Chris's+page

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Phase 1
Yang Hui "Pascals Triangle" Yang Hui triangle (Pascal's triangle) using rod numerals, as depicted in a publication of Zhu Shijie in 1303 AD. " //The men of old changed the name of their methods from problem to problem, so that as no specific explanation was given, there is no way of telling their theoretical origin or basis" -Yang Hui// __Yang Hui__ -[] Yang Hui lived from 1238-1298AD the name that everybody called him was Qianguang. Being a chinese mathematician from the what is now modern day Hangzhou, Zhejiang province. He was tourted by Qin Jiushao and many other famous chinese mathematicians and some of his knowledge was just practical math facts and terms he created and help create during his life's work. Some of the things he created or helped create are the quadratic formula which he helped by being the first one to use the negative coefficient's of x in the quadratic formula. He also worked on the magic squares, magic circles, and the binomial theorem. Then again he also helped touter many other famous mathematicians.

In around 1275AD Yang Hui Finally got two books published they were known as Xugu Zhaiqi and Suanfa Tongbian Benmo and this helped his fame tremendously. After this things settled down for him. Until he started to change his methods from problem to problem and this made stuff just a bit more harder for him because it would make the time it took to work each problem out signifintaly longer than usual. This made him display a modern yet sort of laid back approach to mathematics and this just seemed to make him become more famous in the world spotlight. Then again he didn't need any help becoming more famous he just needed to take things a lot slower for his sake and his family's.

= __Abraham bar Hiyya__ = http://www-history.mcs.st-andrews.ac.uk/Biographies/Abraham.html **Born: 1070 in Barcelona, Spain**
 * Died: 1136 in Provence, France**

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A jewish mathematician and astronomer he was already famous because of his book Hibbur ha-Meshihah ve-ha-Tishboret. He was also very fmaous in other ways but we wont get into that right now so we will just continue on about the Quadratic formula and how he helped in its creation or helped develope it. But Then again he also had an impact on jewish and spanish math and this helped him a lot. In the world spot light he was a little famous beause of his many books and writings that he created. He needed some help as an astronomer because he was more of a mathematician than astronomer. But this made him good at calculations and things like that were deeply involved in math. But then again he was a good astronomer he wasn't the best but he was good.======

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When he helped with the really didn't help that much with quadratic formula so he didn't get that much fame from helping with it. He helped wit hthe positive coefficient's and that is really only thing that he helped with. He also helped many kings and queens of his day mostly the king and queen of spain. He also had major controbutions to arabic mathematics but this was not the only thing he had impact on in mathematics. He also had an impact on astronomy.======

Phase 2 1. Suppose a farmer has 3500 yards of fencing to enclose a rectangular field. What is the largest area that the farmer can enclose? (10 points)

Square has the largest area so 3500 / 4 = 875 875 x 875 = 765,625 square yards

2. The owner of a ranch decides to enclose a rectangular region with 380 feet of fencing. To help the fencing cover more land, he plans to use one side of his barn as part of the enclosed region. What is the maximum area the rancher can enclose? (10 points)

2y + x = 380 = x = 380 – 2y A = xy

A = (380 – 2y)(y) A = 380y – 2y² A = -2y² + 380y

Complete the square ax² + bx + c = 0 a(x + d)² + e = 0 d = b/2a e = c – (b²/4a)

a = -2, b = 380, c = 0, d = 380/2(-2) = -95, e = 0 – (380² / 4(-2)) = 18050

-2(y – 95)² + 18050 y = 95, 2(95) + x = 380 190 + x = 380 x = 190 Max Area = 95 x 190 = 18,050 square feet



3. A farmer wishes to enclose a rectangular region bordering a river using 3100 feet of fencing. He wants to divide the region into two equal parts using some of the fence material. What is the maximum area that can be enclosed with the fencing? (10 points)

2y + x = 3100 = x = 3100 – 2y A = xy

A = (3100 – 2y)(y) A = 3100y – 2y² A = -2y² + 3100y

Complete the square ax² + bx + c = 0 a(x + d)² + e = 0 d = b/2a e = c – (b²/4a)

a = -2, b = 3100, c = 0, d = 3100/2(-2) = -775, e = 0 – (3100² / 4(-2)) = 76,880,000

-2(y – 775)² + 76,880,000 y = 775

2y + x = 3100

2(775) + x = 3100 1550 + x = 3100 X = 1550

Max Area = 775 x 1550 = 76,880,000 ft²

Part 2 Part two of Phase 2 is to solve the problems that involve projectile in motion. The formula you will be using to solve these problems is below: s(t) = –gt2 + v0t + h0 Describe what the letters g, t, v and h represent. (5 points)

g = gravity t = time v = velocity h = height

Now use the formula and website to solve the following 2 problems. Copy and paste the problems to your wiki and explain them step by step.

1. Some fireworks are fired vertically into the air from the ground at an initial velocity of 88 feet per second. Find the highest point reached by the projectile just as it explodes. (10 points)

s(t) = -gt² + vt + h (Number for gravity is 16 from the internet) s(t) = -16t² + 88t + 0

Find vertex (axis of symmetry) x = -b/2a = -88/2(-16) -88 / -32 = 2.75

s(2.75) = -16 (2.75)² + 88 (2.75) + 0 -16 (7.5625) + 242 + 0 -121 + 242 = 121

121 feet is the highest point and it will take 2.75 seconds

2. A ball is thrown vertically upward with an initial velocity of 89 feet per second. If the ball started from a height of 10 feet off the ground, determine the time it will take for the ball to hit the ground. (10 points)

s(t) = -16t² + 89t + 10

Vertex = -89 / 2(-16) = -89 / -32 = 2.78

s(2.78) = -16(2.78) ² + 89(2.78) + 10 = -16(7.7284) + 247.42 + 10 = -123.65 + 257.42 = 123.77 123.77 = max height and it will take 2.78 x 2 for total time = 5.56 seconds

Phase 3

Phase 3: Three Days: 30 points

You will be given a quadratic Equation; you will come and get your equation from me when you reach this phase. x² + x – 56 = 0

1. You need to explain what does it mean to say “ solve by graphing” once you have this written on your wiki you will solve the quadratic by graphing. (10 points)

The general technique for graphing quadratics is the same as for graphing linear equations. However, since quadratics graph as curvy lines (called "parabolas"), rather than the straight lines generated by linear equations, there are some additional considerations. When you graphed straight lines, you only needed two points to graph your line, though you generally plotted three or more points just to be on the safe side. However, three points will almost certainly //not// be enough points for graphing a quadratic.

The parabola has a (+ a) coefficient and it opens upwards (U-shaped) the parabola has a (-a) coefficient and it opens downwards (n-shaped)

Find the axis of symmetry x = (-b) / 2a x = (-1)/ 2(1) = -1/2


 * x || x² + x – 56 || y ||
 * 0 || 0^2 + 0 – 56 || -56 ||
 * 4 || 4^2 + 4 - 56 || -36 ||
 * -6 || -6^2 -6 -56 || -26 ||
 * 7 || 7^2 + 7 -56 || 0 ||
 * -8 || -8^2 -8 - 56 || 0 ||
 * -10 || -10^2 -10 -56 || 34 ||

2. You need to explain what it means to say “solve the quadratic by factoring.” Once you have this written on your wiki you will then solve the same equation by factoring. What do the solutions represent after factoring a quadratic (10 points)

A "quadratic" is a polynomial that looks like "a//x//2 + b//x// + c", where "a", "b", and "c" are just numbers. For the easy case of factoring, you will find two numbers that will not only multiply to equal the constant term "c", but also add up to equal "b", the coefficient on the //x//-term x² + x – 56 = 0 (x + 8) (x – 7) = 0 x = -8 or x = 7

3. Last you need to explain what it means to say “solve the equation using the quadratic formula.” Put your explanation on your wiki and then take your equation and use the quadratic formula to solve it. Show the step-by-step process in solving this, on your wiki. (10 points)

INCLUDEPICTURE "http://www.1728.com/quadrtc.gif" \* MERGEFORMATINET When you use the quadratic formula you put the coefficients of the equation into the formula and solve it.

So for x² + x – 56 = 0 then a = 1, b = 1, c = -56

x = (-1± square root 1^2 – 4(1)(-56)) / 2(1) x = (-1 ± square root 1 + 224) / 2 x = (-1 ± square root 225)/ 2 x = (-1 + 15) / 2 or x = (-1 – 15) /2 x = 7 x = -8

Phase 4 