Equipment Investment in Abernethy (1c) 20‚000 30‚000 Goodwill 60‚000 Investment in Abernethy (2a1) Equity in subsidiary earnings 60‚000 74‚000 Investment in Abernethy (2a2) No entry required 74‚000 0 No entry required (2b) Investment in Abernathy 0 10‚000 Dividends paid (3a1) Depreciation expense 10000 10‚000 Building (3a2) Equipment 10‚000 4‚000 Depreciation
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Sociology AS at Knights Unit 1: Families and Households Unit 2: Education with Research Methods Revision pack Haberdashers’ Aske’s Federation Sixth Form Mrs Griffiths: sj-griffiths@hahc.org.uk Mr Roaf: sm-roaf@hahc.org.uk 2012 Unit 1 exam: Thursday 17th May‚ am Unit 2 exam: Friday 25th May‚ pm Easter Revision: tbc AS Syllabus: AQA Sociology GCE (new specification) Unit 1: Families and Households (SCLY1) * Worth 40% of your AS and 20% of
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A Semi-Detailed Lesson Plan in BIOLOGY Date: January 22‚ 2013 Time: 7:30-8:30 Section: BSEd-2B I. Objective(s) At the end of the 60-minute period‚ at least 75% of the student’s should be able to: 1. Differentiate prokaryotic from eukaryotic cells. II. Subject Matter a. Topic: Prokaryotes and Eukaryotes b. References: b.1. Department of Education. 2002. Operations Handbook in Biology: 2002 Basic Education Curriculum Secondary Level. II. 1. 1.4. b.2. Department of Education
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Introduction: Procedure: A.spotting of the TLC plates 1.A TLC plates is obtained from my instructor. 2.the plate is set down on a clean‚dry surface then a line Is drawn lightly across the plate about 1cm from the bottom of the plate with the aid of a 2B pencil as shown in figure 3(i). 3.next five 2-3mm lines make‚about 0.8cm is spaced apart and running perpendicularly through the lines across the bottom of the TLC plate. 4.The plate is spotted with 5 different analgesic.A separate capillary tube
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5‚ 6‚ 7‚ 8‚11‚ 12 1‚ 2‚ 3‚ 4 1‚ 2‚ 3‚ 4‚ 6‚ 7‚ 8‚ 9‚11 1A‚ 2A‚ 3A‚ 5A 1B‚ 2B‚ 3B‚ 5B 3. 9‚ 10‚ 11‚ 12 5 1‚ 2‚ 3‚ 6‚ 7‚ 8‚ 10‚ 12 1A‚ 2A‚ 3A‚ 5A 1B‚ 2B‚ 3B‚ 5B 4. 13‚ 14‚ 15 6‚ 7 2‚ 3‚ 5‚ 6‚ 7‚ 8‚ 11‚12‚ 13 1A‚ 2A‚ 3A‚ 4A‚ 5A 1B‚ 2B‚ 3B‚ 4B‚ 5B 5. 16 8 2‚ 3‚ 4‚ 6‚ 7‚ 8‚ 9‚ 10‚ 11 5‚ 12‚ 13 1A‚ 2A‚ 3A‚ 5A 1A‚ 2A‚ 4A‚ 5A 1B‚ 2B‚ 3B‚ 5B 1B‚ 2B‚ 4B‚ 5B 6. 17‚ 18 9 20-1 ASSIGNMENT CHARACTERISTICS TABLE Problem Number
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* AP Calculus Review Limits‚ Continuity‚ and the Definition of the Derivative Teacher Packet Advanced Placement and AP are registered trademark of the College Entrance Examination Board. The College Board was not involved in the production of‚ and does not endorse‚ this product. Copyright © 2008 Laying the Foundation‚ Inc.‚ Dallas‚ Texas. All rights reserved. These materials may be used for face-to-face teaching with students only. Limits‚ Continuity‚ and the Definition of the Derivative Page
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duopoly) P=a-b(Q1+Q2) TR(firm1)=aQ1-b((Q1*Q2)-b(Q1)^2 MR(firm1)=a-bQ2-2bQ1 MR(firm2)=a-bQ1-2bQ2 Therefore‚ each firms MR depends on its own and its rivals output Find firm’s Best-response function‚ MR=MC a-bQ2-2bQ1=MC1 Q1=((a-Mc1)/2b)-(Q2/2) Similarily Q2=((a-MC2)/2b)-(Q2/2) Substitute Q2 into Q1 and solve for Q1 Q1*=(a-MC)/3b Use Q1* to find Q2* Aggregate output‚Q*=Q1*+Q2* Normal Form Game ( Nash equilibrium) Nash equilibrium occurs when no player finds it profitable to unilaterally deviate
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2a. (5 pts) Which of the following species is a resonance form of compound LA YM NAME: 2b. (3 pts) Identify the atomic orbitals in the C-N sigma bond in the oxime shown below? É A)(zsp2‚2sp2) B)(2Sp3‚2sp3) C)(25p‚2sp) D)(2sp2‚2sp3) E) (25p‚ ls) 2c. (2 pts) Identify the atomic orbital that_ the lone pair electrons on N atom are contained in for the oXime shown in problem 2b? A) 2592 B) 25133 C) 23p D) 2s E) 2p 2d. (3 pts) Which of the following is the strongest base:
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III.8. THE TUNNEL DIODE 1. Theory The Japanese physicist Leo Esaki invented the tunnel diode in 1958. It consists of a p-n junction with highly doped regions. Because of the thinness of the junction‚ the electrons can pass through the potential barrier of the dam layer at a suitable polarization‚ reaching the energy states on the other sides of the junction. The current-voltage characteristic of the diode is represented in Figure 1. In this sketch i p and U p are the peak‚ and iv and U v are
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1a Packaging of a potato chip 6cm Volume of a single = 170cm cubed Volume of cylinder = ∏r squared Height= volume Divided by ( ∏r squared) Height= 170cm divided by (∏ x 3 squared) Height = 6.01 round of to 6cm Allowing 1 extra cm for room Top and bottom of cylinder = 6cm + 1cm = 7cm Height= 6cm + 1cm = 7cm Total surface Area= area of base + area of top
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