Tutor:

New! Get regular updates about newly solved problems via algebra.com's RSS system.

---

Recent problems solved by 'jim_thompson5910'

jim_thompson5910 answered: 28586 problems
Jump to solutions: 0..29 , 30..59 , 60..89 , 90..119 , 120..149 , 150..179 , 180..209 , 210..239 , 240..269 , 270..299 , 300..329 , 330..359 , 360..389 , 390..419 , 420..449 , 450..479 , 480..509 , 510..539 , 540..569 , 570..599 , 600..629 , 630..659 , 660..689 , 690..719 , 720..749 , 750..779 , 780..809 , 810..839 , 840..869 , 870..899 , 900..929 , 930..959 , 960..989 , 990..1019 , 1020..1049 , 1050..1079 , 1080..1109 , 1110..1139 , 1140..1169 , 1170..1199 , 1200..1229 , 1230..1259 , 1260..1289 , 1290..1319 , 1320..1349 , 1350..1379 , 1380..1409 , 1410..1439 , 1440..1469 , 1470..1499 , 1500..1529 , 1530..1559 , 1560..1589 , 1590..1619 , 1620..1649 , 1650..1679 , 1680..1709 , 1710..1739 , 1740..1769 , 1770..1799 , 1800..1829 , 1830..1859 , 1860..1889 , 1890..1919 , 1920..1949 , 1950..1979 , 1980..2009 , 2010..2039 , 2040..2069 , 2070..2099 , 2100..2129 , 2130..2159 , 2160..2189 , 2190..2219 , 2220..2249 , 2250..2279 , 2280..2309 , 2310..2339 , 2340..2369 , 2370..2399 , 2400..2429 , 2430..2459 , 2460..2489 , 2490..2519 , 2520..2549 , 2550..2579 , 2580..2609 , 2610..2639 , 2640..2669 , 2670..2699 , 2700..2729 , 2730..2759 , 2760..2789 , 2790..2819 , 2820..2849 , 2850..2879 , 2880..2909 , 2910..2939 , 2940..2969 , 2970..2999 , 3000..3029 , 3030..3059 , 3060..3089 , 3090..3119 , 3120..3149 , 3150..3179 , 3180..3209 , 3210..3239 , 3240..3269 , 3270..3299 , 3300..3329 , 3330..3359 , 3360..3389 , 3390..3419 , 3420..3449 , 3450..3479 , 3480..3509 , 3510..3539 , 3540..3569 , 3570..3599 , 3600..3629 , 3630..3659 , 3660..3689 , 3690..3719 , 3720..3749 , 3750..3779 , 3780..3809 , 3810..3839 , 3840..3869 , 3870..3899 , 3900..3929 , 3930..3959 , 3960..3989 , 3990..4019 , 4020..4049 , 4050..4079 , 4080..4109 , 4110..4139 , 4140..4169 , 4170..4199 , 4200..4229 , 4230..4259 , 4260..4289 , 4290..4319 , 4320..4349 , 4350..4379 , 4380..4409 , 4410..4439 , 4440..4469 , 4470..4499 , 4500..4529 , 4530..4559 , 4560..4589 , 4590..4619 , 4620..4649 , 4650..4679 , 4680..4709 , 4710..4739 , 4740..4769 , 4770..4799 , 4800..4829 , 4830..4859 , 4860..4889 , 4890..4919 , 4920..4949 , 4950..4979 , 4980..5009 , 5010..5039 , 5040..5069 , 5070..5099 , 5100..5129 , 5130..5159 , 5160..5189 , 5190..5219 , 5220..5249 , 5250..5279 , 5280..5309 , 5310..5339 , 5340..5369 , 5370..5399 , 5400..5429 , 5430..5459 , 5460..5489 , 5490..5519 , 5520..5549 , 5550..5579 , 5580..5609 , 5610..5639 , 5640..5669 , 5670..5699 , 5700..5729 , 5730..5759 , 5760..5789 , 5790..5819 , 5820..5849 , 5850..5879 , 5880..5909 , 5910..5939 , 5940..5969 , 5970..5999 , 6000..6029 , 6030..6059 , 6060..6089 , 6090..6119 , 6120..6149 , 6150..6179 , 6180..6209 , 6210..6239 , 6240..6269 , 6270..6299 , 6300..6329 , 6330..6359 , 6360..6389 , 6390..6419 , 6420..6449 , 6450..6479 , 6480..6509 , 6510..6539 , 6540..6569 , 6570..6599 , 6600..6629 , 6630..6659 , 6660..6689 , 6690..6719 , 6720..6749 , 6750..6779 , 6780..6809 , 6810..6839 , 6840..6869 , 6870..6899 , 6900..6929 , 6930..6959 , 6960..6989 , 6990..7019 , 7020..7049 , 7050..7079 , 7080..7109 , 7110..7139 , 7140..7169 , 7170..7199 , 7200..7229 , 7230..7259 , 7260..7289 , 7290..7319 , 7320..7349 , 7350..7379 , 7380..7409 , 7410..7439 , 7440..7469 , 7470..7499 , 7500..7529 , 7530..7559 , 7560..7589 , 7590..7619 , 7620..7649 , 7650..7679 , 7680..7709 , 7710..7739 , 7740..7769 , 7770..7799 , 7800..7829 , 7830..7859 , 7860..7889 , 7890..7919 , 7920..7949 , 7950..7979 , 7980..8009 , 8010..8039 , 8040..8069 , 8070..8099 , 8100..8129 , 8130..8159 , 8160..8189 , 8190..8219 , 8220..8249 , 8250..8279 , 8280..8309 , 8310..8339 , 8340..8369 , 8370..8399 , 8400..8429 , 8430..8459 , 8460..8489 , 8490..8519 , 8520..8549 , 8550..8579 , 8580..8609 , 8610..8639 , 8640..8669 , 8670..8699 , 8700..8729 , 8730..8759 , 8760..8789 , 8790..8819 , 8820..8849 , 8850..8879 , 8880..8909 , 8910..8939 , 8940..8969 , 8970..8999 , 9000..9029 , 9030..9059 , 9060..9089 , 9090..9119 , 9120..9149 , 9150..9179 , 9180..9209 , 9210..9239 , 9240..9269 , 9270..9299 , 9300..9329 , 9330..9359 , 9360..9389 , 9390..9419 , 9420..9449 , 9450..9479 , 9480..9509 , 9510..9539 , 9540..9569 , 9570..9599 , 9600..9629 , 9630..9659 , 9660..9689 , 9690..9719 , 9720..9749 , 9750..9779 , 9780..9809 , 9810..9839 , 9840..9869 , 9870..9899 , 9900..9929 , 9930..9959 , 9960..9989 , 9990..10019 , 10020..10049 , 10050..10079 , 10080..10109 , 10110..10139 , 10140..10169 , 10170..10199 , 10200..10229 , 10230..10259 , 10260..10289 , 10290..10319 , 10320..10349 , 10350..10379 , 10380..10409 , 10410..10439 , 10440..10469 , 10470..10499 , 10500..10529 , 10530..10559 , 10560..10589 , 10590..10619 , 10620..10649 , 10650..10679 , 10680..10709 , 10710..10739 , 10740..10769 , 10770..10799 , 10800..10829 , 10830..10859 , 10860..10889 , 10890..10919 , 10920..10949 , 10950..10979 , 10980..11009 , 11010..11039 , 11040..11069 , 11070..11099 , 11100..11129 , 11130..11159 , 11160..11189 , 11190..11219 , 11220..11249 , 11250..11279 , 11280..11309 , 11310..11339 , 11340..11369 , 11370..11399 , 11400..11429 , 11430..11459 , 11460..11489 , 11490..11519 , 11520..11549 , 11550..11579 , 11580..11609 , 11610..11639 , 11640..11669 , 11670..11699 , 11700..11729 , 11730..11759 , 11760..11789 , 11790..11819 , 11820..11849 , 11850..11879 , 11880..11909 , 11910..11939 , 11940..11969 , 11970..11999 , 12000..12029 , 12030..12059 , 12060..12089 , 12090..12119 , 12120..12149 , 12150..12179 , 12180..12209 , 12210..12239 , 12240..12269 , 12270..12299 , 12300..12329 , 12330..12359 , 12360..12389 , 12390..12419 , 12420..12449 , 12450..12479 , 12480..12509 , 12510..12539 , 12540..12569 , 12570..12599 , 12600..12629 , 12630..12659 , 12660..12689 , 12690..12719 , 12720..12749 , 12750..12779 , 12780..12809 , 12810..12839 , 12840..12869 , 12870..12899 , 12900..12929 , 12930..12959 , 12960..12989 , 12990..13019 , 13020..13049 , 13050..13079 , 13080..13109 , 13110..13139 , 13140..13169 , 13170..13199 , 13200..13229 , 13230..13259 , 13260..13289 , 13290..13319 , 13320..13349 , 13350..13379 , 13380..13409 , 13410..13439 , 13440..13469 , 13470..13499 , 13500..13529 , 13530..13559 , 13560..13589 , 13590..13619 , 13620..13649 , 13650..13679 , 13680..13709 , 13710..13739 , 13740..13769 , 13770..13799 , 13800..13829 , 13830..13859 , 13860..13889 , 13890..13919 , 13920..13949 , 13950..13979 , 13980..14009 , 14010..14039 , 14040..14069 , 14070..14099 , 14100..14129 , 14130..14159 , 14160..14189 , 14190..14219 , 14220..14249 , 14250..14279 , 14280..14309 , 14310..14339 , 14340..14369 , 14370..14399 , 14400..14429 , 14430..14459 , 14460..14489 , 14490..14519 , 14520..14549 , 14550..14579 , 14580..14609 , 14610..14639 , 14640..14669 , 14670..14699 , 14700..14729 , 14730..14759 , 14760..14789 , 14790..14819 , 14820..14849 , 14850..14879 , 14880..14909 , 14910..14939 , 14940..14969 , 14970..14999 , 15000..15029 , 15030..15059 , 15060..15089 , 15090..15119 , 15120..15149 , 15150..15179 , 15180..15209 , 15210..15239 , 15240..15269 , 15270..15299 , 15300..15329 , 15330..15359 , 15360..15389 , 15390..15419 , 15420..15449 , 15450..15479 , 15480..15509 , 15510..15539 , 15540..15569 , 15570..15599 , 15600..15629 , 15630..15659 , 15660..15689 , 15690..15719 , 15720..15749 , 15750..15779 , 15780..15809 , 15810..15839 , 15840..15869 , 15870..15899 , 15900..15929 , 15930..15959 , 15960..15989 , 15990..16019 , 16020..16049 , 16050..16079 , 16080..16109 , 16110..16139 , 16140..16169 , 16170..16199 , 16200..16229 , 16230..16259 , 16260..16289 , 16290..16319 , 16320..16349 , 16350..16379 , 16380..16409 , 16410..16439 , 16440..16469 , 16470..16499 , 16500..16529 , 16530..16559 , 16560..16589 , 16590..16619 , 16620..16649 , 16650..16679 , 16680..16709 , 16710..16739 , 16740..16769 , 16770..16799 , 16800..16829 , 16830..16859 , 16860..16889 , 16890..16919 , 16920..16949 , 16950..16979 , 16980..17009 , 17010..17039 , 17040..17069 , 17070..17099 , 17100..17129 , 17130..17159 , 17160..17189 , 17190..17219 , 17220..17249 , 17250..17279 , 17280..17309 , 17310..17339 , 17340..17369 , 17370..17399 , 17400..17429 , 17430..17459 , 17460..17489 , 17490..17519 , 17520..17549 , 17550..17579 , 17580..17609 , 17610..17639 , 17640..17669 , 17670..17699 , 17700..17729 , 17730..17759 , 17760..17789 , 17790..17819 , 17820..17849 , 17850..17879 , 17880..17909 , 17910..17939 , 17940..17969 , 17970..17999 , 18000..18029 , 18030..18059 , 18060..18089 , 18090..18119 , 18120..18149 , 18150..18179 , 18180..18209 , 18210..18239 , 18240..18269 , 18270..18299 , 18300..18329 , 18330..18359 , 18360..18389 , 18390..18419 , 18420..18449 , 18450..18479 , 18480..18509 , 18510..18539 , 18540..18569 , 18570..18599 , 18600..18629 , 18630..18659 , 18660..18689 , 18690..18719 , 18720..18749 , 18750..18779 , 18780..18809 , 18810..18839 , 18840..18869 , 18870..18899 , 18900..18929 , 18930..18959 , 18960..18989 , 18990..19019 , 19020..19049 , 19050..19079 , 19080..19109 , 19110..19139 , 19140..19169 , 19170..19199 , 19200..19229 , 19230..19259 , 19260..19289 , 19290..19319 , 19320..19349 , 19350..19379 , 19380..19409 , 19410..19439 , 19440..19469 , 19470..19499 , 19500..19529 , 19530..19559 , 19560..19589 , 19590..19619 , 19620..19649 , 19650..19679 , 19680..19709 , 19710..19739 , 19740..19769 , 19770..19799 , 19800..19829 , 19830..19859 , 19860..19889 , 19890..19919 , 19920..19949 , 19950..19979 , 19980..20009 , 20010..20039 , 20040..20069 , 20070..20099 , 20100..20129 , 20130..20159 , 20160..20189 , 20190..20219 , 20220..20249 , 20250..20279 , 20280..20309 , 20310..20339 , 20340..20369 , 20370..20399 , 20400..20429 , 20430..20459 , 20460..20489 , 20490..20519 , 20520..20549 , 20550..20579 , 20580..20609 , 20610..20639 , 20640..20669 , 20670..20699 , 20700..20729 , 20730..20759 , 20760..20789 , 20790..20819 , 20820..20849 , 20850..20879 , 20880..20909 , 20910..20939 , 20940..20969 , 20970..20999 , 21000..21029 , 21030..21059 , 21060..21089 , 21090..21119 , 21120..21149 , 21150..21179 , 21180..21209 , 21210..21239 , 21240..21269 , 21270..21299 , 21300..21329 , 21330..21359 , 21360..21389 , 21390..21419 , 21420..21449 , 21450..21479 , 21480..21509 , 21510..21539 , 21540..21569 , 21570..21599 , 21600..21629 , 21630..21659 , 21660..21689 , 21690..21719 , 21720..21749 , 21750..21779 , 21780..21809 , 21810..21839 , 21840..21869 , 21870..21899 , 21900..21929 , 21930..21959 , 21960..21989 , 21990..22019 , 22020..22049 , 22050..22079 , 22080..22109 , 22110..22139 , 22140..22169 , 22170..22199 , 22200..22229 , 22230..22259 , 22260..22289 , 22290..22319 , 22320..22349 , 22350..22379 , 22380..22409 , 22410..22439 , 22440..22469 , 22470..22499 , 22500..22529 , 22530..22559 , 22560..22589 , 22590..22619 , 22620..22649 , 22650..22679 , 22680..22709 , 22710..22739 , 22740..22769 , 22770..22799 , 22800..22829 , 22830..22859 , 22860..22889 , 22890..22919 , 22920..22949 , 22950..22979 , 22980..23009 , 23010..23039 , 23040..23069 , 23070..23099 , 23100..23129 , 23130..23159 , 23160..23189 , 23190..23219 , 23220..23249 , 23250..23279 , 23280..23309 , 23310..23339 , 23340..23369 , 23370..23399 , 23400..23429 , 23430..23459 , 23460..23489 , 23490..23519 , 23520..23549 , 23550..23579 , 23580..23609 , 23610..23639 , 23640..23669 , 23670..23699 , 23700..23729 , 23730..23759 , 23760..23789 , 23790..23819 , 23820..23849 , 23850..23879 , 23880..23909 , 23910..23939 , 23940..23969 , 23970..23999 , 24000..24029 , 24030..24059 , 24060..24089 , 24090..24119 , 24120..24149 , 24150..24179 , 24180..24209 , 24210..24239 , 24240..24269 , 24270..24299 , 24300..24329 , 24330..24359 , 24360..24389 , 24390..24419 , 24420..24449 , 24450..24479 , 24480..24509 , 24510..24539 , 24540..24569 , 24570..24599 , 24600..24629 , 24630..24659 , 24660..24689 , 24690..24719 , 24720..24749 , 24750..24779 , 24780..24809 , 24810..24839 , 24840..24869 , 24870..24899 , 24900..24929 , 24930..24959 , 24960..24989 , 24990..25019 , 25020..25049 , 25050..25079 , 25080..25109 , 25110..25139 , 25140..25169 , 25170..25199 , 25200..25229 , 25230..25259 , 25260..25289 , 25290..25319 , 25320..25349 , 25350..25379 , 25380..25409 , 25410..25439 , 25440..25469 , 25470..25499 , 25500..25529 , 25530..25559 , 25560..25589 , 25590..25619 , 25620..25649 , 25650..25679 , 25680..25709 , 25710..25739 , 25740..25769 , 25770..25799 , 25800..25829 , 25830..25859 , 25860..25889 , 25890..25919 , 25920..25949 , 25950..25979 , 25980..26009 , 26010..26039 , 26040..26069 , 26070..26099 , 26100..26129 , 26130..26159 , 26160..26189 , 26190..26219 , 26220..26249 , 26250..26279 , 26280..26309 , 26310..26339 , 26340..26369 , 26370..26399 , 26400..26429 , 26430..26459 , 26460..26489 , 26490..26519 , 26520..26549 , 26550..26579 , 26580..26609 , 26610..26639 , 26640..26669 , 26670..26699 , 26700..26729 , 26730..26759 , 26760..26789 , 26790..26819 , 26820..26849 , 26850..26879 , 26880..26909 , 26910..26939 , 26940..26969 , 26970..26999 , 27000..27029 , 27030..27059 , 27060..27089 , 27090..27119 , 27120..27149 , 27150..27179 , 27180..27209 , 27210..27239 , 27240..27269 , 27270..27299 , 27300..27329 , 27330..27359 , 27360..27389 , 27390..27419 , 27420..27449 , 27450..27479 , 27480..27509 , 27510..27539 , 27540..27569 , 27570..27599 , 27600..27629 , 27630..27659 , 27660..27689 , 27690..27719 , 27720..27749 , 27750..27779 , 27780..27809 , 27810..27839 , 27840..27869 , 27870..27899 , 27900..27929 , 27930..27959 , 27960..27989 , 27990..28019 , 28020..28049 , 28050..28079 , 28080..28109 , 28110..28139 , 28140..28169 , 28170..28199 , 28200..28229 , 28230..28259 , 28260..28289 , 28290..28319 , 28320..28349 , 28350..28379 , 28380..28409 , 28410..28439 , 28440..28469 , 28470..28499 , 28500..28529 , 28530..28559 , 28560..28589, >>Next

Distributive-associative-commutative-properties/183092: 36x²+12xy+y² 1 solutions Answer 137482 by jim_thompson5910(28598)   on 2009-02-21 23:40:03 (Show Source): You can put this solution on YOUR website! I assume that you want to factor. Looking at we can see that the first term is and the last term is where the coefficients are 36 and 1 respectively. Now multiply the first coefficient 36 and the last coefficient 1 to get 36. Now what two numbers multiply to 36 and add to the middle coefficient 12? Let's list all of the factors of 36: Factors of 36: 1,2,3,4,6,9,12,18 -1,-2,-3,-4,-6,-9,-12,-18 ...List the negative factors as well. This will allow us to find all possible combinations These factors pair up and multiply to 36 1*36 2*18 3*12 4*9 6*6 (-1)*(-36) (-2)*(-18) (-3)*(-12) (-4)*(-9) (-6)*(-6) note: remember two negative numbers multiplied together make a positive number Now which of these pairs add to 12? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 12 First Number Second Number Sum 1 36 1+36=37 2 18 2+18=20 3 12 3+12=15 4 9 4+9=13 6 6 6+6=12 -1 -36 -1+(-36)=-37 -2 -18 -2+(-18)=-20 -3 -12 -3+(-12)=-15 -4 -9 -4+(-9)=-13 -6 -6 -6+(-6)=-12 From this list we can see that 6 and 6 add up to 12 and multiply to 36 Now looking at the expression , replace with (notice adds up to . So it is equivalent to ) Now let's factor by grouping: Group like terms Factor out the GCF of out of the first group. Factor out the GCF of out of the second group Since we have a common term of , we can combine like terms So factors to So this also means that factors to (since is equivalent to ) note: is equivalent to since the term occurs twice. So also factors to ------------------------------------------------------------ Answer: So factors to

Quadratic_Equations/183093: 2x²-x=15 1 solutions Answer 137481 by jim_thompson5910(28598)   on 2009-02-21 23:37:41 (Show Source): You can put this solution on YOUR website! Start with the given equation. Subtract 15 from both sides. Notice we have a quadratic equation in the form of where , , and Let's use the quadratic formula to solve for x Start with the quadratic formula Plug in , , and Negate to get . Square to get . Multiply to get Rewrite as Add to to get Multiply and to get . Take the square root of to get . or Break up the expression. or Combine like terms. or Simplify. So the answers are or

Quadratic_Equations/183094: x²+3x-10=0 1 solutions Answer 137480 by jim_thompson5910(28598)   on 2009-02-21 23:36:46 (Show Source): You can put this solution on YOUR website! Start with the given equation. Notice we have a quadratic equation in the form of where , , and Let's use the quadratic formula to solve for x Start with the quadratic formula Plug in , , and Square to get . Multiply to get Rewrite as Add to to get Multiply and to get . Take the square root of to get . or Break up the expression. or Combine like terms. or Simplify. So the answers are or

Polynomials-and-rational-expressions/183087: Do the operations and simpify! a+3 ------- 36-a^2 multiplied by 4a-24 ------ 4a+12 a) -1/a-6 b) 1/a+6 c) 1/a-6 d) -1/a+6 1 solutions Answer 137462 by jim_thompson5910(28598)   on 2009-02-21 21:42:48 (Show Source): You can put this solution on YOUR website! Start with the given expression. Factor to get . Factor to get . Factor to get . Combine the fractions. Highlight the common terms. Cancel out the common terms. Simplify. So simplifies to . So the answer is D)

Equations/183086: -7y-8y= -15 I have tried adding 7y to the -7y to cancel that out then add the 7y to -8y which comes out to be -y= -15 so my answer should be 15 but the answer according to the back of my book says it should be 1. What am I doing wrong? 1 solutions Answer 137461 by jim_thompson5910(28598)   on 2009-02-21 21:37:52 (Show Source): You can put this solution on YOUR website! Start with the given equation. Combine like terms on the left side. Note: -7y-8y combines to -15y NOT -y Divide both sides by to isolate . Reduce. ---------------------------------------------------------------------- Answer: So the answer is

Miscellaneous_Word_Problems/183085: This question is from textbook Precalculus Page 132 (problem 95): A length of wire 16 inches is to be cut into two pieces, and then each piece will be bent to form a square. Find the length of the two pieces if the sum of the areas of the two squares is 10 square inches. (x/4)^2 + ((16-x)/4)^2 = 10 I've gotten this far but can't figure out the next steps to solve it. Thank you, Lynn Scott 1 solutions Answer 137460 by jim_thompson5910(28598)   on 2009-02-21 21:34:30 (Show Source): You can put this solution on YOUR website! You're on the right track, you just need to solve the equation: Start with the given equation. Square to get Square to get Combine the fractions. Multiply both sides by 16. Multiply FOIL Subtract 160 from both sides. Combine like terms. Notice we have a quadratic equation in the form of where , , and Let's use the quadratic formula to solve for x Start with the quadratic formula Plug in , , and Negate to get . Square to get . Multiply to get Subtract from to get Multiply and to get . Take the square root of to get . or Break up the expression. or Combine like terms. or Simplify. So the answers are or This means that the second length of the square is either or Note: either way, the two side lengths are 12 and 4 =============================================================== Answer: So the length of the two pieces are 12 and 4 inches.

Probability-and-statistics/183084: use binomal thearom to write the binomial expression (x-2)^3 1 solutions Answer 137457 by jim_thompson5910(28598)   on 2009-02-21 21:06:17 (Show Source): You can put this solution on YOUR website! Start with the given expression To expand this, we're going to use binomial expansion. So let's look at Pascal's triangle: 1    1   1    1   2   1    1   3   3   1    Looking at the row that starts with 1,3, etc, we can see that this row has the numbers: 1, 3, 3, and 1 These numbers will be the coefficients of our expansion. So to expand , simply follow this procedure: Write the first coefficient. Multiply that coefficient with the first binomial term and then the second binomial term . Repeat this until all of the coefficients have been written. Once that has been done, add up the terms like this: Notice how the coefficients are in front of each term. However, we're not done yet. Looking at the first term , raise to the 3rd power and raise to the 0th power. Looking at the second term raise to the 2nd power and raise to the 1st power. Continue this until you reach the final term. Notice how the exponents of are stepping down and the exponents of are stepping up. So the fully expanded expression should now look like this: Distribute the exponents Multiply Multiply the terms with their coefficients So expands and simplifies to . In other words,

Probability-and-statistics/183083: This question is from textbook algbra 2 (x-3y)^6 i need to wirte this as a binomal expansion using pascal triangle 1 solutions Answer 137455 by jim_thompson5910(28598)   on 2009-02-21 20:48:11 (Show Source): You can put this solution on YOUR website! Start with the given expression To expand this, we're going to use binomial expansion. So let's look at Pascal's triangle: 1    1   1    1   2   1    1   3   3   1    1   4   6   4   1    1   5   10   10   5   1    1   6   15   20   15   6   1    Looking at the row that starts with 1,6, etc, we can see that this row has the numbers: 1, 6, 15, 20, 15, 6, and 1 These numbers will be the coefficients of our expansion. So to expand , simply follow this procedure: Write the first coefficient. Multiply that coefficient with the first binomial term and then the second binomial term . Repeat this until all of the coefficients have been written. Once that has been done, add up the terms like this: Notice how the coefficients are in front of each term. However, we're not done yet. Looking at the first term , raise to the 6th power and raise to the 0th power. Looking at the second term raise to the 5th power and raise to the 1st power. Continue this until you reach the final term. Notice how the exponents of are stepping down and the exponents of are stepping up. So the fully expanded expression should now look like this: Distribute the exponents Multiply Multiply the terms with their coefficients So expands and simplifies to . In other words,

Radicals/183076: Show steps: [98]-[50]-[72] Hint: Where the [] are it demonstrates a square sign. 1 solutions Answer 137448 by jim_thompson5910(28598)   on 2009-02-21 19:13:07 (Show Source): You can put this solution on YOUR website! Start with the given expression Simplify to get . Note: If you need help with simplifying square roots, check out this solver. Simplify to get . Simplify to get . Since we have the common term , we can combine like terms Combine like terms. Remember, Now simplify to get So simplifies to . In other words,

Equations/183072: 1. -8(x+4)-(-9x-3)=-8 2. 3(y-2)/5=1-3y 1 solutions Answer 137443 by jim_thompson5910(28598)   on 2009-02-21 18:22:00 (Show Source): You can put this solution on YOUR website! I'll do the first one to get you started # 1 Start with the given equation. Distribute. Combine like terms on the left side. Add to both sides. Combine like terms on the right side. ---------------------------------------------------------------------- Answer: So the answer is

Linear_Algebra/183071: This question is from textbook Contemporary Linear Algebra Let u = (1,-1,3,5) and v = (2,1,0,-3). Find scalars a and b so that au + bv = (1,-4,9,18). I would like to know the steps on how to solve for scalars a and b. 1 solutions Answer 137442 by jim_thompson5910(28598)   on 2009-02-21 18:20:21 (Show Source): You can put this solution on YOUR website! au+bv = (1,-4,9,18) ... Start with the given equation a(1,-1,3,5)+b(2,1,0,-3) = (1,-4,9,18) ... Plug in the given vectors (a,-a,3a,5a)+(2b,b,0,-3b) = (1,-4,9,18) ... Multiply the scalars by EVERY element in the vectors (a+2b,-a+b,3a,5a-3b) = (1,-4,9,18) ... Add the vectors by adding the corresponding components. Now because both vectors on both sides are equal in dimension, this tells us that the left components correspond to the right components. This means that , , , and Which gives us the system of equations: We can use any method to solve this system (substitution, elimination, matrix, etc), but notice how the third equation is composed of only one variable in which we can easily solve for. Let's now solve the third equation Start with the third equation. Divide both sides by 3 to isolate "a". Reduce. ----------------------------------- Go back to the first equation. Plug in Subtract from both sides. Combine like terms on the right side. Divide both sides by to isolate . Reduce. ======================================================== Answer: So the scalars are and In other words, 3(1,-1,3,5)-1*(2,1,0,-3) = (1,-4,9,18)

Linear_Equations_And_Systems_Word_Problems/183069: This question is from textbook Intermediate Algebra HELP!!!! Ive been trying to solve this most irritating word problem and I cant manage to find the 3rd equation. The Problem is .... COFFEE. A coffee manufacturer sells a 10-pound package of coffee that consists of three flavors of coffee, vanilla flavored coffee costs $2/pound,Hazelnut costs $2.50/pound, and french roast cost $3/pound. The package contains the same amount of Hazelnut coffee as French roast coffee. The cost of the 10-pound package is $26. How many pounds of each type of coffee are in the package? I labeled the equation as: x- pounds of Vanilla flavored coffee y- pounds of Hazelnut flavored coffee z-pounds of French Roast flavored coffee x + y + Z = 10 2x + 2.50y + 3z = $26 -Now if I can only manage to get the third variable Id be in good shape, BUT I just dont get it! What am I doing wrong??? 1 solutions Answer 137436 by jim_thompson5910(28598)   on 2009-02-21 17:50:50 (Show Source): You can put this solution on YOUR website! Since the "package contains the same amount of Hazelnut coffee as French roast coffee", this means that is your third equation. Since you have 3 equations in 3 unknowns, you can find a unique solution (if there is one) to the system. Here's the Updated Solution: Start with the first equation. Plug in . In other words, replace each "y" with "z" Combine like terms. Let's call this equation 4. --------------------------------------------------- Move onto the second equation Multiply EVERY term by 10 to make every number a whole number. Plug in Combine like terms. Let's call this equation 5. ----------------------------------------------------------- So we have the system of equations 4 and 5: Let's solve this smaller system by substitution Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to solve for z. So let's isolate z in the first equation Start with the first equation Subtract from both sides Rearrange the equation Divide both sides by Break up the fraction Reduce --------------------- Since , we can now replace each in the second equation with to solve for Plug in into the second equation. In other words, replace each with . Notice we've eliminated the variables. So we now have a simple equation with one unknown. Distribute to Multiply Multiply both sides by the LCM of 2. This will eliminate the fractions. Distribute and multiply the LCM to each side Combine like terms on the left side Subtract 550 from both sides Combine like terms on the right side Divide both sides by -15 to isolate x Divide Since we know that we can plug it into the equation (remember we previously solved for in the first equation). Start with the equation where was previously isolated. Plug in Multiply Combine like terms and reduce. Now because we know that and , this tells us that also ========================= Answer ============================= So the solutions are , and which form the ordered triple (2,4,4) This means that there are 2 lbs of Vanilla flavored coffee, 4 lbs of Hazelnut flavored coffee, and 4 lbs of French Roast flavored coffee

Square-cubic-other-roots/183059: This question is from textbook Algebra 2 Divide using synthetic division: (4x^2-13x+5) divided by (x-3) 1 solutions Answer 137430 by jim_thompson5910(28598)   on 2009-02-21 16:56:14 (Show Source): You can put this solution on YOUR website! Let's simplify this expression using synthetic division Start with the given expression First lets find our test zero: Set the denominator equal to zero Solve for x. so our test zero is 3 Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero. 3 | 4 -13 5 | Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 4) 3 | 4 -13 5 | 4 Multiply 3 by 4 and place the product (which is 12) right underneath the second coefficient (which is -13) 3 | 4 -13 5 | 12 4 Add 12 and -13 to get -1. Place the sum right underneath 12. 3 | 4 -13 5 | 12 4 -1 Multiply 3 by -1 and place the product (which is -3) right underneath the third coefficient (which is 5) 3 | 4 -13 5 | 12 -3 4 -1 Add -3 and 5 to get 2. Place the sum right underneath -3. 3 | 4 -13 5 | 12 -3 4 -1 2 Since the last column adds to 2, we have a remainder of 2. This means is not a factor of Now lets look at the bottom row of coefficients: The first 2 coefficients (4,-1) form the quotient and the last coefficient 2, is the remainder, which is placed over like this Putting this altogether, we get: So which looks like this in remainder form: remainder 2

Polynomials-and-rational-expressions/183052: please help me with this problem find the quotient 6y6-9y4+12y2 3y2 1 solutions Answer 137426 by jim_thompson5910(28598)   on 2009-02-21 16:39:30 (Show Source): You can put this solution on YOUR website! Start with the given expression. Factor out the GCF Highlight the common terms. Cancel out the common terms. Simplify ========================================= Answer: So simplifies to In other words, where

Square-cubic-other-roots/183050: This question is from textbook Algebra 2 evaluate 5^2 over 5^5 1 solutions Answer 137425 by jim_thompson5910(28598)   on 2009-02-21 16:36:17 (Show Source): You can put this solution on YOUR website! Here's the long way to do it: Start with the given expression. Expand to get Expand to get (note: there are five 5's) Highlight the common terms. Cancel out the common terms. Simplify Multiply to get So ==================================================== Here's the short way to do it: Start with the given expression. Divide the terms by subtracting the exponents. Subtract Flip the base to make the exponent positive. Expand to get Multiply to get So

Square-cubic-other-roots/183049: This question is from textbook Algebra 2 write (8.4x10^3)(6.5x10^5) in scientific notation 1 solutions Answer 137423 by jim_thompson5910(28598)   on 2009-02-21 16:30:09 (Show Source): You can put this solution on YOUR website! Step 1) Multiply the coefficients 8.4 and 6.5 to get Step 2) Convert the previous result to scientific notation (if it is not already). So So this means that the expression now becomes Step 3) Now multiply out by adding the exponents: Now putting everything together, we get the result ===================================================== Answer: So

Average/183041: This question is from textbook Please help. Question from text. On three examinations, you have grades of 88,78, and 86. There is still a final examination, which counts as one grade. In order to get an A, your average must be at least 90. If you get 100 on the final, compute your average and dertermine if an A int he course is possible. Thank you.... 1 solutions Answer 137420 by jim_thompson5910(28598)   on 2009-02-21 16:25:29 (Show Source): You can put this solution on YOUR website! Remember, the average of "x" scores is simply the sum of those scores over "x". In this case, we have the scores: 88,78,86, and 100 So add them up and divide them by 4 (the number of scores present) So if you receive a perfect score on the final exam, you will get an 88. Since , this means that it is NOT possible to get an A.

Linear-systems/183042: Solve the equations for x and y or state that there is no solution. x + y = 4 2x + 3y = -7 1 solutions Answer 137419 by jim_thompson5910(28598)   on 2009-02-21 16:21:24 (Show Source): You can put this solution on YOUR website! Start with the given system of equations: Multiply the both sides of the first equation by -2. Distribute and multiply. So we have the new system of equations: Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this: Group like terms. Combine like terms. Simplify. ------------------------------------------------------------------ Now go back to the first equation. Plug in . Multiply. Subtract from both sides. Combine like terms on the right side. Divide both sides by to isolate . Reduce. So our answer is and . Which form the ordered pair . This means that the system is consistent and independent.

Equations/183045: Solve the equations for x and y or state that there is no solution. 5x - 4y = 2 4x - 5y = -1 1 solutions Answer 137418 by jim_thompson5910(28598)   on 2009-02-21 16:19:49 (Show Source): You can put this solution on YOUR website! Start with the given system of equations: Multiply the both sides of the first equation by 4. Distribute and multiply. Multiply the both sides of the second equation by -5. Distribute and multiply. So we have the new system of equations: Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this: Group like terms. Combine like terms. Simplify. Divide both sides by to isolate . ------------------------------------------------------------------ Now go back to the first equation. Plug in . Multiply. Multiply EVERY term by the LCD to clear any fractions. Distribute and multiply. Add to both sides. Combine like terms on the right side. Divide both sides by to isolate . Reduce. So the solutions are and . Which form the ordered pair . This means that the system is consistent and independent.

Graphs/183046: Solve the following by either elimination or substitution method: {x = 9/4y + 7 {4x - 9y = 30 Is the system inconsistent, are the equations dependent, or is the solution an ordered pair? 1 solutions Answer 137416 by jim_thompson5910(28598)   on 2009-02-21 16:16:39 (Show Source): You can put this solution on YOUR website! I'm going to use the substitution method (as "x" is already isolated) Start with the second equation. Plug in Distribute Multiply and reduce Combine like terms on the left side. Subtract from both sides. Combine like terms on the right side. Simplify. Since this equation is NEVER true for any "y" value, this means that there are no solutions. So the system is inconsistent.

Equations/183047: Solve the following equations for x and y or state that there is no solution. 5x – 3y = 5 x/2 + y/10 = 3 1 solutions Answer 137415 by jim_thompson5910(28598)   on 2009-02-21 16:13:30 (Show Source): You can put this solution on YOUR website! Start with the second equation. Multiply EVERY term by the LCD to clear any fractions. Distribute and multiply. So we have the system of equations: Multiply the both sides of the second equation by -1. Distribute and multiply. So we have the new system of equations: Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this: Group like terms. Combine like terms. Simplify. Divide both sides by to isolate . Reduce. ------------------------------------------------------------------ Now go back to the first equation. Plug in . Multiply. Multiply EVERY term by the LCD to clear any fractions. Multiply. Add 75 to both sides. Combine like terms. Divide both sides by to isolate . Reduce. So the solutions are and Which form the ordered pair . This means that the system is consistent and independent.

Linear-equations/183043: Find the equation, in standard form, of the line perpendicular to 2x - 3y = -5 and passing through (3,-2). Write the equation in standard form, with all integer coefficients. 1 solutions Answer 137414 by jim_thompson5910(28598)   on 2009-02-21 16:06:03 (Show Source): You can put this solution on YOUR website! Start with the given equation. Subtract from both sides. Rearrange the terms. Divide both sides by to isolate y. Break up the fraction. Reduce. We can see that the equation has a slope and a y-intercept . Now to find the slope of the perpendicular line, simply flip the slope to get . Now change the sign to get . So the perpendicular slope is . Now let's use the point slope formula to find the equation of the perpendicular line by plugging in the slope and the coordinates of the given point . Start with the point slope formula Plug in , , and Rewrite as Multiply both sides by 2. Distribute Add 3x to both sides. Subtract 4 from both sides. Combine like terms. =============================================== Answer: So the equation of the line perpendicular to that goes through the point in standard form is . So the answer you're looking for is: Here's a graph to visually verify our answer: Graph of the original equation (red) and the perpendicular line (green) through the point .

Equations/183044: Solve the equations for x and y or state that there is no solution. 3x + 7y = 5 2x - y = 1 1 solutions Answer 137413 by jim_thompson5910(28598)   on 2009-02-21 16:00:01 (Show Source): You can put this solution on YOUR website! Start with the given system of equations: Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to solve for y. So let's isolate y in the second equation Start with the second equation Subtract from both sides Rearrange the equation Divide both sides by Break up the fraction Reduce --------------------- Since , we can now replace each in the first equation with to solve for Plug in into the first equation. In other words, replace each with . Notice we've eliminated the variables. So we now have a simple equation with one unknown. Distribute to Multiply Combine like terms on the left side Add 7 to both sides Combine like terms on the right side Divide both sides by 17 to isolate x -----------------First Answer------------------------------ So the first part of our answer is: Since we know that we can plug it into the equation (remember we previously solved for in the first equation). Start with the equation where was previously isolated. Plug in Multiply Combine like terms (note: if you need help with fractions, check out this solver) -----------------Second Answer------------------------------ So the second part of our answer is: -----------------Summary------------------------------ So our answers are: and which form the ordered pair

Proportions/183038: karen wants to bake a cake. The recipe calls for 2 cups of flour. karen only has 28 tablespoons of flour. What percent of a cake can she bake? Ok so ive figured out that 1 cup has 16 tablespoons so 2 cups is 32 tablespoons, but since she only has 28, thats where i get confused because i dont know how to find the percentage of the cake she could bake. please help me 1 solutions Answer 137412 by jim_thompson5910(28598)   on 2009-02-21 15:56:58 (Show Source): You can put this solution on YOUR website! You're on the right track. You just need to know the percentage of 28 out of 32. We can find the percentage using the following proportion: where the "part" is smaller than the "whole" In this case, the "part" is 28 and the "whole" is 32. Let the "percent" be "x" to get Divide 28 by 32 to get Multiply both sides by 100 to isolate "x". Multiply 0.875 and 100 to get 87.5 So the answer is which tells us that the percent of the cake that she can bake is 87.5%

Linear-systems/183032: HELP, I think the answer is no solution. Is that right, if not please show my how to do it. Use elimination to solve each system of equations. X=4Y+8 2X-8Y=-3 1 solutions Answer 137409 by jim_thompson5910(28598)   on 2009-02-21 15:50:06 (Show Source): You can put this solution on YOUR website! Start with the first equation. Subtract from both sides. So we have the system of equations: Multiply the both sides of the first equation by -2. Distribute and multiply. So we have the new system of equations: Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this: Group like terms. Combine like terms. Simplify. Since is NEVER true, this means that there are no solutions. So the system is inconsistent.

Linear-systems/183033: HELP, I think the answer is x=7 and y=7 but how do you set this up. Use elimination to solve each system of equations. 4X-2Y=14 Y=X 1 solutions Answer 137407 by jim_thompson5910(28598)   on 2009-02-21 15:47:23 (Show Source): You can put this solution on YOUR website! Start with the second equation. Subtract "x" from both sides (the goal is to get all the variable terms to the left side). So we have the system of equations: Multiply the both sides of the second equation by 2. Distribute and multiply. So we have the new system of equations: Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this: Group like terms. Combine like terms. Simplify. Divide both sides by to isolate . Reduce. ------------------------------------------------------------------ Now go back to the first equation. Plug in . Multiply. Subtract from both sides. Combine like terms on the right side. Divide both sides by to isolate . Reduce. So our answer is and (so you are correct). Which form the ordered pair . This means that the system is consistent and independent. Notice when we graph the equations, we see that they intersect at . So this visually verifies our answer. Graph of (red) and (green)

Linear-systems/183031: HELP. Use elimination to solve each system of equations. 2X+2/3Y =4 X-1/2Y=7 1 solutions Answer 137403 by jim_thompson5910(28598)   on 2009-02-21 15:22:57 (Show Source): You can put this solution on YOUR website! Start with the first equation. Multiply EVERY term by the LCD to clear any fractions. Distribute and multiply. ----------------------------------------------- Move onto the second equation. Multiply EVERY term by the LCD to clear any fractions. Distribute and multiply. So we have the system of equations: Multiply the both sides of the second equation by 2. Distribute and multiply. So we have the new system of equations: Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this: Group like terms. Combine like terms. Simplify. Divide both sides by to isolate . Reduce. ------------------------------------------------------------------ Now go back to the first equation. Plug in . Multiply. Subtract from both sides. Combine like terms on the right side. Divide both sides by to isolate . Reduce. So our answer is and . Which form the ordered pair . This means that the system is consistent and independent. Notice when we graph the equations, we see that they intersect at . So this visually verifies our answer. Graph of (red) and (green)

Polynomials-and-rational-expressions/183030: This question is from textbook College Algebra Help with Find all rational zeros of the polynomial (using synthetic division) 22. p(x)= What I did: multiples of 1= +-1 multiples of 4= +-1 +-4 +-2 possible zeros: +-1 +-4 +-2 (after dividing constant over leading coeficient) I used synthetic division to test which are zeros. +1 worked, +4 did not, +2 using the quotient of +1 (1 -1 -4 4) did not work gave me remainder of 8, but when I used the original coeficients (1 -2 -3 8 -4)it gave me a zero. I was told by my professor that either way it should work. Any information is much appreciated, have at test coming up. Thank you 1 solutions Answer 137401 by jim_thompson5910(28598)   on 2009-02-21 15:14:20 (Show Source): You can put this solution on YOUR website! Any rational zero can be found through this equation where p and q are the factors of the last and first coefficients So let's list the factors of -4 (the last coefficient): Now let's list the factors of 1 (the first coefficient): Now let's divide each factor of the last coefficient by each factor of the first coefficient Now simplify These are all the distinct rational zeros of the function that could occur (ie some of these values are NOT zeros, but could be) =========================================================================== Let's see if the possible zero is really a root for the function So let's make the synthetic division table for the function given the possible zero : 1 | 1 -2 -3 8 -4 | 1 -1 -4 4 1 -1 -4 4 0 Since the remainder (the right most entry in the last row) is equal to zero, this means that is a zero of ------------------------------------------------------ Let's see if the possible zero is really a root for the function So let's make the synthetic division table for the function given the possible zero : 2 | 1 -2 -3 8 -4 | 2 0 -6 4 1 0 -3 2 0 Since the remainder (the right most entry in the last row) is equal to zero, this means that is a zero of ------------------------------------------------------ Let's see if the possible zero is really a root for the function So let's make the synthetic division table for the function given the possible zero : 4 | 1 -2 -3 8 -4 | 4 8 20 112 1 2 5 28 108 Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of ------------------------------------------------------ Let's see if the possible zero is really a root for the function So let's make the synthetic division table for the function given the possible zero : -1 | 1 -2 -3 8 -4 | -1 3 0 -8 1 -3 0 8 -12 Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of ------------------------------------------------------ Let's see if the possible zero is really a root for the function So let's make the synthetic division table for the function given the possible zero : -2 | 1 -2 -3 8 -4 | -2 8 -10 4 1 -4 5 -2 0 Since the remainder (the right most entry in the last row) is equal to zero, this means that is a zero of ------------------------------------------------------ Let's see if the possible zero is really a root for the function So let's make the synthetic division table for the function given the possible zero : -4 | 1 -2 -3 8 -4 | -4 24 -84 304 1 -6 21 -76 300 Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of ====================================================================== Answer: So the rational zeros of are: 1,2,-2 In other words, if we plug in , or into , we'll get 0 as a result (try it out if you aren't sure) Note: the zero 1 has a multiplicity of 2 (ie it is counted twice)

Linear-equations/183029: This question is from textbook Introductory Algebra Graph the equation and identify the y-intercept y=x+3 Thank you for your time. 1 solutions Answer 137398 by jim_thompson5910(28598)   on 2009-02-21 14:59:52 (Show Source): You can put this solution on YOUR website! Looking at we can see that the equation is in slope-intercept form where the slope is and the y-intercept is Since this tells us that the y-intercept is .Remember the y-intercept is the point where the graph intersects with the y-axis So we have one point Now since the slope is comprised of the "rise" over the "run" this means Also, because the slope is , this means: which shows us that the rise is 1 and the run is 1. This means that to go from point to point, we can go up 1 and over 1 So starting at , go up 1 unit and to the right 1 unit to get to the next point Now draw a line through these points to graph So this is the graph of through the points and

Rational-functions/183027: I need to find the domain of this given function. f(x)= |4x+6| 1 solutions Answer 137396 by jim_thompson5910(28598)   on 2009-02-21 14:47:31 (Show Source): You can put this solution on YOUR website! Since you can plug in ANY number in for "x", this means that the domain is all real numbers. Note: there are no restrictions like dividing by zero or taking the square root of a negative number to worry about here. So the domain of the function in set-builder notation is: In plain English, this reads: x is the set of all real numbers (In other words, x can be any number) Also, in interval notation, the domain is: Note: if you graph , you'll see that the graph extends infinitely in both directions along the x-axis. So this shows you that any "x" value will work.

Radicals/183026: Multiply and show steps: (5[2]+2[3])([2]-2[3]) Hint: where there is a [] it is also a square root sign as well. I am limited to what I am demonstrate! Thanks 1 solutions Answer 137395 by jim_thompson5910(28598)   on 2009-02-21 14:36:17 (Show Source): You can put this solution on YOUR website! Start with the given expression. Now let's FOIL the expression. Remember, when you FOIL an expression, you follow this procedure: Multiply the First terms:. Multiply the Outer terms:. Multiply the Inner terms:. Multiply the Last terms:. --------------------------------------------------- So we have the terms: , , , and Now collect every term listed above to make a single expression. Now combine like terms. So FOILs to . In other words, .