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S  0P֍ "^ `  >*  0\ۍ "^   @*  0T "^ `  @*H  0޽h ? 3380___PPT10.]R Default Design0 p\*(  \ \ 0  P     X*  \ 0D       Z* d \ c $ ?    \ 0P   0   RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S \ 6\  _P    X*  \ 6d  _    Z* H \ 0޽h ? 3380___PPT10.I!)0^N0 IAD (  D D  fl5 o?"6@`NNN?N<`,$D 0  Write the equation in the form ax + b = 0. Write the related function y = ax + b. Graph the equation y = ax + b. The solution of ax + b = 0 is the x-intercept of y = ax + b. 2q 2= 2      F   D  T  p` D#   p`" D 0v:  0 D 0G"` P <GOAL 2`2 D 0p` D 0,K 91 2  D 6Op X$SOLVING LINEAR EQUATIONS GRAPHICALLY% 2%F @ D @ D 6(SD(P   ;4.7(2( D 6W@ ]%Solving Linear Equations Using Graphs& 2& D(n D  fZ33o?"6@ NNN?NK`Q,$@0 f.STEPS FOR SOLVING LINEAR EQUATIONS GRAPHICALLY/(2/ D # l o?"6@ NNN?NB`,$D0 @  D#  ;,$D 0N   0 D   0N   0 D   0n D 0"`> 0n2 D 0"` 00n2 D 0"` 0 D 0x`k  A EXAMPLE 1 2 Z D C *Aj0311784@ H D 0!޽h ? 33 ___PPT10.]R+o D' = @B D@' = @BA?%,( < +O%,( < +D' =%(D' =%(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*D%(D' =-g6B fade*<3<*DD+' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*D%(D' =-g6B fade*<3<*DD' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*D-%(D' =-s6Bwipe(left)*<3<*D-D' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*D-T%(D' =-s6Bwipe(left)*<3<*D-TD' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*DTs%(D' =-s6Bwipe(left)*<3<*DTsD' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*Ds%(D' =-s6Bwipe(left)*<3<*DsD' =%(D@' =%(D' =4@BBBB%()))D' =1:Bvisible*o3>+B#style.visibility<*D%(D' =+4 8?RCBBCB#ppt_wB*Y3>B ppt_w<*DD' =+4 8?RCBBCB#ppt_hB*Y3>B ppt_h<*DD' =+K4 8?CBB#ppt_x+(cos(-2*pi*(1-$))*-#ppt_x-sin(-2*pi*(1-$))*(1-#ppt_y))*(1-$)CB?B*Y3>B ppt_x<*DD' =+K4 8?CBB#ppt_y+(sin(-2*pi*(1-$))*-#ppt_x+cos(-2*pi*(1-$))*(1-#ppt_y))*(1-$)CB?B*Y3>B ppt_y<*D+8+0+D0 +hj 0^N0 0-(-0? @ `,(      0e0e    B*CDEF Ao 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||*@ "0e@     @ABC DEEFGHIJK5%LMNOPQRSTUWYZ[ \]^_ `abN E5%  N E5%  N F   5%    !"?N@ABC DEFFGHIJK5%LMNOPQRSTUWYZ[ \]^_ `ab+ B ,$D 0 P 6  p CExtra Example 1 @  0 # @@?,$D 0N   0 1    0N   0 2    0n 3  0"`> 0n2 4  0"` 00n2 5  0"` 0 6  0ڈk  A EXAMPLE 2 2 Z 7  C *Aj0311784@  : C xވ o?"6@ NNN?N  lSolve  x  3 = 0.5x algebraically. Check graphically.@7 $ <  f|x o?"6@ NNN?NW ` ,$D 0 RSolve  x  3 = 0.5x for x: x =  2 2   x >  f o?"6@ NNN?N @ K,$D 0 pTo check graphically: Rewrite as ax + b = 0:  1.5x  3 = 0 Write the related function: y =  1.5x  3 Graph the equation: b =  3 m =  1.5 =6- 2  2  2 !   F        `B ] 0Do33 `B ^  0Do U$T [, m9 _ #  ) ZB `  s *D1[y- y-ZB a  s *D1[4 4ZB b  s *D1[55 55ZB c  s *D1[5 5ZB d  s *D1[6 6ZB e  s *D1[e2 e2ZB f  s *D1[Q7 Q7ZB g  s *D1[m9 m9ZB h  s *D1[8 8ZB i  s *D1[8 8ZB j  s *D1[I0 I0ZB k  s *D1[0 0ZB l  s *D1[1 1ZB m  s *D1[. .ZB n  s *D1[/ /ZB o  s *D1[-. -.ZB p  s *D13,3m9ZB q  s *D1,m9ZB r  s *D1,m9ZB s  s *D1x,xm9ZB t  s *D1G,Gm9ZB u  s *D1,m9ZB v  s *D1,m9ZB w  s *D1,m9ZB x  s *D1,m9ZB y  s *D1,m9ZB z  s *D1,m9ZB {  s *D1d,dm9ZB |  s *D1,m9ZB }  s *D1,m9ZB ~  s *D1P,Pm9ZB   s *D1 , m9ZB   s *D1[3 3ZB   s *D1[, ,ZB   s *D1[,[m9ZB   s *D1 , m9   # l o?"6@ NNN?N`P 5y 2   # l  o?"6@ NNN?N 5x 2zl 0 P   0 P ,$D 02   # lo?"6@ NNN?N J.     f o?"6@ NNN?NP0 P  @(0,  3) 22  # lo?"6@ NNN?ND,$D0  C x`G H`q}o?"6@`NNN?Np0,$D0 fThe x-intercept is  2, which confirms the solution.,4.   c $A ?? 8 $D  02  # lo?"6@ NNN?N P ,$D0H  0޽h ?  33<<___PPT10<._`+{D<<' = @B D;' = @BA?%,( < +O%,( < +D' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*< %(D' =-s6Bwipe(left)*<3<*< D' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*< *%(D' =-s6Bwipe(left)*<3<*< *D' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*> %(D' =-s6Bwipe(left)*<3<*> D' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*> -%(D' =-s6Bwipe(left)*<3<*> -D' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*> -<%(D' =-s6Bwipe(left)*<3<*> -<D' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*> <X%(D' =-s6Bwipe(left)*<3<*> <XD' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*> Xh%(D' =-s6Bwipe(left)*<3<*> XhD' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*> h|%(D' =-s6Bwipe(left)*<3<*> h|Dr' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*> |%(D' =-s6Bwipe(left)*<3<*> |D' =%(D+' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-g6B fade*<3<* D' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*> %(D' =-s6Bwipe(left)*<3<*> D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-s6Bwipe(left)*<3<* D' =%(D+' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-g6B fade*<3<* D' =%(D=' =4@BB1BB%()?)?)D' =.I7 BBBBB[M -3.33333E-6 -4.44444E-6 L 0.04584 0.09445 *3>*B ppt_xB ppt_y=@0BBAApBB<BlA=<* D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-s6Bwipe(left)*<3<* D' =%(DI' =4@BB5BB%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =+4 8?RCBBCB#ppt_wB*Y3>B ppt_w<* D' =+4 8?RCBBCB#ppt_hB*Y3>B ppt_h<* D' =-g6B fade*<3<* D' =%(DV' =A@BB5BB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =+4 8?RCBBCB#ppt_wB*Y3>B ppt_w<* D' =+4 8?RCBBCB#ppt_hB*Y3>B ppt_h<* D' =-g6B fade*<3<* D' =%(D@' =%(D' =4@BBBB%()))D' =1:Bvisible*o3>+B#style.visibility<*0 %(D' =+4 8?RCBBCB#ppt_wB*Y3>B ppt_w<*0 D' =+4 8?RCBBCB#ppt_hB*Y3>B ppt_h<*0 D' =+K4 8?CBB#ppt_x+(cos(-2*pi*(1-$))*-#ppt_x-sin(-2*pi*(1-$))*(1-#ppt_y))*(1-$)CB?B*Y3>B ppt_x<*0 D' =+K4 8?CBB#ppt_y+(sin(-2*pi*(1-$))*-#ppt_x+cos(-2*pi*(1-$))*(1-#ppt_y))*(1-$)CB?B*Y3>B ppt_y<*0 +8+0+ 0 +eQ 0^N0 !!+@  !(    6(6  p CExtra Example 2`  c $A ?? H   8 @ 0p\  40 ` ZB  s *D@ ##ZB  s *D\ RT a(*#8 #  D "N %%(*#8  %%(*#8TB  c $D%%(%%#8TB  c $D&(&#8TB  c $D(((#8TB  c $D)()#8TB  c $D*(*#8TB   c $Da(a#8TB ! c $D(#8TB " c $DM(M#8TB # c $D ( #8TB $ c $D9"(9"#8T )s,#8 &#  b TB ' c $D_/s,_/TB ( c $D-s,-TB ) c $Ds,s,s,TB * c $D*s,*TB + c $D)s,)TB , c $D#8s,#8TB - c $D6s,6TB . c $D75s,75TB / c $D3s,3TB 0 c $DK2s,K2 2  fA o?"6@ NNN?N0P 5y 2 3  f$E o?"6@ NNN?NP p@ 5x 2 5  fH o?"6@ NNN?N @ C ,$D 0 `To solve graphically: Rewrite as ax + b = 0: Write the related function: Graph the equation:bK 2 ! ! 6 c $A ??H8 $D 0 7 c $A ??(  8 $D 0 8 c $A ??Y  ) 8 $D 02 9 # lo?"6@ NNN?NZN,$D 0H :  f@S o?"6@ NNN?NP,$ 0 @(0,  1) 22 ; # lo?"6@ NNN?NZN,$D 0 <  0e0e    BC DEF Ao 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E|| @ "0e@     @ABC DEEFGHIJK5%LMNOPQRSTUWYZ[ \]^_ `abN E5%  N E5%  N F   5%    !"?N@ABC DEFFGHIJK5%LMNOPQRSTUWYZ[ \]^_ `abU ,$D  0p =  f # lo?"6@ NNN?Nj.,$D0 ? c $A ??( p 8 $D 0 @  f^ o?"6@ NNN?N ,$ 0 The solution can be checked algebraically by solving the original equation. Note: The same solution will be found even if you rewrite the equation as 2x + 1 = 0.R 2MK H  0޽h ? 33U/M/___PPT10-/._`+nGelDY.' s= @B D.' = @BA?%,( < +O%,( < +D' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*5%(D' =-s6Bwipe(left)*<3<*5D' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*5-%(D' =-s6Bwipe(left)*<3<*5-D' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*6%(D' =-s6Bwipe(left)*<3<*6D' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*5/K%(D' =-s6Bwipe(left)*<3<*5/KD' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*7%(D' =-s6Bwipe(left)*<3<*7D' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*5Ma%(D' =-s6Bwipe(left)*<3<*5MaD ' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*8%(D' =-s6Bwipe(left)*<3<*8D' =%(D+' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*9%(D' =-g6B fade*<3<*9D' =%(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*:%(D' =-g6B fade*<3<*:D' =%(D+' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*;%(D' =-g6B fade*<3<*;D' =%(D' =%(D[' =4@BBBB%()?)?D' =.w7 BBBBBM 0.00018 -0.00092 C 0.00538 0.0132 0.02639 0.06922 0.0316 0.08334 *3>*B ppt_xB ppt_y=@0BBffpBB<B,=<*;D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*<%(D' =-s6Bwipe(down)*<3<*<D' =%(D' =%(DD' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*=%(D' =-s6Bwipe(left)*<3<*=D' =%(D+' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*>%(D' =-g6B fade*<3<*>D' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*?%(D' =-s6Bwipe(left)*<3<*?D' =%(D' =%(DD' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*@%(D' =-s6Bwipe(left)*<3<*@++0+:0 ++0+=0 ++0+@0 +h0^N0   00 z (  0F  Z: y0 Z :T  p` 0#  Z:`" 0 0v:  0  0 0T~"` P <GOAL 2`2  0 0p`  0 0 92 2   0 6p X$APPROXIMATING SOLUTIONS IN REAL LIFE% 2% @  0#  j,$D 0N   0 0   0N   0 0   0n 0 0"`> 0n2 0 0"` 00n2 0 0"` 0 0 0؋k  A EXAMPLE 4 2 Z 0 C *Aj0311784@ F @ 0 @ 0 0D(P   ;4.7(2( 0 0Д@ ]%Solving Linear Equations Using Graphs& 2& D(H 0 0!޽h ? 33___PPT10.]R+D' = @B Da' = @BA?%,( < +O%,( < +D' =%(D@' =%(D' =4@BBBB%()))D' =1:Bvisible*o3>+B#style.visibility<*0%(D' =+4 8?RCBBCB#ppt_wB*Y3>B ppt_w<*0D' =+4 8?RCBBCB#ppt_hB*Y3>B ppt_h<*0D' =+K4 8?CBB#ppt_x+(cos(-2*pi*(1-$))*-#ppt_x-sin(-2*pi*(1-$))*(1-#ppt_y))*(1-$)CB?B*Y3>B ppt_x<*0D' =+K4 8?CBB#ppt_y+(sin(-2*pi*(1-$))*-#ppt_x+cos(-2*pi*(1-$))*(1-#ppt_y))*(1-$)CB?B*Y3>B ppt_y<*0+ 0^N0 ~v|0 (  | | H  `    | # l o?"6@ NNN?N `N GBased on census data from 1987 to 1995, a model for the hourly wage w of people employed in the production of computers and related goods in the United States is w = 0.409t + 10.74, where t is the number of years since 1987. According to this model, in what year will the hourly wage for these workers be approximately $18.10?VHD^ |  f֜ o?"6@ NNN?N@`F  QTo solve, graph the equation and find the value of t when the value of w = 18.10.BR 23 H | 0޽h ? 33___PPT10i.@+D='  = @B +&? 0^N0 --8< ,(    H8  `   >  # l o?"6@ NNN?N`  dw = 0.409t + 10.746  @  #  [,$D 0N   0    0N   0    0n  0"`> 0n2   0"` 00n2   0"` 0   0 k  A EXAMPLE 5 2 Z   C *Aj0311784@ F    :   NB  S Y HNB  S Y HNB  S Y /H/NB  S Y HNB  S Y $ H$ NB  S Y FHFNB  S Y H NB  S Y H NB  S Y  H NB  S Y HNB  S Y QHQNB  S Y HNB  S Y HNB  S Y ]H]NB  S Y hHhNB  S QsQ NB  S  s NB  S S sS NB   S  s NB ! S U sU NB " S  s NB # S W sW NB $ S  s NB % S OsO NB & S s NB ' S MsM NB ( S s NB ) S s NB * S LsL NB + S s NB , S JsJ NB - S Y :H: .  ~@AB CMDEF M@#" s  /  ~@AB[C DEF [@#" X EHB 0B C jJP pX   1 # BXC DEFjJX @#" X  H  2 # l o?"6@ NNN?NP   5t 2 3 # lH  o?"6@ NNN?N`   5w 2. 5 # ld  o?"6@ NNN?N P  L0 4 8 10 12 14 16 18 20 2\  6  0e0e    B|CDEF8 A@  Ao 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||""EA9-*# $.~5y=sEqNjUa`_jZsU|NvEr9i5c-RIF<<@           s " 0e@        @ABC DEEFGHIJK5%LMNOPQRSTUWYZ[ \]^_ `abN E5%  N E5%  N F   5%    !"?N@ABC DEFFGHIJK5%LMNOPQRSTUWYZ[ \]^_ `ab   % 7 # l  o?"6@ NNN?N P *  C20 18 16 14 12 10  9  0e0e    BLC7DEF Ao 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||7L@ "0e@     @ABC DEEFGHIJK5%LMNOPQRSTUWYZ[ \]^_ `abN E5%  N E5%  N F   5%    !"?N@ABC DEFFGHIJK5%LMNOPQRSTUWYZ[ \]^_ `abM G ,$D 0 ;  f  o?"6@ NNN?NPp ,$0 zWhen w = 18.10, t H" 18, which corresponds to the year 2005.D> *"2 < # lo?"6@ NNN?N6f,$D0H  0޽h ? 33___PPT10t.@+V~D' & = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*9%(D' =-s6Bwipe(left)*<3<*9Ds' =%(D' =%(D+' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*<%(D' =-g6B fade*<3<*<D' =%(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*;%(D' =-g6B fade*<3<*;D' =%(D@' =%(D' =4@BBBB%()))D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?RCBBCB#ppt_wB*Y3>B ppt_w<*D' =+4 8?RCBBCB#ppt_hB*Y3>B ppt_h<*D' =+K4 8?CBB#ppt_x+(cos(-2*pi*(1-$))*-#ppt_x-sin(-2*pi*(1-$))*(1-#ppt_y))*(1-$)CB?B*Y3>B ppt_x<*D' =+K4 8?CBB#ppt_y+(sin(-2*pi*(1-$))*-#ppt_x+cos(-2*pi*(1-$))*(1-#ppt_y))*(1-$)CB?B*Y3>B ppt_y<*+8+0+;0 +KG 0^N0 |3t3p3h3(  h h H8/   `    kh # lD3  o?"6@ NNN?N ` Write and graph a function for each side of the equation 18.10 = 0.409t + 10.74 from Extra Example 4. What is the t-coordinate of the point where the two graphs intersect?@F,9F    h  HB h C Y HHB h C Y HHB h C Y /H/HB h C Y HHB h C Y $ H$ HB h C Y FHFHB h C Y H HB h C Y H HB h C Y  H HB h C Y HHB h C Y QHQHB h C Y HHB h C Y HHB h C Y ]H]HB h C Y hHhHB h C QsQ HB h C  s HB h C S sS HB h C  s HB h C U sU HB h C  s HB h C W sW HB h C  s HB h C OsO HB h C s HB h C MsM HB h C s HB h C s HB h C LsL HB h C s HB h C JsJ HB h C Y :H: h  ~@AB CMDEF M@#" s  h  ~@AB[C DEF [@#" X EBB hB 3 jJP pX   h # BXC DEFjJX @#" X  H  h  f?  o?"6@ NNN?NP   5t 2 h  fC  o?"6@ NNN?N`   5w 2( h  fG  o?"6@ NNN?N P  L0 4 8 10 12 14 16 18 20 2\  h  0e0e    B|CDEF8 @  o 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||""EA9-*# $.~5y=sEqNjUa`_jZsU|NvEr9i5c-RIF<<@           s " 0e@        @ABC DEFGHIJK5%LMNOPQRSTUWYZ[ \]^_ `abN 5%  N 5%  N    5%    !"?N@ABC DEFGHIJK5%LMNOPQRSTUWYZ[ \]^_ `ab    h  fK  o?"6@ NNN?N P *  C20 18 16 14 12 10 l h  fP  o?"6@ NNN?N0  P,$ 0 dGraph w = 18.10.8 2  h  0e0e    BCDEF o 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||@ "0e@     @ABC DEEFGHIJK5%LMNOPQRSTUWYZ[ \]^_ `abN E5%  N E5%  N F   5%    !"?N@ABC DEFFGHIJK5%LMNOPQRSTUWYZ[ \]^_ `abX lZ,$D 0 h  fU  o?"6@ NNN?NP P p,$ 0 Graph w = 0.409t + 10.74 V 2  h  0e0e    BLC7DEF o 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||7L@ "0e@     @ABC DEFGHIJK5%LMNOPQRSTUWYZ[ \]^_ `abN 5%  N 5%  N    5%    !"?N@ABC DEFGHIJK5%LMNOPQRSTUWYZ[ \]^_ `ab  ,$D 02 h # lo?"6@ NNN?NBr,$D0 h  ft]  o?"6@ NNN?N ,$ 0 |4The t coordinate of the point of intersection is 18.,5 20H h 0޽h ? 33_W___PPT107.@+Dc'  = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(DD' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*h%(D' =-s6Bwipe(left)*<3<*hD' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*h%(D' =-s6Bwipe(left)*<3<*hD' =%(D' =%(DD' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*h%(D' =-s6Bwipe(left)*<3<*hDr' =%(D' =%(D7' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*h%(D' =-s6Bwipe(left)*<3<*hD' =%(D+' =4@BB BB%(D' =1:Bvisible*o3>+B#style.visibility<*h%(D' =-g6B fade*<3<*hD' =%(D' =%(DD' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*h%(D' =-s6Bwipe(left)*<3<*h++0+h0 ++0+h0 ++0+h0 +  0^N0  m(    Hx   `      # l`y  o?"6@ NNN?N `~ hThe average speeds in miles per hour of the first six winners (1911 1916) of the Indianapolis 500 can be modeled by the equation y = 1.84x + 80, where x is the number of years since 1911 and y is the average speed in miles per hour. Use this model to find the average speed of Ralph DePalma, the 1915 winner.l5 'u    f(  o?"6@ NNN?Np ,$0 ]Regardless of the way you choose to solve this problem, the solution is 87.36 miles per hour.^(2^H  0޽h ? 33z___PPT10Z.@+2SD'  = @B D' = @BA?%,( < +O%,( < +D' =%(D' =%(D8' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =-g6B fade*<3<* +8+0+ 0 +  0^N0 pA(    N o?"0@NNN?N C QUESTION: (2 q   fp  o?"6@ NNN?N@@ ;What linear function is related to the equation 4x + 3 = 5?F0(2 1   N o?"0@NNN?N  AANSWER:(2   f  o?"6@ NNN?N @@ ,$0 y = 4x  2X H  03޽h ? 33___PPT10.{+,DD'  = @B D' = @BA?%,( < +O%,( < +D6' =%(D' =%(D' =A@BB5BB0B%()))D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?RCBBCB#ppt_wB*Y3>B ppt_w<*D' =+4 8?RCBBCB#ppt_hB*Y3>B ppt_h<*D' =-g6B fade*<3<*+8+0+0 +  0 d(  dX d C \    d S l \ 0    H d 0޽h ? 3380___PPT10.J@PxYMl[E$@QUF R$pԊSqt_#1cq~,񓂄B#B@!ʨHU$\UPE\Jk38yN-w͛ݷ?m@= NXyŢ/ОQH/*` iV– Ҟ7-79Q}meζEO85,'pF|sK)`fAÙppF̷9KRoy+X7`& p^>-}l,Y]+z0syO9eQz_oFqۦ\3 %ǿrq(HS&ˆ|fYGvXد'e2vP8`B0t^_Ƈ?$^߸c:2媡O |w(J =Bc3%C<>8BXt(ӿ:2#Yu!lִ628*UP;yAPT\b*^j}54=eb)ɥ)p㻬 z|G?N_[7 ]|P?S{/TLyǔi tn< t6XІ1888[ѳlC_}Z_} &^4_'G+Xپy3Ew.14NffSڝ$Q*nYNpQ1sR<="rPENjN˗ |r͌?vbC;l׻{ƎVP|$sz\Hkظ[緼Og~1B,I,NlQE;0^cUZ@?ٍ$zMKӈM/UK,K,` g#iyx6W2VXX _ ظaZA^V-GGVXDb,,[yxоxLfjK]i)o=i jwbۙI/+F g$;!qZ$ktE}nmQ hlxXMlUv4T,ZP(^ζj\QNf /C9 TJEz@ z*#R\P8 T/3`';;g͛of|]?: 4~d= e~GNA?UPA^P >7܇WܖEqة_/ɂ>ur`lKV2>l{^yXwE;IeOOs zPB;Lb;п[,7p\{=I7zm?eE:I^tvn9|(pM{8 NczX~R|*0n;?DWcNwv~;(S 3aϹR]$^q7~D v.1-/ٿ)"8N1#r\:DP LH#2.>FB0W_LJeْ#)=}l0SB<;W&)QsUjEdB9kդwlި*D}iU{Z1 VHD&ᇐ ̃ʱ rp.+91Wl^ Lg4pGBTĸvMQid?z- 'p9Dܵ.GÝ@ՠx j!w^[߶?mv@tn{C?iBM|L+2h9w{_ݣUlk@bGLו)D~E%>)Oᶘ~A$X?gJ??ǦsB{3MA@}#`[}=\m6mۿtL4u\BCK_Gwzl{^ r8?)~?nE#4LWe~OQ[FI4U{jUF,N#Ys+MoU׿;ud֫lZb#V=jvDMF=Aq˭x?ݐnxXoE3kv *HT  #dK9čTR#$CSZ !\J=pzzF9V*q FBڼ7l8]Ϟ73;7W{ lhlǰp̾fբ. 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