Grade 11 – Manroop's Blog

The Taking Tree – Poetry Project

manroopt2015 on June 21, 2018June 21, 2018






The Taking Tree - Manroop Thandi 

The maple tree is dying,

Once red and beautiful

With a strong thick trunk,  

Took a sip of the bittersweet

Flight away from life,

In the midst of removing every dried-up leaf,

Giving the other roots grief

Disregarding its surroundings.



The maple tree is dying,

This tree is drowned in an anesthetic

That is reactive but not proactive,

The drink that causes sad happiness

Trusting a true trick.  

creating loneliness.



The maple tree is dying,

The dark day has come,

Now unable to see

It will never breathe

The dagger strikes the tree ,

as it falls down gloomily

Quietly and quickly as fast as

when it started blooming,

it finally acknowledges the world it didn’t get to see.


the maple tree is dying.

Composition:

People often rely on drugs and alcohol as a vanity away from problems but fail to acknowledge the repercussions of such actions. In the poem The Taking Tree, Manroop Thandi displays the theme of using drugs/alcohol to hide pain. The Taking Tree is a free verse and open poem. The speaker is someone who is close with someone who has experienced drug abuse before. The poem is an extended metaphor about someone who abuses substances, this is shown in line 10 “this tree is drowned in an anesthetic”. The anesthetic represents a numbing agent that limits pain but only lasts for a short amount of time, this could represent a drug or alcohol.

Thandi’s choice of using maple trees as a comparison to a person could be because a maple tree is very vibrant but after constantly misusing substances it is dead and has lost its life. In line three Manroop says “with a strong thick trunk” this could be representing the strong mind it began with and how substances started destroying its solid roots (mind). The “dried-up leaves” is a metaphor for problems. The tree wants to remove all of the things that bring down the vibrancy of the tree. The line “of removing every dried-up leaf” is also a hyperbole because removing all problems from a one’s life is impossible. Line 8 shows that drug abusers do not know the trail that substance abuse causes and how much it emotionally affects others. The line, “this tree is drowned in an anesthetic” is a hyperbole because the tree is over using this drug for tranquility. Line 11 “that is reactive but not proactive” represents that the drugs the person is taking is only a solution for a limited time but in the long run it will never go away. The poet also shows that the ultimate result in abusing drugs instead of facing problems will result in in loneliness, “trusting a true trick. Creating loneliness”. In lines 21-22 the poet compares the day the tree bloomed to the day it is dying by using a simile. This comparison is made because when the tree bloomed it was excited to grow, similarly, when it dies it will finally have peace. Thandi also uses the symbol of darkness and a dagger to represent death. There are also a couple of examples of personification in the poem. In line 3 Thandi says, “took a sip of the bittersweet” this is personification because trees cannot sip.

Thandi also uses sound devices such as repetition, alliteration, cacophony and oxymorons. Throughout the poem, Thandi repeats the line “the maple tree is dying” to emphasize the effect of the drug abuse on the people surrounding the abuser. In line 13 Thandi says, “trusting a true trick” this is alliteration because of the repeating “tr” sounds. In line 12, Thandi uses an oxymoron “the drink that causes sad happiness”, this represents a faux experience. In the hopes of reviving everything, the person has become depressed. The person fails to realize that this drug is not causing happiness but affecting everything around it. Line 13 is also an oxymoron because “true trick” contradict each other, this is because the person using the drugs believes that this is a real solution but later figures out it was all a trick.The person abusing drugs believes that drugs will solve all problems but later realizes that life with the drugs/alcohol created many more problems rather than solving them. The poet also uses cacophony in line 3 by repeating the “k” sound and in line 21 by repeating the “q” sound.

The poem The Taking Tree use many figurative devices to show the path someone going through drug abuse. This poem is insightful to the human condition because many people deal with substance abuse in attempts to hide pain but fail to recognize the effects of these actions.

Week 18 – Final Review

manroopt2015 on June 18, 2018June 18, 2018

This week in Pre-Calculus 11 we are reviewing for the final. The top 5 things I learned in Pre-Calculus 11 are calculating the discriminant, CDPEU, solving equations, finding the vertex of a parabola, and special triangles.

Calculating the Discriminant: The discriminant has been helpful throughout Pre-Calculus 11 because with the discriminant you can find out how many solutions/roots/x-intercepts an equation has or whether it is an extraneous solution. The discriminant is derived from the quadratic formula $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ . The discriminant is the part that is under the square root sign $b^2-4ac$ . If the discriminant is negative that means that there are no solutions, it does not have a x-intercept and/or it is an extraneous solution because there cannot be a square root of a negative number. If the discriminant is equal to zero that means that the equation has one solution. If the discriminant is positive that means there are two solutions.

Ex. $x^2+3x+30$

$b^2-4ac$

$3^2-4(1)(30)$

$9-120$

$-111$

CDPEU: CDPEU was introduced in the quadratics unit. The acronym helps you remember the steps of factoring. The letter C represents “Is there anything in common?”.

Ex. $2x^2+8x-10$

$2(x^2+4-5)$

$2(x+5)(x-1)$

In the example above all of the terms shared a common multiple of 2, so I factored it out. After factoring the 2 out I continued on and factored.

The letter D represents “Is there a difference of squares”. A difference of squares only occurs in binomials. A difference of squares requires two square roots where one is negative.

Ex. $x^2-49$

$(x-7)(x+7)$

The example above is a difference of squares. Both the first term and the second term are perfect squares which means I could factor the expression easily. You have to make sure it is a DIFFERENCE of squares and not a SUM of squares, otherwise you will not be able to use this strategy.

The P represents the word pattern. The pattern that we look for when factoring is $a^2+bx+c$ . If the expression or equation does not have this pattern it is linear. If it is a quadratic then you have to categorize it using the next two letters.

$9x-3$

$3(x-1)$

The example above is not a quadratic expression but a linear expression because there is no $x^2$ . Even though this is a linear expression you can still factor it.

The letter E represents the word easy. This is when there is not a coefficient in front of the $x^2$ and it is easily factorable.

Ex. $x^2+x-20$

$(x+5)(x-4)$

The example above, is easy to solve because there are two numbers that multiply to 20 and add to 1.

The letter U represents the world ugly. An ugly quadratic equation is when there is a coefficient in front of the $x^2$ . This causes the quadratic expression to be hard to factor.

Solving Inequality Equations: To solve inequality equations algebraically we used the method of substitution. After finding each of the values of x and y it is important to always verify the solution by plugging it back in.

Ex. $y=x^2+6x+2$ and $y=2x-1$

$y=2x-1$

STEP 1: $2x-1=x^2+2x+4$

$0=x^2+6x-2x+2+1$

$0=x^2+4x+3$

$0=(x+3)(x+1)$

$x+3=0$

$x=-3$

$x+1=0$

$x=-1$

STEP 2: $y=2x-1$ $x=-3$

$y=2(-3)-1$

$y=-6-1$

$y=-7$

$(-3,-7)$

STEP 3: $y=2x-1$ $x=-1$

$y=2(-1)-1$

$y=-2-1$

$y=-3$

$(-1,-3)$

STEP 4: $y=2x-1$ $(-1,-3)$

$-3=2(-1)-1$

$-3=-2-1$

$-3=-3$

STEP 5: $y=2x-1$ $(-3,-7)$

$-7=2(-3)-1$

$-7=-6-1$

$-7=-7$

In the example above, the first step was to isolate one of the variables from one of the equations but that was already done. The next step was to plug in the value we found in the previous step into the other equation. After that I brought all of the terms onto one side of the equation and then factored the equation to find out the $x$ values. Then I plugged in the x values into one of the equations to find the $y$ value. To make sure that both of the points that I found were solutions I plugged those values back into one of the original equations and made sure both of the side were equal to each other.

Finding the Vertex: Finding the vertex of a parabola is one of the most important things I’ve learned this year because we have used parabolas throughout the course. To find the vertex of a parabola you have to complete the square. When you complete the square you divide the middle term by two and then square it.

$y=x^2+8x-23$

$\frac{8}{2}^2=\frac{64}{4}=16$

$y=x^2+4x+16-16-23$

$y=(x+4)^2-39$

Vertex: $(-4,-39)$

In the example above I used completing the square to find the vertex. There was not a coefficient in front of the $x^2$ which made the equation easier to factor.

Special Triangles: One of the most important things I learned was the special triangles. Special triangles prevent the use of calculators because they always have the same patterns. One of the special triangles has a $1-\sqrt{3}-2$ pattern which means that the angles are 90-60-30. Another one of the special triangles is the $1-1-\sqrt{2}$ which has the angles of 45-45-90.

Ex. What is the exact cosine ratio for $225$ degres?

$225-180=45$

$\cos45=\frac{1}{\sqrt{2}}$

Week 17 – Trigonometry

manroopt2015 on June 11, 2018

This week in Pre-Calculus 11 we started the trigonometry unit. This week we learned about the sine law and the cosine law.

What is the Sine Law? The sine law is typically used when you are not dealing with right triangles. Although the sine law could be used with right triangles it involves more steps. When solving a triangle that doesn’t have a right angle you cannot use SOH CAH TOA, you have to use the sine law or cosine law. The formula for finding an angle is $\frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{sin(C)}{c}$ . The formula for finding a side length is the reciprocal $\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}$ .

Ex. In a triangle $\angle(B)=34$ , side $c$ is 14.0cm and $\angle(C)=65$ . What is the side length of $b$ to the nearest tenth?

$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}$ .

$\frac{a}{\sin(A)}=\frac{b}{\sin(34)}=\frac{14}{\sin(65)}$

$\frac{b}{\sin(34)}=\frac{14}{\sin(65)}$

$b=\frac{14\times\sin(34)}{\sin65}$

$b=\frac{7.82}{\sin65}$

$b=8.638$

$b=8.6cm$

In the example above, the first step I did was plug in the known values into the formula. I knew this triangle was not a right triangle because it does not have a $\angle90$ . The second step was to isolate the variable and find it’s value.

Cosine Law: The Cosine Law is used when the Sine Law cannot be used. The cosine law can calculate a missing side length or angle. The formula for finding a missing side length is $a^2=b^2+c^2-2bc\cos(A)$ . The formula for finding an angle is $\cos(A)=\frac{b^2+c^2-a^2}{2bc}$ .

Ex. $\triangle{MKT}$ , side m is 12 cm long, side k is 7 cm and side t is 13 cm long. What is $\angle{K}$ to the nearest degree?

$\cos{K}=\frac{m^2+t^2-k^2}{2(mt})$

$\cos{K}=\frac{12^2+13^2-7^2}{2(12)(13)}$

$\cos{K}=\frac{144+169-49}{312}$

$\cos{K}=\frac{264}{312}$

$\cos{K}=0.846$

$K=\cos^{-1}{0.846}$

$K=32.2$

$K=32$ degrees

In the example above I used the cosine law to solve for $\angle{K}$ . The first step was to fill in the values. After that, I solved for $\angle{K}$ by calculating one side of the equation and then using the inverse function of cosine to solve for $\angle{K}$

Week 16 – Pre Calc 11

manroopt2015 on June 3, 2018

This week in Pre-Calculus 11 we finished the Rational Expressions unit. This week we learned about the application of rational equations in real life situations.

Word Problem Basics: Word problems always contain valuable information for solving the problem, so it is always useful to read the question slowly and carefully. To understand the problem better you can create a chart or organize the information in a format you like. Being able to identify key words like greater than (+), less than (-), the sum (x+x), the difference (x-x), the quotient ( $\frac{x}{y}$ ), the product of ( $x*y=$ ), etc. These words are helpful to remember when solving a word problem. When giving the answer to a word problem always answer with a sentence and with the units.

Ex. Sally can do all the house’s chores in 55 minutes. When Sally and Jake work together, they can do all the chores in 32 minutes. How long does it take Jake to do the chores?

What We Know:

From the clues in the word problem we know the time it takes Sally to do all the chores and how long it takes Jake and Sally together to do all the chores. Because they complete all of the chores in that amount of time we know that they complete 100% of the chores, which can be represented as $\frac{1}{1}$ .

The Equation: $\frac{32}{55}+\frac{32}{x}=\frac{1}{1}$

The first fraction represents the portion which it takes Sally and Jake to do all the chores over the amount of time Sally takes by herself. The second fraction represents the amount of time Sally and Jake take to do all the chores over the time Jake can do the chores. The time that Jake takes to do the chores is represented by $x$ because we are solving for that variable. The sum of both the rational numbers is $1/1$ because they complete all of the chores.

Solving: $\frac{32}{55}+\frac{32}{x}=\frac{1}{1}$

$(55x)(\frac{32}{55}+\frac{32}{x})=(55x)(\frac{1}{1})$

$32x+1760=55x$

$1760=55x-32x$

$1760=23x$

$\frac{1760}{23}=x$

$76\frac{12}{23}=x$

Final Answer: It takes Jake around 76.5 minutes to do all the chores by himself.

Week 15 – Pre Calc 11

manroopt2015 on May 26, 2018

This week in Pre-Calculus 11 we continued the rational expressions unit. This week we learned how to add and subtract rational expressions and how to solve equations.

How To Add and Subtract Rational Expressions: When you add rational expressions you always have to find a common denominator. To find a common denominator it is best to figure out the lowest common multiple by factoring both of the denominators.

Ex. $\frac{3}{5}+\frac{7}{8}$

$(\frac{3}{5}\times\frac{8}{8})+(\frac{7}{8}\times\frac{5}{5})$

$\frac{24}{40}+\frac{35}{40}$

$\frac{59}{40}$ or $1\frac{19}{40}$

In the example above, I had to add two fractions together. The first step was to find a common denominator, the lowest common multiple between 5 and 8 is 40. Then I added the numerators together. After adding the fractions together you must check if the fractions simplify any further.

Ex. $\frac{13}{9}-\frac{4}{3}$

$(\frac{13}{9}\times\frac{2}{2})-(\frac{4}{3}\times\frac{6}{6})$

$\frac{26}{18}-\frac{24}{18}$

$\frac{2}{18}$

$\frac{1}{9}$

In the example above I subtracted two fractions together. The steps were the exact same as when I added. The first step was to find a common denominator, the lowest common multiple between 9 and 3 is 18. After changing the denominators I subtracted the numerators. Then I simplified the fraction.

How To Add and Subtract Rational Expressions with Variables: Similarly, when dealing with variables while adding and subtracting you still need to find a common denominator.

Ex. $\frac{12}{5x}+\frac{7x}{2}$

$(\frac{12}{5x}\times\frac{2}{2})+(\frac{7x}{2}\times\frac{5x}{5x})$

$\frac{24}{10x}+\frac{35x^2}{10x}$

$\frac{35x^2+24}{10x}$

$x\neq0$

In the example above, I found a common denominator of 10x and then I added the fractions together.

Ex. $\frac{8}{6x+9}+\frac{3}{4x-4}$

$(\frac{8}{6x+9}\times\frac{4(x-1)}{4(x-1))}+(\frac{3}{4x-4}\times\frac{3(2x+3)}{3(2x+3)})$

$\frac{32x-32}{3(2x+3)(4(x-1))}+\frac{18x+27}{3(2x+3)(4(x-1))}$

$\frac{32x-32+18x+27}{3(2x+3)(4(x-1))}$

$\frac{50x-5}{12(2x+3)(x-1)}$

$\frac{5(10x-1)}{12(2x+3)(x-1)}$

$2x+3\neq0$

$2x\neq-3$

$x\neq\frac{-3}{2}$

$x-1\neq0$

$x\neq1$

In the example above, the first thing I did was factor both of the fractions. After, I found a common denominator by multiplying both of the denominators together. Then, I made a big fraction and added the like terms together and simplified. One key step is to remember to find the non-permissible values.

How To Solve Fractional Equations: There are a number of different ways to solve fractional equations that involve variables. If the equation has one fraction on both sides of the equals sign then you can cross multiply the fractions. Because we are solving, you have to isolate the variable and find out it’s value.

Ex. $\frac{x+2}{x-3}=\frac{x-1}{x-2}$

$(x+2)(x-2)=(x-1)(x-3)$

$x^2-2x+2x-4=x^2-3x-x+3$

$x^2-4=x^2-4x+3$

$x^2-x^2+4x-4-3=0$

$4x-7=0$

$4x=7$

$x=\frac{7}{4}$

$x-3\neq0$
$x\neq3$

$x-2\neq0$

$x\neq2$

In the example above, the first step was to cross multiply. After cross multiplying I collected all the like terms. Then, I isolated x and found x’s value.

True Happiness – Fahrenheit 451 Essay

manroopt2015 on May 22, 2018

To be happy is anticipated by many, but does our society know the meaning of true happiness? Often, happiness is thought to be achieved by the amount one consumes and the entertainment one has. In the novel “Fahrenheit 451”, Ray Bradbury exaggerates the overuse of technology in society. The society in Fahrenheit 451 relies on parlour walls and violence for entertainment. Due to the distraction of technology the society was oblivious to the intense warfare surrounding it. This society burns books because they believe books contain useless information. In his heroic journey the main character; Guy Montag, realizes that happiness cannot be attained through technology but with knowledge and books. Without the help of Clarisse, Montag would have never seen the beauty of the world. Ray Bradbury is prophetic by emphasizing the addiction to technology people face in the dystopian society, which is very similar to today.

People tend to seek fill their lives with reactive happiness. In Fahrenheit 451, the society fails to care about things that matter and invests more time in entertainment. The society in the book, believes that life should strictly be composed of entertainment in order to possess happiness. While talking to Montag, Beatty says, “bring on your clubs and parties, your acrobats and magicians, your daredevils, your jet cars, motorcycle helicopters, your sex and heroin, more of everything to do with automatic reflex… I just like solid entertainment” (pg. 58). The society seeks glorification to avoid thinking about political issues. The society seeks to be entertained and has faux experiences. Rather than participating in activities that generate true happiness they are stuck believing that by undertaking drugs you will be in contentment. Whereas it is paradoxical, the parties and entertainment will only be able to grant happiness for a limited amount of time. After the time is up, they will continue to be in melancholy which results in suicidal behavior. In the book, true happiness is only experienced by Clarisse, who neglects technology. Clarisse is able to communicate with others and she has insight into the world. The rest of the society does not challenge their minds, which diminishes their conscience. Similarly, in our society we seek for glorification. One lyric from Donald Glover’s video, This Is America is “we just wanna party”, then Donald Glover continues to dance and later shoots someone. We believe that doing such revolting acts are entertaining, but the adrenaline and happiness lasts for a restricted amount of time. We experience faux happiness and continuously repeat those actions. Once we feel malaise, we quickly react instead of being proactive and doing something that will truly enrich our lives. Both Fahrenheit 451 and society today, lack the feeling of authentic enjoyment.

The society in Fahrenheit 451 and society today are both reliant on technology and disregard the ability to critically think. In Fahrenheit 451, technology is the prominent influence in society. The parlour walls in the society display useless information, this makes the society incompetent at critically thinking. Instead of having the parlour walls show cartoons, it could have documentaries that educate the people. The citizens do not have a conscience and do not use the parlour walls for educational purposes. While talking about politics and the presidential election, one of the parlour ladies says this: “I voted last election, same as everyone, and I laid it on the line for President Noble. I think he’s one of the nicest looking men to ever became president” (pg. 93). The parlour women voted for a president based on his looks and did not consider his campaign. They have the privilege to vote, but they abuse this honor. The parlour walls are undervalued for the effective information they could spread. Due to the distraction of the parlour walls, the society is unaware of the warfare happening in its own country. They have no insight into the world. Similarly, in the society’s schools the students learn absurd facts that are irrational. The purpose of school is exploited and used improperly. In our society, we equivalently do not use contextual information. In Time To Do Everything But Think, David Brooks says, “the problem with all this speed, and the frantic energy that is spent using time efficiently, is that it undermines creativity”. Time To Do Everything But Think is a very universal and interpretive article. The internet can dispense everything, and this takes away from our conscience. We can find pointless information online, but we do not use the tool for its full extent. We can easily find educational material online and this takes away from our learning skills. Our society has become lazy and dependent on technology to critically think for ourselves. We do not dive deep with the information we receive but we only stay at a surface level understanding. With the tool of technology, we have forgotten our own ability to think. We are oblivious to events happening around the world because of our hubris. We only care about things happening around the countries we live in and neglect our responsibility in this world.

People tend to believe consumption leads to happiness. In Fahrenheit 451, Mildred is exposed to technology all around her. Mildred faces problems with depression yet, she still claims to be happy. At the beginning of the book, Montag was pleased to have a life filled with technology. But at the end of the book he is in despair because of technology and urges to change society’s functions. In the book Mildred is willing to spend “one third of [Montag’s] yearly pay” (pg. 18). Mildred is deeply obsessed with the technology because she believes it is her family. She cannot maintain a relationship with real people but a virtual reality world. This very closely relates to society today. We are indulged in buying the newest products available. By consuming we are tricked into thinking we are happy. We tend to buy things even if they will lead us into debt. After looking at the Happiness Index, countries such as the USA and Canada are claimed to be the saddest countries in the world. The USA is the top consumer in the world, yet, it is the saddest. Mexico and Colombia are the happiest countries in the world because they find joy without consuming useless products. Happiness cannot be measured with entertainment and technology, but it is measured in from within.

Fahrenheit 451 was written over 50 years ago, yet the themes still relate immensely to society today. Happiness does not come from faux experiences and technology, but it comes from knowledge. Happiness is a tangible feeling and with intangible items, no one will be able to experience happiness. Technology is taking over the ability to critically think and contextualize information. Happiness needs to be redefined to fit both the society in Fahrenheit 451 and society today. Sadly, Ray Bradbury’s dystopian fiction society in Fahrenheit 451 will soon become a reality due to our hubris.

Sources:
Bradbury, Ray. Fahrenheit 451. Simon & Schuster, 2013.

Brooks, David. “Time To Do Everything Except Think.” Newsweek, 13 Mar. 2010, www.newsweek.com/time-do-everything-except-think-150597.

“Childish Gambino – This Is America (Official Video).” YouTube, 5 May 2018, youtu.be/VYOjWnS4cMY.

NEF. Happy Planet Index. Mono, 2016, http://happyplanetindex.org/countries.

Week 14 – Rational Expressions

manroopt2015 on May 20, 2018

This week in Pre-Calculus 11 we started the rational expressions unit. This week we learned about equivalent rational expressions and how to multiply and divide rational expressions.

What is a Rational Expression: A rational expression is a number that can be expressed as a fraction. For example, the number 3 can be represented as $\frac{3}{1}$ , therefore it is a rational number. Irrational numbers are numbers that form an endless amount of decimals and are non-repeating.

Non Permissible Values: When there is a variable in the denominator of a fraction you have to give it a restriction. This is an important step because the denominator of a fraction can never be zero because that would make the fraction undefined.

Ex. $\frac{1}{x-4}$

$x-4\neq0$

$x\neq4$

In the example above, because there was a variable in the denominator I had to create a restriction. The first step is to make the denominator equal to zero. Then, I isolated the variable to figure out a value that would make the fraction undefined.

What are Equivalent Rational Expressions? Equivalent rational expressions are fractions where the numerator and denominator have something in common, so that they can simplify.

Ex. $\frac{10}{25}$

$\frac{10}{25}=\frac{5\times2}{5\times5}$

$\frac{10}{25}=\frac{2}{5}$

In the example above, I simplified a fraction by taking something out that both the numerator and denominator had in common. I took something in common out by factoring the fraction first, then cancelling out equal values. Similarly, in fractions that involve variables you have to factor the numerator and denominator first and then take out common terms.

Ex. $\frac{3x-12}{x^2+x-20}$

$\frac{3(x-4)}{(x+5)(x-4)}$

$\frac{3}{x+5}$

$x-4\neq0$

$x\neq4$

$x+5\neq0$

$x\neq-5$

In the example, the first thing I did was factor the numerator and denominator. Then, I took out the factors that they had in common. After, I figured out the non permissible values for the original expression and the factored expression.

How To Multiply and Divide Rational Expressions: To multiply and divide rational expressions you just have to multiply straight across. But it is important to factor first because it will be easier to simply the expression later.

Ex. $\frac{2x+2}{3x}\times\frac{x^2-2x}{4x+4}$

$\frac{2(x+1)}{3x}\times\frac{x(x-2)}{4(x+1)}$

$\frac{2(x+1)(x)(x-2)}{3x(4)(x+1)}$

$\frac{2x(x-2)}{3x(4)}$

$\frac{2x(x-2)}{12x}$

$\frac{x-2}{6}$

$3x\neq0$

$x\neq0$

$x-2\neq0$

$x\neq2$

The first thing I did was factor both of the rational expressions and then I took out the factors that were the same. After that, I simplified the fraction because 2 and 12 share a common factor of 2. Then, I found the non permissible values for the original and factored rational expression.

How To Divide Rational Expressions: To divide rational expressions you do the exact same as multiplying, by factoring and then taking out common factors. But when dividing fractions you have to remember to multiply by the reciprocal of the second fraction. You also have to find the non permissible values for the variable.

Ex. $\frac{x+5}{x-4}\div\frac{x^2-25}{3x-12}$

$\frac{x+5}{(x-4)}\div\frac{(x-5)(x+5)}{3(x-4)}$

$\frac{x+5}{(x-4)}\times\frac{3(x-4)}{(x-5)(x+5)}$

$\frac{(x+5)(3)(x-4)}{(x-4)(x-5)(x+5)}$

$\frac{3}{x-5}$

$x-4\neq0$

$x\neq4$

$x+5\neq0$

$x\neq-5$

$x-5\neq0$

$x\neq5$

For dividing the rational expression, the first step I did was to factor both of the fractions. Then, I reciprocated the second fraction and multiplied them together. After, I took out the common factors and everything that was left over become the new rational expression.

Amazon is Taking Over

manroopt2015 on May 14, 2018May 14, 2018

The Amazon-ification of Whole Foods

This article is about Amazon’s new branch of delivering grocery items to homes. This article attracted to me because the title simply amazed me. Amazon is the world’s biggest retailer and it is growing at an exponential rate. I had never thought that you would be able to sit at home and not only order your groceries online but also get them delivered to you. Although this new program sounds convenient, I would never use this tool. I do not support this movement because I believe this is going to make the human race lazier and it will prevent face-to-face interactions from occurring. This reminds me of the society in Fahrenheit 451 because they rarely had conversations with others and this movement is a step towards that society. The author explains how Amazon had once before ran “Amazon Fresh” but it went downhill. This gets me wondering how Amazon will be able to pull off this new campaign without shutting it down like “Amazon Fresh”. I do not think that the online grocery store will have the same quality as stores like Superstore or Costco. When you go into a grocery store you get to pick the vegetables that seem fresh and that are in good quality but online you will only be able to select the quantity of food you want. I like how the author, Derek Thompson, has a clear explanation on Amazons plan and includes the future ambitions of the company. At the end of the article, the author talks about how Amazon is planning on creating an online pharmacy and offer healthcare services. By Amazon expanding it’s empire, I think that many businesses will decide to close down and transfer online as well.

Week 13 – Absolute Value And Reciprocal Functions

manroopt2015 on May 14, 2018

This week in Pre-Calculus 11 we started the Absolute Value and Reciprocal Functions unit. This week we learned how to graph an absolute value linear function/quadratic function and we learned about reciprocal functions.

What is an Absolute Value Function? An absolute value function is always in the top half of the graph because the y value has to be positive. Linear absolute value functions make a v-shape and quadratic absolute value functions make a w shape.

Critical Point: The critical point or the point of inflection is where the line or parabola changes direction. The critical point is always the x-intercept.

How To Graph an Absolute Value Function: When graphing an absolute value function it is always useful to graph the original function first. For linear functions, after graphing the original function, you have to look where the line hits the x-axis and then you flip the line by making the slope the opposite.

Ex. $y=\mid2x+9\mid$

Original Function: $y=2x+9$

Absolute Value Function: $y=\mid2x+9\mid$

y-intercept: (0,9)

x-intercept: (-4.5,0)

Domain: $x\varepsilon\mathbb{R}$

Range: $y\geq0$

Piecewise Notation: $f(x)=2x+9,x\geq-4.5$

$f(x)=-(2x+9),x<-4.5$

In the example above, I graphed a linear absolute value function. In the example, as soon as the line hit the x-axis it changed directions. The slope of the original function is $\frac{2}{1}$ but once the line hit the x-axis the slope became $\frac{-2}{1}$ .

Ex. $y=\mid$ $x^2-x-6\mid$

Original Function: $y=x^2-x-6$

Absolute Value Function: $y=\mid$ $x^2-x-6\mid$

y-intercept: (0,6)

x-intercept: (-2,0) and (3,0)

Domain: $x\varepsilon\mathbb{R}$

Range: $y\geq0$

Piecewise Notation: $f(x):x^2-x-6,x\leq-2$ or $x\geq3$

$f(x):-(x^2-x-6),-2<x<3$

In the example above, I graphed a quadratic absolute value function. The first step was to graph the original function. After graphing the original function, I saw where the parabola hit the x-axis and then flipped everything below the x-axis. The original vertex was (0.5,-6.25) but with the absolute value function it turned into (0.5, 6.25) because the y value had to become positive.

What is a Reciprocal Function? The reciprocal function of a quadratic or linear function is one over the original function.

Ex. $y=x+2$

Reciprocal Function: $y=\frac{1}{x+2}$

When a reciprocal function is graphed it creates two curves also known as the hyperbolas. The reciprocal of 1 is $\frac{1}{1}$ which equals 1, so it stays the same. Similarly, the reciprocal of -1 is $\frac{-1}{1}$ which equals -1. The points of 1 and -1 on the y-axis are the invariant points (they do not change) these are used to draw the asymptotes. The asymptote is used to separate the hyperbolas. This year, one of the asymptotes is always going to be the x-axis.

Ex. $f(x)=3x+4$

$f(x)^{-1}=\frac{1}{3x+4}$

In the example above, the first thing I did was graph the original function. After graphing the original function I found where the line meets at (_, 1) and (_ , -1) to draw my asymptotes. Then I drew my hyperbolas accordingly.

Week 12 – Pre Calc 11

manroopt2015 on May 7, 2018

This week in Pre-Calculus 11 we finished the solving quadratic inequalities unit. This week we learned how to solve linear-quadratic systems and quadratic-quadratic systems by using algebra.

How To Solve Systems Algebraically: To solve systems algebraically you have to use a method called substitution. Substitution is a concept that we learned in the systems unit from Math 10.

Substitution: Substitution is when you isolate a variable of an equation and then plug it into the other equation. After that you use the value that you solved for and plug it in into one of the equations to figure out the second variables value.

Linear-Quadratic Systems: Linear-Quadratic Systems can have 0,1, and 2 possible points of intersection. You can figure the points of intersection by graphing or using substitution algebraically.

Ex. $y=-2x^2+8$ / $3x-y=-3$

STEP 1: $3x-y=-3$

$3x+3=y$

STEP 2: $3x+3=-2x^2+8$

$2x^2-8+3x+3$

$2x^2+3x-5$

$(2x+5)(x-1)$

STEP 3: $2x+5=0$

$2x=-5$

$x=\frac{-5}{2}$

$x-1=0$

$x=1$

STEP 4: ( $\frac{-5}{2}$ ,y)

$3x+3=y$

$3(\frac{-5}{2})+3=y$

$\frac{-15}{2}+3=y$

$\frac{-15}{2}+\frac{6}{2}=y$

$\frac{-9}{2}=y$

$(\frac{-5}{2}.\frac{-9}{2})$

STEP 5: ( $1$ , y)

$3x+3=y$

$3(1)+3=y$

$3+3=y$

$6=y$

(1,6)

In the example above I solved the linear-quadratic system by breaking up the process in steps. In Step 1, I isolated $y$ because it was the easiest variable to isolate. By isolating $y$ I was able to figure out it’s value. In Step 2, I plugged in the value we found in step 1 into the other equation. After plugging in the value I found for $y$ I factored the equation to find out the x-intercepts (the values for x). In Step 3, because we factored the equation in the previous step I used those factors to isolate x and find the roots of the equation (the values for x). In Step 4, I took the first value I found for x and plugged it into one of the equations to find the value of y, that gave me a point where the linear-quadratic system intersects. In Step 5, I did the same thing as step 4 except I used the other value of x we found and plugged it into one of the equations to find the other point of intersection.

Quadratic-Quadratic Systems: Quadratic-Quadratic Systems can have 0,1,2, and an infinite amount of points of intersection. In order to figure out the number of points of intersection you can graph the parabolas or solve it algebraically by using substitution.

Ex. $y-4=x^2$ / $y=-x^2+12$

STEP 1: $y-4=x^2$

$y=x^2+4$

STEP 2: $y=-x^2+12$

$x^2+4=-x^2+12$

$x^2+x^2+4-12=0$

$2x^2-8=0$

$2(x^2-4)=0$

$2(x-2)(x+2)=0$

STEP 3: $x-2=0$

$x=2$

$x+2=0$

$x=-2$

STEP 4: $(2,y)$

$y=x^2+4$

$y=(2)^2+4$

$y=4+4$
$y=8$

(2,8)

STEP 5: $(-2,y)$

$y=x^2+4$

$y=(-2)^+4$

$y=4+4$

$y=8$

(-2,8)

In the example above, I did the exact same steps as I did in the linear-quadratic system example. The first step was to isolate a variable, then I plugged that value into the other equation, after that I factored the equation and found the values for x, and finally I plugged those values for x into one of the equations to figure out the values for y.

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