HS Pre-Alger Notes Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Roots nd Cue Roots Syllus Ojetive 5.4: The student will find or pproximte squre roots of numers to 4. CCSS 8.EE.-: Evlute squre roots of smll perfet squres nd ue roots of smll perfet ues. The squre root of numer n is numer m suh tht m n. The rdil sign,, represents the nonnegtive squre root. The symol ±, red plus or minus, refers to oth the positive nd negtive squre root. Therefore, the squre roots of 6 re 6 nd 6, euse 6 6 nd ( 6) 6. Also, 6 6, 6 6, nd ± 6 ± 6. We understnd tht will e the positive vlue. We refer to this s the prinipl squre root. Students should memorize the vlues of the squres for 1 through. These perfet squres (squres of integers) re listed elow: Perfet s: 1 1 4 9 4 16 5 5 6 6 7 49 8 64 9 81 1 1 11 11 1 144 1 169 14 196 15 5 16 56 17 89 18 4 19 61 4 Simplifying expressions suh s the 5 nd 64 re pretty stright forwrd. 5 5 nd 64 8 5 Even expressions like re esy to simplify one we see n exmple. 81 5 5 5 81 81 9 But wht if the expression is lrge, like 576? We will onsider method for determining this nswer whih extends eyond wht is required y the CCSS. Consider the perfet squres of few multiples of 1: 1 1 9 4 4 16 nd so on... MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge 1 of 19 Revised 1 - CCSS
Now to find 576 : 4 < 576 < 9 Identify perfet squres losest to 576. 4 < 576 < 9 Tke positive squre root of eh numer. < 576 < Evlute squre root of eh perfe t squre tht I know. Vlues we now need to onsider re 1,,, 4, 5, 6, 7, 8, nd 9. But wit! Sine we know n even even even, our nswer must e even. So now our nswer hoies re redued to, 4, 6, nd 8. We now need to sk ourselves whih of those nswer hoies will give us 6 in the one s ple ( 576 ). There re only two: 4 (sine 4 16 ) nd 6 (sine 6 6). Sine 576 is loser to 4 thn it is to 9, we quikly know tht the nswer is 4. A quik hek will identify 4 s the nswer: 576 4. Let s look t nother exmple, finding 15. 9 < 15 < 16 Identify perfet squres losest to 15. 9 < 15 < 16 Tke positive squre root of eh numer. < 15 < 4 Evlute squre root of e h perfet squre tht I know. The vlues we need to onsider re 1,,, 4, 5, 6, 7, 8, nd 9. Sine our rdind is odd, we ondense those onsidertions to 1,, 5, 7, nd 9. However, if we look t the rdind 15, we n quikly reognize tht the only numer whose squre will give us 5 is 5. We hve our nswer very quikly! 15 5. In review, evlute the following: 16, 16, ± 16 16 denotes the prinipl (positive) squre root, so the nswer is 4 euse 16 denotes the negtive squre root, so the nswer is 4 euse 4 16. 4 16. ± 16 denotes the positive nd negtive squre roots. It is red plus or minus the squre root of 16. The nswer is ± 4 euse (4)(4) 16 nd ( 4)( 4) 16. Speil Cses: root of :. Sine is neither positive nor negtive, only hs one squre root. There should e mention of the squre root of negtive numer: negtive numers hve no rel squre roots, euse the squre of every rel numer is nonnegtive. Emphsize tht negtive numers hve no REAL squre roots (they hve imginry roots) not tht negtive numers hve no squre roots. MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge of 19 Revised 1 - CCSS ( )
The ue root of numer n is numer m suh tht m n. The rdil sign,, represents the ue root. Therefore, the ue root of 8 is, euse 8. Students should memorize the vlues of the ues for 1 through 5. These perfet ues (ues of integers) re listed elow: Perfet Cues: 1 1 8 7 4 64 5 15 6 16 7 4 8 51 9 79 1 1 Therefore, 64 4 (sine 4 64). NOTE: Whenever there is n even numer of identil ftors, the produt will positive euse every pir of identil ftors hs positive produt. 5 5 5 positive positive positive ( 5)( 5) 5 negtive negtive positive However, when there is n odd numer of n identil ftor, the produt will hve the sme sign s the ftors. So, while negtive squre root is imginry, negtive ue root is possile. 5 5 5 15 positive positive positive positive ( 5)( 5)( 5) 15 negtive negtive negtive negtive So 8 nd 7. (Note: onsidertion of negtive numers is not inluded in the CCSS for this grde level.) Simplifying Roots To simplify squre root, you rewrite the rdind s produt of the lrgest perfet squre tht is ftor nd some other numer. You then tke the squre root of the perfet squre. Exmple: Simplify 5. Now 5 n e written s produt of 5 nd 1. Should I use those ftors? Hopefully, you sid no. We wnt to rewrite the rdind s produt of the lrgest perfet squre, nd neither 5 nor 1 re perfet squres. So, looking t my list of perfet squres, whih, if ny, re ftors of 5? MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge of 19 Revised 1 - CCSS
Tht s right, 5 is ftor of 5 nd it is perfet squre. Simplifying, I now hve 5 5 5 5 Exmple: Simplify 98x y. 98 49 xy x y 7 x y 7x y Exmple: Simplify 6 49. 6 6 49 49 4 15 7 15 7 Solving Equtions using Root nd Cue Root CCSS 8.EE.-1: Use squre root nd ue root symols to represent solutions to equtions of the form x p nd x p, where p is positive rtionl numer. We re now redy to solve equtions in the form x. Exmple: Solve x 9. Solve 5 p x x 9 9 x ± 5 5 ± 15 p p p MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge 4 of 19 Revised 1 - CCSS
Exmple: Solve x 8 Solve 15 r x x 8 x 8 15 r 15 5 r r Rel Numers CCSS 8.NS.1-1: Know tht numers tht re not rtionl re lled irrtionl. CCSS 8.EE.-: Know tht is irrtionl. The set of rel numers onsists of ll rtionl nd irrtionl numers. This reltionship n e shown in Venn digrm. Rel Numers Counting/Nturl Numers {1,, } Integers Whole Numers {, 1,, } Rtionl Numers {, 1,,1,, } 1.4 11.5 5 Irrtionl Numers π.17447 A rtionl numer is numer tht n e written s quotient of two integers. The deiml form repets or termintes. An irrtionl numer is numer tht nnot e written s quotient of two integers. The deiml form neither termintes nor repets. You n illustrte n irrtionl numer y hving students try to "Think of numer, when multiplied y itself, equls. Perhps we re finding the length of the side of squre when we know the re is. Or for more visul representtion, use tngrms s disussed elow. MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge 5 of 19 Revised 1 - CCSS
The tsk is to ompre the three squres nd try to estimte the side length of the middle squre. side 1 unit side? side units re 1 squre unit re squre units re 4 squre units You n tell tht the smll squre is 1 y 1, so its re is 1 squre unit. The iggest squre hs units on side, so its re is 4 squre units euse it is y. But the middle squre is mde from 4 hlf-squres, so it must hve n re of squre units. Wht is its side length? 1 1 1 nd 4. But? By mthing the sides to eh other, students n see tht the unknown side is etween 1 nd, perhps 1 nd hlf. Hve students get out lultors nd try 1.5 times 1.5 nd mke tle, with lrge gp etween 1 nd. side length re 1 1 1.5.5 4 Students will find 1.5 to e too lrge (.5) nd 1.4 to e too smll (1.96). Wht numer squred will mke? It must e etween 1.4 nd 1.5. MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge 6 of 19 Revised 1 - CCSS
Further investigtion: side length re 1 1 1.4 1.96 1.41 1.9881 1.415.5 1.4.164 1.5.5 This is n exellent form of interpoltion tht tehers n uild upon to lrify ple vlue in the deimls. Let students go for s long s they wnt within reson, getting ever loser y never rehing extly. Review some repeting deimls. By now students re redy to her tht there re some speil numers tht will never repet s deimls, silly rtios tht go on to infinity with no repeting pttern. We were looking for one of them (1.4141 ) nd it does not repet. In ft, these deiml ples will only round to when squred rised to the seond power. Insted of rememering the deiml nd sying it is when squred, we hve etter wy to del with this. The side length, tht when squred is, we ll the squre root of nd write it. It n never e expressed s repeting deiml euse it is not the rtio of two numers, like 1/6 or /8. Therefore it is lled irrtionl. (You might hve students explore the squre root key on the lultor to see how mny deiml ples they get.) Another wy to model is to wit until you hve introdued the Pythgoren Theorem. Drw right tringle with eh leg mesuring 1 unit. Find the length of the hypotenuse of the right tringle y using the Pythgoren Theorem. 1 1 + 1 + 1 1+ 1 MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge 7 of 19 Revised 1 - CCSS
Approximting Roots CCSS 8.NS.-: Estimte the vlue of irrtionl expressions (e.g., π ) You n use perfet squres to pproximte the squre root of numer. Exmple: Approximte 5 to the nerest integer. First, find the perfet squre tht is losest ut less thn 5. Tht would e 49. The perfet squre losest to 5 ut greter thn 5 is 64. So, 5 is etween 49 nd 64. 49 < 5 < 64 Identify perfet squres losest to 5. 49 < 5 < 64 Tke positive squre root of eh numer. 7 < 5 < 8 Evlute squre root of eh perfet squre. Beuse 5 is loser to 49, 5 is loser to 7. Therefore, 5 7. Students n pproximte squre roots y itertive proesses. Exmple: Approximte the vlue of 5 to the nerest hundredth. Solution: Students strt with rough estimte sed upon perfet squres. 5 flls etween nd euse 5 flls etween 4 nd 9. The vlue will e loser to thn to. Students ontinue the itertive proess with the tenths ple vlue. 5 flls etween. nd. euse 5 flls etween. 4.84 nd. 5.9. The vlue is loser to.. Further itertion shows tht the vlue of 5 is etween. nd.4 sine. is 4.979 nd.4 is 5.176. Another wy to mke the estimte is shown elow: Exmple: Approximte 5. Strt s we did with the previous exmple. 49 < 5 < 64 49 5 64 4 15 Determine the differene etween 49 (smller perfet squre) nd 5 (the rdind). The differene is 4. Then find the differene etween 49 (smller perfet squre) nd 64 (the lrger perfet squre). The differene is 15. We know tht 5 flls etween 7 nd 8. To pproximte the deiml, tke 4 15.6. We would estimte 5 7.7. Using lultor, we find 5 7.8, whih is very lose to our estimte! MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge 8 of 19 Revised 1 - CCSS
CCSS 8.NS.-: Lote irrtionl numers pproximtely on numer line digrm. Exmple: Compre nd y estimting their vlues, plotting them on numer line, nd mking omprtive sttements. Solution: Sttements for the omprison ould inlude: is pproximtely 1.4 is etween the whole numers 1 nd is etween 1.7 nd 1.8 Ordering Rel Numers Syllus Ojetive 5.7: The student will ompre nd order rel numers. CCSS 8.NS.-1: Use rtionl pproximtions of irrtionl numers to ompre the size of irrtionl numers. Now is good time to return to the onept of ordering rel numers, inluding irrtionl numers. Exmple: Order the numers from lest to gretest: 4,, 4.1, 5 One wy to solve this would e to onvert ll terms into deiml form (or pproximte deiml form). 4 6 < 4 < 49 Identify perfet squres losest to 4. 6 < 4 < 49 Tke the positive squre root of ll terms. 6 < 4 < 7 Simplify the perfet squres tht I know. 5 4 will fll etween 6 nd 7, loser to 6. 4 44. 5 5 16 < < 5 Identify perfet squres losest to. 16 < < 5 Tke the positive squre root of ll terms. 4 < < 5 Simplify the perfet squres tht I know. will fll etween 4 nd 5, loser to 5. Thus will fll loser to 5 thn 4. So I would order these numers from lest to gretest: loser to 5, 4.1, 4.4, loser to 6. Tht trnsltes to, 4.1,, 4. 5 MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge 9 of 19 Revised 1 - CCSS
The Pythgoren Theorem Syllus Ojetive 5.6: The student will explin the Pythgoren Theorem. CCSS 8.G.6-1: Explin proof of the Pythgoren Theorem nd its onverse. In right tringle, the side opposite the right ngle is lled the hypotenuse. The legs re the sides tht form the right ngle. We typilly lel the legs nd, while the hypotenuse is. hypotenuse legs If we look t enough right tringles nd experiment little, we will notie reltionship developing. Drw right tringle on piee of grph pper, with leg mesurements of units nd 4 units. Drw squre on eh side of the tringle s shown. Cut out squres A nd B nd ple them side y side. A B Tpe the repositioned tringles. Compre your new squre with squre C. A C Drw two tringles s shown; ut long the hypotenuses nd slide s shown. A B B A B It ppers tht the sum of the res of the squres formed y the legs is equl to the re of the squre formed y the hypotenuse. There is nother wy to look t this reltionship. MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge 1 of 19 Revised 1 - CCSS
Begin with right tringle, legs nd with hypotenuse. Crete squre tht hs mesure + on side, using 4 ongruent tringles s shown. The re of this new squre n e expressed two wys: 1) A ( + ) (length of side is +, so multiply sides to get re) 1 ) A 4 + (4 tringles, eh with re of 1 ; dd to the re of the squre with side ) Setting these two vlues equl to eh other, we get ( ) + + 1 4 + + + + This is n importnt reltionship in mthemtis. Sine it is importnt, we will give this reltionship nme, the Pythgoren Theorem. MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge 11 of 19 Revised 1 - CCSS
And yet nother wy to model the theorem: Drw right tringle on piee of grph pper, with leg mesurements of units nd 4 units. Drw squre on eh side of the tringle s shown. Cut out squres eh, ongruent to A nd B, in two different olors Glue one A nd one B on the originl grph. Lel with the re. B A B A C Are 16 units B A Are 9 units C Hve students ut nd pste the remining squres (9 + 16) to form third squre C, whih represents the re of 5. Hve students ut nd pste the remining squres (9 + 16) to form third squre C, whih represents the re of 5. Are 16 units Are 9 units Are 5 units Finlly, hve students lel the legs of the originl right tringle with nd, hypotenuse. (This represents one possile solution.) It ppers tht the sum of the res of the squres formed y the legs is equl to the re of the squre formed y the hypotenuse, or +. Pythgoren Theorem: If tringle is right tringle, then +. MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge 1 of 19 Revised 1 - CCSS
The onverse of the Pythgoren Theorem sys If tringle. Let s see if this seems to e true. +, then the tringle is right Fill in the first empty olumn in the hrt elow. Then, use grid strips ut to one squre width to rete the tringle with the side lengths given (see exmple elow the hrt). Does it pper to e right tringle? Is + true? Does this pper to e right tringle? 4 5 + 4 5 true Yes, see elow. 5 1 1 7 4 5 8 15 17 5 5-4-5 Looks like right tringle. To model the onverse, use the exmple of the Egyptin rope-strethers. Some nient toms depit sries rrying ropes tied with 1 eqully sped knots. It turns out tht if the rope ws pegged to the ground in the dimensions of -4-5, right tringle would emerge. This enled the Anient Egyptin uilders to ly the foundtions for their uildings urtely. To model this, deide on unit length nd mke 1 knots (representing this unit s mesure) in piee of rope or twine. This wy you n hold the first nd lst knot together. Using people you n streth the rope into -4-5 tringle nd model where/how the right ngle ourred. Syllus Ojetive 5.5: The student will use the Pythgoren Theorem to solve prolems. CCSS 8.G.7-1: Apply the Pythgoren Theorem to determine unknown side lengths in right tringles in rel-world nd mthemtil prolems in two dimensions. Exmple: Find the unknown length in simplest form. + 5 + 1 5 + 144 169 169 1 5 1 MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge 1 of 19 Revised 1 - CCSS
Plese note tht this 5-1-1 is one of our Pythgoren triples. A Pythgoren triple is set of three positive integers,,, nd suh tht +. Another triple worth noting is -4-5. You n rete other Pythgoren triples y multiplying the originl triple y ftor. For instne, -4-5 eomes 6-8-1 y multiplying ll the sides y. One might rete nother, 1-4-6, y multiplying the 5-1-1 y. This skill is very vlule when working with prolems on the Nevd CRT nd NV High Shool Profiieny Exm. It n sve the student time nd frustrtion! Exmple: Find the unknown length. 15 5 If we reognize the ommon ftor of 5, we n think: 5 15 5 5 5 This is leg nd the hypotenuse of the -4-5. So the missing side would e 5 4, whih is. We hd -4-5 tht ws multiplied y ftor of 5 to produe 15--5. So. Exmple: A 1-foot ldder is pled ginst uilding. The foot of the ldder is 5 feet from the se of the uilding. How high up the side of the uilding does the ldder reh? Give your nswer to the nerest foot. + 5 + 1 5 + 144 119 119 1 < 119 < 11 1 < 119 < 11 1 < 119 < 11 Sine 119 is loser to 11, 119 is loser to 11. Therefore, the ldder rehes pproximtely 11 feet up the side of the uilding. 1 ft 5 ft MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge 14 of 19 Revised 1 - CCSS
Exmple: You use four stkes nd string to mrk the foundtion of house. You wnt to mke sure the foundtion is retngulr. The four sides mesure ft, 7 ft, ft nd 7 ft. You then mesure digonl to e 78 ft. Is your foundtion retngulr? The digonl divides the foundtion into tringles. If the tringles re right tringles, + will e true sttement. Beuse + 7 78 (9 + 5184 684), you n onlude tht the tringles re right nd your foundtion is retngulr. CCSS 8.G.7-: Apply the Pythgoren Theorem to determine unknown side lengths in right tringles in rel-world nd mthemtil prolems in three dimensions. Exmple: Find the length of the longest line segment (lled the min digonl) in the retngulr prism shown on the piture elow. Solution: We will pply the Pythgoren Theorem twie. First, lel the points nd sides tht will used on the piture. We will find x using the Pythgoren Theorem in tringle ABC. Then we n find y using the Pythgoren Theorem in tringle ACD. For ABC, + 4 + 5 x 16 + 5 x 41 x We do not hve to simplify x s we need it in tht form in the Pythgoren Theorem: MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge 15 of 19 Revised 1 - CCSS
For ACD, + + 41 y 45 y 45 y 45 y The length of the min digonl would e 45 feet (or pproximtely 6.7 feet). Note: Our result is tully + 4 + 5. Indeed, we n see tht the length of the min digonl in retngulr prism with sides x, y, nd z is L x + y + z. This is sometimes lled the -dimensionl Pythgoren Theorem. Another Retngulr Box prolem: You hve huge retngulr ox with dimensions1ft 15ft 9ft. Wht is the length of the longest rod you ould fit in your ox? 9 ft 15 ft 1 ft Solution: We will pply the Pythgoren Theorem twie. First, lel the points nd sides tht will used on the piture. We will find AC using the Pythgoren Theorem in tringle ABC. n find AD using the Pythgoren Theorem in tringle ACD. Then we For ABC, 1 + 15 x D 144 + 5 x 69 x For ACD, 9 + 69 y 9 ft A 1 ft B 15 ft C 45 y 45 y 45 y The length of the longest rod would e 45 feet (or pproximtely 1. feet). MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge 16 of 19 Revised 1 - CCSS
The Distne nd Midpoint Formuls Syllus Ojetive 5.8: The student will use distne nd midpoint formuls. CCSS 8.G.8: Apply the Pythgoren Theorem to find the distne etween two points in oordinte system. Tke the time to link the distne formul with the Pythgoren Theorem. Here s wy to show it. You re given two points, (, 1) nd (6, 5). You re sked to find the distne etween them. Plot the points. 6 4 4 6 You n drw right tringle, using these points s two of the verties, s shown. 6 4 4 6 Find the lengths of the horizontl nd vertil sides of the right tringle y sutrting the x- nd y- vlues. or You ould simply ount the units from one point to the next. 6 4 (, 1) 4 6 (6, 5) 5 1 4 We now will use the Pythgoren Theorem to find the missing length, whih is the hypotenuse. + ( 6 ) ( 5 1) + 4 9 + 16 5 5 + ± 5; however, sine we re looking t mesurement, only the positive vlue will e onsidered. 6 MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge 17 of 19 Revised 1 - CCSS
Now, insted of using the ordered pirs (, 1) nd (6, 5), we use pir of points, x, y nd x, y. Sustituting into the Pythgoren Theorem one gin: ( ) ( ) 1 1 + ( ) ( ) x x1 + y + y1 ( ) ( ) 1 1 x x + y + y Sine we re determining distne, we reple the with d, nd we will only onsider the positive vlue. So we now hve the distne formul: ( ) ( ) d x x + y y 1 1 Exmple: Find the distne etween the points A(, 1 ) nd B(, 5) ( ) ( ) d x x + y y 1 1 ( ) 5 ( 1) d + ( ) 4 6 d +. d d 16 + 6 5 ( 5 is etween 7 nd 8, ut loser to 7) d 4 1 d 1 The distne etween the points A(, 1 ) nd B(, 5) is 5 or 1 units. Sine 49 < 5 < 64 7 < 5 < 8 nd. then 5 7. units 15 Students need to know the midpoint formul lso. Consider the line segment with endpoints (, 1 ) nd ( 6, 5 ). 6 4 4 6 MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge 18 of 19 Revised 1 - CCSS
The midpoint is the point on the segment tht is equidistnt from the endpoints. We re looking for the middle! If we think of this s finding the verge, we ould dd the two x vlues nd divide y. We ould likewise verge the y vlues. Finding the verge for our exmple, it ppers the midpoint is t (4, ). x1 y1 x y, we n rrive t the formul for finding the midpoint: x 1+ x 1, y + M y If we use endpoints (, ) nd (, ) Exmple: Find the midpoint of the segment with endpoints ( 5, 7 ) nd ( 1, 5). x1+ x y1+ y M, 5+ ( 1) 7+ ( 5) M, 5+ 1 M (, 6) The midpoint of the segment with endpoints ( 5, 7 ) nd ( 1, 5) is (, 6) 6 4. 4 6 6+ MDougl Littell, Chpter 9, Setions 1-5 HS Pre-Alger, Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Pge 19 of 19 Revised 1 - CCSS