Exercise before English Examination

Last week, on 2nd April 2009 Mr. Marsigit gave English examination in my class. I and my friends very surprised because Mr.Marsigit didn’t say anything in before meeting. So we do it during thirty minutes and finally Mr.Marsigit said that it is just exercise to do English examination next week. Immediately, our profile change becomes very happy and we felt relieved.

Exercise I:
1. Explain how to prove that square root of 2 is irrational number!
2. Explain how to show or to indicate that the some angel of triangle is equal to 180 degree!
3. Explain how you are able to get phi!
4. Explain how you are able to find of the area of region foundered by the graph of y equal x square and y equal x plus 2!
5. Explain how you are able to determine the intersection point between the circle x square plus y square equal 20 and y equal x plus 1!

Answer:
1. To prove that the square root of 2 is irrational number so for example we have assumption that the square root of 2 is rational number. It means the square root of 2 equal a over b, which a and b as prime so the square root of 2 equal a over b. a over b can be writing with a equal b times the square root of 2 or a square equal 2 times b square because a square equal 2 times of integer so a square even so a also odd. Example equal 2c so the equation because four c square equal 2b. So, be square even and b also even. But, it's impossible because a and b impossible even because they are relative prime. So, assumption that square root of 2 is rational number is impossible. So, the conclusion that the square root of 2 is irrational number.
2. To show or to indicate that the some angels of triangle are equal to 180 degree so we can cut all of angel of triangle. Then, we adhere become one so three angel that will configured straight supplementary angel. We also prove it with we draw a line pass one of point angel from triangle which parallel with side in front of it. For example we have triangle ABC and we draw MN line pass point angel B and parallel with AC. Attention that size supplementary angel on B equal size sides of triangle ABC are a degree plus b degree plus c degree equal 180 degree. Every pair angel congruent is pair angel inside opposite on lines parallel.
3. To get phi we must do exercise. We can make a circle has radius of one cm from wire. Then, we wrap wool accord form of that wire. Then, we measure the length of wire with rule. For example, really we get 6 coma 28 it means 6 coma 28 is round of the circle. We know that the round of circle is phi time’s diameter or phi times two times of radius so phi equal round divide two times of radius. So, we can write that 3 coma 14 equal phi times two times of radius. And know phi equal 6 coma 28 divide two equal 3 coma 14.
4. To find of the area of region foundered by the graph of y equal x square and y equal x plus 2 so we must draw of each curve. After draw it, so we look for intersection point between two equations above. So, we get value of x and x as interval of integral to account the area of region it. If, y equal x square and y equal x plus 2 so we can write that x square equal a plus 2 and we get that x square minus x minus two equal zero. So, x minus two in bracket times x plus one in bracket equal zero. So, we get value of x with x equal two and x equal negative one as interval of integral it. Now, we can account the area of region it with integral x plus two minus x square in bracket to x with an equal negative one until x equal two. If we integral the equation so we get half x square plus two x minus one over three x cube with interval x equal negative one until a equal two. So we must substitution x with negative one and two and we get four half. So, the area of region foundered by the graph of y equal x square and y equal x plus two is four half.
5. To determine the intersection point between he circle x square plus y square equal 20 and y equal x plus 1 so we must substitution y equal x plus 1 inside x square plus y square equal 20. And we get x square plus open bracket x plus 1 close bracket square equal 20, so x square plus x square plus two x plus 1 equal 20, so two x square plus two x minus 19 equal zero. To get value of x, we must use the formula abc that x one or x two equal negative b plus minus the root of b square minus four times a times c divide angel negative 4a. So, x one or x two equal negative 2 plus minus the root of 2 square minus 4 times 2 times negative 19 divide negative 4 times 2. After we account it so we get that value of x one equal negative 1 coma 291 or x two equal 1 coma 791.

Exercise II:
1. High line
High line from a triangle is segment line which perpendicular correlates a point angle with a point on side in front of point angle. The theorem high line on hypotenuse a right angle establishes two triangles which uniform and also uniform with right angles.
2. Percentile
Percentile is value which divide batch of data becomes 100 part same.
Formula of percentile:
Pi equal TB Pi plus open bracket (i divide 100) N minus fk divide f P10 close bracket times p.
TB is limit under percentile i
i is number of percentile
N is sum all percentile
f is frequentation class percentile i
fk is frequentation cumulative class percentile i
p is interval class

Exercise to percentile:
Table frequentation and frequentation cumulative score
Score (x) Frequentation (fi) Frequentation cumulative
0-9 3 3
10-19 67 70
20-29 205 275
30-39 245 520
40-49 213 733
50-59 147 880
60-69 77 957
70-79 34 991
80-89 8 999
90-99 1 10000
sum 10000

P10? To get it, so we look for class which contains P10 is we account 10 divide100 in braked N equal 10 divide 100 times 1000 equal 100.
P10 equal 19 coma 5 plus open bracket 100 minus 70 divide 205 close bracket times 10 so we get that P10 equal 20 coma 96.
3. Relation
• Definition of relation is making a pair from element of compilation A to element of compilation B.
• Domain relation R from A to B is subset compilation from A while range E is subset compilation from B. Example A={1,2,3,4,5} and B={a,b,c,d}
Relation R= {(1,b), (3,b), (4,c), (4,d)}
Domain from relation R from A to B={1,3,4}
Range from relation R from A to B={b,c,d}
• Kind of relation
a. Reflective relation
Relation R called reflective relation if to every a element S, (a,a) element R is every element S is related with their self. Example relation R inside G is same and uniform relation
b. Symmetry relation
Relation R called symmetry relation if to every a,b element S (a,b) element R so (b,a) element R. Example G is parallel relation.
c. Transitive relation
Relation R called transitive relation if to every a,b element S if (a,b) element R and (b,c) element R. Example Q is relation minus from.
d. Equivalent relation
Relation R called equivalent relation if relations R are reflective relation, symmetry relation, and transitive relation.