June 2018 – 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.