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Operating instructions HYUNDAI ACCENT - mypdfmanuals.com

Inner strength Accent's solid feel starts with the use of a safe, rigid body shell Your safety is our mission Hyundai engineers have raised standards in their quest to protect everyone in the event of a crash and that includes any pedestrians unfortunate enough to be caught in the incident. For a start, the safety blueprint for Accent requires impeccable handling, which is now accompanied by larger anti-lock disc brakes. The reinforced body shell is complemented by solid steel side impact beams, six air bags and seat belts equipped with pre-tensioners. Scientifically monitored tests in our high technology laboratories confirm computer calculations that the Accent exceeds the toughest crash protection legislation from around the world. Cushioning the impact Occupants have the benefit of dual front air bags, another on the side of each front seat, and roof-mounted bags that form protective curtains along each side of the interior. Safety first Sure handling and anti-lock brakes help to keep the Accent on-line. 1.6-litre DOHC CVVT petrol Variable valve timing and extra displacement produce 82.4kW (112 PS) 1.4-litre DOHC petrol A small displacement increase takes power output to 69. 9kW (97 PS) 1.5-litre CRDi turbodiesel Common rail fuel injection helps to produce top-class 81kW (110 PS) The best of everything Accent's trio of engines have it all: economy, power, refinement and durability Stop and go Front brake disc diameters are bigger for more stopping power. Well-proven MacPherson strut-type front suspension gives excellent on-road performance. Accent revives the time when driving was fun Driving today can sometimes be a chore, but Hyundai does not believe that means cars should be dull. Quite the contrary. Our aim is to put the sparkle of enjoyment back into driving. Accent is an example of what we mean. The stiffness and strength of the new platform and body shell make the car solid and secure. Our engineers paid particular attention to reducing what they know as NVH noise, vibration and harshness. The steering is light and precise and the anti-lock disc brakes are world-class.

SDHQ Ford Raptor A-Pillar Light Mount Parts List ... - SDHQ Off Road

SDHQ Ford Raptor A-Pillar Light Mount SDHQ-13-1206 Parts List 1 – Passenger Bracket 1 – Driver Bracket Tools Required 10MM RATCHET WRENCH / BOX WRENCH SMALL FLATHEAD SCREWDRIVER ESTIMATED 40-60 MINUTES INSTALL TIME FOR BRACKETS ONLY STEP 1: Raise hood, locate hood hinge bolts near base of windshield. With 10mm wrench (ratchet wrench preferred) remove both hinge bolts on one side of the truck. To ease installation, also remove the gas strut that lifts the hood on the side that you are working on (Be sure to brace center of hood with a prop). A small flathead screwdriver will pop the lower retaining clip enabling you to remove the lower end of the strut off the ball stud STEP 2: Feed the A-Pillar bracket into the cavity above the hood hinge. You may have to lift the hood slightly to feed the bracket into place. The vertical of the A-Pillar bracket sits next to the vertical of the hood hinge for reference. Reinstall 10mm bolts and final tighten. STEP 3: Reinstall strut, repeat step 1 and 2 for opposite side.

Subaru Forester SJG new - cusco usa
by kahn 0 Comments favorite 9 Viewed Download 0 Times

2012 Subaru Forester 2.0XT 2.0L Turbocharged FA20 Engine (SJG Chassis) Street ZERO-S (Fixed Damper Rate) Part No: 697 60P CB MSRP: $1,627.00 Sports Accel Pedal Part No: 965 766 A MSRP: $51.00 Folding Tow Hook (Front) Part No: 693 017 R MSRP: $131.00 Street ZERO-A (40 Way Damper Adjust) Part No: 697 60N CB MSRP: $1,974.00 Powerbrace Rear Member Side Part No: 687 492 RS MSRP: $174.00 Front Sway Bar Part No: 697 311 A26 MSRP: $347.00 Rear Sway Bar Part No: 692 311 B20 MSRP: $280.00 Powerbrace Front Side Part No: 697 492 FS MSRP: $227.00 Front Strut Bar Type OS Part No: 697 540 A MSRP: $200.00 Rear Strut Bar Type OS Part No: 692 541 A MSRP: $200.00 Powerbrace Rear End Part No: 677 492 RE MSRP: $227.00 CARROSSER Co. Ltd 1664-1 Shinbo-cho Takasaki-shi Gunma-ken 370-0018 JAPAN Tel: (+81) 27-352-3578 • Fax: (+81) 27-352-1919 http://www.cusco.co.jp oversea@cusco.co.jp CUSCO USA Inc. 16635 Gemini Lane Huntington Beach, CA 92647 USA Tel: (714) 907-0033 • Fax: (714) 369-8142 http://www.cuscousainc.com info@cuscousainc.com *CUSCO PRODUCTS ARE FOR OFFROAD & COMPETITION USE ONLY *Genuine 100% Made in Japan

moog® coNTRol aRmS - FMe360
by Fabio21 0 Comments favorite 15 Viewed Download 0 Times

moog R-SERIES™ control arms for value-driven repairs Total undercar solutions When it comes to undercar solutions, MOOG® offers the complete steering and suspension package. MOOG provides advanced engineering, problemsolving innovation, industry-leading technical expertise, superior products and the broadest range of foreign and domestic coverage that today’s technicians need. That is why MOOG is the choice of professional technicians and top NASCAR® Crew Chiefs. Upper Strut Mount Upper Control Arm Coil Spring Control Arm Bushing Upper Ball Joint For the cost-conscious consumer, the new R-Series™ line of control arms is the right choice. With a broad range of applications, and with the quality construction and design you expect from MOOG, R-Series provides technicians the perfect solution for the value-driven repair. INDUSTRY EXPERTISE Technical Training & Support Today’s vehicle technologies create new challenges for technicians. That’s why Federal-Mogul offers an array of industry-leading training and support options: • Technical Education Center • ASE-certified call center • In-market seminars • Live webinars • 24/7 parts lookup • Technical web-based library It’s superior training and support you only get with MOOG. ww PR B w.m OB oo Ba LE to ll joi M: inf nt eri fai gp ro w Ball em w w so lv er U LL ET Years 19971999- 2003 1990- 2003 1995- 2002 1996- 1998 1999 disin Pla tegr stic ated bea uPP ring er R O B .m LE : 21 10 02 ll fa join t il ure Ac ys Hon ura se da CL LU y, Ac MO TI Isuz co & TL Lo OG ON we ® K9 : u Oard , r Ba 64 To ensu & ll Jo3 Fro catas re sis parts troph depe int nt whic like ic ndab failur MOO stren h le Od SO...

Impreza Rear Brace Installation Instructions - Oswald Performance

Note: The following instructions were written for a 2005 WRX STi. The procedure may be slightly different for different models or years. This installation should be done with the suspension loaded and the wheels on level ground. (Notes for Wagon Installation added in red) 1. Remove seat bottom by removing the two bolts (one on each side). 2. Remove seat back by removing the three bolts (left, right, and center). (Wagon: Remove seat backs by folding seats down and unbolting center bracket from the seat backs. Then pull seat backs out of the side mount/pivot points by pulling toward the center of the car.) 3. Remove the two vertical bars (left and right) by removing the four bolts each (two at the top and two at the bottom). Remove the trunk trim panel by removing the four plastic clips (circled). (Wagon: This step does not apply to Wagon Installations) 4. Remove the two side seatbelts by removing the bolt shown. (Wagon: This step does not apply to Wagon Installations) 5. Remove the three nuts on each strut top.

2012 Nissan | Warranty Information Booklet | Nissan USA

TABLE OF CONTENTS 1 WARRANTY COVERAGE AT A GLANCE 2 NISSAN’S CUSTOMER CARE PROGRAM 4 NISSAN’S COMMITMENT TO CUSTOMER SATISFACTION CONTINENTAL/GENERAL TIRE LIMITED WARRANTY 33 MICHELIN TIRE LIMITED WARRANTY 35 TOYO TIRE LIMITED WARRANTY 40 ORIGINAL EQUIPMENT TIRE LIMITED WARRANTIES 41 IMPORTANT TIRE SAFETY INFORMATION 2012 NEW VEHICLE LIMITED WARRANTY FEDERAL VEHICLE EMISSION CONTROL LIMITED WARRANTIES GOODYEAR/DUNLOP TIRE LIMITED WARRANTY 30 9 BRIDGESTONE FIRESTONE TIRE LIMITED WARRANTY 24 5 21 12 CALIFORNIA VEHICLE EMISSION CONTROL WARRANTIES 18 SEAT BELT LIMITED WARRANTY 19 BFGOODRICH TIRE LIMITED WARRANTY TABLE OF CONTENTS 47 49 50 LIMITED WARRANTY ON GENUINE NISSAN REPLACEMENT PARTS, GENUINE NISMO S-TUNE PARTS, AND GENUINE NISSAN ACCESSORIES 52 REPLACEMENT BATTERY LIMITED WARRANTY 54 GENUINE NISSAN PARTS AND ACCESSORIES NISSAN LIFETIME REPLACEMENT PANEL CORROSION LIMITED WARRANTY 55 CORROSION PROTECTION GUIDELINES 56 NISSAN’S SECURITY+PLUSா VEHICLE PROTECTION PLAN GENUINE NISSAN ORIGINAL EQUIPMENT MUFFLER, GENUINE NISSAN SHOCK ABSORBER AND STRUT LIFETIME LIMITED WARRANTY

2010 Nissan Warranty Information Booklet

TABLE OF CONTENTS 1 WARRANTY COVERAGE AT A GLANCE 2 NISSAN’S CUSTOMER CARE PROGRAM 3 NISSAN’S COMMITMENT TO CUSTOMER SATISFACTION 4 2010 NEW VEHICLE LIMITED WARRANTY 7 FEDERAL VEHICLE EMISSION CONTROL LIMITED WARRANTIES 10 CALIFORNIA VEHICLE EMISSION CONTROL WARRANTIES 14 SEAT BELT LIMITED WARRANTY 15 BFGOODRICH TIRE LIMITED WARRANTY 17 BRIDGESTONE FIRESTONE TIRE LIMITED WARRANTY 19 GOODYEAR/DUNLOP TIRE LIMITED WARRANTY 24 CONTINENTAL/GENERAL TIRE LIMITED WARRANTY TABLE OF CONTENTS 27 MICHELIN TIRE LIMITED WARRANTY 29 TOYO TIRE LIMITED WARRANTY 32 ORIGINAL EQUIPMENT TIRE LIMITED WARRANTIES 33 IMPORTANT TIRE SAFETY INFORMATION 38 LIMITED WARRANTY ON GENUINE NISSAN REPLACEMENT PARTS, GENUINE NISMO S-TUNE PARTS, AND GENUINE NISSAN ACCESSORIES SUMMARY OF THE NISSAN LIFETIME REPLACEMENT PANEL CORROSION WARRANTY GENUINE NISSAN ORIGINAL EQUIPMENT MUFFLER, GENUINE NISSAN SHOCK ABSORBER AND STRUT LIFETIME LIMITED WARRANTY 43 REPLACEMENT BATTERY LIMITED WARRANTY 44 GENUINE NISSAN PARTS AND ACCESSORIES 45 CORROSION PROTECTION GUIDELINES 46 40 41 NISSAN’S SECURITY+PLUS® VEHICLE PROTECTION PLAN...

SNMPTN 2012 Matematika - zenius.net

SNMPTN 2012 Matematika Doc. Name: SNMPTN2012MATDAS999 Version : 2013-04 halaman 1 01. Jika a dan b adalah bilangan bulat positif yang memenuhi ab = 220 - 219, maka nilai a+b adalah …. (A) 3 (B) 7 (C) 19 (D) 21 (E) 23 02. Jika 4log3 = k , maka 2log27 adalah … (A) k 6 (B) (C) (D) (E) k 6k 6 k6 k 03. Jika p+1 dan p-1 adalah akar-akar persamaan x2 - 4x + a = 0, maka nilai a adalah …. (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 04. Jika f adalah fungsi kuadrat yang grafiknya melalui titik (1,0), (4,0), dan (0,-4), maka nilai f(7) adalah …. (A) -16 (B) -17 (C) -18 (D) -19 (E) -20 Kunci dan pembahasan soal ini bisa dilihat di www.zenius.net dengan memasukkan kode 2429 ke menu search. Copyright © 2012 Zenius Education SNMPTN 2012 Matematika, Kode Soal doc. Name: SNMPTN2011MATDAS999 version : 2013-04 | halaman 2 05. Semua nilai x yang memenuhi (x + 3)(x - 1) ≥ (x - 1) adalah (A) 1 ≤ x ≤ 3 (B) x ≤ -2 atau x ≥ 1 (C) -3 ≤ x ≤ -1 (D) -2 ≥ x atau x ≥ 3 (E) -1 ≥ x atau x ≥ 3 06. Jika 2x - z = 2, x + 2y = 4, dan y + z = 1, maka nilai 3x + 4y + z adalah …. (A) 4 (B) 5 (C) 6 (D) 7 (E) 8 07. Jika diagram batang di bawah ini memperlihatkan frekuensi kumulatif hasil tes matematika siswa kelas XII, maka persentase siswa yang memperoleh nilai 8 adalah…. (A) (B) (C) (D) (E) 12 % 15 % 20 % 22 % 80 % Kunci dan pembahasan soal ini bisa dilihat di www.zenius.net dengan memasukkan kode 2429 ke menu search. Copyright © 2012 Zenius Education SNMPTN 2012 Matematika, Kode Soal doc. Name: SNMPTN2011MATDAS999 version : 2013-04 | halaman 3 08. Ani telah mengikuti tes matematika sebanyak n kali. Pada tes berikutnya ai memperoleh nilai 83 sehingga nilai rata-rata Ani aalah 80, tetapi jika nilai tes tersebut adalah 67, maka rata-ratanya adalah 76. Nilai n adalah …. (A) 2 (B) 3 (C) 4 (D) 5 (E) 6 09. Nilai maksimum fungsi objektif (tujuan) f(x,y) = 3x + 2y dengan kendala x + 2y ≤ 12, x ≥ 2, dan y ≥ 1 adalah …. (A) 16 (B) 18 (C) 32 (D) 36 (E) 38 10. Jika dan , maka determinan matriks AB - C adalah …. (A) -5 (B) -4 (C) 5 (D) 6 (E) 7 11. Agar tiga bilangan a + 2, a - 3, a - 4 merupakan barisan aritmatika, maka suku ke dua harus ditambah dengan …. (A) -3 (B) -2 (C) -1 (D) 1 (E) 2 Kunci dan pembahasan soal ini bisa dilihat di www.zenius.net dengan memasukkan kode 2429 ke menu search. Copyright © 2012 Zenius Education SNMPTN 2012 Matematika, Kode Soal doc. Name: SNMPTN2011MATDAS999 version : 2013-04 | halaman 4 12. Jika suku pertama barisan aritmatika adalah -2 dengan beda 3, Sn adalah jumlah n suku pertama deret aritmatika tersebut, dan Sn+2 - Sn = 65, maka nilai n adalah …. (A) 11 (B) 12 (C) 13 (D) 14 (E) 15 13. Jika suatu persegi dengan sisi satu satuan dibagi menjadi 5 persegi panjang dengan luas yang sama seperti ditunjukkan pada gambar di bawah ini, maka panjang ruas garis AB adalah … (A) 3 5 (B) 2 3 (C) 2 5 (D) (E) 1 5 1 5 14. Di suatu kandang tedapat 40 ekor ayam, 15 ekor diantaranya jantan. Di antara ayam jantan tersebut, 7 ekor berwarna putih. Jika banyak ayam berwarna putih adalah 22 ekor, maka banyak ayam betina yang tidak berwarna putih adalah … (A) 5 (B) 7 (C) 8 (D) 10 (E) 15 Kunci dan pembahasan soal ini bisa dilihat di www.zenius.net dengan memasukkan kode 2429 ke menu search. Copyright © 2012 Zenius Education SNMPTN 2012 Matematika, Kode Soal doc. Name: SNMPTN2011MATDAS999 version : 2013-04 | halaman 5 15. Jika f(x) = ax + 3, a ≠ 0 dan f-1 (f-1(9)) = 3, maka nilai a2 + a + 1 adalah … (A) 11 (B) 9 (C) 7 (D) 5 (E) 3 Kunci dan pembahasan soal ini bisa dilihat di www.zenius.net dengan memasukkan kode 2429 ke menu search. Copyright © 2012 Zenius Education

Soal-Soal dan Pembahasan Matematika IPA SNMPTN 2012 ...

Soal-Soal dan Pembahasan Matematika IPA SNMPTN 2012 Tanggal Ujian: 13 Juni 2012 1. Lingkaran (x + 6)2 + (y + 1)2 = 25 menyinggung garis y = 4 di titik... A. ( -6, 4 ) B. ( 6 , 4) C. ( -1, 4 ) D. ( 1, 4 ) E. ( 5 , 4 ) Jawab: BAB XI Lingkaran Masukkan nilai y=4 pada persamaan (x + 6)2 + (4 + 1)2 = 25 (x + 6)2 = 25 – 25 = 0 x = -6 Didapat titik x = -6 dan y = 4  (-6,4) Jawabannya A 2. Jika 2x3 – 5x2 – kx + 18 dibagi x - 1 mempunyai sisa 5, maka nilai k adalah... A. -15 B. -10 C. 0 D. 5 E. 10 Jawab: BAB XII Suku Banyak Metoda Horner x3 x= 1 2 x2 x -k 18 2 2 -5 -3 -3 - k -3 ( -3- k) + = kalikan dengan x =1 (15 – k)  sisa =5 15 – k = 5 k = 15 – 5 = 10 Jawabannya E www.belajar-matematika.com 1 3. Luas daerah yang dibatasi oleh kurva y = x2, y = 1, dan x = 2 adalah... A. ∫ (1 − B. ∫ ( ) C. ∫ ( − 1) − 1) D. ∫ (1 − Jawab BAB XVI Integral E. ∫ ( ) − 1) Buat sketsa gambar untuk mengetahui batas luas: terlihat bahwa bidang luasnya (arsiran) bagian atasnya adalah y = x 2 dan bagian bawahnya y = 1 dengan dibatasi oleh batas atas x = 2 dan batas bawah x =1. Dalam notasi integralnya : b ∫ ( b b a a a L =  y2 dx -  y1 dx =  ( y 2  y1) dx − 1) Jawabannya C 4. ( ( A. B. ) ) = .... C. E. D. www.belajar-matematika.com 2 Jawab: BAB VII Trigonometri ( ( + 2 sin cos ) ) = = = =1 = 2 Jawabannya E 5. Lingkaran (x - 3)2 + (y - 4)2 = 25 memotong sumbu –x di titik A dan B. Jika P adalah titik pusat lingkaran tersebut, maka cos ∠APB = ... A. C. B. E. D. Jawab: BAB XI Lingkaran dan BAB VII Trigonometri Sketsa gambar: Lingkaran dengan pusat (3,4) APB merupakan segitiga. www.belajar-matematika.com 3 Untuk menjawab soal ini digunakan teorema di bawah ini: Aturan sinus dan cosinus C  b  a  A c B Aturan cosinus 1. a 2 = b 2 + c 2 - 2bc cos  2. b 2 = a 2 + c 2 - 2ac cos  3. c 2 = a 2 + b 2 - 2ab cos  Kita pakai rumus (3) c = AB = 6 a = b = AP = PB = √3 + 4 = √25 = 5 c 2 = a 2 + b 2 - 2ab cos P 2ab cos P = + − cos P = = = = . . . Jawabannya A 6. Grafik fungsi f(x) = ax3 – bx2 + cx + 12 naik jika.... A. b2 – 4ac < 0 dan a > 0 B. b2 – 4ac < 0 dan a < 0 C. b2 – 3ac > 0 dan a < 0 D. b2 – 3ac < 0 dan a > 0 E. b2 – 3ac < 0 dan a < 0 Jawab: BAB XV Differensial www.belajar-matematika.com 4 Syarat fungsi naik ( )>0 3ax2 - 2bx + c > 0  fungsi naik ( - , 0, + ) * variabel x2 > 0 3a > 0 a>0 *D<0 ( ) > 0 , maka tidak ada titik potong dan singgung di sb x sehingga D < 0  karena (-2b)2 – 4.3a.c < 0 4b2 – 12.a.c < 0 b2 – 3 ac < 0 didapat a > 0 dan b2 – 3 ac < 0 Jawabannya D 7. →0 = .... E. √3 √ A. -1 C. 1 B. -0 D. Jawab: XIV Limit Fungsi →0 = →0 = = = →0 →0 1 . 1. = = =1 Jawabannya C www.belajar-matematika.com

Soal dan Pembahasan Matematika IPA SNMPTN 2011

Soal-Soal dan Pembahasan SNMPTN Matematika IPA Tahun Pelajaran 2010/2011 Tanggal Ujian: 01 Juni 2011 1. Diketahui vektor u = (a, -2, -1) dan v = (a, a, -1). Jika vektor u tegak lurus pada v , maka nilai a adalah ... A. -1 B. 0 C. 1 D. 2 E. 3 Jawab: Vektor: vektor u tegak lurus pada v maka u . v = 0 u = −2 , v = −1 −2 . −1 −1 (a – 1) (a-1) = 0 maka a = 1 −1 = a2 – 2a + 1 = 0 (a - 1)2 = 0 Jawabannya adalah C 2. Pernyataan berikut yang benar adalah ... A. Jika sin x = sin y maka x = y B. Untuk setiap vektor u , v dan w berlaku u . ( v . w ) = ( u . v ). w C. Jika b  f ( x) dx = 0, maka a D. Ada fungsi f sehingga E. 1 – cos 2x = 2 cos2 x f ( x )= 0 Lim f(x) ≠ f(c) untuk suatu c xc www.belajar-matematika.com - 1 Jawab: Trigonometri, vektor, integral, limit A. Ambil nilai dimana sin x = sin y  sin α = sin (1800 – α ) ambil nilai α = 600  sin 600 = sin 1200 ; tetapi 600 ≠ 1200 Pernyataan SALAH B. Operasi u . ( v . w ) tak terdefinisi karena v . w = skalar, sedangkan u = vektor vektor . skalar = tak terdefinisi Pernyataan SALAH C. Ambil contoh cari cepat hasil dimana b  f ( x) dx = 0 ; a 1 Didapat b = 1 dan a = -1 maka f(x)= x   x dx = 0  1 terbukti : f(x) = x bukan f(x) = 0 x2 | Pernyataan SALAH D. Ambil contoh f(x) = Lim xc f(x) = Lim x 1 ( ( = ( ( ) ( )( ) = ) ( ) Lim f(x) ≠ f(c)  2 ≠ 1 xc ) ( )( ) = ) ( ) =2 Pernyataan BENAR E. 1 – cos 2x = 1 – ( 2cos2 x – 1) = 1 + 1 - 2cos2 x = 2 - 2cos2 x = 2 ( 1 – cos2 x) Pernyataan SALAH Jawabannya adalah D www.belajar-matematika.com - 2 = (1 – 1) = 0 3. Luas daerah di bawah y = -x2 +8x dan di atas y = 6x - 24 dan terletak di kuadran I adalah.... a. ∫ (− b. ∫ (− c. ∫ (− +8 ) +8 ) +8 ) d. ∫ (6 − 24) e. ∫ (6 − 24) Jawab: Integral: +∫ ( + ∫ (− + ∫ (− + ∫ (− + ∫ (− − 2 − 24) + 2 + 24) + 2 + 24) +8 ) +8 ) kuadran I titik potong kedua persamaan : y1 = y2 -x2 +8x = 6x-24 -x2 +8x - 6x+24 = 0 -x2 +2x + 24 = 0 x2 -2x - 24 = 0 (x - 6) (x+4)0 x = 6 atau x = -4  karena di kuadran I maka yang berlaku adalah x = 6  y = 6.6 – 24= 12 berada di titik (6,12) www.belajar-matematika.com - 3 L = ∫ (− = ∫ (− +8 ) +8 ) + ∫ ((− + ∫ (− Jawabannya adalah B + 8 ) − (6 − 24)) + 2 + 24) 4. sin 350 cos 400 - cos 35 sin 400 = A. cos 50 B. sin 50 C. cos 950 D. cos 750 E. sin 750 Jawab: Trigonometri: Pakai rumus: sin (A - B) = sin A cos B - cos A Sin B A= 350 ; B = 400 = sin (350 - 400) = sin -50 Cos (90 0 -  ) = sin   rumus Cos (90 0 - (-50) ) = sin -50   = -50 Cos 950 = sin -50 Jawabannya adalah C 5. Diketahui g(x) = ax2 – bx + a – b habis dibagi x – 1. Jika f(x) adalah suku banyak yang bersisa a ketika dibagi x – 1 dan bersisa 3ax + b2 + 1 ketika dibagi g(x), maka nilai a adalah...... A. -1 B. -2 C. 1 D. 2 Jawab: Suku Banyak: g(x) = ax2 – bx + a – b habis dibagi x – 1  g(1) = 0 g(1) = a . 1 – b .1 + a – b = 0 =a–b+a–b=0 2a – 2b = 0 2a = 2b  a = b karena a = b maka: g(x) = ax2 – ax + a – a = ax2 – ax www.belajar-matematika.com - 4 E. 3 f(x) dibagi dengan f(x-1) sisa a  f(1) = a f(x) dibagi dengan g(x) sisa 3ax + b2 + 1 f(x) dibagi dengan ax2 – ax sisa 3ax + b2 + 1 f(x) dibagi dengan ax(x – 1) sisa 3ax + b2 + 1 teorema suku banyak: Jika suatu banyak f(x) dibagi oleh (x- k) akan diperoleh hasil bagi H(x) dan sisa pembagian S  f(x) = (x- k) H(x) + S f(x) dibagi dengan ax(x – 1) sisa 3ax + b2 + 1 f(x) = ax (x - 1) H(x) + (3ax + b2 + 1) substitusikan nilai nol dari pembagi yaitu x = 0 dan x = 1  dari ax (x - 1) ambil x = 1  untuk x = 1 f(1) = a . 1 (1 – 1) H(0) + 3a.1 + b2 + 1 a = 0 + 3a + b2 + 1  diketahu a = b, masukkan nilai a = b a = 3a + a2 + 1 a2 + 2a + 1 = 0 (a+1)(a+1) = (a+1)2 = 0 a = -1 Jawabannya adalah A 6. Rotasi sebesar 450 terhadap titik asal diikuti dengan pencerminan terhadap y = -x memetakan titik (3,4) ke .... A. √ B. − Jawab: ,√ √ ,√ C. D. √ √ ,−√ ,−√ E. − Transformasi Geometri:  cos  Rotasi sebesar 450 terhadap titik asal =   sin    sin    cos     0  1 pencerminan terhadap y = -x    1 0     www.belajar-matematika.com - 5 √ ,√

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