Which Graph Represents the Compound Inequality - Edubirdie

University:
Bradley University
Course:
MTH 109 | College Algebra
Academic year:

2019
Views:

223

Pages:

8
Author:

Gudrunecg9m

GRAPH THE INEQUALITY * I Unit 5 ; x< —3 y A !! 7 4 >— ;J c / » a Solving & Graphing Linear Inequalities i J ~H ' 3 ' * J 3 —f . j 1 8 if A ALGEBRA 1 r .t r t , « y .iC o- - r 2 1 f ‘V . * i g Keywords OPENCIRCLE < or > \ Write the Inequality for the Graph CLOSED CIRCLE V J “ « i i i or - # i Graph the ”T io 6i ) i i 1111 i qfcflio —>* • 8 1 H > -3 Wav the Arrow Points afiCfiEBE . . y 'V . '* -' ** U£ ./r .‘T • /1 ct. - 8 l BBHWBBMMMB r> ' — ~ — * ——— r * 1 1 1 I I O1 1t 1Q M J H 'S ' " _ 9/3/2019 ^ (- I ' P X ,/ <2 - Jn <> -4* # (r ;< • Solving Inequalities Is the Same as Solving Equations -7* > 21 T -» -M • If you MULTIPLY OR DIVIDE by a NEGATIVE NUMBER, you MUST FLIP THE INEQUALITY * / - 1C 2 7 J K>. " 6 xJty < -j G> lC /1 Solving Inequalities in (, * F5 5 (/c - 6) > 15 Sir ' KJ 4> l > ^ Solving Inequalities is the Same as Solving Equations • If you MULTIPLY OR /5 Solving Inequalities Solving Inequalities * t >. - « 20 lity DIVIDE by a NEGATIVE NUMBER, you MUST FLIP THE INEQUALITY • Solving Inequalities Is >’ i -/w- cs * Same as Solving Equations J ». > A + 10 ( j!(/> - 14) > 0 >0 hi 10 7 k < £ the : • If you MULTIPLY OP DIVIDE by a NEGATIVE ! NUMBER, you MUST FLIP { THE INEQUALITY j 9/3/ 2 9p _15/ < (12pj+ 3 ^ I • Solving Inequalities is the ir 6 I 'g Unit 5 Linear Inequalities with Variables on Both Sides Solving Inequalities i Same as Solving Equations • If you MULTIPLY OR 3 ' DIVIDE by a NEGATIVE NUMBER, you MUST FLIP THE INEQUALITY 53 ALGEBRA 1 • pp - 1 + 9w > 8 (w + 1) O '? ' ' g^ * E±2 \ \ J 1 + 17n > 16 ( n + 1) )IIA > (pT tii, ^I I - /1 2: /C I | * If you MULTIPLY OR { DIVIDE by a NEGATIVE j NUMBER, you MUST FLIP !j THE INEQUALITY J ~«r. ig «wi 5 (5* ^X ' »«>i * - 16 6.23 ~~' TT 1 ' Solving Inequalities • If TRUE: The solution is ALL REAL NUMBERS • If FALSE: NO SOLUTION 2 STC > - lUK 1 rT~ either a true or false statement. — - 3) > 25 . - - i • If the variables cancel ; out, you will be left with 0 > - -N * ! ! "tf ; * Solving Inequalities is the Same as Solving ; Equations ' i —4 3 (3% 4- 2) > 9 x I ~ v> > (1 Solving Inequalities :- r i I 1 + 5 - WX IQX . - 9 x H- 7 L lyjn PV -'irttys M ® c 'j ** * -lx 6 + 12% . 5 x + 9 t iff » tfjmmu ' / - 472 ^ / *> r s y, * **» A* I Solving Inequalities s ms ^ I - VC 1 ** . - * If the variables cancel out, you will be left with either a true or false f statement . • If TRUE; The solution Is AM REAL NUMBERS • If FALSE; NO SOLUTION 7/ tfjQt * ttt / K * IWfHHM jt f if A M el M Unit 5 r, <» < Vi f IHifi a COMPOUND INEQUALITIES 4 YJ * 1 * i {) 44M tmJ * 11* f / f *y *! v/ /- * f ^1 1 1 *1 t t A 1 i * / ~)y 7 /. M AlCil ItUA 1 HiiHI HSKIB i, ', dy I- I #< x<# I I Ww4t . .. . : \ :,. Mw M'/yWVy tWPM Ufm tlvfW Uu: u 0 » 4 > • 4 J • I / W e a r e shading to the + left and right * -< » " " J .TiZi ' 4F £ ^o^Wy yiyon ***'* £ » hH ; « « * ; i ‘ . » | | 4 . jtutn Lj II tu u jifjwjpJ wx * or And o J& ' COMPOUND INEQUALITIES We’re shading hetween the two points, L .; L:S . , BE # -10 ‘ fr tt # ° 4 * « - 4 ; • -• 4 « ? > ' /0 ~ / r 4. 142 1 1 1 1» 4 9 4 / \ 0(Ur \ 1 9/3/2019 SoN* the fotomoQ compound inequality A. Jf B - C. *> Soivo th* - ^ 3* ' 7 > - 19 OR 5 s 4 * >8 A. - . 4 0. * « *4 OR r > 8 ^ -vQl .« OR *<1 4 i* < 3 * *a 4 < 15 8 OIL. compound (noquMlY - 4 OR i>3 C >f 5 8 OR x ^ 4 0 *5 4 Absolute Value Inequalities 10 - “biw or flofvn Ihn following Inmiunlily Bdvt Ifte foN« >wtng inequality 5 f ?i * * £ / 7.5 - „ * C. * D 10 or r s 10 or r * 5 4 -2 £) -> ) >X i 3 £ 4 M M 0 < '< c *< r < 4 S- i -Jr .: * MM <• i T ^ «*y»,* V'i l i4 i j JV ’ V>. • :*KV W V \\ ' - W \\ •> ^ 4 - or « > 4 j - 13 < r < 4 - \^ / -•» 4 J y .« .8 c- •*- } /- - » —— D • * - J .? i i i i awftf u * ' • • 4 I I I * : « » u * * « V ^ 0j -1 or £ 4 -1-3 * 4Y £ ID y M ^ a 0 i 8 * U t i -7 1« /\ Hv* I- I I I -l> ‘. r* ’ ; i Cio • IMV4 ^ ): I ? L l y * UU BilWPBlia ; ILWUWB. v i. ; * »v.w - X 2 -H Wi 10 v2 j2 -l 'I I - I t > IO %0\ K - - lo v y > io < Or -3|3 ifn £j) i - - *| ( Q* > 7 or x < -1 2H B. x > 5 or x < 1 c x < 7 or x > TP C 4 D. Oftr x B£&2 < 5 or x • V*). H -1 rii iv -p Of T I 3 "4 > H $-* i q > 1 Y 4 - I or 7 2 ~) 7 ? -\ o r - 9 -J-2 U H. r , • I « It > i -\ 4 “ I 9 %> f Solve the following inequality. M i i- 1 4 - I ^ L t y4 ; *14 Which of the fofcawng iMmber knes shows the solution to the inequality given below? 5* 3 < 7 OR 4* 9 > t 5 - -Ilf «)c * <1 Vi , - • I -4 - V\ L 'r&i . ** * ****** A mU A M Awtf keen « 24 « « d to * tt MUkM AUo htoTH Of # kA/St trn 21% f /jf J#ti oA'# » > WA # «K , ( % concenontoft 'A 5% 'f£0frH* / , EtowAonTt# xAftPA / Mow toomMJ* Aw# / 1 Kmor* * •* *** * Mm •// * ***** * *** * 1tWi ' * ** *#* neetff 0,21 * Unit 5 ^ - *J 2 f 0/X/ l ** t#rA Mtoo t#i # t' rtkeutfft A 15% fcooMffi# - ' * - / v. A if # 21% fctiofl to m«to 24 ««'/ *^ ** to* 15 ounem S \# A UAOn Mflft # * A 10% U 0 * ** C Itofor / r> 24' ? #%' A Hr nmo f? 21% tofton to ** * ^ # mod 15 ** * ^ h nd kJton W4h 15% ( ^ p ftwjtetJtjtn f / 7- ' . WlticfcjjoM&tify repr^mto #ie 2 L ( . -^ 0 , / j, 2>,/£ - ftorfc ba$inef # wewps %3f/ X) px mortti in intorn«l mvAhes VAT/ ) vv v&v&wi rs on the- fneqfokfy Mon, ton meny MetpeoAe, 9, o 5 to te nortorto to oro r tor s roorth, ^ to qen&efe rt \eeiA $4,100 towonyy twerm7 *r ^ ** $ (AV&LJ Y/ffii 2 C^n A ^ C, W Z 2 / + 44 ZW) . . 4 O Q? . qo lv - HV^QS * # f #n o ,' jy +. /, y ~ o,or 2 C 3 ** 2 Q. ic 5 3> 2 ZC o*. fi' X, 2 ? C - /. 2 2/ ^^ * * :, :i O ' /\ y, Keton b f/fctori# bread tor o torn#/ tonfwrilurr He knows fid yeeet, a key togredtord, fhtwee/Mtti Ahe tempx&UMe farvp,of f #¥ f to 35* F 3o to frsMrte yeesi , 3 vtoc/ jrq C Ben mfrfv an rno& 13 D Eton oedt d mod 3 safetpeopte v/or ng worktop ^ AO ^ i > L/1Oo ^ 2TO JX> S 2oo 3 2 C 3 > ^ ' ::;,w; ... nr • "f * y 1 9/3/ 2019 \ B The mileage must be greater than 2.200 miles. . » 1 - ° * boU*f - — is less ^ than iho «1» M 25 3 M M 14 U * & B. a a ta 4 13 IO i—« i ) * 1 1 1 1 »1 1 «1• so1 1 .1 5 a x N The mileage must be greater than 1.100 miles c. The mileage must be less than 1,100 miles. o. “ «i«S fV»? w rS A % ;v \ / *> $3.50 The junkx dass would need to sell a! most 293 roses. B. ihe junior C. i he junior ihe junior dass would need to sell at least 293 roses nS&Sr-r —— ^ %> .* ? r< M * «w> ? *2 / 0 2T1 <> i .ro . —I—I—\—I—l-H—1—Y* - H I I— * 3 53 \ ) - -»• Which graph represents the time when the height of the oak tree is greater than or equal to the height of the pecan tree? -2 X£ 293 *5l «9 O 0 > W PO M 4 . class would need to sell at most 866 roses. dass would need to sell at most 573 roses. * * * V S Marcus planted two trees in his backyard, an oak tree and a pecan tree. The height, in feel, of the oak 9 The height, in feet, of the pecan tree after tree after x years con be modeled by the expression 4 x x years can be modeled by the expression 3x 12 A $3,030.72 A. S 4 ' The junior dass has been seSing roses as an ongoing fundraiser Roses sell for $3.50 each and. to date, they have rased $2,005.22 They would like to raise a total of $3,030 72 by the end of the year If thes sdua&on cs modeled by trie inequality below how many more roses, x, do they need to sell to raise a* leas!S3.030.72? 2,005.22 3 * I »I I I I 1 soI I I I« sI* M ccaa • > I—I—I—I—I—1 1 1 1 1 4—<> <— > > < «! T" , 1 1 -J i> I I ::I -I I t—H ) « -2o : The mileage must be less than 5.560 miles C A. < $299.70 A .* bottle Which graph lepresonts tho time when the amount remaining »n the diet sodo amount remaining in tho regular soda bottlo? . + $0.09x , , n going P T rtown hc? rtr #l n soda The amount ftmdning in the regular BodmbaMmmn in the diet soda bottle remaining amount The x 4 . 5 x seconds can be modeled by the expression 1 2 0 after x seconds can bo modeled by the expression 240 12r Stephanie has two bottles of liquid that she »oda and the other bottle *contain* diet . An insurance company is considering implementing a plan based on mileage. The current plan Plan A, charges a Rat rate of $299 70 per month. Plan 6 charges a flat rate of $200 70 per month plus an additional $0.09 per mile driven the previous month. Using the inequality below, find the number of mJes, y where the cost of Plan 6 is less than the cost of Plan A » i 9 —i— c. 4 ) o D. * - A \0 12 1« « 15 25 i i 1 i i ) i i i i i 5 12 15 20 24 25 32 35 i i i i i i i i i i» « S 12 15 20 24 7 52 VI 4l> 44 I I I l °H— I I I l l l l i « — mm — . ^ > > • •2 O 2 4 6 V 9 • ID i v 12 14 r- v T : I * 1

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