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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 √ ,√

A New Direction for Streetwear Apparel

Many designers have tried their hand at taking urban fashion in new directions. It isn’t easy to harness everything into a single fashion expression, but the L.A.T.H.C. approach may come closest to that ideal.

NYS Mathematics Glossary* – Algebra 2/Trig - Regents Exam Prep ...

NYS Mathematics Glossary* – Algebra 2/Trig *This glossary has been amended from the full SED Commencement Level Glossary of Mathematical Terms (available at http://www.emsc.nysed.gov/ciai/mst/math/glossary/home.html) to list only terms indicated to be at the Algebra 2/Trig level.) This Glossary, intended for teacher use only, provides an understanding of the mathematical terms used in the Regents-approved course entitled Algebra 2/Trig (as reflected in the NYS Mathematics Core Curriculum). A a + bi form The form of a complex number where a and b are real numbers, and i = −1 . abscissa The horizontal or x-coordinate of a two-dimensional coordinate system. absolute value The distance from 0 to a number n on a number line. The absolute value of a number n is indicated by n . Example: −3 = 3 , +3 = 3 , and 0 = 0 . absolute value equation An equation containing the absolute value of a variable. Example: x+3 = 9 absolute value function A function containing the absolute function of a variable. ⎧ x, x ≥ 0 ⎫ Example: f ( x) = x = ⎨ ⎬ ⎩ − x, x < 0 ⎭ absolute value inequality An inequality containing the absolute value of a variable. Example: x + 3 < 9 adjacent angles Two coplanar angles that share a common vertex and a common side but have no common interior points. Example: In the figure, ∠AOB and ∠BOC are a pair of adjacent angles, but ∠AOC and ∠BOD are not adjacent. A B C O D 2 adjacent sides Two sides of any polygon that share a common vertex. algebraic equation A mathematical statement that is written using one or more variables and constants which contains an equal sign. Examples: 3y + 5 = 1 2 x − 5 = 11 log 5 ( x − 3) = 2 2x = 1 8 algebraic expression A mathematical phrase that is written using one or more variables and constants, but which does not contain a relation symbol ( <, >, ≤, ≥, =, ≠ )...

The Role of Packaging in Building Brand Identity - Franke+Fiorella

commercial paint, the packaging offers a huge opportunity to positively impact a brand’s image. But what we frequently hear from many business–to–business companies is, “the packaging doesn’t really matter, our customers place their orders via catalog”. Or in other words, they don’t need the package to sell the product on a retail shelf. But selling is only one function that a package offers. In addition to selling, packaging serves four other purposes, all of which influence how a brand is perceived: 1. Brand Identity Expression If a product or brand is a leader in the industry or category, the design of the packaging — from sleek health & beauty products to utilitarian chemical drums — should reflect that position. High quality printing, distinctive design that supports the brand strategy and unique materials can help engage users and differentiate a brand in today’s crowded markets. 2. Relationship Building Packaging is a core part of building a relationship between a brand and the end user, in some cases, long after the sale has been made. For example, packaging for products like toothpaste or commercial office supply products will impact how the brand is perceived every time the product is used. Is it aesthetically pleasing? Does the package make life easier (or more difficult)? A positive user experience can encourage loyalty and even increase the amount end users are willing to pay for a product. 3. Communication Good package design makes information easy to find. If necessary information is difficult to locate, read or otherwise use, the package is likely not living up to the expectation of the end user. Consider commercial cleaning products — the packaging doesn’t need to sell the product on a retail shelf. But it absolutely must communicate (frequently in more than one language) what the product is, how it’s different from other products in the line and how it should be used. Creating an experience with your brand that makes life easier and minimizes frustration will go a long way toward building loyalty. 4. Selling If your products are marketed through consumer channels, you know the package is your last opportunity to convince someone to buy your product. But when it comes to commercial packaging, the role of selling is more subtle — you’re selling the next purchase. If the primary package is one that end users interact with regularly, you have an opportunity to build favorability between your brand and the user every time they use your product. 5. Protection The appearance and condition of a product when it reaches the customer impacts their identityWise® is published by Franke+Fiorella 401 North 3rd Street Suite 380 Minneapolis, MN 55401 612.338.1700 www.frankefiorella.com ©2006, Franke+Fiorella identityWise® is a trademark of Franke+Fiorella.

Laporan lengkap
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Potensi bencana alam yang tinggi pada dasarnya tidak lebih dari sekedar refleksi fenomena alam yang secara geografis sangat khas untuk wilayah tanah air kita. Indonesia merupakan Negara kepulauan tempat dimana tiga lempeng besar dunia bertemu, yaitu: lempeng Indo-Australia, Eurasia dan Pasifik. Interaksi antar lempeng-lempeng tersebut lebih lanjut menempatkan Indonesia sebagai wilayah yang memiliki aktivitas kegunungapian dan kegempaan yang cukup tinggi. Lebih dari itu, proses dinamika lempeng yang cukup intensif juga telah membentuk relief permukaan bumi yang khas dan cukup bervariasi, dari wilayah pegunungan dengan lereng-lerengnya yang curam dan seakan menyiratkan potensi longsor yang tinggi hingga wilayah yang landai sepanjang pantai dengan potensi ancaman banjir, penurunan tanah dan tsunaminya (Sadisun, 2005-2006). Berbagai potensi bencana alam yang mungkin timbul sudah sebaiknya harus kita kenal agar karakter bahaya alam tersebut dapat kita minimalkan dampaknya. Selain itu, potensi bencana alam ini telah diperparah oleh beberapa permasalahan lain yang muncul di tanah air kita yang memicu peningkatan kerentanannya. Laju pertumbuhan penduduk yang sangat tinggi merupakan salah satu contoh nyata, sehingga akan banyak membutuhkan kawasan-kawasan hunian baru yang pada akhirnya kawasan hunian tersebut akan terus berkembang dan menyebar hingga mencapai wilayah-wilayah marginal yang tidak selayaknya dihuni. Tidak tertib dan tepatnya perencanaan tata guna lahan, sebagai inti dari permasalahan ini merupakan faktor utama yang menyebabkan adanya peningkatan kerentanan. Peningkatan kerentanan ini akan lebih diperparah bila masyarakat sama sekali tidak menyadari dan tanggap terhadap adanya potensi bencana alam di daerahnya. Pengalaman memperlihatkan bahwa kejadian-kejadian bencana alam selama ini telah banyak menimbulkan kerugian dan penderitaan yang cukup berat sebagai akibat dari perpaduan bahaya alam dan kompleksitas permasalahan lainnya. Untuk itu diperlukan upaya-upaya yang komprehensif untuk mengurangi resiko bencana alam, antara lain yaitu dengan melakukan kegiatan migitasi. Bencana (disaster) merupakan fenomena sosial akibat kolektif atas komponen bahaya (hazard) yang berupa fenomena alam/buatan di satu pihak, dengan kerentanan (vulnerability) komunitas di pihak lain. Bencana terjadi apabila komunitas mempunyai tingkat kapasitas/kemampuan yang lebih rendah dibanding dengan tingkat bahaya yang mungkin terjadi padanya. Misalnya, letusan G. Merapi dan bahaya lainnya gempa bumi, banjir, gerakan tanah, dan lainnya tidak akan sertamerta menjadi bencana apabila komunitas memiliki kapasitas mengelola bahaya. Bencana cenderung terjadi pada komunitas yang rentan, dan akan membuat komunitas semakin rentan. Kerentanan komunitas diawali oleh kondisi lingkungan fisik, sosial, dan ekonomi yang tidak aman (unsave conditions) yang melekat padanya. Kondisi tidak aman tersebut terjadi oleh tekanan dinamis internal maupun eksternal (dynamic pressures), misalnya di komunitas institusi lokal tidak berkembang dan ketrampilan tepat guna tidak dimiliki. Tekanan dinamis terjadi karena terdapat akar permasalahan (root causes) yang menyertainya. Akar permasalahan internal umumnya karena komunitas tidak mempunyai akses sumberdaya, struktur dan kekuasaan, sedang secara eksternal karena sistem politik dan ekonomi yang tidak tepat. Oleh karenanya penanganan bencana perlu dilakukan secara menyeluruh dengan meningkatkan kapasitas dan menangani akar permasalahan untuk mereduksi resiko secara total. Siklus penanggulangan bencana yang perlu dilakukan secara utuh. Upaya pencegahan (prevention) terhadap munculnya dampak adalah perlakuan utama. Tsunami tidak dapat dicegah. Pencegahan dapat dilakukan pada bahaya yang manusia terlibat langsung maupun tidak langsung. Pada tsunami misalnya. Pencegahan dapat dilakukan rakyat dengan membuat bendung penahan ombak, bangunan panggung tahan ombak, penataan ruang dan sebagainya. Agar tidak terjadi jebolnya tanggul, maka perlu disusun save procedure dan kontrol terhadap kepatuhan perlakuan. Walaupun pencegahan sudah dilakukan, sementara peluang adanya kejadian masih ada, maka perlu dilakukan upaya-upaya mitigasi...

MIPS green sheet - EECS Instruction
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MIPS Reference Data Card (“Green Card”) 1. Pull along perforation to separate card 2. Fold bottom side (columns 3 and 4) together M I P S Reference Data CORE INSTRUCTION SET FORNAME, MNEMONIC MAT OPERATION (in Verilog) add Add R R[rd] = R[rs] + R[rt] Add Immediate addi ARITHMETIC CORE INSTRUCTION SET 1 OPCODE / FUNCT (Hex) (1) 0 / 20hex I Add Imm. Unsigned addiu R[rt] = R[rs] + SignExtImm (1,2) I R[rt] = R[rs] + SignExtImm (2) Add Unsigned addu and R R[rd] = R[rs] & R[rt] And Immediate andi I Branch On Equal beq I Branch On Not Equal bne I 0 / 24hex Jump j J R[rt] = R[rs] & ZeroExtImm if(R[rs]==R[rt]) PC=PC+4+BranchAddr if(R[rs]!=R[rt]) PC=PC+4+BranchAddr PC=JumpAddr Jump And Link jal J R[31]=PC+8;PC=JumpAddr Jump Register jr ll R PC=R[rs] R[rt]={24’b0,M[R[rs] I +SignExtImm](7:0)} R[rt]={16’b0,M[R[rs] I +SignExtImm](15:0)} I R[rt] = M[R[rs]+SignExtImm] Load Upper Imm. lui I Load Word lw I R[rt] = M[R[rs]+SignExtImm] Nor nor or ori I Set Less Than slt (4) 4hex R R[rd] = R[rs] | R[rt] Or Immediate chex R R[rd] = ~ (R[rs] | R[rt]) Or (3) R[rt] = {imm, 16’b0} R R[rd] = (R[rs] < R[rt]) ? 1 : 0 Load Byte Unsigned lbu Load Halfword Unsigned Load Linked lhu Set Less Than Imm. slti Set Less Than Imm. sltiu Unsigned Set Less Than Unsig. sltu 9hex 0 / 21hex R R[rd] = R[rs] + R[rt] And 8hex (4) (5) (5) 5hex 2hex 3hex 0 / 08hex (2) (2) (2,7) 24hex 25hex 30hex fhex (2) 23hex 0 / 27hex 0 / 25hex R[rt] = R[rs] | ZeroExtImm (3) dhex 0 / 2ahex FLOATING-POINT INSTRUCTION FORMATS R[rt] = (R[rs] < SignExtImm)? 1 : 0 (2) ahex R[rt] = (R[rs] < SignExtImm) bhex I ?1:0 (2,6) R R[rd] = (R[rs] < R[rt]) ? 1 : 0 (6) 0 / 2bhex 0 / 00hex R R[rd] = R[rt] << shamt I Shift Left Logical sll Shift Right Logical srl Store Byte sb Store Conditional sc Store Halfword sh Store Word sw R R[rd] = R[rt] >> shamt M[R[rs]+SignExtImm](7:0) = I R[rt](7:0) M[R[rs]+SignExtImm] = R[rt]; I R[rt] = (atomic) ? 1 : 0 M[R[rs]+SignExtImm](15:0) = I R[rt](15:0) I M[R[rs]+SignExtImm] = R[rt] Subtract sub R R[rd] = R[rs] - R[rt] Subtract Unsigned subu 31 31 29hex 2bhex (1) 0 / 22hex 0 / 23hex R R[rd] = R[rs] - R[rt] (1) May cause overflow exception (2) SignExtImm = { 16{immediate[15]}, immediate } (3) ZeroExtImm = { 16{1b’0}, immediate } (4) BranchAddr = { 14{immediate[15]}, immediate, 2’b0 } (5) JumpAddr = { PC+4[31:28], address, 2’b0 } (6) Operands considered unsigned numbers (vs. 2’s comp.) (7) Atomic test&set pair; R[rt] = 1 if pair atomic, 0 if not atomic BASIC INSTRUCTION FORMATS opcode R 31 rs 26 25 opcode I 31 rs 26 25 opcode J 31 rt 21 20 rd 16 15 shamt 11 10 rt 21 20 funct 6 5 0 immediate 16 15 0 address 26 25 ft 21 20 fmt 26 25 fs 16 15 ft 21 20 fd 11 10 funct 6 5 16 15 REGISTER NAME, NUMBER, USE, CALL CONVENTION PRESERVED ACROSS NAME NUMBER USE A CALL? $zero 0 The Constant Value 0 N.A. $at 1 Assembler Temporary No Values for Function Results $v0-$v1 2-3 No and Expression Evaluation $a0-$a3 4-7 Arguments No $t0-$t7 8-15 Temporaries No $s0-$s7 16-23 Saved Temporaries Yes $t8-$t9 24-25 Temporaries No $k0-$k1 26-27 Reserved for OS Kernel No $gp 28 Global Pointer Yes $sp 29 Stack Pointer Yes $fp 30 Frame Pointer Yes $ra 31 Return Address Yes 0 Copyright 2009 by Elsevier, Inc., All rights reserved. From Patterson and Hennessy, Computer Organization and Design, 4th ed. 0 immediate PSEUDOINSTRUCTION SET NAME MNEMONIC OPERATION blt if(R[rs]Branch Less Than bgt if(R[rs]>R[rt]) PC = Label Branch Greater Than ble if(R[rs]<=R[rt]) PC = Label Branch Less Than or Equal bge if(R[rs]>=R[rt]) PC = Label Branch Greater Than or Equal li R[rd] = immediate Load Immediate move R[rd] = R[rs] Move 38hex (2) (2) fmt 26 25 opcode FI 28hex (2,7) opcode FR 0 / 02hex (2) OPCODE / FMT /FT FOR/ FUNCT NAME, MNEMONIC MAT OPERATION (Hex) bc1t FI if(FPcond)PC=PC+4+BranchAddr (4) 11/8/1/-Branch On FP True Branch On FP False bc1f FI if(!FPcond)PC=PC+4+BranchAddr(4) 11/8/0/-div R Lo=R[rs]/R[rt]; Hi=R[rs]%R[rt] 0/--/--/1a Divide divu Divide Unsigned R Lo=R[rs]/R[rt]; Hi=R[rs]%R[rt] (6) 0/--/--/1b add.s FR F[fd ]= F[fs] + F[ft] 11/10/--/0 FP Add Single FP Add {F[fd],F[fd+1]} = {F[fs],F[fs+1]} + add.d FR 11/11/--/0 Double {F[ft],F[ft+1]} 11/10/--/y FP Compare Single c.x.s* FR FPcond = (F[fs] op F[ft]) ? 1 : 0 FP Compare FPcond = ({F[fs],F[fs+1]} op c.x.d* FR 11/11/--/y Double {F[ft],F[ft+1]}) ? 1 : 0 * (x is eq, lt, or le) (op is ==, <, or <=) ( y is 32, 3c, or 3e) FP Divide Single div.s FR F[fd] = F[fs] / F[ft] 11/10/--/3 FP Divide {F[fd],F[fd+1]} = {F[fs],F[fs+1]} / div.d FR 11/11/--/3 Double {F[ft],F[ft+1]} mul.s FR F[fd] = F[fs] * F[ft] 11/10/--/2 FP Multiply Single FP Multiply {F[fd],F[fd+1]} = {F[fs],F[fs+1]} * mul.d FR 11/11/--/2 Double {F[ft],F[ft+1]} 11/10/--/1 FP Subtract Single sub.s FR F[fd]=F[fs] - F[ft] FP Subtract {F[fd],F[fd+1]} = {F[fs],F[fs+1]} sub.d FR 11/11/--/1 Double {F[ft],F[ft+1]} lwc1 I F[rt]=M[R[rs]+SignExtImm] Load FP Single (2) 31/--/--/-Load FP F[rt]=M[R[rs]+SignExtImm]; (2) ldc1 I 35/--/--/-Double F[rt+1]=M[R[rs]+SignExtImm+4] mfhi R R[rd] = Hi 0 /--/--/10 Move From Hi mflo R R[rd] = Lo 0 /--/--/12 Move From Lo 10 /0/--/0 Move From Control mfc0 R R[rd] = CR[rs] mult R {Hi,Lo} = R[rs] * R[rt] 0/--/--/18 Multiply Multiply Unsigned multu R {Hi,Lo} = R[rs] * R[rt] (6) 0/--/--/19 sra R R[rd] = R[rt] >>> shamt 0/--/--/3 Shift Right Arith. swc1 I M[R[rs]+SignExtImm] = F[rt] Store FP Single (2) 39/--/--/-Store FP M[R[rs]+SignExtImm] = F[rt]; (2) sdc1 I 3d/--/--/-Double M[R[rs]+SignExtImm+4] = F[rt+1] 2 0 IEEE 754 FLOATING-POINT STANDARD 4 IEEE 754 Symbols Exponent Fraction Object 0 0 ±0

Interactive Furniture Layout Using Interior Design Guidelines

You are moving into a new home and need to arrange the living room furniture. You have a sofa, armchairs, coffee table, end tables, ottomans, and a media center. What arrangement will create the most comfortable and visually pleasing setting for your home? Furniture placement is challenging because it requires jointly optimizing a variety of functional and visual criteria. Skilled interior designers follow numerous high-level guidelines in producing furniture layouts [Lyons 2008; Ward 1999]. In a living room for example, the furniture should support comfortable conversation, align with prominent features of the space, and collectively form a visually balanced composition. In practice these guidelines are often imprecise and sometimes contradictory. Experienced designers learn to balance the tradeoffs between the guidelines through an iterative trial-and-error process. Yet most people responsible for furnishing a new home have no training in interior design. They may not be aware of interior design guidelines and they are unlikely to have the tacit knowledge and experience required to optimally balance the tradeoffs. Instead such amateur designers rely on intuitive rules such as pushing large furniture items against the walls. These intuitive rules often lead to functionally ineffective and visually imbalanced arrangements [Lyons 2008]. The resulting furniture layouts “simply don’t look or feel right,” and even worse the amateur designer “can’t pinpoint what the problems are” [Ward 1999]. In this paper, we identify a set of interior design guidelines for furniture layout and develop an interactive system based on these guidelines. In our system, the user begins by specifying the shape of a room and the set of furniture that must be arranged within it. The user then interactively moves furniture pieces. In response, the system suggests a small set of furniture layouts that follow the interior design guidelines. The user can interactively select a suggestion and move any piece of furniture to modify the layout. Thus, ...

Expert Voice Instruction & Coaching

Robert Wendell offers the Charlotte area music training services that range from teaching beginning singers to coaching accomplished professionals in multiple styles. Expert guidance in Italian, French, and German repertoire, style, expression and diction! Improve vocal agility for baroque ornamentation, fast, clean, complex runs, melismas, etc.

Graphic Standard Style Guide - Gibson
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The objective of this Graphic Standards Style Guide is to provide standards of usage for the Gibson Corporate Logo. A corporate logo symbolizes a company’s strength, confidence, promise of quality and competence. Proper treatment of a corporate logo ensures that it will be recognized amid communications clutter. Graphic standards are a visual expression of our Company, and serve as the foundation for our overall communications strategy. A well-managed Graphic Standards program provides a consistent message and image. It is extremely important that these standards be followed precisely to maintain the uniformity of presentation necessary for successful promotional and sales efforts. Do not deviate, under any conditions, from the standards set forth in this Guide. This Guide outlines the core elements of logo usage and how to apply them to printed and promotional materials. Possession of this Graphic Standards Style Guide does not relieve the holder from the responsibility of obtaining corporate approval. Items of any kind that bear a Gibson logo may not be produced prior to approval by the Creative Services Department at creativeservices@gibson.com. There are no exceptions to this procedure. Every employee is responsible for helping to make the Company’s communications strategy a success. Please contact the Creative Services Department if you have questions about the policies or standards in this manual. ...

Some Calculus Problems - Penn Math
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—... A por whi™h re—l x does this improper integr—l ™onvergec ˜A ƒhow th—t G(x + I) = xG(x) —nd dedu™e th—t G(n + I) = n3 for —ny integer n ! HF QIF ƒ—y g(t ) X R 3 R2 de¢nes — smooth ™urve in the pl—neF —A sf g(H) = H —nd kgH (t )k ™ D show th—t for —ny „ ! HD kg(„ )k ™„ F woreoverD show th—t equ—lity ™—n o™™ur if —nd only if one h—s g(t ) = ™vt where v is — unit ve™tor th—t does not depend on t F ˜A sf g(H) = HD gH (H) = H —nd kgHH (t )k IPD give —n upper ˜ound estim—te for kg(P)k F ‡hen ™—n this upper ˜ound ˜e —™hievedc QPF vet r(t ) de¢ne — smooth ™urve th—t does not p—ss through the originF —A sf the point a = r(t0 ) is — point on the ™urve th—t is ™losest to the origin @—nd not —n end point of the ™urveAD show th—t the position ve™tor r(t0 ) is perpendi™ul—r to the t—ngent ve™tor rH (t0 ) F ˜A ‡h—t ™—n you s—y —˜out — point b = r(t1 ) th—t is furthest from the originc QQF gonsider two smooth pl—ne ™urves g1 ; g2 X (H; I) 3 R2 th—t do not interse™tF ƒuppose €1 —nd €2 —re points on g1 —nd g2 D respe™tivelyD su™h th—t the dist—n™e j€1 €2 j is miniE m—lF €rove th—t the str—ight line €1 €2 is norm—l to ˜oth ™urvesF R QRF vet h(x; y; z) = H de¢ne — smooth surf—™e in R3 —nd let € X= (—; ˜; ™) ˜e — point not on the surf—™eF sf  X (x; y; z) is — point on the surf—™e th—t is ™losest to € D show th—t the line € is perpendi™ul—r to the t—ngent pl—ne to the surf—™e —t  F QSF vet r(t ) des™ri˜e — smooth ™urve —nd let V ˜e — ¢xed ve™torF sf rH (t ) is perpendi™ul—r to V for —ll t —nd if r(H) is perpendi™ul—r to V D show th—t r(t ) is perpendi™ul—r to V for —ll t F QTF vet f (s) ˜e —ny differenti—˜le fun™tion of the re—l v—ri—˜le s F ƒhow th—t u(x; t ) X= f (x + Qt ) h—s the property th—t ut = Qux F ƒhow th—t u —lso s—tis¢es the w—ve equ—tion utt = Wuxx F QUF vet u(x; y) ˜e — smooth fun™tionF —A sf ux = H with u(H; y) = sin(Qy) D ¢nd u(x; y) F ˜A sf ux = Pxy with u(H; y) = sin(Qy) D ¢nd u(x; y) F ™A sf ux + uy = H with u(H; y) = sin(Qy) D ¢nd u(x; y) F ss there more th—n one su™h fun™tionc dA sf ux + uy = Q Pxy with u(H; y) = sin(Qy) D ¢nd u(x; y) F ss eA sf ux Puy = H with u(H; y) = sin(Qy) D ¢nd u(x; y) F ss there more th—n one su™h fun™tionc QVF vet r X= xi + yj —nd V(x; y) X= p(x; y)i + q(x; y)j ˜e @smoothA ve™tor ¢elds —nd g — R smooth ™urve in the pl—neF sn this pro˜lem s is the line integr—l s = C V ¡ d r F por e—™h of the followingD either give — proof or give — ™ounterex—mpleF —A sf g is — verti™—l line segment —nd q(x; y) = HD then s = HF ˜A sf g is — ™ir™le —nd q(x; y) = HD then s = HF ™A sf g is — ™ir™le ™entered —t the origin —nd p(x; y) = q(x; y) D then s = HF dA sf p(x; y) > H —nd q(x; y) > HD then s > HF QWF vet g denote the unit ™ir™le ™entered —t the origin of the pl—neD —nd h denote the ™ir™le of r—dius S ™entered —t (P; I) D ˜oth oriented ™ounter™lo™kwiseF vet  denote the ring region ˜etween these R ™urvesF sf — ve™tor ¢eld V s—tis¢es div V = HD show th—t the line R integr—l C V ¡ N ds = D V ¡ N ds = ‘„his extends immedi—tely to the situ—tion where g —nd h —re more gener—l ™urves —nd  is the region ˜etween themF por £uid £ow it is —n expression of ™onserv—tion of m—ssD sin™e div V = H me—ns there —re no sour™es or sinks in the region  F“...

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