Category: Grade 11

Week17 / Cosine Law

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Before using Cosine Law you have to know when to use it. Normally when we are trying to find a side or an angle with Sine Law we will have a side that is opposite to an angle, but if we dont have that, what should we do; well we use c2 = a2 + b2 – 2ab cos C which incorperates what looks to be pythagoras. What I noticed when trying to find out when trying to find out when to use pythagoras is when the sides are all next to one angle, its a dead givaway that you need to use Cosine law. If you still do not understand try watching this video by Vivid Math which helps explain when to use cosine law.

Now that we know when to use Cosine Law we can use it to solve an equation.

In the photo theres a triangle that shows two sides and one angle, if you tryed to use Sine law it wouldnt have worked, but ising the Cosine law which is c2 = a2 + b2 – 2ab cos C and fill in the blanks you would get c2 = (1.7)2 + (3.1)2 – 2(1.7)(3.1) cos 110 then the rest is algebra like what you see in the photo.

 

 

Week 15/ Rational Equation’s in Word Problems

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Starting off with word problems you also want to read the equation a few times over and underline any important words that with give you an idea of what kind of a word problem you’ll be solving.

Given a word problem underline key words: Jorge rows a boat 24km down stream and back where they began. When the average speed of the current is 2km/h, they can complete the journey in 9 hours. What is Jorge’s average rowing speed in still water.

(Note: If in the word problem you find the word from at any time, reverse the two fractions/ or numbers.)

A key thing to help you solve these equtions is D = S x T, which means Distance = Speed times Time formula.

In the word problem (Jorge rows a boat 24km down stream and back where they began. When the average speed of the current is 2km/h, they can complete the journey in 9 hours. What is Jorge’s average rowing speed in still water.)

We pulled out key words in the sentence that will give us a clue to what will be solved the 24km down stream represents that the equation D \div S = T. And if you fill in the numbers it would have 24km as the distance. The sentence (Back where they began also tells us its for both fractions. So 24 will be the distance all around both up and down stream. Now we need the speed to divide the distance by. In the sentence (average speed of the current is 2km/h) we know that x+2 must be the speed going downstream and x-2 is the speed going up. Because if you think about it going up stream your going against the current and going down the current is pushing you by you overall speed. So now the equation is \frac {24}{x+2} +\frac {24}{x-2} = ?. We still are missing the time it took, but luckily in the sentence we underlined (they can complete the journey in 9 hours.) Meaning that the total time it took is nine hours completing the equation D \div S = T now we just need to solve for x. \frac {24}{x+2} +\frac {24}{x-2} = 9

You can do multiple things with this equation but what I would recommend is doing is common denomonator, and dont forget to right down you non-permissible values which is + or – 2. Then just factor.

The reason why I crossed out -2/3 is because it is a negitive number and the speed connnot be a negitive number.

Week 14/ Multiplying and Dividing Rational Expressions

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When multiplying and dividing rational expressions the steps you want to take are:

Step 1: Check if their are no plus or minus symboles, (this means that if their are no plus or minus symboles you can cancel out terms on the top and bottom)

Step 2: Factor both the numerator and the denominator on the fraction

Step 3: Write the non-permissible values x cannot equal (number) (To find the non-permissible values its what ever is in the denomonator that is after factored form.) Ex. $latex \frac{(-x+1)}{(x-1)} the Non-permissible values are x cannot equal 1, because you cannot have a 0 as your denominator.

Step 4: Simplify the rational exspression

Step 5: Cancel out any of the same simplified exspressions. (Note, if the number is \frac{(-x+1)}{(x-1)} you can take a -1 out of the (-x+1) and make it the same as (x-1) then cancel them out leaving you with -1.)

Step 6: leave it in factored form

 

Week 13/Graphing Reciprocal Functions

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Reciprocal Functions

Reciprocal Functions is taking a function like y= 2x^2+12x-20 and making it y=1/ (2x^2+12x-20) is the functions reciprical.

When finding y= 2x^2+12x-20 and its reciprical y=1/ (2x^2+12x-20), you want to graph the original function first all the time. After you have graphed the original function, you have to find its vertical asymptote which is where the graph touches the x axis and where the reciprocal will never cross over. It acts like a dotted line on going vertically upwords to help you graph the reciprical. There is also a horizontal asymptote which is for now always on the a axis, it acts simular to the vertical asymptote except it sits vertically on the x axis making sure that your reciprical never touches it.

Once you have graph the parabola, now that you have found that there is no x intercept in the equation, so there is no vertical asymptote. This is what you would call a pimple graph, one that does not have a vertical asymptote but does however still have a horizontal asymptote. To find out where it recipricates you have to find the vertex on the graph.

(3,-2) now that the vertex is found, you can recipricate the vertex, but only the -2 or the y, then you get (3,-1/2) and then you draw the bump that carries on through the x axis and down towards the new vertex for the reciprocal function.

Week 12/ Absolute value functions

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Absolute value functions

I’ll be graphing a absolute value equation, graphing it, and showing it in piecewise notation.

y= |-3x+9|  <——- Note: absolute value brackets (|x|)

With a normal line, (-3x+9) it would go below the x-axis, but when there are absolute value lines in the equation it makes it so the line does not go into the negitive not. Which also means it does not go below the x-axis, istead it just rickashays off and comes right back up. It refects the original line and bounces off the x-axis.

 

Now that we haved graphed we must state it peace wise notation. The point that intercecpts with the x axis is where you will be basing the x is < or > off of. For stating the x is < or > is with two equations, becuase the graph shows two line reflecting off the x axis there are two equations. 1. -3x+9 and -(-3x+9) then you state whether  x is < or > for the number on the x axis.

y = {-3x+9  if x < and equal to 3}

y = {-(-3x+9)  if x > 3}

To figure out if x is < or > you must test a point on the x axis to see if its true. Test numbers before and after three.

Week 11/ Solving system of equations graphically

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Solving system of equations

First you must be able to show the equation graphically. By doing so you have to show any linear equations into y=mx+b form; then change any quadratic equations into factored form or vertex form.  After you have algebraically changed all the equations you can then graph the equations and see where they intercept. Then that will be your solution.

There can be three type of solutions when solving for x equations that are linear and quadratic. The first one is no solution, the second is one solution, and the third is two solutions.

Equations: y= 2x^2 + 4x + 4

y= -2x+7

The graph states that there is no solution because the equation does not touch or intercept with each other.