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1. English as Medium of Instruction Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) Smart Solution Smart Solution English as Medium of Instruction Ridho Alfarisi dan Agustin Puspitasari Pendidikan Matematika Pendidikan MIPA Fakultas Keguruan dan Ilmu Pendidikan Universitas Jember 22nd May 2013 Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) FKIP English as Medium of Instruction

2. English as Medium of Instruction Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) Smart Solution Smart Solution Smart Solution 1 Number Theory 2 Answer of Number Theory 3 Algebra 4 Answer of Algebra 5 Geometry 6 Answer of Geometry 7 Probability 8 Answer of Probability Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) FKIP English as Medium of Instruction

10. English as Medium of Instruction Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Number Theory Exercise 1 How many digit of multiplication 22002*52003 ? Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) FKIP English as Medium of Instruction

11. English as Medium of Instruction Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Number Theory Exercise 1 How many digit of multiplication 22002*52003 ? Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) FKIP English as Medium of Instruction

12. English as Medium of Instruction Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Answer of Number Theory Answer 1 22002 ∗ 52003 = 22002 ∗ 52002 ∗ 5 = (2 ∗ 5)2002 ∗ 5 = (10)2002 ∗ 5 so all digits is 2003 Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) FKIP English as Medium of Instruction

13. English as Medium of Instruction Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Answer of Number Theory Answer 1 22002 ∗ 52003 = 22002 ∗ 52002 ∗ 5 = (2 ∗ 5)2002 ∗ 5 = (10)2002 ∗ 5 so all digits is 2003 Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) FKIP English as Medium of Instruction

14. English as Medium of Instruction Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Algebra Exercise 1 Lets a and b is natural number with a > b. if √ (94 + 2 ∗ √ (2013)) = √ (a) + √ (b), then value of a − b is..... 2 Let’s p and q is prims number. if its known equation x2014 − p ∗ x2013 + q = 0 haveing root integers number, then value of p + q is... Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) FKIP English as Medium of Instruction

17. English as Medium of Instruction Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Answer of Algebra Answer 1 for a, b ≥ 0, then ( √ a + √ b)2=a ∗ b + 2 ∗ √ a ∗ b←→ √ a + √ b= (a + b) + 2 ∗ √ a ∗ b.We back that 94 + 2 ∗ √ 2013=(61 + 33) + 2 ∗ √ 61 ∗ 33, therefore 94 + 2 ∗ √ 2013= √ 61 + √ 33 so we get the value a = 61, b = 33, then a − b=61 − 33=28 Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) FKIP English as Medium of Instruction

18. English as Medium of Instruction Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Answer of Algebra Answer 1 for a, b ≥ 0, then ( √ a + √ b)2=a ∗ b + 2 ∗ √ a ∗ b←→ √ a + √ b= (a + b) + 2 ∗ √ a ∗ b.We back that 94 + 2 ∗ √ 2013=(61 + 33) + 2 ∗ √ 61 ∗ 33, therefore 94 + 2 ∗ √ 2013= √ 61 + √ 33 so we get the value a = 61, b = 33, then a − b=61 − 33=28 Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) FKIP English as Medium of Instruction

19. English as Medium of Instruction Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Answer of Algebra Answer 1 Let’s one of root integers of equation x2014 − p ∗ x2013 + q = 0 is t, then we get t2014 − p ∗ t2013 + q = 0 ⇐⇒ q = t2013(p − t). Following that −1 and 0 isn’t root of equation x2014 − p ∗ x2013 + q = 0. So with remember that q is prims number, then we get t = 1 therefore q = p − 1 ⇐⇒ p − q = 1. This information of above can concluded that one of p,q is even and because even prims number only 2, then we get q = 2 and p = 3. So p + q = 5 Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) FKIP English as Medium of Instruction

20. English as Medium of Instruction Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Answer of Algebra Answer 1 Let’s one of root integers of equation x2014 − p ∗ x2013 + q = 0 is t, then we get t2014 − p ∗ t2013 + q = 0 ⇐⇒ q = t2013(p − t). Following that −1 and 0 isn’t root of equation x2014 − p ∗ x2013 + q = 0. So with remember that q is prims number, then we get t = 1 therefore q = p − 1 ⇐⇒ p − q = 1. This information of above can concluded that one of p,q is even and because even prims number only 2, then we get q = 2 and p = 3. So p + q = 5 Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) FKIP English as Medium of Instruction

21. English as Medium of Instruction Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Geometry Exercise 1 Lets P is interior point in the triangle ABC, so value of < PAB = 10◦, < PBA = 20◦,< PCA = 30◦,< PAC = 40◦, value of < ABC = ....... 2 Given a triangle ABC with this area 10. Point D,E, dan F respectively lies on the edge AB, BC, dan CA with AD =2, DB=3. If a triangle ABE and a rectangle DBFE has same the area, then it’s the area is........ Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) FKIP English as Medium of Instruction

24. English as Medium of Instruction Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Answer of Geometry Answer 1 Following a sketch at below: Because the area ∆ABE= DBFE cause the area ∆ADE=∆DEF. We know that DE is partnership edge among ∆ADE and ∆DEF, so distance point A to the edge DE equal to distance point F to the edge DE. in other words AF parallel DE so CE EB = AD DB = 2 3 . Therefore, the area ∆ ABE = 3 5 ∗ 10 = 6. Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) FKIP English as Medium of Instruction

25. English as Medium of Instruction Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Answer of Geometry Answer 1 Following a sketch at below: Because the area ∆ABE= DBFE cause the area ∆ADE=∆DEF. We know that DE is partnership edge among ∆ADE and ∆DEF, so distance point A to the edge DE equal to distance point F to the edge DE. in other words AF parallel DE so CE EB = AD DB = 2 3 . Therefore, the area ∆ ABE = 3 5 ∗ 10 = 6. Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) FKIP English as Medium of Instruction

26. English as Medium of Instruction Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Probability Exercise 1 A dice at toss six times. How many trick for get total dies 28 correctly one dice arises digit 6 is........ Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) FKIP English as Medium of Instruction

27. English as Medium of Instruction Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Probability Exercise 1 A dice at toss six times. How many trick for get total dies 28 correctly one dice arises digit 6 is........ Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) FKIP English as Medium of Instruction

28. English as Medium of Instruction Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Answer of Probability Answer 1 Tanpa mengurangi keumuman misalkan tos pertama muncul angka 6. maka pada tos ke dua sampai tos ke enam hanya boleh muncul angka 1, 2, 3, 4, 5 dan jumlanya 22. kemungkinan hal seperti ini hanya ada 3 kasus yaitu : yang pertama (2, 5, 5, 5, 5) ada 5 cara dari 5! 4!, yang kedua (3, 4, 5, 5, 5) ada 20 cara dari 5! 3!, yang ketiga (4, 4, 4, 5, 5) ada 10 cara dari 5! 2!∗3!. sehingga total ada 35 cara jika pada tos pertama muncul angka 6. Karena keenam tos memiliki peluang yang sama untuk muncul angka 6 berarti total keseluruhan cara yang mungkin yaitu 6 ∗ 35 = 210 cara Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) FKIP English as Medium of Instruction

29. English as Medium of Instruction Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Number Theory Number Theory Algebra Algebra Geometry Algebra Probability Algebra Answer of Probability Answer 1 Tanpa mengurangi keumuman misalkan tos pertama muncul angka 6. maka pada tos ke dua sampai tos ke enam hanya boleh muncul angka 1, 2, 3, 4, 5 dan jumlanya 22. kemungkinan hal seperti ini hanya ada 3 kasus yaitu : yang pertama (2, 5, 5, 5, 5) ada 5 cara dari 5! 4!, yang kedua (3, 4, 5, 5, 5) ada 20 cara dari 5! 3!, yang ketiga (4, 4, 4, 5, 5) ada 10 cara dari 5! 2!∗3!. sehingga total ada 35 cara jika pada tos pertama muncul angka 6. Karena keenam tos memiliki peluang yang sama untuk muncul angka 6 berarti total keseluruhan cara yang mungkin yaitu 6 ∗ 35 = 210 cara Ridho Alfarisi (110210101043) dan Agustin Puspitasari (110210101061) FKIP English as Medium of Instruction

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