Algebra ->  Tutoring on algebra.com -> See tutors' answers!      Log On

 Tutoring Home For Students Tools for Tutors Our Tutors Register Recently Solved
 By Tutor
| By Problem Number |

Tutor:

# Recent problems solved by 'ankor@dixie-net.com'

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, >>Next

 Square-cubic-other-roots/145355: This question is from textbook Beginning Algebra I have to simplify and assume that all variables represent positive numbers...Can you help? Page 526 Number 50 square root of 18x^2y^31 solutions Answer 106046 by ankor@dixie-net.com(15622)   on 2008-06-15 10:12:44 (Show Source): You can put this solution on YOUR website!Simplify: : Factor inside the radical to reveal the perfect squares : Extract the square roots of these perfect squares: or
 Unit_Conversion_Word_Problems/145331: You have two empty containers that water can be placed in. One is 3 gallons in capacity while the other is 5 gallons in capacity. You are standing next to the ocean. Your task is to determine how to measure 4 gallons of water. No estimation!! You can fill, dump, refill, and dump as many times as you want. 1 solutions Answer 106008 by ankor@dixie-net.com(15622)   on 2008-06-14 21:25:36 (Show Source): You can put this solution on YOUR website!You have two empty containers that water can be placed in. One is 3 gallons in capacity while the other is 5 gallons in capacity. You are standing next to the ocean. Your task is to determine how to measure 4 gallons of water. No estimation!! You can fill, dump, refill, and dump as many times as you want. : Fill the 5 gallon container Dump 3 gal into the 3 gallon container Empty the 3 gal container Pour the remaining 2 gal from the 5 gal container to the 3 gallon container Fill the 5 gallon container Pour 1 gal from the 5 gallon container to fill the 3 gallon container 4 gal remains in the 5 gallon container
 logarithm/145281: Graph one exponential equation and one logarithmic equation. Attach your graphs for our review. 1 solutions Answer 106003 by ankor@dixie-net.com(15622)   on 2008-06-14 21:15:56 (Show Source): You can put this solution on YOUR website!Graph one exponential equation and one logarithmic equation. Attach your graphs for our review. : plot these two equations: find y for x=0 to x=4 y = 2^x y = ln(x) :
 Polynomials-and-rational-expressions/145319: Identify the variable and write an equation that describes gwyn sold her car on consignment for x dollars. The salespersons comission was 10% of the selling price and gwen received 6570 1 solutions Answer 106001 by ankor@dixie-net.com(15622)   on 2008-06-14 21:00:41 (Show Source): You can put this solution on YOUR website!Identify the variable and write an equation that describes Gwyn sold her car on consignment for x dollars. The salespersons commission was 10% of the selling price and Gwen received 6570. : Price - commission = \$6570 x - .1x = 6570 : .9x = 6570 x = x = \$7,300
 Numbers_Word_Problems/145327: I am having a hard time breaking down story problems, I can solve once I have the equation. Here is my ? Charlie and Doggie are squirrels that live in the trees in my mother's yard. Before snow covered the fall crop of acorns, Doogie gathered ten pounds more that two-thirds as many acorns as Charlie. Now, since Doogie played in the road and took himself out of the gene pool, Charlie has all thirty-five pounds of acorns they gathered to himself. How may pounds did each of them gather? Thank You! Beth1 solutions Answer 106000 by ankor@dixie-net.com(15622)   on 2008-06-14 20:45:24 (Show Source): You can put this solution on YOUR website!Charlie and Doggie are squirrels that live in the trees in my mother's yard. Before snow covered the fall crop of acorns, : Write an equation for the statements: : " Doogie gathered ten pounds more than two-thirds as many acorns as Charlie." D = C + 10 : " Charlie has all thirty-five pounds of acorns they gathered to himself." C + D = 35 D = (35-C) : How may pounds did each of them gather? Substitute (35-C) for D in the 1st equation, find C ; Do you think you can handle it now?
 expressions/145324: Western State College is 18 years older than Southern State. Western is also 2 1/2 times as old as Southern State. How old is eac?1 solutions Answer 105999 by ankor@dixie-net.com(15622)   on 2008-06-14 20:10:43 (Show Source): You can put this solution on YOUR website!Write an equation for each statement: : "Western State College is 18 years older than Southern State." w = s + 18 : "Western is also 2 1/2 times as old as Southern State." w = 2.5s : How old is each? : Substitute (s+18) for w in the 2nd equation, find s: s + 18 = 2.5s : 18 = 2.5s - s : 18 = 1.5s s = s = 12 yr is S.States age then 2.5(12) = 30 yrs is W's age
 Linear-equations/145264: This question is from textbook Elementary and Intermediate Algebra The function S = 500 + 200X and D = 9500 - 100X express the supply S and the demand D, respectively, for a popular compact disc brand as a function of its price x (in dollars. A) Graph the functions on the same coordinate system. 1 solutions Answer 105985 by ankor@dixie-net.com(15622)   on 2008-06-14 18:13:48 (Show Source): You can put this solution on YOUR website!he function S = 500 + 200X and D = 9500 - 100X express the supply S and the demand D, respectively, for a popular compact disc brand as a function of its price x (in dollars. A) Graph the functions on the same coordinate system. Plot both equations; you can use x=10, x=50; find y for each as follows: : y = 500+200x (Purple) x | y ------- 10 | 2500 50 | 10500 : y = 9500-100x (Green) x | y ------- 10 |8500 50 |4500 : : you can see that D = S at x = 30, y = 6500
 Linear_Equations_And_Systems_Word_Problems/145304: an alloy of tin and copper contains 14 pounds of tin and 26 pounds of copper. a second alloy of tin and copper contains 8 pounds of tin and 24 pounds of copper. How many pounds of each alloy must be taken to form a third alloy containing 15 pounds of tin and 35 pounds of copper.1 solutions Answer 105975 by ankor@dixie-net.com(15622)   on 2008-06-14 15:17:03 (Show Source): You can put this solution on YOUR website!n alloy of tin and copper contains 14 pounds of tin and 26 pounds of copper. a second alloy of tin and copper contains 8 pounds of tin and 24 pounds of copper. How many pounds of each alloy must be taken to form a third alloy containing 15 pounds of tin and 35 pounds of copper. : Solve this using the decimal value of %tin: : Find the per cent tin in the 1st alloy: = .35 : Find the per cent tin in the 2nd alloy = .25 : Find the per cent tin in the resulting alloy = .30 : Let x = amt of the 1st alloy required : The resulting total amt is (15 + 35) = 50 lb, therefore: (50-x) = amt of the 2nd equation : The per cent tin equation: .35x + .25(50-x) = .3(50) : .35x + 12.5 - .25x = 15 : .35x - .25x = 15 - 12.5 : .10x = 2.5 x = x = 25 lb of alloy #1 and 50 - 25 = 25 lb of alloy #2 : : Check solution using %tin equation .35(25) + .25(25) = .30(50) 8.75 + 6.25 =15
 Money_Word_Problems/145297: Tickets for a train ride were \$120 for a sleeping room, \$80 dollars for a berth, and \$50 for a coach seat. The total ticket sales were \$8600. If there were 20 more berth tickets than sleeping room tickets and 3 times as many coach tickets as sleeping room tickets, how many of each type of ticket were sold?1 solutions Answer 105972 by ankor@dixie-net.com(15622)   on 2008-06-14 14:23:16 (Show Source): You can put this solution on YOUR website!Let x = no. of rooms Let y = no. of berths Let z = no. of seats : Write an equation for each statement: : "Tickets for a train ride were \$120 for a sleeping room, \$80 dollars for a berth, and \$50 for a coach seat. The total ticket sales were \$8600." 120x + 80y + 50z = 8600 : "there were 20 more berth tickets than sleeping room tickets" y = x + 20 : " 3 times as many coach tickets as sleeping room tickets," z = 3x : how many of each type of ticket were sold? : Substitute (x+20) for y and 3x for z in the 1st equation: 120x + 80(x+20) + 50(3x) = 8600 : 120x + 80x + 1600 + 150x = 8600; multiplied whats in the brackets : 120x + 80x + 150x = 8600 - 1600; subtracted 1600 from both sides : 350x = 7000 x = x = 20 sleeping room tickets : I'll let you find y and z using the 2nd and 3rd equations : Check your solutions in the \$total equation
 Miscellaneous_Word_Problems/145296: Allen bought 56 stamps at the post office in 37 cent and 20 cent denominations. If the total cost of the stamps was \$18, how many of each denomination did Allen buy?1 solutions Answer 105966 by ankor@dixie-net.com(15622)   on 2008-06-14 14:09:31 (Show Source): You can put this solution on YOUR website!Allen bought 56 stamps at the post office in 37 cent and 20 cent denominations. If the total cost of the stamps was \$18, how many of each denomination did Allen buy? : Let x = no. of 37 cent stamps Let y = no. of 20 cent stamps : Total no. of stamps given as 56, therefore: x + y = 56 or y = (56-x) : Total cost given as \$18, therefore: .37x+ 2.0y = 18 : Substitute (56-x) for y and find x: .37x + .20(56-x) = 18 : .37x + 11.2 - .20x = 18 : .37x - .20x = 18 - 11.2 : .17x = 6.8 x = x = 40 ea 37 cent stamps and 56 - 40 = 16 ea 20 cent stamps : : Check solution: .37(40) + .20(16) = 18
 Equations/145200: BENNET CUSTOM FLOORING HAS 0.5 GAL OF STAIN THAT IS 20% BROWN AND 80% NEUTRAL. A CUSTOMER ORDERS 1.5 GAL OF A STAIN THAT IS 60% BRONW AND 40% NEWUTRAL. HOW MUCH PURE BROWN STAIN AND HOW MUCH NEUTRAL STAIN SHOULD BE ADDED TO THE ORIGINAL 0.5 GAL IN ORDER TO MAKE UP THE ORDER?1 solutions Answer 105873 by ankor@dixie-net.com(15622)   on 2008-06-13 13:44:58 (Show Source): You can put this solution on YOUR website!BENNET CUSTOM FLOORING HAS 0.5 GAL OF STAIN THAT IS 20% BROWN AND 80% NEUTRAL. A CUSTOMER ORDERS 1.5 GAL OF A STAIN THAT IS 60% BROWN AND 40% NEUTRAL. HOW MUCH PURE BROWN STAIN AND HOW MUCH NEUTRAL STAIN SHOULD BE ADDED TO THE ORIGINAL 0.5 GAL IN ORDER TO MAKE UP THE ORDER? : Let x = amt of Pure Brown stain required to accomplish this : It says the total has to be 1.5 gal, therefore: (1.5-x) = amt of neutral : Write an equation based on the % of Brown stain .2(.5) + 1.0(x) = .6(1.5) : .10 + x = .9 ; x = .9 - .1 ; x = .8 gal of pure brown required Therefore: 1.5 - .8 = .7 gal of neutral stain required : : Check solution in original equation: .2(.5) + 1(.8) = .1 + .8 = .6(1.5)
 Linear-systems/129871: Please help me solove this problem: 2X-1/3y=8 3X-1/2y=6 I'm not sure how to start this, do I need to get a common multiple to remove the fraction or can I use the substitution method? Or am I on the right tract if I do the following 3(2x-1/3y=8) 2(3x-1/2y=6) 6x-y=24 6x-y=12 -(6x-y=24) 6X-y=12 1 solutions Answer 105848 by ankor@dixie-net.com(15622)   on 2008-06-13 07:53:03 (Show Source): You can put this solution on YOUR website!You are on the right track
 Linear_Equations_And_Systems_Word_Problems/145147: 24 is divided into two parts such that 7 times the first part added to 5 times the second part makes 146.find each part1 solutions Answer 105836 by ankor@dixie-net.com(15622)   on 2008-06-12 21:37:37 (Show Source): You can put this solution on YOUR website!24 is divided into two parts such that 7 times the first part added to 5 times the second part makes 146.find each part : Let x = 1 part which makes up 24 Then (24-x) = 2nd part which makes up 24 : Write an equation for the statement: " 7 times the first part added to 5 times the second part makes 146." 7x + 5(24-x) = 146 : 7x + 120 - 5x = 146 : 7x - 5x = 146 - 120 : 2x = 26 x = x = 13, one part of 24 then 24 - 13 = 11; the other part of 24 : Check solution in the statement: "7 times the first part added to 5 times the second part makes 146." 7(13) + 5(11) = 146
 Linear_Equations_And_Systems_Word_Problems/145149: if 10 be added to 4 times a certain number , the result is 5 less than five times the no.. find the no.1 solutions Answer 105835 by ankor@dixie-net.com(15622)   on 2008-06-12 21:28:52 (Show Source): You can put this solution on YOUR website!Let x = "a certain number" : Write an equation for what it says. "10 be added to 4 times a certain number, the result is 5 less than five times the no." 4x + 10 = 5x - 5 : 10 + 5 = 5x - 4x : 15 = x ; : Check solution in the given statement 10 + 4(15) = -5 + 5(15)
 Quadratic_Equations/144909: I need some help with this word problem. Any help would be greatly appreciated. Thanks. A photo is 3 inches longer that it is wide. A 2-inch border is placed around the photo making the total area of the photo and border 108in2. What are the dimensions of the photo?1 solutions Answer 105597 by ankor@dixie-net.com(15622)   on 2008-06-10 21:48:41 (Show Source): You can put this solution on YOUR website!A photo is 3 inches longer that it is wide. A 2-inch border is placed around the photo making the total area of the photo and border 108in2. What are the dimensions of the photo? : Let x = the width of the photo : It says,"A photo is 3 inches longer that it is wide. ", therefore: (x+3) = the length of the photo : The dimensions of the whole thing: (x+3)+ 4, x + 4, which is (x+7) by (x+4) : Given the area of the whole thing: (x+7)(x+4) = 108 FOIL x^2 + 11x + 28 - 108 = 0 : x^2 + 11x - 80 = 0 Factor (x+16)(x-5) = 0 The positive solution: x = 5" is the width of the photo and 5 + 3 = 8" is the length of the photo : : Check solution) (8+4) * (5+4) = 108
 Linear-systems/144881: This question is from textbook prentice hall algebra 1 Find the equation of the line containing the point (4,1) and having a slope of -1/3. How do I solve this?1 solutions Answer 105596 by ankor@dixie-net.com(15622)   on 2008-06-10 21:37:38 (Show Source): You can put this solution on YOUR website!Find the equation of the line containing the point (4,1) and having a slope of -1/3. : Use the point/slope formula: y - y1 = m(x - x1) : In this problem x1=4; y1=1 and m= : y - 1 = (x - 4) : y - 1 = x + : y = x + + 1 : y = x + + : y = x +
 Quadratic_Equations/144782: Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle. 1 solutions Answer 105554 by ankor@dixie-net.com(15622)   on 2008-06-10 14:58:46 (Show Source): You can put this solution on YOUR website!Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle : Let s = speed of his vehicle then (s+10) = the faster speed : Write a time equation: Time = : Actual time - 1 hr = Faster speed time - 1 = : Multiply equation by s(s+10) s(s+10)* - s(s+10)(1) = s(s+10)*: Cancel the denominators: : 200(s+10) - s(s+10) = 200s : 200s + 2000 - s^2 - 10s = 200s : 0 = s^2 + 10s + 200s - 200s - 2000 : A quadratic equation s^2 + 10s - 2000 = 0 : Factors to: (s+50)(s-40) = 0 : Positive solution s = 40 mph : : Check solution by finding the times: 200/40 = 5 hrs 200/50 = 4 hrs
 Travel_Word_Problems/144790: A freight train with 75 cars is travelling 90km/h in a tunnel 150m long. How long will it take the train to travel completely through the tunnel if each car is 15m long?1 solutions Answer 105536 by ankor@dixie-net.com(15622)   on 2008-06-10 14:03:05 (Show Source): You can put this solution on YOUR website!A freight train with 75 cars is traveling 90km/h in a tunnel 150m long. How long will it take the train to travel completely through the tunnel if each car is 15m long? : Find the length of the train. 75 * 15 = 1125 meters, however, it says, "75 cars" so it may not include the engine which, for example, a 3100 hp locomotive is about 21 meters long, anyway ignoring that, and assuming the train is 1125 meters long> : The total distance from the end of the tunnel to the back of the train as the train just enters the tunnel: 1125 + 150 = 1275 meters = 1.275 km : Time = = .014167 * 60 = .85 min or about .85*60 = 51 seconds
 Miscellaneous_Word_Problems/144688: Solve the following simultaneous equation 32x^2+15y^2=2112 7x^2-3y^2=60 1 solutions Answer 105437 by ankor@dixie-net.com(15622)   on 2008-06-09 14:18:12 (Show Source): You can put this solution on YOUR website!Solve the following simultaneous equation 32x^2+15y^2 = 2112 7x^2 - 3y^2 = 60 : Multiply the 2nd equation by 5 and add to the 1st equation Solve the following simultaneous equation 32x^2 + 15y^2 = 2112 35x^2 - 15y^2 = 300 ----------------------adding eliminates y 67x^2 = 2412 divide both sides by67 x^2 = 36 x = +/- x = +/-6 : Find y using the 2nd original equation: 7(36) - 3y^2 = 60 252 - 3y^2 = 60 -3y^2 = 60 - 252 -3y^2 = -192 3y^2 = +192 y^2 = y^2 = 64 y = +/-8 ; : Check solutions in 1st original equation : 32(6^2) + 15(8^2) = 32(36) + 15(64) = 1152 + 960 = 2112
 Linear_Equations_And_Systems_Word_Problems/144667: This question is from textbook Prealgebra & Introductory Algebra The sum of the legs of a right triangle is 17 in. The longer leg is 2 more than twice the shorter. Th hypotenuse is 13 in. Find the length of each leg. Isn't the formula for a right triangle a^2 +b^2=c^2? From what I understand this is the way I would write to solve, but it doesn't make sense to me. 17=a^2+b^2+2 a^2+b^2= 13^2 Thanks for your help.1 solutions Answer 105434 by ankor@dixie-net.com(15622)   on 2008-06-09 13:35:45 (Show Source): You can put this solution on YOUR website!The sum of the legs of a right triangle is 17 in. The longer leg is 2 more than twice the shorter. The hypotenuse is 13 in. Find the length of each leg. : Isn't the formula for a right triangle a^2 +b^2 = c^2? Yes it is, but we don't need it here. : It says,"The hypotenuse is 13 in." therefore c = 13 : Let a = shorter leg : then it says,"The longer leg is 2 more than twice the shorter."therefore b = 2a+2 : "the sum of the legs of a right triangle is 17": a + b = 17 : Substitute (2a+2) for b and 13 for c, find a: a + (2a+2) = 17 : 3a + 2 = 17 : 3a = 17 - 2 : 3a = 15 a = a = 5 : Find the longer leg (b): b = 2(5) + 2 b = 12 : We can confirm our solution using the pythag Does 5^2 + 12^2 = 13^2?
 Polynomials-and-rational-expressions/144622: Some students planned for a get-together. The budget for food was \$500. Five of the students failed to come because of the distance and therefore the cost of food for each member increased by \$5. How many students attended the get-together? 1 solutions Answer 105390 by ankor@dixie-net.com(15622)   on 2008-06-08 21:40:13 (Show Source): You can put this solution on YOUR website!Some students planned for a get-together. The budget for food was \$500. Five of the students failed to come because of the distance and therefore the cost of food for each member increased by \$5. How many students attended the get-together? : Let x = no. of students that attended then (x+5) = no. of planned students : Planned cost per student = Actual cost per student = : Actual cost = planned cost + \$5 = + 5 Multiply equation by x(x+5), resulting in: 500(x+5) = 500x + 5x(x+5) : 500x + 2500 = 500x + 5x^2 + 25x : Arrange as a quadratic equation: 5x^2 + 25x - 2500 = 0 Simplify x^2 + 5x - 500 = 0 Factor: (x+25)(x-20) = 0 positive solution x = 20 students actually attended : : Check solution by finding the cost of each 500/20 = 25 500/25 = 20
 Trigonometry-basics/144643: Hi, if anyone can tell me if I am in the right direction with this problem please. Two boats leave a marina at the same time. Boat A travels at 20km/hr in a direction of 65 degrees. Boat B travels at 12.5km/hr in a direction of 145 degrees. How far apart are the boats after 2 hours. (2) In what direction would the skipper of Boat A have to look to see Boat B. I have this. Boat A 20km x 2 hrs = 40 km Boat B 12.5 km x 2 hrs = 25 km. They are offset by an angle of 145-60 = 86 degrees. Using the Law of Cosines I get x^2 = 40^2 + 25^2 - 2(40)(25)cos(85 degrees) = 2050.688. so x^2 = 45.28. The boats are 45.28 km apart after 2 hours. However this just doesnt seem right to me. I have no idea how to answer part b. Thanks.1 solutions Answer 105388 by ankor@dixie-net.com(15622)   on 2008-06-08 21:20:28 (Show Source): You can put this solution on YOUR website!Two boats leave a marina at the same time. Boat A travels at 20km/hr in a direction of 65 degrees. Boat B travels at 12.5km/hr in a direction of 145 degrees. How far apart are the boats after 2 hours. (2) In what direction would the skipper of Boat A have to look to see Boat B. I have this. : I think you have the right idea here: : Boat A 20km x 2 hrs = 40 km Boat B 12.5 km x 2 hrs = 25 km. However the angle should be They are offset by an angle of 145-65 = 80 degrees. : Using the Law of Cosines you should get: : x^2 = 40^2 + 25^2 - 2(40)(25)cos(80 degrees) = 1877.70 so x = 43.33. The boats are 43.33 km apart after 2 hours. : Draw a diagram of this and you will see that Skipper A would have to look south-west to see boat B
 Polynomials-and-rational-expressions/144613: Jane invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. She received yearly interest of \$130. Randy also invested in the same scheme, but he interchanged the amounts invested, and received \$4 more as interest. How much amount did each of them invest at different rates? 1 solutions Answer 105379 by ankor@dixie-net.com(15622)   on 2008-06-08 17:14:46 (Show Source): You can put this solution on YOUR website!Jane invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. She received yearly interest of \$130. Randy also invested in the same scheme, but he interchanged the amounts invested, and received \$4 more as interest. How much amount did each of them invest at different rates? : Let x = amt J invested at 12% Let y = amt J invested at 10% J's total interest = \$130 Also x = amt R invested at 10% y = amt R invested at 12% R's total interest = \$134 : Two equations .12x + .10y = 130 (J's investments) .10x + .12y = 134 (R's investments) : Multiply the 1st equation by 1.2 and subtract the 2nd equation, find x .144x + .12y = 156 .10x + .12y = 134 ---------------------subtracting eliminates y .044x = 22 x = x = \$500; J has invested at 12% and R has invested at 10% : Find y using (.12x + .10y = 130) .12(500) + .10y = 130 60 = .10y = 130 .10y = 130 - 60 .1y = 70 y = y = \$700; J has invested at 10% and R has invested at 12% : : Check solution using (.10x + .12y = 134) .10(500) + .12(700) = 50 + 84 = 134; confirms our solutions
 Equations/144592: Preform the indicated operations and simplify: (v-5/v-7)-(v+1/v+7)+(v-35/v^2-49) Thank you I really need help with this!!!1 solutions Answer 105369 by ankor@dixie-net.com(15622)   on 2008-06-08 15:31:35 (Show Source): You can put this solution on YOUR website! - + : Factor the difference of squares: - + : The common denominator is (v-7)(v+7) : FOIL : : ; : Factor: : Cancel (v-7)
 Polynomials-and-rational-expressions/144611: An art dealer sold two artworks for \$1520 thereby making a profit of 25% on the first work and 10% profit on the other, whereas if he had approached any exhibition he would have sold them together for \$1535 with a profit of 10% on the first and 25% on the other artwork. Find the actual cost of each artwork. 1 solutions Answer 105367 by ankor@dixie-net.com(15622)   on 2008-06-08 15:07:16 (Show Source): You can put this solution on YOUR website!An art dealer sold two artworks for \$1520 thereby making a profit of 25% on the first work and 10% on the other, whereas if he had approached any exhibition he would have sold them together for \$1535 with a profit of 10% on the first and 25% on the other artwork. Find the actual cost of each artwork. : Let x = cost of the 1st artwork, Let y = cost of 2nd artwork : Art dealer: Retail price 1st piece = 1.25(x); Retail price of 2nd = 1.10(y) 1.25x + 1.10y = 1520 : Exhibition: Retail price 1st piece = 1.10(x); Retail price of 2nd = 1.25(y) 1.10x + 1.25y = 1535 : Use two equation in the elimination method: multiply the 1st equation by 100, and the 2nd equation by 88,resulting in: 125x + 110y = 152000 96.8x +110y = 135080 ----------------------subtracting eliminates y 28.2x + 0y = 16920 x = 16920/28.2 x = \$600 is the cost of the 1st piece : 2nd piece cost using the 1st equation: 1.25(600) + 1.1y = 1520 750 + 1.1y - 1520 1.1y = 1520 - 750 1.1y = 770 y = 770/1.1 y = \$700 is the cost of the 2nd piece: : : Check our solutions in the 2nd equation: 1.1(600) + 1.25(700) = 660 + 875 = 1535 as given : Did this make sense to you? In case you wondered how I came up 88 as multiplier in the 2nd equation. I multiplied the 1st equation by 100 to get rid of the decimal. Then divided 1.25 into 110 and saw it came out to an even 88.
 Polynomials-and-rational-expressions/144612: A professor started a project with a group of students, 60% of whom were boys. Due to some unavoidable reasons, six girls couldn't turn up, so the professor had to make some changes in the group. He admitted six boys. In doing so, the percentage of boys in the project increased to 75%. Find the number of boys and girls involved in the project initially. 1 solutions Answer 105363 by ankor@dixie-net.com(15622)   on 2008-06-08 14:54:15 (Show Source): You can put this solution on YOUR website!A professor started a project with a group of students, 60% of whom were boys. Due to some unavoidable reasons, six girls couldn't turn up, so the professor had to make some changes in the group. He admitted six boys. In doing so, the percentage of boys in the project increased to 75%. Find the number of boys and girls involved in the project initially. : Let t = total number of students, .6t = number of boys originally : (.6t + 6) = actual number of boys (which the problem says became 75%) : From all this we have one simple equation: .6t + 6 = .75t .6t - .75t = -6 -.15t = -6 t = -6/-.15 t = +40 kids : .6 * 40 = 24 boys originally .4 * 40 = 16 girls "
 logarithm/144422: Tutors I submitted this once showing my work but I can't find my notes. If you sear your email you will see how I worked the problem. I asked if you could check my work. Here is the word problem again..Thank you \$5000 is invested at interest rate K, compounded continuously and grows to \$6954.84 in 6 years. Find the interest rate; the exponential growth function; the balance after 10 years and find the doubling time. If I remember my notes I think I had the doubling time as 12 years, the interest rate at 6%. Thank you1 solutions Answer 105358 by ankor@dixie-net.com(15622)   on 2008-06-08 14:04:01 (Show Source): You can put this solution on YOUR website!\$5000 is invested at interest rate K, compounded continuously and grows to \$6954.84 in 6 years. Find the interest rate; the exponential growth function; the balance after 10 years and find the doubling time. : P*e^rt = A : 5000*e^6r = 6954.84 e^6r = Divide both sides by 5000 e^6r = 1.390968 : Find the nat log of both sides: ln(e^6r) = 1.390968 : log equiv of exponents 6r*ln(e) = ln(1.390698) : Find the ln of both sides (ln(e) = 1) 6r = .33000 r = r = .055, 5.5% : Growth function f(t) = 5000*e^.055t : use this to check above solution on a calc: enter: 5000*e^(.055*6) = 6954.84 : : the balance after 10 years A = 5000*e^(.055*10) A = 8666.26 : and find the doubling time. 1*e^.055t = 2 .055t*ln(e) = ln(2) .055t = .693147 t = t = 12.6 yrs : Check this using the value of 5000 on calc: enter 5000*e^(.055*12.6) = 9998.53 ~ 10000
 Radicals/144588: Simplify by taking roots of the numerator and denominator. Assume that all expressions under radicals represent positive numbers. ^3sqrt((1000x^5)/(y^3)) Thank you so very much~1 solutions Answer 105354 by ankor@dixie-net.com(15622)   on 2008-06-08 13:24:58 (Show Source): You can put this solution on YOUR website!^3sqrt((1000x^5)/(y^3)) : Assume you mean cuberoot() ; write it to reveal the cubes cuberoot() : *cuberoot(x^2)
 Equations/144593: Multiply and Simplify ((4c^2)/(3c^2-12c+12))*((3c-6)/2c)) Thank you I really need help with this!!!1 solutions Answer 105352 by ankor@dixie-net.com(15622)   on 2008-06-08 12:52:54 (Show Source): You can put this solution on YOUR website! * : Note that 2c cancels into 4c^2, 2c times, * (3c-6) : Factor * (3c-6) : I'm sure you have noted now, that the (3c-6)'s will cancel, leaving:
 Linear_Equations_And_Systems_Word_Problems/144606: The perimeter of a rectangle is 54cm. Two times the height is 3cm more than the width. Find the length of the height and the length of the width. this is the way i've been trying to solve this problem: P=2l+2w, right? we have the perimeter. 54=(2L+3)+2W. Is this right? I'm stuck!1 solutions Answer 105348 by ankor@dixie-net.com(15622)   on 2008-06-08 12:33:49 (Show Source): You can put this solution on YOUR website!The perimeter of a rectangle is 54cm. Two times the height is 3cm more than the width. Find the length of the height and the length of the width. : Let x = the width : It says,"Two times the height is 3cm more than the width." so we can say: 2h = (x + 3} : Given perimeter: 2h + 2w = 54 : Substitute (x+3) for 2h and x for w and we have: (x+3) + 2x = 54 : 3x = 54 - 3 x = x = 17 cm is the width. : Find the length from the equation 2h = (x+3) 2h = 17 + 3 2h = 20 h = 10 cm : : Check solution by finding the perimeter using these values
 Human-and-algebraic-language/144581: A designer attempts to arrange the characters of his artwork in the form of a square grid with equal numbers of rows and columns, but finds that 24 characters are left out. When he tries to add one more row and column, he finds that he has 25 too few characters. Find the number of characters used by the designer.1 solutions Answer 105344 by ankor@dixie-net.com(15622)   on 2008-06-08 10:21:41 (Show Source): You can put this solution on YOUR website!A designer, attempting to arrange the characters of his artwork in the form of a square grid with equal number of row and columns, found that 24 characters were left out." : If x = number of characters in each row we translate this statement as x*x + 24 = total number of characters : "When he tried to add one more row and column, he found that he was short of 25 characters." : Now, the number of rows is "x + 1", and the number of columns is "x + 1", so we have (x + 1)(x + 1) - 25 = total number of characters. : continuing: large grid - 25 = small grid + 24 x^2 + 2x + 1 -25 = x^2 + 24 : x^2 + 2x -24 = x^2 + 24 : x^2 - x^2 + 2x - 24 - 24 = 0 : 2x - 48 = 0 : 2x = +48 : x = 48/2 : x = 24 : Grid is 24 by 24 which contains 576 character + 24 left out = 600 Grid of 25 by 25 = 625 character so he would have 25 blanks