This week in math 11 we learned about the Sine law. They’re two different laws we use to find different things on a triangle: we use one of them to find a missing side and we use the other law to find a missing angle. We have 3 different ratios for each law but when solving for either a side or angle we only use two of these ratios at a time. All of the ratios are equivalent. We know we’re using the Sine law because we are given 3 clues on the triangle: we are given an angle with its corresponding side along with either an angle or side on a different part of the triangle. The Sine law is used when the triangle is not a right triangle.

Sine Laws:

• This law is used to find a missing side

• This law is used to fin a missing angle

Today I’m just going to be talking about solving for a missing angle

Example:

• in this triangle we are trying to find angle “B”
• We’ll say that angle 92 is “A” and the other point is “C”
• Now the first thing we do is figure out which law we need to use
• In this case we’re trying to find an angle

• so we would use this one
• so we can fill in the information we are given by the triangle

2.

• we see that “C” isn’t used so we can get rid of it because we are only suppose to use two of the ratios.

3.

• because we want to find “B” we have to isolate it

4.

• we multiply by 28 to cancel it out on one side but what you do to one side you must do to the other

5.

• Now we have to move sine to the other side and when we do this sine becomes inverse sine

6.

• Now we can plug all this into our calculator and get our answer

7. B = 43°

This week in Math 11 we learned about we learned how to add and subtract rational expressions that have binomial along with trinomial denominators. To do this we have to first simplify each rational expression. Then find the lowest common denominator between the rational expressions you are add or subtracting, once you find the LCD make sure you multiply the numerator by what you multiplied to get the LCD. Then put the numerators of both rational expressions over the LCD. What you do from here is solve. Once you finish solving you reduce if you can and make sure you don’t forget to write your non-permissible values.

Example:

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• First thing we do is simplify

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• now that we have simplified our expressions we can also write our non-permissible values.

• also note that you can simplify as long as it’s within the same expression.

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• here we can simplify the (x – 2) because it is within the same fraction

– 

• now we find our LCD which is (x – 3)(x – 5)(x – 6)
• and whatever we multiplied the bottom by we have to multiply the top by

• Now we can solve the numerator

• and we can see that we can factor out a -6 in our numerator

• so our final answer would be

This week in Math 11 we learned how to simplify rational expressions. We do this by cancelling out what is equivalent. The first thing we do is figure out our non-permissible values, these are values that we cannot use because if we do our denominator will equal 0 and a math rule is that we can’t have a 0 in our denominator. Then we use the different math skills we have learned to first simplify the numerator and denominator, then once everything is simplified then we can cancel out what is the same or something common.

Example:

• right now due to the way our equation is written we can’t say what our non-permissible values are
• so first thing we can do is factor

• now that it is factored we can see that x cannot equal -9 and 9 because our rule is that we cannot have a zero inn our denominator and we we use those numbers as x’s we will get a zero

• we see that we can something in the numerator and denominator which is the (x+9)
• so we would cancel both of those out and be left with our final answer because we cannot reduce anymore

And as our final answer we would have our simplified expression and what our non-permissible values are.

x cannot equal -9 and 9

This week in Math 11 we learned how to graph reciprocal linear functions. We do this by first reciprocating our function, then we graph the original function, then we find our invariant points on our graph which are always 1 and -1 and we graph our hyperbola that intersects through our invariant points. The hyperbola never touches our asymptotes which is y=0 and it can vary for our vertical asymptote depending on our graph.

Example:

y=-3x + 6

• the first thing we do is reciprocate our graph which is pretty easy, we know that technically it is over 1 so we just flip it.

y =

now that we have our reciprocal we can graph our original function

• we see that our y-intercept is 6 and our slope is -3

Now we find our invariant points on our graph which is -1 and 1

Now that we know our invariant points we can graph our hyperbola through those points

• In this case our vertical asymptote would be x=2 because our vertical asymptote is always right between our invariant points and and horizontal asymptote is always going to be y=0(in grade 11 at least)
• so our hyperbola intersects our invariant point but never touches our asymptotes

This week in Math 11 we learned how to solve absolute value equations algebraically. When solving with an absolute value and or a square root you may come across extraneous solutions which are solutions that are not true solutions of the equation given. Algebraically we can find out whether a solution is extraneous by verifying and inserting the “x” value back into the original equation to see if the equation is true or to see whether both sides equal each other.

3 = |2x + 5|

• first thing we do is separate the equation
• because they original equation uses absolute value we times everything by a negative to make it to make it positive.

3 = 2x + 5

3 = -(2x + 5)

• now that we have our equations we solve each of them

3 = 2x +5

3 – 5 = 2x

-2 = 2x

x = -1

Now for the second equation:

3 = -(2x +5)

3= -2x – 5

3 + 5 = -2x

8 = -2x

x = -4

Now we have to make sure these solutions are not extraneous by verifying:

• to do this we insert each solution back into the original equation

x = -1      x = -4

X = -1: Verify

3 = |2x + 5|

3 = |2(-1) + 5|

3 = |-2 + 5|

3 = |3|

3 = 3

X = -4: Verify

3 = |2x + 5|

3 = |2(-4) + 5|

3 = |-8 + 5|

3 = |-3|

3 = 3

We see that both solutions make the equation true because both sides are equal.

This week in Math 11 we learned about how to graph linear inequalities with two variables. By graphing the linear inequality we can use a test point to find out which side of the line has points that will work as a solution for our inequality to make it true. When graphing inequalities we need to know if the line we are graphing is a broken line(dotted line) or a solid line. A broken line tells us that the points on the line are not solutions, so every set of coordinates on the true side excluding the line is a solution. A solid line tells us that the coordinates on the line are also solutions, so every set of coordinates on the true side including the line is a solution. To know when the linear inequality is a solid or broken line we look at the sign of the inequality given; for example, greater than(>) and less than(<) signs are both broken lines. Greater than or equal to() and less than or equal to() signs are solid lines. Now to graph out inequality we have to see what we know first just by looking at the inequality without doing anything to it.

Ex.)

First thing is that we can see that the sign is less than or equal to so we know our line is going to be solid. Just by looking at this we can see that it is linear because the highest degree of our variables is 1, also knowing the information from grade 10 we can see that this inequality is in the form of            “y = mx + b”, where “m” is equal to the slope and “b” is equal to the y-intercept. So knowing this we can graph our y-intercept which would be (0, 5)

Now we would use our “m” which tells us the slope to finish the line which is 2.

We know that the slope =

So to make 2 and fraction we just put it over 1 and use that as our rise and run.

Now that we have our line we can use a test point to figure out which side has the solutions

• To make it easy we can use the most easy test point which is (0, 0)
• Now we substitute the zeros into our original inequality
• 
• 
• 
• Now we see that the solution is true which means that everything including the line on the side of the test point we used is a solution for this inequality.
• to show this we shade in that side

This is how we show our answer for these types of questions.