# Top 5 things I learned in Precalculus 11

Top 5 things I learned in Precalculus 11

Analyzing Quadratic Functions and Inequalities Unit

Key Learning：

Every quadratic function can be written in the form y = ax^2+bx+c, where a, b, and c are real numbers and a is not zero.

1. I learned the properties of quadratic functions and how to analyze the quadratic equation. And I learned that when we are given the quadratic equation in vertex form then we can graph the equation.

Vertex Form $a(x-p)^2 + q = y$

General Form $ax^2 + bx +c = y$

Factored Form $a(x-x_1)(x-x_2) = y$

2. I learned equivalent forms in order to solve equations by using 3 different formulas.

vertex form- $y = a (x-p)^2 + q$, where (p, q) is the vertex of the parabola, and a is the stretch value. It’s the easiest equation to graph because we are given a starting point and the pattern that it goes up by.

general form- $y = ax^2 + bx + c$,  a, b, and c are three real numbers. Once these are given, the values for x and y that make the statement true express a set of (x, y) points which form a parabola when graphed. It’s the most useless equation to graph because we are only given the y-intercept and can’t do much else unless it’s converted to one of the other forms.

factored form– $y = a(x-x1)(x-x2)$,  x1 and x2 are the opposite of the x-intercepts of a graph. This equation is useful only when given the x-intercepts and is used mostly when trying to search for a value as long as there is another point given along the parabola to substitute y and x for to find a.

Rational Expressions And Equations Unit

3. I learned how to simplify rational expressions.

For example, to simplify this expression, we need to factor first. $\frac {(2x+2)}{(x^2+3x+2)}$

Now we can begin with factoring. $\frac{2(x+1)}{(x+2)(x+1)}$

The next step is to look at the denominator and the numerator and cancel out any of them that are identical. This has to be monomial and monomial, binomial and binomial, variable with variable, etc. $\frac {2}{(x+2)}$

Now we can simplify the expression and state the non-permissible values. The non-permissible values are the restrictions and are for the denominator. In this expression, we need to also state the NPVs of the expression before and after we converted it into an all multiplication since it had the division in it.

Therefore x≠-2 or -1.

4. I learned how to solve rational equations.

For example : $\frac{5}{x+4}=\frac {3}{x-3}$

First, you would want to multiply each side by the common denominator $(x+4)$ and $(x-2)$ , which would make the equation become: $(x-2)(5) = (x+4)(3)$

then you would multiply the brackets which would turn the equation into: $5x-10 = 3x+12$

after that, you would just solve from here moving the 3x to one side then moving the 10 too the other side which would have the equation end at:

x = 11

Trigonometry Unit

4. I learned about special triangles.

We use these special triangles when the angle is 30, 60, 45 degrees and also the Multiple of 90 degrees. For these special triangles, we will not have to use a triangle and they give us a more exact answer.

Ex: 5. I learned that Sine law is a way that we can use to find a certain angle or length of something.

For example, for a Triangle ABC from A to B it’s 8.6; from A to C it’s 6.1; Angle C is 65°. Now, we are trying to find the length of BC.

You would use the equation of $\frac{A}{SinA} = \frac{B}{SinB} = \frac{C}{SinC}$ , this is for when you’re looking a missing side but for a missing angle,  and the equation $\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}$ it’s the same equation just flipped. And what we need to do right now is to find the angle of B, so it can help us find the side length of BC so you would use the equation of $\frac{SinA}{A} = \frac{SinB}{6.1} = \frac{Sin65}{8.6}$ . Now we would use the two fractions that are filled in the most so now the equation would be $\frac{SinB}{6.1} = \frac{Sin65}{8.6}$ .  To solve this equation is to basically isolate the B and put everything else on one side so the equation and solve from there which would give the angle B 40°, and from there you would keep using the same equation to find side BC.

# Thermos Challenge

Prototype testing procedure:

1. Heat up 150ml of water in a beaker on a hot plate set to high and heat it up to around 100 °c.
2. Take the initial temperature of the hot water as soon as it poured into the prototype thermos. Record the temperature and quickly place the cap on the thermos.
3. Leave the water for 20 minutes and observe any changes in the heat on the outside of the thermos.
4. Remove the cap and record the temperature of the water. Calculate the final and initial temperature difference.
5. Note any flaws and possible improvements that could be made to the design.
6. Make improvements based on the notes and repeat the procedure with the next prototype.

The idea behind this prototype is to reduce heat loss through heat transfer, so we use a lot of materials that have less thermal conductively, so the conduction of heat through the contact between the hot water and the material becomes weaker and slower.

Prototype #1   Materials list and cost: Data: Pros:

1.Pretty cheap. Cheapest one in all three prototypes.

Cons:

1. The appearance is simple, not attractive.

2. The lid (ballon) is not easy to take off, it takes time and will lose the heat by Conduction.

Progression:

– Improve external insulation to reduce heat loss.

– Try another material to make the lid.

– Add decoration to make themes look better.

Because sand has strong thermal conductivity and is easy to absorb thermal energy, we add sand between the cups, so that the sand absorbs the heat of water, and plays a role of heat insulation through the sand. On the other hand, due to the heat loss caused by convection inside the water, a large amount of heat of water molecules with higher molecular kinetic energy will move to the surface. If there is no cover, it will cause a large amount of heat loss, so we use a cut balloon As a cover.

Prototype#2

Data: Pros:

1.Cheap.

2.General insulation effect.

Cons:

1. The appearance is simple, not attractive.

Progression:

– Try to add aluminum foil inside the foam cup, to absorb heat.

– Add decoration to make themes look better.

We made the second idea based on the first one to reduce heat transfer and enhance the heat reflection in the inner layer. Since aluminum foil has very strong thermal conductively, if used in inner layer, it can strengthen thermal reflection and achieve the effect of ” reheating ”to keep heat. With a lid made of foam cup, it can reduce the heat loss caused by convection inside the water.

Prototype#3  Materials list and cost: Data: Pros:

1.Better insulation effect.

2.Beautiful appearance.

Cons:

The coffee cup lid has a hole on the top, which will lose heat through the hole.

Progression:

-Find something to cover the hole on the lid to reduce or slow down the heat dissipation.

-Use aluminum to cover the lid, to achieve better insulation effect.

The third major is a combination and improvement of ideas. We have used the aluminum foil in the second experiment to achieve the internal thermal insulation effect of “reheating the water.” It can better achieve the effect of external insulation.

Final Product Materials list and cost: Data: Our final product evolved from the third prototype, we proposed to add improvement of aluminum foil to the lid, and our final product is to test our hypothesis and improve. Since aluminum foil has very strong thermal conductively, if used in the inner layer, it can strengthen thermal reflection and achieve the effect of “reheating” to keep heat. The lid is made of aluminum foil, on the one hand, because of material price, on the other hand, because in the interior of the water can produce heat loss by convection, high quantity of heat of water molecules have higher molecular kinetic energy will move to the surface, if there is no lid can cause a lot of heat loss, and we want to use radiation of aluminum foil to reduce heat loss.

Overall, water in the prototype lose heat main is by conduction, so we are doing the project’s main idea is to use less thermal conductivity of materials to reduce heat conduction, so each of our design in outer do have “shell”, and through the results of the first we knew, the “shell” is not enough to reduce conduction, we need to strengthen inside, so we adopt with radiation heat material for “heating water again”.And at the top of the design is similar to the cup body, which is also using thermal conductivity less to reduce heat loss. But the second main consideration is the use of radiation, which reflects heat, to conserve it.

# Week 14 – Math 11

This week in Precal 11 we learned about special triangles. We use these special triangles when the angle is 30, 60, 45 degrees and also the Multiple of 90 degrees. For these special triangles, we will not have to use a triangle and they give us a more exact answer.      # Week 15 – Math 11

For the fourteenth week of Preacul 11, we started a new chapter —Trigonometry.  In grade 11 trig there are way more things to learn and the simplicity of SOH CAH TOA won’t cut it for grade 11 trig.

In this new trig we learn that there are 4 quadrants, it’s different from what we learned in grade 10. In grade 10 we were doing trig in quadrant 1 for most of the time, and we know this because Sine, Cosine and tangent were all positive in quadrant 1. Another thing that we learned was Sine Law, Sine law is a way that we can use to find a certain angle or length of something.

For example, for a Triangle ABC from A to B it’s 8.6; from A to C it’s 6.1; Angle C is 65°. Now, we are trying to find the length of BC.

You would use the equation of $\frac{A}{SinA} = \frac{B}{SinB} = \frac{C}{SinC}$ , this is for when you’re looking a missing side but for a missing angle,  and the equation $\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}$ it’s the same equation just flipped. And what we need to do right now is to find the angle of B, so it can help us find the side length of BC so you would use the equation of $\frac{SinA}{A} = \frac{SinB}{6.1} = \frac{Sin65}{8.6}$ . Now we would use the two fractions that are filled in the most so now the equation would be $\frac{SinB}{6.1} = \frac{Sin65}{8.6}$ .  To solve this equation is to basically isolate the B and put everything else on one side so the equation and solve from there which would give the angle B 40°, and from there you would keep using the same equation to find side BC.  This is how you use the Sine Law.

# Week 13 – Precalc 11

For the thirteenth week of Precalc 11,  we continued learning about rational expressions and equations.

For example: $\frac{5}{x+4}=\frac {3}{x-3}$

First, you would want to multiply each side by the common denominator $(x+4)$ and $(x-2)$ , which would make the equation become: $(x-2)(5) = (x+4)(3)$

then you would multiply the brackets which would turn the equation into: $5x-10 = 3x+12$

after that, you would just solve from here moving the 3x to one side then moving the 10 too the other side which would have the equation end at:

x = 11

# Week 12 – Precalc 11

For the twelfth week of Precalc 11, we started a new chapter about rational expressions and equations. In this chapter, we have so far covered how to multiply/divide and add/subtract.

To simplify this expression, we need to factor first. $\frac {(2x+2)}{(x^2+3x+2)}$

Now we can begin with factoring. $\frac{2(x+1)}{(x+2)(x+1)}$

The next step is to look at the denominator and the numerator and cancel out any of them that are identical. This has to be monomial and monomial, binomial and binomial, variable with variable, etc. $\frac {2}{(x+2)}$

Now we can simplify the expression and state the non-permissible values. The non-permissible values are the restrictions and are for the denominator. In this expression, we need to also state the NPVs of the expression before and after we converted it into an all multiplication since it had the division in it.

Therefore x≠-2 or -1.

# Week 9 – Precalc 11

For the ninth week of Precalc 11, we learned more about the Analyzing Quadratic Functions and inequalities unit.

Key Learning：

1. Vertex Form $a(x-p)^2 + q = y$
2. General Form $ax^2 + bx +c = y$
3. Factored Form $a(x-x_1)(x-x_2) = y$

We learned equivalent forms in order to solve equations by using 3 different formulas.

vertex form- $y = a (x-p)^2 + q$, where (p, q) is the vertex of the parabola, and a is the stretch value. It’s the easiest equation to graph because we are given a starting point and the pattern that it goes up by.

general form- $y = ax^2 + bx + c$,  a, b, and c are three real numbers. Once these are given, the values for x and y that make the statement true express a set of (x, y) points which form a parabola when graphed. It’s the most useless equation to graph because we are only given the y-intercept and can’t do much else unless it’s converted to one of the other forms.

factored form– $y = a(x-x1)(x-x2)$,  x1 and x2 are the opposite of the x-intercepts of a graph. This equation is useful only when given the x-intercepts and is used mostly when trying to search for a value as long as there is another point given along the parabola to substitute y and x for to find a.