Mu Alpha Theta Personal Statement

The next thing we have to do is find the perimeter of the small triangle & that comes out to be 58 in.

I only prove that $E\hat{H}G = 90^{\circ}$, for the reason that the prove of the other angles will be similar. When I was asked to prove if the inscribed rhombus is a square, I simply say to myself, ``it means I have to prove that the corners of the rhombus are right angled ($90^{\circ}$)" since the properties of a rhombus are such that opposite angles are equal, all sides are equal, and opposite sides are parallel, the diagonals bisect the angles, and the diagonals are perpendicular bisectors of each other. Therefore, the inscribed rhombus can only be a square if the angles are right angles. I observe that all four triangles are congruent by SSS, therefore from $A\hat{E}H + E\hat{H}A = 90^{\circ}$ and $E\hat{H}G + A\hat{H}E + D\hat{H}G = 180^{\circ}$ and $G\hat{H}D = A\hat{E}H$, it follows that $A\hat{H}E + D\hat{H}G = 90^{\circ}$, hence $E\hat{H}G = 90^{\circ}$. There is a need to draw the picture since it gives more logical reasoning and the understanding of how the angles and sides are defined, without drawing the picture we may not understand what is happening, therefore it is very crucial that as teachers we make use of pictures when we dealing with this kind of

So we start of with x^2-4x+y^2+8y=-4 you would then take 1/2 of -4 and square it, then take 1/2 of 8 and square it. You would then get x^2-4x+4-4+(y^2+8y+16-16=-4. After all of these steps you would factor the square, x-2^2-4+y+42-16=-4, x-2^2+y+4^2-20=-4. Then you would add 20 to both sides, x-)^2+y+4^2=16

5 x 2 + -6 x 4 + 5 x 8 + -6 x 1= 10 -24 + 40 -6= 20

When you are multiplying fractions you multiply the numbers on top as well as the numbers on the bottom. For example:

5 = a + b*0 //from this we see that a=5, and we plug that value into the second equation

Running with this information can now write out the equation AB2 + BC2 = AC2. One important thing is that we must note that AB is equal to “X” and the line segment of BC is equal to that of 2x+4, and that AC will be equal to that of 2x+6. So we will now input this information to create (x)2 + (2x + 4)2 = (2x + 6)2 and begin factoring each term into two sections. These two sections will be as x*x + (2x + 4)(2x + 4) = (2x + 6)(2x + 6). x times x is x2. An important tool to use now would be the FOIL method, so we will take (2x + 4)(2x + 4) and create 4x2 + 16x + 16. Right off the bat we notice that we have like terms. So we will add x2 to 4x2 to get 5x2. This will create 5x2 + 16x + 16 = 4x2 + 24x+ 36. Now we will use the subtraction property to get 5x2 – 4x2 + 16x – 24x + 16 – 36 = 0, however we still have like terms, so because 5x2 is a like term with -4x2 we will add them together to get x2. We will also combine 16x and –24x and also 16 and –36 which are also like terms and create –8x and –20, our equation should now look like x2 – 8x -- 20 = 0.We will now factor the equation from left to right, first factoring x2 which has 1 coefficient so the fact will be 1 and -1. The other term will be 20 which have no coefficient so we will do 5x4 and then 4 still can be divided so 2x2. This will create 20=225.

You must show all steps and provide any evidence needed in your solution to receive full credit.

H(2+h)= -5(2+h)2 + 20(2+h) +1 = -5(4+4h+h2) + 40+20h+1 = -20-20h-5h2+40+20h+1= -5h2+21 lim(h->0) H(2+h)-H(2)/h = -5h2+21-21/h = -5h = -5(0) = 0 √

Also I still do not quit understand how this amounts to 8 ------> (1+1)**(5-2) = 8 why 2 * ? and what is the value?

For my community service, I have been going to Lyndon B. Johnson Head Start Program, or LBJ. At LBJ, it is my responsibility to look after children and clean up after them. The teachers there set up many activities for the kids to work on, most of the time they finger paint. The teachers also set up small field trips. For example, we once went on a field trip to a house that had set up a pumpkin patch and many other activities for the kids. The house we went to wasn’t very far from LBJ. The children had so much fun choosing a pumpkin and playing different games, and I had fun helping them and taking care of them.

During my Junior year at Colfax High School, I was recruited into a new program called Link Crew which is a freshman transition program that welcomes incoming freshmen and makes them feel comfortable and supported throughout the first year of high school. This program was not only a club but also a class called Wings in which we would precisely plan Academic Follow Up's (AFU) for the freshmen, and organize events such as a Haunted House and a Blood Drive for the whole school. Also during my Junior year, I joined the Leo's Club which is a student run club that's purpose is to perform service projects within our school and community. I was able to become a certified volunteer of Colfax and Meadow Vista once I had not only met, but exceeded the minimum of 35 hours of volunteer work. Outside of school, I became interested in snowboarding which allowed me to get a seasonal job at a Ski

Throughout my four years of high school, I have dedicated quite a bit of my time to Mu Alpha Theta, between practices after school, Saturday competitions, and practicing on my own. It has always been my primary extracurricular. I’ve attended almost every competition GHS has gone to. Since our team relies on student coaching, I have helped with that as needed. My sophomore year I helped coach Geometry, my junior year I helped with Pre-Calculus, and this year I coach Calculus to the first year students since this is my second year. On top of that, I have been a stand in coach for every division I’m capable in for the past three years. I’m almost always available after school, so if another student coach was unavailable a certain day I would

In short, these three trig. ratios are the reciprocal of sine (sin.), cosine (cos.), and tangent (tan.). Although many may assume that cosecant is the reciprocal of cosine, it is actually that of sine, which means that cosecant is hypotenuse over opposite. Thereafter, that leaves us with secant, which is in fact the reciprocal of cosine, demonstrating that it is hypotenuse over adjacent. Nonetheless, cotangent is the last trig. ratio, meaning that it is the reciprocal of tangent, being rather adjacent over opposite. Now, we know the formulas for these trigonometric functions being: Csc=HypotenuseOpposite, Sec=HypotenuseAdjacent, and Cot=AdjacentOpposite . For example, the triangle on the next page is a 7-24-25 right triangle, and we must determine the six trigonometric ratios for angle C of the right triangle. Based off this information, we can determine that the adjacent is 7, with the opposite being 24 and hypotenuse as 25. In order to find the sine of angle C, we must write out Sin C =OppositeHypotenuse and plug in the numbers to equal Sin C=2425. Next, in order to find the cosine of angle C, by writing out Cos C=AdjacentHypotenuse and plugging in the adjacent and hypotenuse number to equal Cos C=725. After solving for the sine and cosine of the right triangle, we must find the tangent being Tan C=OppositeAdjacent.

My motto is God first, husband and then family and friends. I have the successfully balanced family, my career, and school for many years. I also maintain an outstanding GPA and am an inspiration to many adults who decide to continue to pursue education as a traditional student. My ability to overcome obstacles sets me aside from many others. When it comes to my success I’m willing to put in extra work to ensure that I will achieve my goals.