Category: Grade 11 (Page 1 of 3)
This week in PreCalc 11, we started to learn about Trigonometry. We learned many concepts regarding trigonometry, but the one thing that stood out to me was the Sine Law. The reason why it stood out to me was because it cuts down the work we had to do last year to solve the same things that the Sine Law does. Last year if we were introduced to triangles that weren’t typical 90 degrees angles, we had to make a line from the top point perpendicular to the bottom line to solve each angle and solve for each sides measurements. Fortunately, the Sine Law helps us do that with less amount of work needed to be done.
The Sine Law is a/sin A = b/sin B = c/sin C, where the lower case letters are the measurements of the side lengths and and the upper case letters are the lower case letters corresponding angle. By corresponding angle i mean that if angle B describing one point of the triangle, then side b would be the side of the triangle that is directly opposite of angle B. Because the Sine Law works with the angles of triangles and the measurements of the triangles side then it’s useful in finding a missing angle or side.
Example:
Work:
Step 1: draw triangle
When given a problem that describes the angles and the side lengths of a particular triangle, the first step is to always draw it. Drawing the triangle gives us a better understanding of what we have and what we’re looking for with little room for error. In this example, we we’re given two side lengths, KN and MK , and angle M. I drew the triangle, labelled the points, and then labelled the characteristic given. That’s when I found out what I was looking for, angle N, angle K, and side MN.
Step 2: find the equation to use first
The easiest way to do this is to label the sides first. In this example, opposite of angle K would be side k, opposite angle M would be side m, and opposite angle N would be side n. Once we have labelled those we should write out the full equation and plug in everything we know. The equation comes with three different ones, but we can only use two at a time. So the important thing is to use two equations that we can solve for one variable meaning all other variables we should already have. In this example, we have Sin M, m, and n, meaning we could find Sin N.
Step 3: solve the equation
Once we have the first equation we are going to use to solve for something, then all we have to do is solve for it. To solve for it, I used cross-multiplication and solved from there. When i moved Sin to the other side, I inverted it.
Step 4: find the last angle
Once we found angle M, we can easily find the last unknown angle with little work. We know that the 3 angles of one triangle have to add up to 180 degrees. So we know that one angle is 70 degrees and another angle is 75, so that means if we take 180 and subtract 70 and subtract another 75, it will give us the final angle. In this case, the final angle is 35 degrees. That means we have angle M as 70 degrees, angle N as 75 degrees, and angle K as 35 degrees.
Step 5: solve for the last side length
The only thing we have left to solve is the final side length. We can use the Sine Law to find the last side length. Since we have all variables of the equations except for k then we can use any two of the equations to find it. The final step is to solve for it. In this example the final side length was 8.6 cm.
Now we know all the triangles angles and side lengths all thanks to the Sine Law!
This week in PreCalc 11 we learned how to solve rational equations. Rational equations are equations containing at least one fraction whose numerator or denominator is a variable.
There are two ways to solve rational equations, one of them is multiplying every term by each of the denominators or cross multiplication. Cross multiplication only works when there are two fractions and one is on each side of the equal sign whereas multiplying by the denominator is a strategy that will work with every type of rational equation.
Example:
Explanation:
Step 1: Factor
Factor any polynomials that aren’t already factored. To factor, find two numbers that when multiplied together equal the last term and when added together equal the middle term.
Step 2: Multiply by the Denominator
Multiplying by the denominator is just doing the opposite operation to take the fraction away to make the equation easier to solve. When you have a fraction it means that its the numerator divided by the denominator, so to get rid of the denominator, we should multiply the term by the denominator to get rid of it. If we multiply the term by it, then the same term that is on the denominator will also be apart of the numerator making them cancel out. But what we do to one term, we do to every term, so we have to multiply each term in the equation by the denominator. Typically there is more than one denominator, so we follow the same procedure for the rest of the fractions.
Step 3: Non-Permissible Values
Non permissible values are numbers in which x cannot equal because it will make the denominators zero which can never happen. So stating non permissible values is important. In this case, the denominators consists of (b – 2), (b +2), and (b – 3) so b cannot equal 2, -2, and 3 because it will make the denominators equal zero.
Step 4: Solve
The final step is to solve. Multiply each term by the denominators, cancel out similar factors that are on the numerator and denominator and solve for x.
This week in PreCalc 11 we learned how to multiply and divide rational expressions.
Rational Expressions are fractions whose numerator and denominator are both polynomials and this week we learned how to multiply two rational expressions together. Typically when we multiply two regular fractions together, we multiply the top numbers which turns equals the numerator and multiply the bottom numbers which equals the denominator. Then at the end we simplify the remaining fraction. The same idea goes for multiplying two or more rational expressions together.
Example:
Explanation:
Step 1: completely factor
Factor all the polynomials present in the rational expressions. All factors should be completely factored.
Step 2: name non permissible values
Typically with regular factors, the denominator can never equal zero. The same concept goes for rational expressions, the denominator can’t equal zero. That is why if there is ever a variable on the denominator it can’t equal any numbers that would make the denominator zero. In this case, if x equals 1, -1, 9, -9, 0 then the denominator would equal zero so that’s why we must put that x cannot equal these numbers.
Step 3: simplify
To simplify, cross out any factors that are the same on the top and the bottom.
Final Answer:
The final answer will equal all the numbers that remain on top multiplied together to equal the numerator, and the same goes for the denominator.
This picture represents the light and darkness that Montag associates with his culture. The dark half of the page is nighttime and the bright side is the sunrise. Montag knows that his society is heading towards it’s demise, but he knows that there is going to be a new beginning. Muh like in the world, nighttime comes, but people always know that the sun will rise the next day.
This week in PreCalc 11 we learned about rational expressions. Rational Expressions are fractions whose numerator and denominator are both polynomials. They are usually given in unfactored form, and the point is to simplify them, just like a regular fraction, and determine which numbers x cannot equal.
Example:
Explanation:
Step 1: Factor Numerator
In this example, the factoring is simple. All that needs to be done is find two numbers that add to 11 and the product of those two numbers equal 30.
Step 2: Factor Denominator
In this example, the denominator is a difference of squares so to factor is two figure out the conjugates of the square roots of both terms of the polynomial.
Step 3: find what x can’t equal
For fractions, in general, the denominator can never equal zero. Same rule applies for rational expressions, the denominator cannot equal zero. In this example, if x was 5 or -5 then the denominator will equal zero which can never happen. So because of that, we must write that x cannot equal 5 or -5.
Step 4: simplify
Once the fraction is completely factored then cross out factors that are the same on the top and the bottom. In this example, (x +5) is on the top and the bottom so we can cross those out. We can only cross out complete factors that are the same, not single terms. For example, if there was a 3 on the top and 3 on the bottom then we couldn’t cross those out. We could cross them out if it was (x + 3).
Step 5: final answer
After simplifying the fraction, then that leaves you with the final answer.
This week in PreCalc 11 we learned how to graph absolute value functions.
Absolute Value Functions is a function that contains an algebraic expression within absolute value symbols. When we learned about Absolute Values before, we learned that whatever number is between the absolute value symbols must come out a positive. For example, | -3 | = 3 . Absolute Value Functions are similar in the sense that the y-values for Absolute Value Functions must be equal to or greater than zero.
Example:
y = | -5x + 10 |
Step 1: Graph the Parent Function
The first step to graphing this Absolute Value Function is to graph the parent function. The parent function is the same function but without the absolute value symbols. In this case, the parent function is y = -5x + 10.
Step 2: Move the Negative Y-Values
As I mentioned before, y-values in the parent function can’t be less than zero which is where the absolute value function comes in. As I mentioned earlier, | -3 | equals 3 because they’re both the same distance from zero on the number line. This relates back to absolute value functions because the negative y-values for the parent function, when put in the absolute value symbols, is a positive. So all we have to do is change the negative y-values into a positive and then graph it again.
That is how you graph Absolute Value Functions
This week in PreCalc 11 we learned how to solve systems of equations algebraically. A system of equations is a collection of two or more equations with a same set of unknowns. In solving a system of equations, we try to find values for each of the unknowns that will satisfy every equation in the system. The systems can either be both linear, be both quadratic, or be one linear and one quadratic. The reason for finding the unknowns (typically x and y) is because it’s the point that, when both the equations are graphed, they intersect. If the 
Example:
Step 1: Isolate One Variable
The first step to solving a system is to make one of the two equation equal either one of the two variables. In this example, one of the equations is equal to y. To make things easier you can make the simpler equation equal one of the variables, too.
Step 2: Substitute
Now that we have one equation equal y we can place that equation into the other equation. At any place of the equation that I see y I’m going to place the expression that y equals
Step 3: Solve for X
Step 4: Substitute
Now we have two x- coordinates that could lead to the solution. There is possible to have two solutions, but even if we have 2 x-coordinates we could only end up with 1 solution. With the 2 x-coordinates that we have, we put them back into the other equation and solve for y
Step 5: Verify
To make sure 
Since both solutions make both equations true then that means they are real solutions and that is where the two equations intersect on a graph.
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Open in new tab My group did pages 59- 65 from the big book. My role in this project was the literary luminary meaning I had to pick out quotes that contained figurative language and talk about their meanings and why they are significant to the theme of the book. I completed by role with care and detail and used notes that Ms. Shong gave us to complete it. After I completed the quotes, I moved on to putting it on the powerpoint. Closer to the due date I messaged group members outside of class to remind them to stick to the timeline we discussed prior. I also added group members work that they only put on the collaborative google docs and not the powerpoint to make sure it was completed on time.
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This week in PreCalc 11, we learned about graphing quadratic inequalities with two variables. Inequalities state whether two values are equal, not equal, greater than or less than the other. The point of the inequality is to make it true, If the inequality states that 
Inequalities can be expressed by quadratic equations. For example 
Step 1: Graph
To start this process, we have to graph the quadratic. We can start graphing this quadratic by one of two ways. One way is to convert this inequality from general form to vertex form. We can do this by the completing the square method. This form tells us where the parabola is located and what the stretch pattern is so we know the pattern to go up by. Another way we can graph this is to factor it. To factor we have to find two numbers that when multiplied equal -8 and when added equal -2:
In this case the numbers were -4 and 2.
Now we know what the zeros of the equation is so that means the opposites are the x-intercepts. The opposite of -4 and 2 are 4 and -2 which means that those are the x-intercepts of the graph. Since we know where the x-intercepts are, we can find out the vertex. We can do this by adding the x intercepts together and dividing them by 2.
Now we know that 1 is the x value of the vertex. To find the y value, we can put the x value into the original equation to solve for y.
Now we have the y-value which means we know the vertex
We also know that the stretch value, the value that determines the pattern the parabola follows, is 1 because there is no number attached to the 
The graph should look like the following:
Step 2: Determine the Solution
Now that we know what the parabola looks like, we have to figure out which coordinates on the graph make the inequality statement true. The points that make the statement true are either going to be all the ones inside the parabola or all the ones outside the parabola. To figure this out we can start by picking one point of the graph and dropping the x value and the y value in the original equation. If, when the inequality is solved, the statement is true, then all the points, where its located (i.e. outside the parabola) then all the points outside the parabola are true, so we would shade that area. If it was located inside the parabola then we would shade the inside of the parabola.
When the coordinate
All the coordinates inside the parabola make the original inequality true.