Numerical methods for engineers 5th e-solution ch 04
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Numerical methods for engineers 5th e-solution ch 04 Document Transcript

  • 1. 1CHAPTER 44.1 (a) For this case xi = 0 and h = x. Thus, x2 f ( x ) = ) + 0) x + (0) f (0 f ( f " + ⋅⋅ ⋅ 2 f (0) = f (0) = f " (0) =e 0 =1 x2 f ( x) = 1 + x + +⋅⋅⋅ 2 (b) h2 h3 f ( xi +1 ) = e − xi − e − xi h + e − xi − e − xi + ⋅⋅⋅ 2 6 for xi = 0.2, xi+1 = 1 and h = 0.8. True value = e–1 = 0.367879. zero order: f (1) =e −0.2 =0.818731 0.367879 − 0.818731 εt = ×100% =122.55% 0.367879 first order: f (1) = 0.818731 − 0.818731(0.8) = 0.163746 0.367879 − 0.163746 εt = ×100% = 55.49% 0.367879 second order: 0 .8 2 f (1) = 0.818731 − 0.818731(0.8) + 0.818731 = 0.42574 2 0.367879 − 0.42574 εt = ×100% =15.73% 0.367879 third order: 0.8 2 0.8 3 f (1) = 0.818731 − 0.818731(0.8) + 0.818731 − 0.818731 = 0.355875 2 6 0.367879 − 0.355875 εt = ×100% = 3.26% 0.367879PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 2. 24.2 Use the stopping criterion ε s = 0.5 ×10 2 − 2 % = 0.5% True value: cos(π/3) = 0.5 zero order: π  cos  = 1 3 0.5 − 1 εt = ×100% = 100% 0.5 first order: π  cos  = 1 − (π / 3) = 0.451689 2 3 2 0.451689 −1 ε t = 9.66% εa = ×100% =121.4% 0.451689 second order: π  cos  = 0.451689 + ( π / 3) = 0.501796 4 3 24 0.501796 − 0.451689 ε t = 0.359% εa = ×100% = 9.986% 0.501796 third order: π  cos  = 0.501796 − (π / 3) = 0.499965 6 3 720 0.499965 − 0.501796 ε t = 0.00709% εa = ×100% = 0.366% 0.499965 Since the approximate error is below 0.5%, the computation can be terminated.4.3 Use the stopping criterion: ε s = 0.5 × 10 2−2 % = 0.5% True value: sin(π/3) = 0.866025… zero order:PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 3. 3 π  π sin   = = 1.047198 3 3 0.866025 −1.047198 εt = ×100% = 20.92% 0.866025 first order: π  sin   = 1.047198 − ( π / 3) = 0.855801 3 3 6 0.855801 −1.047198 ε t = 1.18% εa = ×100% = 22.36% 0.855801 second order: π  sin   = 0.855801 + (π / 3) = 0.866295 5 3 120 0.866295 − 0.855801 ε t = 0.031% εa = ×100% =1.211% 0.866295 third order: π  sin   = 0.866295 − ( π / 3) = 0.866021 7 3 5040 0.866021 − 0.866295 ε t = 0.000477% εa = ×100% = 0.0316% 0.866021 Since the approximate error is below 0.5%, the computation can be terminated.4.4 True value: f(3) = 554. zero order: f (3) = f (1) = −62 554 − ( −62) εt = ×100% = 111.191% 554 first order: f (3) = −62 + f (1)(3 −1) = −62 + 70( 2) = 78 ε t = 85.921%PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 4. 4 second order: f " (1) 138 f (3) =78 + (3 −1) 2 =78 + 4 =354 2 2 ε t = 36.101% third order: ( 3) f (1) 150 ε t = 0% f (3) = 354 + (3 − 1) 3 = 354 + 8 = 554 6 6 Thus, the third-order result is perfect because the original function is a third-order polynomial.4.5 True value: f(2.5) = ln(2.5) = 0.916291... zero order: f ( 2.5) = f (1) = 0 0.916291 − 0 εt = ×100% = 100% 0.916291 first order: f ( 2.5) = f (1) + f (1)(2.5 −1) = 0 +1(1.5) =1.5 0.916291 −1.5 εt = ×100% = 63.704% 0.916291 second order: f " (1) −1 f ( 2.5) = .5 + 1 ( 2.5 − ) 2 = .5 + 1 1 1.5 2 2 2 0.916291 − 0.375 εt = ×100% = 59.074% 0.916291 third order: ( 3) f (1) 2 f (2.5) = 0.375 + (2.5 − 1) 3 = 0.375 + 1.5 3 = 1.5 6 6PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 5. 5 0.916291 −1.5 εt = ×100% = 63.704% 0.916291 fourth order: ( 4) f (1) −6 f ( 2.5) = 1.5 + (2.5 − 1) 4 = 1.5 + 1.5 4 = 0.234375 24 24 0.916291 − 0.234375 εt = ×100% = 74.421% 0.916291 Thus, the process seems to be diverging suggesting that a smaller step would be required for convergence.4.6 True value: f ( x) = 75 x 2 − 12 x + 7 f ( 2) = 75( 2) 2 − 12(2) + 7 = 283 function values: xi–1 = 1.8 f(xi–1) = 50.96 xi = 2 f(xi) = 102 xi+1 = 2.2 f(xi+1) = 164.56 forward: 164.56 −102 f ( 2) = = 312.8 0.2 283 − 312.8 εt = ×100% = 10.53% 283 backward: 102 − 50.96 f ( 2) = = 255.2 0.2 283 − 255.2 εt = ×100% = 9.823% 283 centered: 164.56 − 50.96 f ( 2) = = 284 2(0.2)PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 6. 6 283 − 284 εt = ×100% = 0.353% 283 Both the forward and backward have errors that can be approximated by (recall Eq. 4.15), f " ( xi ) Et ≈ h 2 f " (2) = 150 x − 12 = 150(2) − 12 = 288 288 Et ≈ 0.2 = 28.8 2 This is very close to the actual error that occurred in the approximations forward: E t ≈ 283 − 312.8 = 29.8 backward: E t ≈ 283 − 255.2 = 27.8 The centered approximation has an error that can be approximated by, ( 3) f ( xi ) 2 150 Et ≈ − h =− 0.2 2 = −1 6 6 which is exact: Et = 283 – 284 = –1. This result occurs because the original function is a cubic equation which has zero fourth and higher derivatives.4.7 True value: f " ( x) = 150 x − 12 f " ( 2) = 150( 2) − 12 = 288 h = 0.25: f ( 2.25) − ( 2) + .75) 2 f f (1 182 f " ( 2) = 2 = 0.25 h = 0.125: f ( 2.125) − ( 2) + .875) 2 f f (1 13 f " ( 2) = = 0.125 2 Both results are exact because the errors are a function of 4 th and higher derivatives which are zero for a 3rd-order polynomial.4.8PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 7. 7 ( ∂v cgte −( c / m ) t − gm 1 − e −( c / m )t = ) = −1.38666 ∂c c2 ∂v ~ v ~ ∆ (c ) = ∆ =1.38666(1.5) = 2.079989 c ∂c v (12.5) = 9.8(50) 12.5 ( ) 1 − e −12.5( 6) / 50 = 30.4533 v = 30.4533 ± 2.079989 Thus, the bounds computed with the first-order analysis range from 28.3733 to 32.5333. This result can be verified by computing the exact values as v (c − ∆c) = 9.8(50) 11 ( ) 1 − e −(11 / 50 ) 6 = 32.6458 v (c + ∆c ) = 9.8(50) 14 ( ) 1 − e −(14 / 50) 6 = 28.4769 Thus, the range of ±2.0844 is close to the first-order estimate.4.9 v (12.5) = 9.8(50) 12.5 ( ) 1 − e −12.5( 6) / 50 = 30.4533 ∂v ~ ∂v v ~ ~ ∆ ( c , m) = ∆ + c ∆~ m ∂c ∂m ( ∂v cgte −( c / m ) t − gm 1 − e −( c / m )t = ) = −1.38666 ∂c c2 ∂v = ∂m m gt −( c / m ) t e g ( ) + 1 − e −( c / m ) t = 0.871467 c v ~ ~ ∆ (c , m) = −1.38666 (1.5) + 0.871467 ( 2) =2.079989 +1.742934 =3.822923 v = 30.4533 ± 3.8229234.10 For ∆~ =20, T ~ ∂H ~ ∆ (T ) = H ∆T ∂TPROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 8. 8 ∂H = 4 AeσT 3 = 4(0.15)0.9(5.67 ×10 −8 )650 3 = 8.408 ∂T ~ ∆ (T ) =8.408( 20) = H 168.169 Exact error: H (670) − H (630) 1542.468 − 1205.81 ∆H true = = = 168.3286 2 2 Thus, the first-order approximation is close to the exact result. For ∆~ =40, T ~ ∆ (T ) =8.408( 40) =336.3387 H Exact error: H (690) − H (610) 1735.055 − 1059.83 ∆H true = = = 337.6124 2 2 Again, the first-order approximation is close to the exact result. The results are good because H(T) is nearly linear over the ranges we are examining. This is illustrated by the following plot. 2000 1600 1200 800 550 600 650 700 7504.11 For a sphere, A = 4πr2. Therefore, H = 4πr 2 eσT 4 At the mean values of the parameters, H (0.15,0.9,550) = 4π(0.15) 2 0.90(5.67 ×10 −8 )(550) 4 =1320.288 ∂H ~ ∂H ~ ∂H ~ ∆ = H ∆ + r ∆ + e ∆T ∂ r ∂ e ∂TPROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 9. 9 ∂H = 8πreσT 4 =17,603.84 ∂r ∂H = 4πr 2σT 4 = 1466.987 ∂e ∂H =16πr 2 eσT 3 = 9.6021 ∂T ∆H =17603.84(0.01) +1466.987(0.05) + 9.6021( 20) = 441.4297 To check this result, we can compute H (0.14,0.85,530) = 4π(0.14) 2 0.85(5.67 ×10 −8 )(530) 4 =936.6372 H (0.16,0.95,570) = 4π(0.16) 2 0.95(5.67 ×10 −8 )(570) 4 =1829.178 1829.178 − 936.6372 ∆H true = = 446.2703 24.12 The condition number is computed as ~f ( ~ ) x x CN = ~) f (x  1  1.00001  (a) CN =  2 1.00001 −1  = 1.00001(158.1139) =157.617 1.00001 −1 +1 1.003162 The result is ill-conditioned because the derivative is large near x = 1. 10( −e −10 ) 10( −4.54 ×10 −5 ) (b) CN = = = −10 e −10 4.54 ×10 −5 The result is ill-conditioned because x is large.  300  300  −1 (c)  300 2 +1   300( −5.555556 ×10 −6 ) = CN = = −0.99999444 2 300 +1 − 300 0.0016667 The result is well-conditioned.PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 10. 10 − xe − x − e − x + 1 x x2 0.001(0.499667) (d) CN = = − 0.9995 = −0.0005  e −1  −x    x    The result is well-conditioned. (1 + cos x ) cos x + sin x(sin x) x (1 + cos x) 2 3.141907(20,264,237) (e) CN = = = −10,001 sin x − 6366.2 1 + cos x The result is ill-conditioned because, as in the following plot, the function has a singularity at x = π. 2000 1000 0 3.08 3.1 3.12 3.14 3.16 3.18 3.2 -1000 -20004.13 Addition and subtraction: f (u , v ) = u + v ∂f ~ ∂f ~ ∆ = f ∆ + u ∆v ∂u ∂v ∂f ∂f =1 =1 ∂u ∂v ~ ~ ~ ~ f (u , v ) = ∆u + ∆v Multiplication: f (u , v) = u ⋅ v ∂f ∂f =v =u ∂u ∂v ~ ~ ~ u~ ~ v f (u , v ) = v ∆ + u ∆~PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 11. 11 Division: f (u , v) = u / v ∂f 1 ∂f u = = 2 ∂u v ∂v v ~ ~ 1 ~ u ~ f (u , v ) = ∆u + 2 ∆v v v ~ v ∆ +u ∆ u ~ v ~ ~ f (u , v ) = v24.14 f ( x) = ax 2 + bx + c f ( x) = 2ax + b f " ( x ) = 2a Substitute these relationships into Eq. (4.4), 2a 2 ax i2+1 + bxi +1 + c = axi2 + bx i + c + ( 2ax i + b)( x i +1 − x i ) + ( x i +1 − 2 x i +1 x i + x i2 ) 2! Collect terms ax i2+1 + bx i +1 + c = ax i2 + 2ax i ( x i +1 − x i ) + a ( xi2+1 − 2 x i +1 xi + x i2 ) + bx i + b( xi +1 − xi ) + c ax i2+1 + bx i +1 + c = ax i2 + 2ax i xi +1 − 2ax i2 + ax i2+1 − 2axi +1 x i + axi2 + bxi + bx i +1 − bxi + c axi2+1 + bx i +1 + c = (ax i2 − 2axi2 + axi2 ) + ax i2+1 + ( 2ax i xi +1 − 2axi +1 xi ) + (bxi − bxi ) + bx i +1 + c ax i2+1 + bx i +1 + c = axi2+1 + bxi +1 + c4.15 The first-order error analysis can be written as ∂Q ∂Q ∆ = Q ∆ + n ∆S ∂n ∂S ∂Q 1 ( BH ) 5 / 3 ∂Q 1 ( BH ) 5 / 3 1 =− 2 S 0.5 = −50.74 = = 2536.9 ∂n n ( B + 2H ) 2/3 ∂S n (B + 2H ) 2/3 2 S 0.5PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 12. 12 ∆ = −50.74 0.003 + 2536.9 0.00003 =0.152 +0.076 =0.228 Q The error from the roughness is about 2 times the error caused by the uncertainty in the slope. Thus, improving the precision of the roughness measurement would be the best strategy.4.16 Use the stopping criterion ε s = 0.5 ×10 2 − 2 % = 0.5% True value: 1/(1 – 0.1) = 1.111111…. zero order: 1 =1 1−x 1.11111 −1 εt = ×100% = 10% 1.11111 first order: 1 =1 + 0.1 =1.1 1−x 1.1 −1 ε t = 1% εa = ×100% = 9.0909% 1.1 second order: 1 =1 + 0.1 + 0.01 =1.11 1−x 1.11 −1.1 ε t = 0.1% εa = ×100% = 0.9009009% 1.11 third order: 1 =1 + 0.1 + 0.01 + 0.001 =1.111 1− x 1.111 −1.11 ε t = 0.01% εa = ×100% = 0.090009% 1.111 Since the approximate error is below 0.5%, the computation can be terminated.4.17PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 13. 13 d sin φ0 ∆( sin φ0 ) = ∆α dα d sin φ0 −β αβ = + 1− dα 2 (1 + α )(1 + α − αβ ) 1 +α where β = (ve/v0)2 = 4 and α = 0.25 to give, d sin φ0 −4 0.25( 4) = + 1− = −3.1305 dα 2 (1 + 0.25)(1 + 0.25 − 0.25( 4)) 1 + 0.25 ∆( sin φ 0 ) = 3.1305∆α For ∆α = 0.25(0.02) = 0.005, ∆( sin φ0 ) = 3.1305(0.005) = 0.015652 0.25 sin φ0 = (1 + 0.25) 1 − 4 = 0.559017 1 + 0.25 Therefore, max sin φ0 = 0.559017 + 0.015652 = 0.574669 min sin φ0 = 0.559017 − 0.015652 = 0.543365 180 max φ0 = arcsin(0.574669) × = 35.076° π 180 min φ0 = arcsin(0.543365) × = 32.913° π4.18 The derivatives can be computed as f ( x ) = x −1 − 0.5 sin x f ( x ) =1 − 0.5 cos x f " ( x) = 0.5 sin x ( 3) f ( x ) = 0.5 cos xPROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 14. 14 ( 4) f ( x) = −0.5 sin x The first through fourth-order Taylor series expansions can be computed based on Eq. 4.5 as First-order: f 1 ( x ) = f ( a ) + f ( a )( x − a ) π π  π  π f 1 ( x) = − 1 − 0.5 sin + 1 − 0.5 cos  x −  = x − 1.5 2 2  2  2 Second-order: f " (a) f 2 ( x) = f 1 ( x) + ( x − a) 2 2 f 2 ( x) = x − 1.5 + 0.25 sin(π / 2)( x − π / 2) 2 Third-order: ( 3) f (a) f 3 ( x ) = f 2 ( x) + ( x − a) 3 6 0.5 cos(π / 2) f 3 ( x) = x − 1.5 + 0.25 sin(π / 2)( x − π / 2) 2 + ( x − a) 3 6 f 3 ( x ) = x −1.5 + 0.25 sin(π / 2)( x −π / 2) 2 Fourth-order: ( 4) f (a) f 4 ( x) = f 3 ( x ) + ( x − a) 4 24 0.5 sin(π / 2) f 4 ( x ) = x − 1.5 + 0.25 sin(π / 2)( x − π / 2) 2 − ( x − π / 2) 4 24 1 f 4 ( x) = x − 1.5 + 0.25 sin(π / 2)( x − π / 2) 2 − ( x − π / 2) 4 48 Note the 2nd and 3rd Order Taylor Series functions are the same. The following MATLAB script file which implements and plots each of the functions indicates that the 4 th-order expansion satisfies the problem requirements. x=0:0.001:3.2;PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 15. 15 f=x-1-0.5*sin(x); subplot(2,2,1); plot(x,f);grid;title(f(x)=x-1-0.5*sin(x));hold on f1=x-1.5; e1=abs(f-f1); %Calculates the absolute value of the difference/error subplot(2,2,2); plot(x,e1);grid;title(1st Order Taylor Series Error); f2=x-1.5+0.25.*((x-0.5*pi).^2); e2=abs(f-f2); subplot(2,2,3); plot(x,e2);grid;title(2nd/3rd Order Taylor Series Error); f4=x-1.5+0.25.*((x-0.5*pi).^2)-(1/48)*((x-0.5*pi).^4); e4=abs(f4-f); subplot(2,2,4); plot(x,e4);grid;title(4th Order Taylor Series Error);hold off 4.19 Here are Excel worksheets and charts that have been set up to solve this problem:PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 16. 16 12 8 exact backward 4 centered forward 0 -2 0 2 -4PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 17. 17 12 8 exact backward 4 centered forward 0 -2 0 2 -4PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 18. 17 12 8 exact backward 4 centered forward 0 -2 0 2 -4PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 19. 17 12 8 exact backward 4 centered forward 0 -2 0 2 -4PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.
  • 20. 17 12 8 exact backward 4 centered forward 0 -2 0 2 -4PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manualmay be displayed, reproduced or distributed in any form or by any means, without the prior written permission of thepublisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for theirindividual course preparation. If you are a student using this Manual, you are using it without permission.