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kaidenh2016 on October 15, 2019

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Week 15/ Rational Equation’s in Word Problems

kaidenh2016 on December 17, 2018

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 13/Graphing Reciprocal Functions

kaidenh2016 on December 3, 2018

Reciprocal Functions

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

When finding y= 2 $x^2$ +12x-20 and its reciprical y=1/ (2 $x^2$ +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

kaidenh2016 on November 25, 2018

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 3 Square roots

kaidenh2016 on September 24, 2018October 3, 2018

Changing entire radicals to a mixed radical is fairly simple, but is a very important concept.

Your first step is to find out what the number in side the radical divides into. Fifty four can become $\sqrt{2}$ & $\sqrt{27}$ . (Which was from $\sqrt{54}$ )

Since two can’t be made smaller it is left alone. Next is to make $\sqrt{27}$ smaller, $\sqrt{27}$ can be turned into square root 3 and 9. So now you have $\sqrt{2}$ , $\sqrt{3}$ , $\sqrt{9}$ , nine can become smaller again and turn into 3. Nine is a special number called a prime factor which are numbers that can be multiplied by the same number evenly twice. If you have a prime factor like nine it just becomes 3 instead of square root three.

Now that your left with 3 $\sqrt{2}$ , $\sqrt{3}$ you have to multiple the square root 2 and the 3 to make it one number. After words your left with the answer 3 $\sqrt{6}$