This week we learned more about triangles and the ratio (i.e. Sine, Cosine, Tangent) in the triangle.

For me, the hardest part is to solve problems using special triangles, because when the reference angle is a special angle– for example, 45, 60, etc.– there’s always 4 kinds of degrees for each special angle, you need to figure out which is belong to which and also need to know the sign of them (positive or negative).
And there’s another kind of questions made me feel hard to deal with, which is letting you determine the degree of some angle whose sin/cos/tan equals others.

For example, if sin 245= sin A, determine sin A.
There’re a few steps to solve this problem.
First, we need to draw an angle to know which quadrant this angle is in, 245 is in quadrant three, according to “All Students Take Calculus” rule, the tangent of this angle should be negative. So sin A is also negative.
Second, we need to know the reference angle of this angle, it’s 245-180=65, so now we know that the sin of an angle who has a reference angle 65 is negative, so the other angle is in quadrant4, which is 360-65=295. So a=295

I also learned some new words, such as rotation angles, quadrant, opposite, adjacent, hypotenuse, terminal arms, clockwise…… I found that math is really an interesting subject.

This week for precalc11 we learned how to solve rational equations and use them to deal with some problems in reality. There’re a lot of different kinds of problems, such as problems involving distance, time and speed; problems involving work; and those others involving proportions…… For me, this is the part that I found challenging because there’re so many different items in every single question that I don’t know which items should I use to build the equations. For example, this time you should use speed, but the other time you should use the working hours. It’s also hard for me to understand the meanings of those questions, so I may take more time than others on this kind of questions.

Here’s a question that I got wrong on our book: The average speed of an airplane is 10 times that of a car. It takes the aorplane 18 h less thanthe car to travel 1000km. Determind the average speeds of the aorplane and the car.

I assumed the average speed of car as x, so the speed of airplane is 10x, so the equation is: 1000/x-18=1000/10x. The common denominator is 10x, I forgot to cancel the denominator, so I got x=1/18. That’s a silly mistake on calculating, but I always do it…… After asking for help, I got the right answer finally.

I hope I can get good result fo this chapter test. To do that, I must be careful with my calculating.

this week we learned chapter7: how to simplify, multiply and divide, add and substruct rational expressions. I learned it when I was in China so it was not hard for me to understand. All I need to do was to review and to memorize the math languages.
For example,

To solve this question, first, we need to find the lowest common denominator, which is 9x; then multiply 3 to both the top and bottom of the first equation. So it becomes:

We can add the tops together after this step, so we get:

Problem solved.
If the answer is reducible, we must finish the reducing step too.

It’s been a very busy but full week. We finished our chapter5, had a test and started to learn chapter8. We learned how to draw an absolute value function and how to solve the absolute value equations graphically and algebraically.
I want to remind myself of something that needs to be paid attention in about chapter5. I always confused myself when solving inequalities, do I need to sketch the whole graph or just draw the number line when solving the problems? But now I think I understand, when it only has one variable, for example, x, I can just use the number line; when it has both x and y, I should graph the function first, that way I can know the scale visually to solve it. I think that’s all I need to let myself notice for this unit.
Then for chapter8, we should solve the absolute value equation by separating it into two equations to get 1/2/3/4 solutions (every equation has a different amount of solutions from each other, depending on when two lines or parabolas bump into each other.)
For instance, x+8= the absolute value of4x+6, so we can regard it as” x+8=4x+6″ and “x+8=-(4x+6)”.Then we solve these two equations and we get that x=2/3 orx=-14/5.

This week we mainly learned how to solve quadratic, linear inequalities and quadratic systems of equations by graphing. Graphing plays an important role in solving equations and inequalities, especially in quadratic ones.
For the linear and quadratic inequalities, we need to graph the boundary line or parabola first, then test some points in different scales to find the solutions.
To solve the systems, we only need to graph every equation on the graphing paper and find the crossing point of them. Those points are the solutions of the systems.
I need to remember that x belongs to real numbers. I always forgot about adding it in the end.

This week we had two tests and two classes. About the tests, I think I need to know how to use the calculator better because I was not allowed to use it in China, I made a lot of mistakes with it. Also, I’m not good at calculating myself at all, that’s my weakness and I need to improve it.

We also learned 5.1 Solving Quadratic Inequalities in One Variable. To learn this, we must be very familiar with the graphics of the quadratic function, that would help us to think of the solutions of the inequalities.
For example, the inequality is related to the quadratic function ,
then we can draw the graph and find its two zeros, which are x=-4 and x=2,
we need to look at the part of the graph which is above zero,
then we finally are able to find the solution through this graph, which is x<-4 and x>2.

Next week I need to keep working on my weakness and learn new things as well.

This week has passed quickly, we learnt more about quadratic equation such as the equivalent forms, the translations and some modelling problem.
A quadratic function has three equivalent forms:
1.) Standard (Vertex) Form:
y = a (x-p)²+ q

We can find the vertex from this one, it should be (p,q)

2.) General Form: y = ax² + bx + c

This is usually the hard one to figure out, we can find the y-intercept of a function, which equals c in this form.

3.) Factored Form: y = a ( x – x1 ) (x – y2 )

you can tell the X-intercepts, which are X1 and X2 in this case.

They’re really helpful when we solve the problems.
One thing I’m interested in is that sometimes the graph of the function doesn’t cross with the x-axis, so it turns out that the function of it doesn’t have a factored form, it’s unfactorable.
For example, y=2x²-12x+27, you can’t factor it anyway, so it means the graph doesn’t cross the x-axis.

I think I did not good in the last test, so I hope to do better next week.

This week we learned how to graph quadratic functions, and how to interpret the graphs into the functions.
We learnt about many new math words, including the vertex, lines of symmetry, x/y-intercept, vertical/ horizontal translation, congruent, stretch, compression, etc.
The most magical thing, in my opinion, is analyzing the vertex form , you can tell what “q” and “p” is if you see a graph with vertex, and you can also tell what “a” is if the size is changed from the original one(1-3-5) to some other forms. For example, there is a parabola, whose vertex is (-4,-1), and it also passes through the point (-3,-4), what’s the function of it? With the original function form (), you will not be able to know the function, but with vertex form, you can easily know that p=-4,q=-1, and then you can put the point (-3,-4) into the function, which you just figured out, , then you can get that a=-2. Problem solved.
Or sometimes we can tell the value of “a” by observing the change of 1-3-5 form. If it became 2-6-10, then we can say that a=2. I may not very good at explaining what I learned but I don’t think I have any problem with it.
This is a week that I learned a lot. I do my homework every day and I have to keep it this way.
River