Archives: Solution Manual

Chapter 3 Linear Equations in Two Variables 3.1 Check Points 1. 2. (4,2)E (2,0)F (6,0)G 3. a. 39 33(2) 9 36 9 99, true xy    (3, 2) is a solution.    02(0)220,2 12(1)201,0 22(2)222,2 y […]

19-34 Intermediate Accounting, 8/e STOCK APPRECIATION RIGHTS  When an SAR is considered to be equity (because the employer can elect to settle in shares), the amount of compensation is estimated at the grant date as the fair value of […]

Chapter 2 Linear Equations and Inequalities in One Variable 109. Let x = number of miles driven. 80 0.25 400 80 0.25 80 400 80 0.25 320 0.25 320 0.25 0.25 1280 x x x x     […]

REACQUIRED SHARES  The number of reacquired shares is time-weighted for the fraction of the year they were not outstanding, prior to being subtracted from the number of shares outstanding. Sovran Financial Corporation reported net income of $154 million in […]

Section 2.6 Problem Solving in Geometry 32. The sum of the measures of the three angles of a triangle is 180°.   3 40 180 5 40 180 5140 28 384 40 68 x x x x x  […]

C CH HA AP PT TE ER R 1 19 9 S Sh ha ar re e– –B Ba as se ed d C Co om mp pe en ns sa at ti io on n a an nd d […]

Section 2.4 Formulas and Percents 23.  1 for 2 A ha b b   1 22 2 2 2 2 2 2 22 or Ahahb Aha hahbha Aha hb Aha hb hh Aha A bb a hh  […]

Case 18–4 (continued) This is the measure of comprehensive income Cisco reported in the disclosure note. Notice that each component is reported net of its related income tax expense or income tax benefit. The second measure—the comprehensive income accumulated over […]

Section 2.2 The Multiplication Property of Equality 72. a. The bar graph indicates the median weekly earnings, in 2013, for women with some college or an associate’s degree is $657. Since 2013 is 33 years after 1980, substitute 33 into […]

Problem 18–10 Transactions N 1. Sale of common stock N 2. Purchase of treasury stock at a cost less than the original issue price N 3. Purchase of treasury stock at a cost greater than the original issue price D […]

Chapter 2 Linear Equations and Inequalities in One Variable 2.1 Check Points 1. 512 55125 017 17 x  Check: 512 17 5 12 12 12 x   The solution set is   17 . 13 y  […]

CPA / CMA REVIEW QUESTIONS CPA Exam Questions 1. b. The entries to record the stock issuance and subsequent acquisition and retirement (per share) are as follows: Issuance Cash ……………………………………………………………………. 25 Common stock ……………………………………………….. 10 Paid-in capital—excess of par …………………………… […]

Release to Production — Print Supplement Supplement Information ISBN 0-13-418066-6 / 978-0-13-418066-3 Title Instructor’s Solutions Manual for INTRODUCTORY AND INTERMEDIATE ALGEBRA FOR COLLEGE STUDENTS Schedule RTP date 3/1/16 File-out date 3/8/16 Bound book date NA THIS MANUAL IS ONLINE ONLY […]

Exercise 18–7 Requirement 1 ($ in millions) Cash ($424 million – 2 million) …………………………………………… 422 Common stock (15 million shares at $1 par per share) …………… 15 Paid-in capital—excess of par (difference) ……………………… 407 Requirement 2 In recording the sale […]

Chapter 1 Variables, Real Numbers, and Mathematical Models 54. 00 8  55. 7 0 is undefined. 56. 8 0  is undefined. 57. 1 15 3 15 5 3  58. 1 80 8 80 10 8  59. […]

Chapter 18 Shareholders’ Equity QUESTIONS FOR REVIEW OF KEY TOPICS Question 18–1 The two primary sources of shareholders’ equity are amounts invested by shareholders in the corporation and amounts earned by the corporation on behalf of its shareholders. Invested capital […]

Section 1.4 Basic Rules of Algebra 95. a. b. Rhode Island, Georgia, Louisiana, Florida, Hawaii 96. a. b. Wyoming, Wisconsin, Washington, West Virginia, Virginia 121. 12 3.464  and should be graphed between -4 and -3. 122. 12 0.414  […]

18-18 Intermediate Accounting, 8/e RETAINED EARNINGS ❖ In general, retained earnings represents a corporation’s accumulated, undistributed, or reinvested net income (or net loss). It also is called “reinvested capital” or “earned capital.” ❖ Distributions of earned assets are dividends. ❖ […]

Chapter 1 Variables, Real Numbers, and Mathematical Models 1.1 Check Points 1. a. 62 62(10) 26x  b. 2( 6) 2(10 6) 32x  2. a. 7 2 7 3 2 8 21 16 37xy b. 663810 2 2828385 xy […]

C CH HA AP PT TE ER R 1 18 8 S Sh ha ar re eh ho ol ld de er rs s’ ’ E Eq qu ui it ty y Overview We turn our attention in this chapter […]

Case 17–3 (concluded) Requirement 4 Two of the other amounts reported in the disclosure note are reported in the balance sheet. The net gain and prior service cost are reported as components of accumulated other comprehensive income. This is a […]

Solution 19.93 We apply the same formulas derived in the previous problem. E Eie R h )Rh( 1−+ + = = +×+ = ++ = )3800200(101 )RR(h1 A5– ELoe i 5.144 )RR(h1 )hR1)(RhR(h RhZ ELoe oeELreEfe Eiein ++ +− ++= […]

Solution 19.86 (a) By definition, g11 = 0I 1 1 2 V I = , g21 = 0I 2 1 2 V V = . We let V1 = 1 V and open-circuit the output port. The schematic is shown […]

indicates that 0 2=I and that =][y S 5j75.01 105.121050 -6-6       + ×× -6-6 y10)250j65(10)5.12.25j5.77( ×+=×−+=∆ =             ∆ = 11 y 11 […]

Solution 19.50 To get a and c, consider the circuit below. I1=0 2 s I2 + + V1 4/s V2 – – 2 1 2 2 2 21 s25.01 V V aV 4s 4 V s/4s s/4 V+==→ + = […]

Checking using nodal analysis we get, –10 + 2Ia + 2(Ia–Ib) = 0 and 2(Ib–Ia) + 1Ib + 5Ib = 0. From the second equation we get Ia = 4Ib and then from the first equation we get 8Ib + […]

Question 11–1 The terms depreciation, depletion, and amortization all refer to the process of allocating the cost of property, plant, and equipment and finite-life intangible assets to periods of use. The only difference between the terms is that they refer […]

Alternatively, from the given circuit, 211 1.04 VIV −= 212 1.020 VII += Comparing these with the equations for the h parameters show that 4 11 =h , -0.1 12 =h , 20 21 =h , 1.0 22 =h Using […]

Substituting these into (1) and (2), s11 s 1111s1s Zz V IIzZIV + =→=− s11 s21 1212 Zz Vz IzV + == == 2Th VV s11 s21 Zz Vz + Copyright © 2017 McGraw–Hill Education. All rights reserved. No reproduction […]

AMORTIZATION OF INTANGIBLE ASSETS ➢ The cost of an intangible asset with a finite useful life is amortized. ▶ Intangibles typically have no residual value, so the amortization base is simply cost. ▶ The cost of an intangible asset usually […]

Solution 19.1 Obtain the z parameters for the network in Fig. 19.65. 10 Ω 10 Ω 10 Ω Figure 19.65 For Prob. 19.1. Solution Step 1. Label the circuit to allow us to determine the z–parameters. The z–parameter equations are, […]

Solution 18.35 (a) x(t) = f[3(t-1/3)]. Using the scaling and time shifting properties, )j6( e e 3/j2 1 3 1 )(X 3/j 3/j ω+ = ω+ =ω ω− ω− (b) Using the modulation property,       […]

Solution 18.15 (a) =−=ω ω−ω 3j3j ee)(F ω3sinj2 (b) Let ω− =ω−δ= j e2)(G),1t(2)t(g =ω)(F F      ∫∞− tdt)t(g )()0(F j )(G ωδπ+ ω ω = )()1(2 j e2 j ωδ−πδ+ ω = ω− = ω […]

Solution 18.1 Obtain the Fourier transform of the function in Fig. 18.26. Figure 18.26 For Prob. 18.1. Solution f(t) = 7u(t+2) – 7u(t+1) – 7u(t–1) + 7u(t–2) f’(t) = 7δ(t+2) – 7δ(t+1) – 7δ(t–1) + 7δ(t–2) jωF(ω) = 7(ej2ω – […]

Solution 17.28 Obtain the trigonometric Fourier series for the voltage waveform shown in Fig. 17.66. Figure 17.66 For Prob. 17.28. Solution This is half–wave symmetric since f(t − T/2) = −f(t). ao = 0, T = 2, ωo = 2π/2 […]

Solution 17.15 (a) Dcos ωt + Esin ωt = A cos(ωt – θ) where A = 22 ED + , θ = tan-1(E/D) A = 622 n 1 )1n( 16 + + , θ = tan-1((n2+1)/(4n3)) f(t) = ∑ ∞ […]

Solution 17.1 (a) This is periodic with ω = π which leads to T = 2π/ω = 2. (b) y(t) is not periodic although sin t and 4 cos 2πt are independently periodic. (c) Since sin A cos B = […]

Solution 16.89 First select the inductor current iL (left to right) and the capacitor voltage vC to be the state variables. Letting vo = vC and applying KCL we get: sCL sLCCs C CL vvi iivvori v vi +−= ++−==−++− […]

Solution 16.74 Design a problem to help other students to better understand how to find outputs when given a transfer function and an input. Although there are many ways to solve this problem, this is an example based on the […]

Solution 16.1 The current in an RLC circuit is described by 2 2 10 25 0 d i di i dt dt + += If i(0) = 7 A and di(0)/dt = 0, find i(t) for t > 0. Solution […]

Therefore, =)t(y       <<−+ << otherwise,0 6t5,12t5)2t(– 5t3,21 2  <<+− 3t2,2t2)2t( 2 Copyright © 2017 McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw–Hill Education. Solution 15.45 […]

Solution 15.35 (a) Let 2s B 1s A )2s)(1s( 3s )s(G + + + = ++ + = 2A = , -1B = 2t–t– ee2)t(g 2s 1 1s 2 )s(G −=→ + − + = )6t(u)6t(g)t(f)s(Ge)s(F -6s −−=→= =)t(f [ […]

Solution 15.19 Since [ ] 1)t( =δL and 2T = , =)s(F s2– e1 1 − Copyright © 2017 McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw–Hill Education. Solution 15.20 Using Fig. […]

Solution 14.96 22 2 L1 CsR1 sC ||R + ==Z       + ++ =+= 2L 2L 2 L 1 1 1 2CsR1 LCRsRsL || sC 1 )sL(|| sC 1ZZ L R 1 2L 2L 2 […]

Solution 14.88 The schematic is shown below. We insert a voltage marker to measure Vo. In the AC Sweep box, we set Total Points = 101, Start Frequency = 1, and End Frequency = 100. After simulation, we obtain the […]

Solution 14.57 (a) Consider the circuit below. Z V I s = )sRC2(RsC2 sC1 s 1 + = + =Z V II 222 s 1o CRssRC31 )sRC2(sC sRC2 R R++ + ⋅ + == V IV 222 s o CRssRC31 […]

Solution 4.39 222 LR LjR CjCj LjR 1 Yω+ ω− +ω=ω+ ω+ = At resonance, 0)Im( =Y , i.e. 0 LR L C22 0 2 0 0= ω+ ω −ω C L LR 22 0 2 =ω+ 2 -36–-32 2 […]

Solution 14.20 Design a more complex problem than given in Prob. 14.10, to help other students to better understand how to determine the Bode magnitude and phase plots of a given transfer function in terms of jω. Include at least […]

Solution 14.1 Find the transfer function Io/Ii of the RL circuit in Fig. 14.68. Express the transfer function using ωo = R/L. Figure 14.68 For Prob. 14.1. Solution H(ω) = Io/Ii = [RjωL/(R+jωL)]/(jωL) = 1/(1+jωL/R) If we let ωo = […]

Solution 13.87 ZTh = ZL/n2 or n = 300/75Z/Z ThL= = 0.5 Copyright © 2017 McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw–Hill Education. Solution 13.88 n = V2/V1 = I1/I2 or […]