1.
1. We will prove that the square root of 2 is irrational number.
Answer: Two is not a perfect square number. So, the result of square root of two of course is irrational number.
2. We will show that the sum angles of triangle is equal to 180 degree.
Answer:
· Create a triangle. Then, make a line which is parallel with the triangle’s base on the top of triangle.
· Assume the triangle is ABC triangle. Then the angle’s name is A,B and C. And the angle C is on the triangle’s top. The angle A on the right corner and the angle B on the left corner.
· There is an angle between the line and the right side of triangle which has the same size with angle B because they are intern passing angles. We call it angle B’. there is also an angle between the line and the left side of triangle which has the same size with angle A. Then, we call it angle A’.
· So, we can see that A’ plus B’ plus C equals to one hundred and eighty degrees because it does a complete rotation from one side of the straight line to the other.
· Because A equals A’ and B equals B’. so, A plus B plus C equals one hundred and eighty degrees.
· It is proven that the sum angles of the triangle is one hundred and eighty degrees.
3. We will show how to get phi.
· First, we wrap a thread around a cylinder. Then, we measure the thread’s length.
· To have good accuracy of measurement, we can do it several times.
· After we find the length of the thread, we can assume that the length is the perimeter of a circle.
· We know that the perimeter of a circle equals two times phi times the radius. With the same radius, we can count that the value of phi approximately equals to three times one four.
4. We will show how to find out the area of region bounded by the graph of y equals x square and y equals x plus one
Answer:
· First, we must sketch the graph. But before that we must find the intersect points between y equals x square and y equals x plus two. Substitute y equals x square to y equals x plus two.
Then, we will find x square equals x plus two. Then, add both sides with negative x minus two. So, we will find x square minus x minus two equals zero.
After that, we can find the factor of the equation above. We find x minus two in bracket times x plus one in bracket equals zero. So, we find x equals two and x equals negative one. Then, we can find the value of y. If x equals two, so y equals four and if x equals negative one, so y equals one.
The intersect points are (2,4) and (-1,1).
· The area of the region bounded by the equations has lower boundary x equals negative one and upper boundary x equals two.
· After we know the region, we can count the area. Assume A is the area of the region. Make a partition on the graph, then we will find delta A equals x plus two minus x square in bracket times delta x (delta A is found from the subtraction of upper graph y equals x plus two and the lower graph y equals x square in bracket times delta x).
· A equals define integral of x plus two minus x square from lower boundary x equals negative one to upper boundary x equals two.
· A equals a half times x square plus two times x minus one third times x cubeb in bracket from x equals negative one to x equals two.
· A equals a half times four plus two times two minus one third times eight in bracket minus open bracket a half times one plus two times negative one minus one third times negative one close bracket.
· A equals two plus four minus eight third minus a half plus two minus one third.
· A equals four and a half.
· So, the area of the region bounded by y equals x square and y equals x plus two is four and a half.
5. We will show how to determine the intersection point between the circle x square plus y square equals twenty and y equals x plus one.
Answer:
· Substitute y equals x plus one to x square plus y square equals twenty.
· We will find x square plus x plus one in bracket square equals twenty.
· X square plus x square plus two times x plus one equals twenty.
· Add both sides by negative twenty. So, we find x square plus two times x minus nineteen equals zero.
· Then, to find the value of x, we can use ABC formula
First x or second x equals negative two plus minus the square root of open bracket four minus four times two times negative nineteen close bracket over four.
· First x or second x equals negative two plus minus the square root of one hundred and fifty six in bracket over four.
· First x equals negative two plus two times the square root of thirty nine in bracket over four. So, x equals negative one plus the square root of thirty nine in bracket over two. So, the value of y equals negative one plus the square root of thirty nine in bracket over two plus one. Y equals one plus the square root of thirty nine in bracket over two.
· Second x equals negative two minus two times the square root of thirty nine in bracket over four. So, x equals negative one minus the square root of thirty nine in bracket over two. So, the value of y equals x equals negative one minus the square root of thirty nine in bracket over two plus one. Y equals one minus the square root of thirty nine in bracket over two.
· The intersection points are x equals negative one minus the square root of thirty nine in bracket over two, y equals one plus the square root of thirty nine in bracket over two. And x equals negative one minus the square root of thirty nine in bracket over two, y equals one minus the square root of thirty nine in bracket over two.
Differential Mean Value Theorem
Theorem
If function f continue on the closed interval [a,b] and is differentiable at the points on the interval (a,b) that there will be at least a number c on the interval (a,b) with differential of function f with x is c equal to function f with x is b minus function f with x is a in bracket over b minus a on the bracket (f’(c)=f(b)-f(a)/b-a).
How can we find this theorem?
There is a proof to prove that the theorem is right.
For example, there is an equation y equal to function f. That function is on the interval [a,b], so, we will find that the corner points are (a,f(a)) and (b,f(b)). There is a line lies an those two points with the equation y equal to function g and has a gradient equal to f(b) minus f(a) over open bracket b minus a close bracket (m=f(b)-f(a)/b-a).
The line equation of y equal to function g through the point (a,f(a)) is
· y-y1=m(x-x1) (y minus y one equal to gradient times open bracket x minus x one close bracket)
· g(x)-f(a)=f(b)-f(a) over b-a in bracket times (x-a) (function g minus function f with x is a equal to function f with x is b minus function f with x is a over b minus a in bracket times open bracket x minus a close bracket)
· g(x) equal to f(b) minus f(a) over b minus a in bracket times open bracket x minus a close bracket plus f(a) (g(x)=(f(b)-f(a)/b-a) times (x-a) +f(a))
There is a function S. function s is found from f(x) minus g(x).
Substitute g(x)=(f(b)-f(a)/b-a)times (x-a)+f(a) to s(x)=f(x)-g(x). So, we find
S(x)=f(x)-(f(b)-f(a)/b-a)(x-a)-f(a) (s(x) equal to f(x) minus open bracket f(b)minus f(a) over b minus a close bracket times open bracket x minus a close bracket minus f(x).
S’(x)= f’(x)-(f(b)-f(a)/b-a) (differential of function s equal to differential of function f minus f(b) minus f(a) over b minus a in bracket)
S(x) continue at (a,b) because s(x)=f(x)-g(x) which function f and function g continue on the interval [a,b]
We will show that s’(c)=0 which c is element of(a,b).
S(x) continue on the interval a and b, so function s reach maximum and minimum point at (a,b).
a. If the value of s(x)maximum and s(x)minimum equal to 0 that s(x) equal to 0 that s(x) equal to 0 which every x is element [a,b]. If s(x) equal to 0 that s’(x) equal to 0 which every x is element of(a,b). It means if c is element (a,b) that s’(x) equal to 0.
b. If the value of s(x)maximum or s(x)minimum is not equal to 0 that s(x)maximum or s(x)minimum reached in the point c on the opened interval [a,b]. It can’t be reached on point a and b because s(a) and s(b) equal 0.
Because s(x) has maximum or minimum value on (a,b), it means s(x) have differential in every point at interval (a,b).
So, if c in (a,b) produces s(c) maximum or s(c) minimum that c is the stationery point from function S.
S’(c) equals 0……………(1)
Then look again the formula
S’(x)= f’(x)-(f(b)-f(a)/b-a)
S’(c)= f’(c)-( f(b)-f(a)/b-a)……………..(2)
From 1 and 2, we can find
f’(c)-( f(b)-f(a)/b-a)=0
f’(c)= f(b)-f(a)/(b-a) The theorem is proven.
The problem related to Differential Mean Value Theorem
F(X)=X square plus X on the interval [-2,2]. Find the value of C!
Solution:
F’(X) equals two times x plus one
f’(c)= f(b)-f(a)/(b-a)
2c+1= f(2)-f(-2)/2-(-2) (two times c plus one equals f(2) minus f(-2) over two minus negative two)
2c+1=4+2-4+2/4 (two times c plus one equals four plus two minus four plus two over four)
2c+1=1 (two times c plus one equals one)
C=0 (c equals zero)
How To Collect Data
To find data which are valid and guaranteed, we must collect them with right ways. There are many methods which are used to collect the data:
1. Interview
Interview is a way to collect data by holding direct interface. The interview must be done with interview’s guide which contains questions list based on the goal.
There are 2 kinds of interview:
a. Stuctured Interview is one kind of interview that the varieties and orders from the questions are arranged before.
b. Unstuctured Interview is an interview which is not secured tightly. It is more flexible because the questions can be increased.
The characteristics of good questions are:
· Based on the goals or the problems of the experiment.
· Clear and undoubtful.
· Fit in the knowledge and the experiences of interviewee.
· The questions can’t be related to someone’s personal.
The positive side from interview is the needed data can be received directly, so the data are more accurate and can be responsibled.
The negative side is it can’t be done in big scale and is not easy to find the personal informations.
2. Quesioner ( ) is a way to collect data by sending or using quesioner which contains many questions. The advantage of using this method is it can be done at big scale and can find personal informations. The negative side is the results are not accurate, not all questions are answered. Even, not all quesioners are not given back.
3. Observation (Supervision) is collecting data’s method by observe the experiment’s object or the events( human, no_living creatures or natural indication). The data can be found to know the habits and behaviours of human, no_living creatures or natural indication. The advantage of the observation is the data can be trusted. But, it can be found false interpretation to the observed events.
4. Test and Objective Scale is a way to find data by giving test to observed objects. This way is used mostly in psychology to measure the characteristics of human personality. There are some examples of objective scale test:
a. Intelligence and Talent Test
b. Personality Test
c. Attitude Test
d. Test about Value
e. Academic Test etc
5. Projective Method is observe or analyze an object by external expression from that object in drawing or writing. This method is used in psychology to know the emotion and personality of someone. The weakness of this method is the same object can be concluded variously by different observers.
Rectangle
1. Definition
Rectangle is a two dimensional shape which has two pairs parallel lines and one of its angles is 90 degrees. It is also defined as parallelogram which has four right angles. Rectangle is not always equilateral. A rectangle with vertices ABCD can be written ABCD.
2. The Characteristics of Rectangle
a. It has the same diagonals.
b. Its diagonals intercept in the middle.
c. Its angle has the sum 360 degree.
d. Each angle is ninety degree.
e. Opposite sides has the same length.
f. Opposite sides are parallel.
3. How to find the perimeter and the area of a rectangle
Assume that the length of a rectangle is L and its width is T
· Its perimeter is from the sum of all sides’ size. So, P(perimeter) equals L plus T plus L plus T. Then, we can find P equals two times open bracket L plus T close bracket [2(L+T)].
· Its area A equals L times T (A=L.T)
· Its diagonal’s length equals the square root of open bracket L square plus T square close bracket.
4. Kinds of Rectangle
§ Square is rectangle which has the same length in four sides.
§ Oblong is used to call the non square rectangle.
5. Problems Related to Rectangle
· Find the perimeter and the area of a rectangle, if the length equals 12 cm and its diagonal line is 13 cm!
Solution:
First, we must find the width of rectangle. T equals the square root of open bracket D square minus L square close bracket. T equals the square root of open bracket thirteen square minus twelve square close bracket. T equals the square root of twenty five. T equals five. So, the width of the rectangle is 5 cm.
Then, we can find its perimeter and area.
P=2(L+T)
P=2(12+5) (P equals two times open bracket twelve plus five close bracket)
P=34 (P equals thirty four)
So, the perimeter is 34 cm.
A=L.T
A=12.5 (A equals twelve times five)
A=60 (A equals sixty)
So, the area is 60 cm.
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