the employment rate of this major is extremely high, and UBC can offer a PhD for this major. In addition, becoming a professor in university is my childhood dream. Before I finish high school, the most intuitive advantage of seeing so many choices is to help me choose a post-secondary school that suits me best by constantly comparing the pros and cons of each choice. My favourite part about seeing these different post-secondary options is to keep me thinking about what I need because I know this choice is important for my future life. Constantly comparing the strengths and weaknesses of different universities and thinking about what courses I really need will, undoubtedly, translate into my later-life  academic success.

# Week 5- Precalc

The new chapter began in this week is “Factoring Polynomial Expressions”, the steps to decompose a polynomial can be divided into five: CDPEU.

C: common

D: difference of squares( binomials)

P: pattern(trinomials)

E: easy( the coefficient of highest power is equal to 1)

U: ugly( the coefficient of highest power is not equal to 1)

EX: factor each polynomial expression

1) $2x^2-50y^2=2(x^2-25y^2)=2(x+5y)(x-5y)$

In this case, find the common is 2 so that take the 2 out and put it on the outside.Each of the terms of the original equation is divided by two and it’s just the expression of difference of squares, and the result can be simplified directly to is equal to $2(x+5y)(x-5y)$.

2)$(x+1)^2+4(x+1)-32=(x+1+8)(x+1-4)=(x+9)(x-3)$

In this case, view (x+1) as “A”, so this term become to$A^2+4A-32$.You can simplify it to (A+8)(A-4) by multiplying the cross, then replace (x+1) back into the formation, so that we can find the answer is (x+9)(x-3).

# Week 4-Precalc 11

In this week, I learn how to add, subtract, multiply and divide radicals ,actually all of the radicals is similar to the way that expressions calculated. An expression is composed of coefficient, variables and terms. The calculate of expression you just need going to evaluate the coefficient of the expression, as the same way to do radical. Nonetheless simplifying the radical expressions have been done at the beginning makes calculation more easier.

Add:$\sqrt[n]{a}+3\sqrt[n]{a}=4\sqrt[n]{a}$, (a≥0)

(2) Group the radical expressions that have same radical,such as $\sqrt{2},2\sqrt{2},3\sqrt{2}$

(3) Calculate the coefficient have same radical only

EX:$\sqrt{36x}+\sqrt{16x}+\sqrt{4x}=6\sqrt{x}+4\sqrt{x}+2\sqrt{x}=12\sqrt{x}$. (x≥0)

Subtract:$\sqrt[n]{a}-3\sqrt[n]{a}=-2\sqrt[n]{a}$, (a≥0)

(2) Group the radical expressions that have same radical,such as $\sqrt{3},2\sqrt{3},3\sqrt{3}$

(3) Calculate coefficients have same radical only

EX:$\sqrt{8x}-\sqrt{32x}-\sqrt{27x}=2\sqrt{2x}-4\sqrt{2x}-3\sqrt{3x}=-2\sqrt{2x}-3\sqrt{3x}$, (x≥0)

Multiply：$\sqrt[n]{a}\cdot\sqrt[n]{b}=\sqrt[n]{ab}$, (a≥0 & b≥0)

The biggest difference between multiplication and addition and subtraction is not just multiplying coefficients, but also radicand.

(2) Calculate coefficients and radicals respectively

(3)Simplify the product

EX:$\sqrt{8x}(\sqrt{2x}-\sqrt{3x})=2\sqrt{2x}(\sqrt{2x}-\sqrt{3x})=2\sqrt{4x^2}-2\sqrt{6x^2}=4x-2x\sqrt{6}$, (x≥0)

Divide:$\frac{a}{\sqrt{b}}=\frac{a}{\sqrt{b}}\cdot\frac{\sqrt{b}}{\sqrt{b}}=\frac{a\sqrt{b}}{b}$, (b＞0)

(2) Calculate coefficients and radicals respectively

(3)Simplify the product

EX:$\frac{2}{\sqrt{3}}=\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$

# Maya‘s Goals

Use the space below to write your personal goal statement. Colour-code it as indicated above to ensure you have all the necessary components for a good goal statement.

In this semester, I will get A or A+ Level of my biology and chemistry classes. I plan to complete all of the assignments on time and write down all of the notes I have learned in these two classes specifically and carefully, also get A or A+ in each quizzes and tests. When I get anything wrong or not learned a certain knowledge point in the learning process clearly，I will going to ask questions for my teacher and classmates until I understand the knowledge point. Teachers who have more experiences can help me to figure out the issue effective and efficiently.

Use the space below to write your academic goal statement. Colour-code it as indicated above to ensure you have all the necessary components for a good goal statement.

Become a biology professor or biomedical engineer before 35 years old. I plan to get A or A+ grade of all the subjects that are related to these two career and attend the study of my professor have doing in university. There are many obstacles to becoming a biology professor and biomedical engineer, such as questions about learning or not fully understanding what the professor has to say. I will communicate with my professor and ask him to make sure that I understand the context. All the people who have helped me, no matter how much they have helped me, are people who deserve my gratitude. Like teachers, parents, friends, etc.

# Week 3-Precalc 11

In this week, we concluded the chapter: The Arithmetic Sequence & Geometric Sequence and started a new chapter :Absolute Value of a Real Number. The defination is that the principal square root of the square of a number, as the example , absolute value of -4.

$\sqrt{(-4)^2}=\sqrt{16}=4$

If we want to find the absolute value of a number, we can use the sign of the absolute value and this sign is called absolute value bars. We can represent it by two vertical lines.

|-4|=$\sqrt{(-4)^2}=\sqrt{16}=4$

EX:Find the absolute value of 15 , -7.6 , 0 ,$\frac{-7}{2}$

a)  |15|=$\sqrt{15^2}=\sqrt{225}=15$

b)  |-7.6|=$\sqrt{-7.6^2}=\sqrt{57.76}=7.6$

c)  |0|=$\sqrt{0^2}=\sqrt{0}=0$

d)  |$\frac{-7}{2}$|=$\sqrt{(\frac{-7}{2})^2}=\frac {7}{2}$

# Week 2- Precalc 11

This week we learned the geometric series, and I think the most interesting thing for me is to find the sum of the geometric series of infinite terms.There’s a formula for summation:$s_{infinite}=\frac{a}{1-r}$.However, before we use this formula we need to determine whether the geometric series is convergent or diverging.If the common ratio (r) of this series are greater than 1 or less than negative 1, it is diverging.We can not calculate the sum of diverging series.When the common ratio is between one and zero or between zero and negative one, the sum of this sequence can be calculated by the formula.

EX：

1)Find the sum of this geometric series :1+$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}$+…

$r=\frac{1}{2}$,

$s_{infinite}=\frac{a}{1-r}=\frac{1}{1-\frac{1}{2}}=2$

2)Find the sum of this geometric series:1+5+10+50+…

r=5

So this sequence is diverging, you can’t find the sum of the infinite terms.

# Week 1- Precalc 11

In the first week of Pre-calculus class I learned the arithmetic sequence.

It is worth mentioning that the difference between consecutive terms is constant and this constant value is called the common difference. We can use letter  “d”  to represent it and  “t” to reflect terms in a sequence.

For example: 2 , 4 , 6 , 8 , 10 , …. is a arithmetic sequence. Our goal is to find the common difference ” d ” of this arithmetic sequence. The simplest way is using $t_2$ mins $t_1$ , at the same time , using $t_3$minus $t_ 2$. If the two differences are the same, we can determine that the common difference is equal to this value. In this arithmetic sequence ” d ” is positive 2. We can figure out any term in this arithmetic sequence by looking at the first term and the common difference. $t_n=t_1+d(n-1)$

In addition, how to calculate the sum of a arithmetic sequence is just as important as $t_n$. When we compute the sum of the first n terms of the arithmetic progression, we need to use a formula $S_n=\frac{n}{2}\cdot (t_1+t_n)$.

EX: There have a arithmetic sequence: 3 , 7 , 11 , 15 ,  19 , …

a) Find $t_{50}$ of this arithmetic sequence.

step 1: Find the common difference ” d ” and the fist term of this sequence “$t_1$“.

step 2: Substitute the value of ” d ” , ” $t_1$ ” & ” n ” into the formula.

d= 4 , $t_1=3$ , n= 50

$t_{50}=t_1+(n-1)d=3+(49\cdot4)=199$

b)Find $s_{50}$ of this arithmetic sequence.

step 1: Find the ” d ” and ” $t_{50}$ “.

step 2: Substitute the value of ” d ” , ” n ” , ” $t_1$ ” & ” $t_ {50}$ ” into the formula.

d=4 , $t_1=3$ , $t_{50}=199$ , n=50

$S_{50}=\frac{n}{2}\cdot (t_1+t_n) =\frac{50}{2}\cdot(3+199)=5050$