Tutor:

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

---

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

ankor@dixie-net.com answered: 15634 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, >>Next

Linear_Equations_And_Systems_Word_Problems/117258: This question is from textbook A company makes two chemicals:type1 and type 2. Due to storage problems, a miximum of 100 pounds of type 1 and 150 pounds of type 2 can be mixed and packaged each year. One pound of type 1 takes 60 hours to mix and 70 hours to package; one pound of type 2 takes 40 hours to mix and 40hours tp package.The mixing department has at most 7200 hours available each year and packaging has at most 7800 hours available. If the profit for one pound of type 1 is $62and for one pound type 2 is $40, what is the maximum profit possible each year? 1 solutions Answer 85399 by ankor@dixie-net.com(15649)   on 2007-12-17 22:00:53 (Show Source): You can put this solution on YOUR website! A company makes two chemicals: type 1 and type 2. Due to storage problems, a maximum of 100 pounds of type 1 and 150 pounds of type 2 can be mixed and packaged each year. One pound of type 1 takes 60 hours to mix and 70 hours to package; one pound of type 2 takes 40 hours to mix and 40 hours to package.The mixing department has at most 7200 hours available each year and packaging has at most 7800 hours available. If the profit for one pound of type 1 is $62 and for one pound type 2 is $40, what is the maximum profit possible each year? : Let x = amt of type 1; y = amt of type 2 : Storage constraints x =<100 y =<150 : Mixing constraint: 60x + 40y =<7200 40y = 7200-60x; arrange in the y=mx +b form y = 180 - 1.5x; divided equation by 40 : Packaging constraint: 70x + 40y =< 7800 40y = 7800 - 70x y = 195 - 1.75x : Plot all 4 of these equations: : : Area of feasibility: At or below the horizontal line At or left of the vertical line At or below the lowest slanted line : The objective functions for each corner of the region solve: 60x+40y = 7200 70x+40y = 7800 Results: x=60,y=90 62(60) + 40(90) = 3720 + 3600 = 7320 : Solve: 70x+40y = 7800 x = 100 Results: x=100,y=20 62(100) + 40(20) = 6200 + 800 = 7000 : solve: 60x+40y = 7200 y = 150 results x=20,y=150 62(20) + 40(150) = 1240 + 6000 = 7240 : Max profit would be 60 type 1 and 90 type 2

Miscellaneous_Word_Problems/117319: Mother made cupcakes for the church picnic."If I had made 6 more chocolate cupcakes, I would have as many chocolate as I do white and pepermint combined,"said Mother."I have 4 more white than peppermint."Altogether she baked 50 cupcakes.How many of each kind did she have? 1 solutions Answer 85367 by ankor@dixie-net.com(15649)   on 2007-12-17 16:38:59 (Show Source): You can put this solution on YOUR website! Mother made cupcakes for the church picnic."If I had made 6 more chocolate cupcakes, I would have as many chocolate as I do white and pepermint combined,"said Mother."I have 4 more white than peppermint."Altogether she baked 50 cupcakes.How many of each kind did she have? : Let c = no. of chocolate c.c. Let w = no. of whites Let p = no. of peppermints : Write an equation from the statement: "If I had made 6 more chocolate cupcakes, I would have as many chocolate as I do white and pepermint combined," c + 6 = w + p : "I have 4 more white than peppermint." w = p + 4 : Altogether she baked 50 cupcakes. c + w + p = 50 : How many of each kind did she have? : Take the 1st equation and replace w with (P+4) c + 6 = (p+4) + p Simplify c + 6 = 2p + 4 c = 2p + 4 - 6 c = 2p - 2 : In the 3rd equation, replace c with(2p-2) and replace w with (p+4); find p (2p-2) + (p+4) + p = 50 2p + p + p -2 + 4 = 50 4p + 2 = 50 4p = 50 - 2 p = 48/4 p = 12 peppermint c.c. : You can find the other amts using the equations we have here : Check solutions in the 3rd equation

Length-and-distance/117314: This question is from textbook Glencoe Geometry Using graph paper, graph each question and plot the given ordered pair. Then construct a perpendicular segment and find the distance from the point to the line. 22. y = 3, (4, -1) What I'm wondering is what y = 3 meant. I thought perhaps it meant that on the graph, the first ordered pair would be (0, 3). I looked through the section, but I can't find anything that would explain this, nor any sample problems like this. 1 solutions Answer 85343 by ankor@dixie-net.com(15649)   on 2007-12-17 14:49:55 (Show Source): You can put this solution on YOUR website! Using graph paper, graph each question and plot the given ordered pair. Then construct a perpendicular segment and find the distance from the point to the line. 22. y = 3, (4, -1) What I'm wondering is what y = 3 meant. I thought perhaps it meant that on the graph, the first ordered pair would be (0, 3). I looked through the section, but I can't find anything that would explain this, nor any sample problems like this. : A graph of y = 3 is just a horizontal line passing thru the y value of 3 Looks like this: : A line perpendicular to this line would have to be a vertical line, right? : I can't draw a vertical line on this graphing system, however, you can. Find the given point x=4, y= -1 and draw vertical line from there to the horizontal line. : Anytime you see y = to a single value, that is a horizontal line Conversely, Anytime you see x = to a single value, that is a vertical line : Did this answer your question

Quadratic_Equations/117292: A garden area is 30ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400ft^2, what is the width of the path? 1 solutions Answer 85340 by ankor@dixie-net.com(15649)   on 2007-12-17 14:26:37 (Show Source): You can put this solution on YOUR website! A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400 ft^2, what is the width of the path? : Draw diagram of this; label the outside dimensions of the rectangle 30 by 20. Label the width of the path as x, it will be apparent that the dimensions of the garden (inside the path), will be (30-2x) by (20-2x) : The area of the garden is given as 400 sq/ft : A simple area equation: : length times width = 400 sq/ft (30-2x) * (20-2x) = 400 : FOIL: 600 - 60x - 40x + 4x^2 = 400 4x^2 - 100x + 600 = 400 4x^2 - 100x + 600 - 400 = 0 4x^2 - 100x + 200 = 0; a quadratic equation : Simplify, divide by 4 and you have: x^2 - 25x + 50 = 0 : We need to use the quadratic formula to solve this: a=1; b=-25;; c=50 : : : Two solutions: x = 22.8, not a possible solution, obviously and x = 2.19 ft is the width of the path : : Check our solution by finding the area of the garden We have to subtract 2x from the outside dimensions: 2*2.19 = 4.38 : (30-4.38) * (20-4.38) = 25.62 * 15.62 = 400.2 ~ 400 sq/ft : How about this, did it make sense to you, any questions

Polynomials-and-rational-expressions/117006: This question is from textbook x varies directly as the square of s and inversely as t. How does x change when s is doubled? When both s and t are doubled? 1 solutions Answer 85321 by ankor@dixie-net.com(15649)   on 2007-12-17 10:28:57 (Show Source): You can put this solution on YOUR website! x varies directly as the square of s and inversely as t. x = : How does x change when s is doubled? Since s is squared, doubling s is the same as 2^2, hence 4 times : When both s and t are doubled? t is in the denominator, therefore x = half as much, t is 4 times as much therefore: .5 * 4 hence, 2 times

Human-and-algebraic-language/116995: How many gallons of water can be stored in a cylindrical tank 35 feet high and 12 feet in diameter? 1 solutions Answer 85303 by ankor@dixie-net.com(15649)   on 2007-12-17 08:26:23 (Show Source): You can put this solution on YOUR website! How many gallons of water can be stored in a cylindrical tank 35 feet high and 12 feet in diameter? : Find the volume of a cylinder: V = pi*r^2*h : V = pi*6^2*35 V = 3958.4 cu ft : There are 7.481 US gallons in 1 cu ft : 3958.4 * 7.481 = 29612.84 gallons

Graphs/117256: TRANSLATE THE FOLLOWING STATEMENTS INTO INEQUALITIES. LET X REPRESENT THE NUMBER IN EACH CASE. 2 times a number, increased by 28, is less than or equal to 6 times that number. 1 solutions Answer 85289 by ankor@dixie-net.com(15649)   on 2007-12-16 22:06:16 (Show Source): You can put this solution on YOUR website! TRANSLATE THE FOLLOWING STATEMENTS INTO INEQUALITIES. LET X REPRESENT THE NUMBER IN EACH CASE. 2 times a number, increased by 28, is less than or equal to 6 times that number. : 2x + 28 <= 6x

Polynomials-and-rational-expressions/117020: This question is from textbook Complex fractions 1/xy + 2/x over 3/y - 1/xy I'm totally lost on complex fractions. Any assistance would be greatly appreciated. 1 solutions Answer 85173 by ankor@dixie-net.com(15649)   on 2007-12-16 11:35:10 (Show Source): You can put this solution on YOUR website! 1/xy + 2/x over 3/y - 1/xy : Assume this is the problem: : Covert the fractions in the numerator and denominator into single fractions : When you divide fractions, you invert the dividing fraction and multiply: * = ; note that the xy's cancel : Did this help?

Average/116767: This question is from textbook algerba1 how many grams of sugar must be added to 60 grams of a solution that is 32% sugar to obtain a solution that is 50% sugar? 1 solutions Answer 85114 by ankor@dixie-net.com(15649)   on 2007-12-15 21:41:14 (Show Source): You can put this solution on YOUR website! how many grams of sugar must be added to 60 grams of a solution that is 32% sugar to obtain a solution that is 50% sugar? : A typical mixture problem, learn this method and you can do most of them easily : Let x = amt of pure sugar required, the decimal amt of pure sugar 1.0(x) : We know the resulting amt = (x+60) ; A simple equation: .32(60) + 1.0(x) = .50(x+60) : 19.2 + x = .5x + 30 : x - .5x = 30 - 19.2 : .5x = 10.8 : x = 10.8/.5 : x = 21.6 grams of pure sugar required : : We can check our solution by substitution in our original equation: .32(60) + 21.6 = .5(21.6+60) 19.2 + 21.6 = .5(81.6) 40.8 = 40.8

Matrices-and-determiminant/117044: please help me with this question.I am having difficulty eliminating variables. solve the linear system without using matrices or determinants x+y+z=2 x-y+3z=12 2x+5y+2z=-2 1 solutions Answer 85108 by ankor@dixie-net.com(15649)   on 2007-12-15 21:21:48 (Show Source): You can put this solution on YOUR website! solve the linear system without using matrices or determinants x + y + z = 2 x - y +3z =12 2x+5y +2z =-2 : Let's eliminate z, and get it down to two, 2 unknown equations: Multiply the 2nd equation by -1 and add all three. x + y + z = 2 -x + y -3z =-12 2x +5y +2z = -2 ------------------adding eliminate z 2x +7y +0z = -12 2x + 7y = -12; our 2 unknown equation : Here we luck out. Multiply eq1 by -2 and add it to eq3, I should have noticed that at first. -2x - 2y - 2z = -4 2x + 5y + 2z = -2 --------------------adding eliminates x & z, easy to find y 0x + 3y + 0z = -6 3y = -6 y = -2 : Use our two unknown equation to find x: 2x + 7y = -12 Substitute -2 for y 2x + 7(-2) = -12 2x -14 = -12 2x = -12 + 14 2x = +2 x = +1 : Use the 1st equation to find z: x + y + z = 2 1 - 2 + z = 2 z = 2 + 1 z = +3 : Check our solution the eq 2 x - y + 3z = 12 1 -(-2) + 3(3) = 12 1 + 2 + 9 = 12 confirms our solutions : It pays to study these systems and see if there is a single operation that will eliminate 2 unknowns at once, otherwise just do what is necessary to get two, 2 unknown equations to work with. Then you can use elimination or substitution quite easily.

Square-cubic-other-roots/116894: Solve the following algebraically. You must show all your work. 1/4= 3- 2x-1/x+2 1 solutions Answer 85100 by ankor@dixie-net.com(15649)   on 2007-12-15 20:59:03 (Show Source): You can put this solution on YOUR website! Assume you mean: = 3 - : Subtract 3 from both sides: - 3 = - : - = - : = - : Get rid of the negatives multiply both sides by -1 = : Cross multiply 11(x+2) = 4(2x-1) : 11x + 22 = 8x - 4 : 11x - 8x = -4 - 22 : 3x = -26 : x = -26/3 or ~-8.67 : : It's tedious to check this value by substitution, however, did check using a calc and this annoying solution does check out.

Miscellaneous_Word_Problems/116980: Can you please help me!! I am so lost with these two. Thanks! Fungicides account for one tenth of the pesticides used in the United States. Insecticides account for one fourth of all the pesticides used in the United States. The ratio of fungicides to insecticides used in the United States can be written as one tenth divided by one fourth. Write this ratio in simplest form. The combined resistance of two resistors R1 and R2 in a parallel circuit is given by the formula RT = 1 over 1r1 + 1r2.Simplify the formula. 1 solutions Answer 85092 by ankor@dixie-net.com(15649)   on 2007-12-15 20:07:15 (Show Source): You can put this solution on YOUR website! Fungicides account for one tenth of the pesticides used in the United States. Insecticides account for one fourth of all the pesticides used in the United States. The ratio of fungicides to insecticides used in the United States can be written as one tenth divided by one fourth. Write this ratio in simplest form. : Here's what it says: / When you divide fractions you: "Invert the dividing fraction and multiply" * = = : : The combined resistance of two resistors R1 and R2 in a parallel circuit is given by the formula RT = 1 over 1r1 + 1r2. Simplify the formula. : I think you mean: Rt = : Change the denominator portion to be a single fraction over a common denominator = : This is divided into 1 which just consists of inverting the dividing fraction, so we have; ; the old "product over sum" formula for two resistors in parallell : : Did this help you find your way, somewhat?

Graphs/116999: This question is from textbook Solve by graphing. 2x = y + 1 2x - y = 5 I know you are supposed to rearrane it into y-intercept form but I'm not really sure how to go about graphing it at all. 1 solutions Answer 85091 by ankor@dixie-net.com(15649)   on 2007-12-15 19:40:16 (Show Source): You can put this solution on YOUR website! Solve by graphing. Actually these two equations have the same slope, therefore they are parallel lines and can't be solved by graphing or any other way. : However this is the procedure and you will see what I mean: Take one equation at a time: : Put in slope/intercept form 2x = y + 1 Subtract 1 from both sides 2x - 1 = y We can also write in the more familiar form: y = 2x - 1 : Make a table to plot this as a graph, you only need 2 points, however here are 3 Let x = -3 Substitute -3 for x in the above equation and find y y = 2x - 1 y = 2(-3)- 1 y = -6 - 1 y = -7 : The table: x | y ------ -3 |-7 do the same with x = 0 y = 2(0) - 1 y = -1 and for x = +3 y = 2(3) - 1 y = +6 -1 y = + 5 the table now x| y ----- -3|-7 0|-1 +3|+5 : Plot these point on a standard +/- 10 graph, should look like this: : : Do exactly the same with this equation and plot on the same graph 2x - y = 5 -y = -2x + 5; subtracted 2x from both sides y = +2x - 5; y always has to be positive, multiplied by -1 : You can see now that the two equations have the same slope (2) : Plot this one using the values for x, -2, 0 +3 Substitute these values for x in the above equation, x| y ------ -2|-9 0|-5 +3|+1 : : Anyway, if the graph of two equation intersect at some point, the x/y value of that point is the solution. : Did this help you?

Miscellaneous_Word_Problems/116877: Solve. Ariana took 2 hours longer to drive 360 miles on the first day of a trip than she took to drive 270 miles on the second day. If her speed was the same on both days, what was the driving time each day? Solve. A plane flies 720 miles against a steady 30 mile per hour headwind and than returns to the same point with the wind. If the entire trip takes 10 hours, what is the plane's speed in still air? 1 solutions Answer 85089 by ankor@dixie-net.com(15649)   on 2007-12-15 19:07:55 (Show Source): You can put this solution on YOUR website! Ariana took 2 hours longer to drive 360 miles on the first day of a trip than she took to drive 270 miles on the second day. If her speed was the same on both days, what was the driving time each day? : Let t = time to drive 270 mi then (t+2) = time to drive 360 mi : Speed of the two days is given as equal, so write speed equation: Speed = distance/time : = : Cross multiply: 360t = 270(t+2) : 360t = 270t + 540 : 360t - 270t = 540 : 90t = 540 : t = 540/90 : t = 6 hrs for the 270 mi day then 6 + 2 = 8 hrs for the 360 mi day : Check solution by finding if the speeds, are indeed equal: 360/8 = 45 mph 270/6 = 45 mph, confirms our solution : : Solve. A plane flies 720 miles against a steady 30 mile per hour headwind and than returns to the same point with the wind. If the entire trip takes 10 hours, what is the plane's speed in still air? : Let s = speed in still air Then (s+30) = speed with the wind and (s-30) = speed against the wind : Write a time equation; Time = Distance/speed : With wind time + against wind time = 10 hrs + = 10 : Multiply equation by (s+30)(s-30) to eliminate the denominators: 720(s-30) + 720(s+30) = 10(s-30)(s+30) : 720s - 21600 + 720s + 21600 = 10(s^2 - 900) : 1440s = 10s^2 - 9000 : 0 = 10s^2 - 1440s - 900; a quadratic equation: : Simplify, divide equation 10 s^2 - 144s - 900 = 0 : Fortunately this will factor: (s-150)(s+6) = 0 : s = +150 mph (the only solution that makes sense) in still air : : Check solution by finding the times. with speed = 180; against speed = 120 : 720/120 = 6 hrs 720/180 = 4 hrs --------------- Total hrs:10 hrs, confirms our solution :

Polynomials-and-rational-expressions/117005: This question is from textbook The distance an object falls is directly proportional to the square of the time it has been falling. After six seconds it has fallen 1296 ft. How long will it take to fall 2304 ft? 1 solutions Answer 85079 by ankor@dixie-net.com(15649)   on 2007-12-15 18:10:29 (Show Source): You can put this solution on YOUR website! The distance an object falls is directly proportional to the square of the time it has been falling. After six seconds it has fallen 1296 ft. How long will it take to fall 2304 ft? : k*t^2 = d Find k, using t=6 sec & d=1296 6^2*k = 1296 36k = 1296 k = 1296/36 k = 36 : Find t when d = 2304 and k = 36 36t^2 = 2304 t^2 = 2304/36 t^2 = 64 t = sqrt(64) t = 8 seconds

Mixture_Word_Problems/116992: This question is from textbook Thanks for your help on this one: On an 18-hole golf course, there are par-3 holes, par-4 holes and par-5 holes. A golfer who shoots par on every hole has a score of 72. The sum of the number of par-3 and the number of par-5 holes is 8. How many of each type of hole are there on the golf course? 1 solutions Answer 85074 by ankor@dixie-net.com(15649)   on 2007-12-15 17:11:31 (Show Source): You can put this solution on YOUR website! On an 18-hole golf course, there are par-3 holes, par-4 holes and par-5 holes. A golfer who shoots par on every hole has a score of 72. The sum of the number of par-3 and the number of par-5 holes is 8. How many of each type of hole are there on the golf course? Let: x = no. of par 3 holes y = no. of par 4 holes z = no. of par 5 holes: : We know: x + y + z = 18 : 3x + 4y + 5z = 72 : It says,"The sum of the number of par-3 and the number of par-5 holes is 8." x + z = 8 or z = (8-x) : In the 1st equation substitute (8-x) for z: x + y + (8-x) = 18 x - x + y = 18 - 8 y = 10 par 4 holes : In the 2nd equation substitute 10 for y and (8-x) for z 3x + 4(10) + 5(8-x) = 72 3x + 40 + 40 - 5x = 72 3x - 5x + 80 = 72 -2x = 72 - 80 -2x = -8 x = -8/-2 x = 4 par 3 holes Then 8 - 4 = 4 par 5 holes : : Check 4(3) + 10(4) + 4(5) = 12 + 40 + 20 = 72 :

Average/116729: This question is from textbook algebra1 An express train travels 90 kilometers per hour from smallville to megatown. A local train takes 2.5 hours longer to travel thensame distance at 50 kilometers per hour. how far apart are Smallville and Megatowm 1 solutions Answer 85073 by ankor@dixie-net.com(15649)   on 2007-12-15 16:48:07 (Show Source): You can put this solution on YOUR website! An express train travels 90 kilometers per hour from smallville to megatown. A local train takes 2.5 hours longer to travel then same distance at 50 kilometers per hour. how far apart are Smallville and Megatown. : Let d = distance from S to M : Write a time equation, Time = Distance/speed: : Local train time - Express train time = 2.5 hrs - = 2.5 : Multiply equation by 450 to get rid of the denominators 450* - 450* = 450(2.5) : Cancel the denominators, we have: 9d - 5d = 1125 : 4d = 1125 : d = 1125/4 : d = 281.25 km from S to M : : Check solution by finding the times for each train: Local: 281.25/50 = 5.625 hrs Exprs: 281.25/90 = 3.125 hrs ------------------------------ Time difference is: 2.5 hrs; confirms our solution : : Did this make sense to you? Any questions? ;

testmodule/116882: This question is from textbook Beginning Algebra 2x2 – x – 3 • 3x2 – 11x – 20 ------------ -------------- 3x2 + 7x + 4 4x2 - 9 1 solutions Answer 85072 by ankor@dixie-net.com(15649)   on 2007-12-15 16:25:25 (Show Source): You can put this solution on YOUR website! * : You can factor everything here: * : Cancel out: (2x-3), (x+1), and (3x+4) leaving us with:

test/116858: x-4 + x-3 _____ ______ x2-4x-5 x2-25 1 solutions Answer 85067 by ankor@dixie-net.com(15649)   on 2007-12-15 15:17:50 (Show Source): You can put this solution on YOUR website! + : Factor where you can: + : The common denominator is (x-5)(x+1)(x+5) FOIL in the numerator: : :

Human-and-algebraic-language/116799: Find A The right triangle is 35 degrees one long side is 23m. Find A the short side. 1 solutions Answer 85063 by ankor@dixie-net.com(15649)   on 2007-12-15 14:56:47 (Show Source): You can put this solution on YOUR website! Find A The right triangle is 35 degrees one long side is 23m. Find A the short side. : a would be the side opposite 35 degrees, Side adjacent is given as 23 m : Tangent = side opposite/side adjacent tan(35) = : .7 = a = .7 * 23 a = 16.1 m is the short side

Evaluation_Word_Problems/116716: Emilio and Rose sell half of their book collection at a garage sale. They donate 1/5 of the remaining books to a resale shop and keep 40 books. How many books did they have to start with? 1 solutions Answer 85048 by ankor@dixie-net.com(15649)   on 2007-12-15 12:09:06 (Show Source): You can put this solution on YOUR website! Emilio and Rose sell half of their book collection at a garage sale. They donate 1/5 of the remaining books to a resale shop and keep 40 books. How many books did they have to start with? : Let x = amt of books initially : x = yard sale + donated + 40 x = .5x + .2(x -.5x) + 40 x = .5x + .2(.5x) x = .5x + .10x + 40 x = .6x + 40 x - .6x = 40 .4x = 40 x = 40/.4 x = 100 books : : Check: .5(100) + .2(50) + 40 = 50 + 10 + 40 = 100

Polynomials-and-rational-expressions/116863: how do you solve this rational expression.... for allr is not equal to 2 r^2-5r+6/r^2-4 the answer is r-3/r+2 i just need to know how to get this answer. 1 solutions Answer 85046 by ankor@dixie-net.com(15649)   on 2007-12-15 11:51:14 (Show Source): You can put this solution on YOUR website! how do you solve this rational expression.... for allr is not equal to 2 : Note that you can factor the numerator and the denominator is the difference of squares, you can factor also, then cancel out (r-2) =

Polynomials-and-rational-expressions/116864: I have to express the following rational expressions in lowest terms Can you please help with the following questions. I am so confused and I don't quiet understand. There are five questions that I tried and can't get. PLEASE HELP ME!!!! 1) x/x+xy 2)3x^2+6x/3x^2+9x 3)x^2+2x+1/x+1 4)x^2+3x-18/2x-6 5)2x^2-4x-6/x-3 1 solutions Answer 85042 by ankor@dixie-net.com(15649)   on 2007-12-15 11:35:01 (Show Source): You can put this solution on YOUR website! I have to express the following rational expressions in lowest terms Can you please help with the following questions. I am so confused and I don't quiet understand. There are five questions that I tried and can't get. PLEASE HELP ME!!!! 1) Factor x out in the denominator, then cancel with the x in the numerator: = : 2) Factor out 3x in both the numerator and denominator; cancel the 3x's = : 3) You can factor the numerator here; cancel one of the (x+1)'s leaving: = x+1 : 4) Factor the numerator and factor out 2 in the denominator, cancel the (x-3)'s: = : 5) Factor the numerator: Notice now that we can factor (2x-6) in the numerator, then we can cancel = 2(x+1) : : Study what we did here diligently, and it will all makes sense, you can email me if you have a question about it.

Polynomials-and-rational-expressions/116947: Can someone Please Help mE WITH THIS QUESTION I tried to get and it is really confusing. PLEASE HELP ME!!!!! 1) 16x^2/y^4*5x^2/y^2 2)6x^5y^3/5z^3*6x^4/5yz^4 1 solutions Answer 85033 by ankor@dixie-net.com(15649)   on 2007-12-15 11:02:49 (Show Source): You can put this solution on YOUR website! 1) You can't cancel anything here, just multiply the numerators together and the denominators, remember to add exponents of like terms when you multiply * = : 2) Do the same with this one: * = ; Did this help? Any questions?

Polynomials-and-rational-expressions/116948: Please help me with this question I tried to figure it out and i just can't get it. PLEASE HELP ME!!!! 1) x^2+7x+12/x+4*1/x+3 2)25xy^2/7z/5x^2y^2/14z^2 1 solutions Answer 85024 by ankor@dixie-net.com(15649)   on 2007-12-15 10:44:43 (Show Source): You can put this solution on YOUR website! 1) * The first thing you want to do in these problems is see what can be factored * = 1 Note that you have (x+3) and (x+4) in the numerator and the denominator cancel these and that leaves 1 : : 2) --------- : Remember, when you divide by fractions, you invert the dividing fraction & multiply * Cancel the following: 5 into 25 7 into 14 x in the numerator into x^2 of the denominator y^2 into y^2 z in the denominator into z^2 in the numerator Leaving us with: 5 * = : How about this, is this the help you need? Any questions?

Rate-of-work-word-problems/116938: Good afternoon ma'am or sir. Will you please help me in solving this worded problems: 1). Ryoma can painttheir house 7 1/2 days, however Sakuno helped him and together they were able to do the painting in just 4 days. If Sakuno work alone how many days would it take to paint the whole house? 2).A pipe can fill the pool in 6 hours. A small pipe can fill the pool in 8 hours. The pool will be empty within 12 hours. How long would it take to fill the pool if both filling pipe and draining pipe are both open? 1 solutions Answer 85018 by ankor@dixie-net.com(15649)   on 2007-12-15 09:24:38 (Show Source): You can put this solution on YOUR website! 1). Ryoma can paint their house 7 1/2 days, however Sakuno helped him and together they were able to do the painting in just 4 days. If Sakuno work alone how many days would it take to paint the whole house? : Let x = S's time alone Let the completed job = 1 : Each will do a fraction of the house: : R's fraction + S's fraction = completed job + = 1 : Multiply equation by 7.5x to get rid of the denominators: 7.5x* + 7.5x* = 7.5x(1) Cancel out the denominators and you have; 4x + 7.5(4) = 7.5x 4x + 30 = 7.5x 30 = 7.5x - 4x 30 = 3.5x x = 30/3.5 x = 8.57 days together or 8 days, 13.7 hrs : : 2).A pipe can fill the pool in 6 hours. A small pipe can fill the pool in 8 hours. The pool will be empty within 12 hours. How long would it take to fill the pool if both filling pipe and draining pipe are both open? : Let x = time required when working together Let the full pool = 1 : Let filling be positive & draining be negative: + - = 1 : Multiply equation by 24 to get rid of the denominators 24* + 24 - 24 = 24(1) cancel out the denominators and you have: 4x + 3x - 2x = 24 : 5x = 24 : x = 24/5 : x = 4.8 hr to fill the pool (or 4 hrs 48 min)

Geometry_Word_Problems/116944: Construction. A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400 , what is the width of the path? 1 solutions Answer 85013 by ankor@dixie-net.com(15649)   on 2007-12-15 08:15:23 (Show Source): You can put this solution on YOUR website! A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400 ft^2, what is the width of the path? : Draw diagram of this; label the outside dimensions of the rectangle 30 by 20. Label the width of the path as x, it will be apparent that the dimensions of the garden (inside the path), will be (30-2x) by (20-2x) : The area of the garden is given as 400 sq/ft : A simple area equation: : length times width = 400 sq/ft (30-2x) * (20-2x) = 400 : FOIL: 600 - 60x - 40x + 4x^2 = 400 4x^2 - 100x + 600 = 400 4x^2 - 100x + 600 - 400 = 0 4x^2 - 100x + 200 = 0; a quadratic equation : Simplify, divide by 4 and you have: x^2 - 25x + 50 = 0 : We need to use the quadratic formula to solve this: a=1; b=-25;; c=50 : : : Two solutions: x = 22.8, not a possible solution, obviously and x = 2.19 ft is the width of the path : : Check our solution by finding the area of the garden We have to subtract 2x from the outside dimensions: 2*2.19 = 4.38 : (30-4.38) * (20-4.38) = 25.62 * 15.62 = 400.2 ~ 400 sq/ft : How about this, did it make sense to you, any questions?

Mixture_Word_Problems/116832: what is the name given to part of a decanted mixture left behind in the vessel? 1 solutions Answer 84943 by ankor@dixie-net.com(15649)   on 2007-12-14 13:38:27 (Show Source): You can put this solution on YOUR website! The Dregs??

Square-cubic-other-roots/116538: Use the quadratic formula to solve the equation. 2x^2+ 12x =-5 1 solutions Answer 84942 by ankor@dixie-net.com(15649)   on 2007-12-14 13:28:43 (Show Source): You can put this solution on YOUR website! Use the quadratic formula to solve the equation. 2x^2+ 12x =-5 : Put the equation in the ax^2 + bx + c = 0 form, add 5 to both sides: 2x^2 + 12x + 5 = 0 : In this equation a = 2; b = 12; c = 5 Substitute: : : : Find the square root of 104: : Two solutions x = -.45 and x = -5.55 : Learn this procedure by heart and you can do any quadratic equation that it's possible to solve. It would be time well spent.

Miscellaneous_Word_Problems/116814: Solve. If one half of one integer is subtracted from three-fifths of the next consecutive integer, the difference is three. What are the two integers? Solve. A passenger train can travel 325 miles in the same time a freight train takes to travel 200 miles. If the speed of the passenger train is 25 miles per hour faster than the speed of the freight train, find the speed of each. 1 solutions Answer 84939 by ankor@dixie-net.com(15649)   on 2007-12-14 13:09:37 (Show Source): You can put this solution on YOUR website! If one half of one integer is subtracted from three-fifths of the next consecutive integer, the difference is three. What are the two integers? : x = one integer; (x+1) = next consecutive integer, change fractions to decimals one-half =.5; and three/fifths = .6 : Write an equation for what it says: .6(x+1) - .5x = 3 .6x + .6 - .5x = 3 .6x - .5x = 3 - .6 .1x = 2.4 x = 2.4/.1 x = 24 is the 1st integer, 25 is the next integer : Check solution in the statement: "one half of one integer is subtracted from three-fifths of the next consecutive integer, the difference is three. .6(25) - .5(24) 15 - 12 = 3 : : Solve. A passenger train can travel 325 miles in the same time a freight train takes to travel 200 miles. If the speed of the passenger train is 25 miles per hour faster than the speed of the freight train, find the speed of each. : Let x = speed of the freight train then (x+25) = speed of the pass train : It says the times for the two is the same. Write a time equation: Time = distance/speed Pass time = Freight time = ; Cross multiply and you have: 325x = 200(x+25) 325x = 200x + 5000 325x - 200x = 5000 125x = 5000 x = x = 40 mph is the freight then 40 + 25 = 65 mph is the pass train : We can check solutions by confirming that the times are, indeed, equal: 325/65 = 5 hr 200/40 = 5 hrs also : Was I able to make this easy for you to understand? Any questions?

Age_Word_Problems/116723: mark is 10 yrs younget than larry.larrys age8 years from now will wxceed twice marks age3 yearsago by 4 years . how old is each boy? 1 solutions Answer 84922 by ankor@dixie-net.com(15649)   on 2007-12-14 10:00:18 (Show Source): You can put this solution on YOUR website! mark is 10 yrs younger than larry. M = L - 10 : larry's age 8 years from now, will exceed twice marks age 3 years ago by 4 years. L + 8 = 2(M-3) + 4 Do some basic algebra on this L + 8 = 2M - 6 + 4 L + 8 = 2M - 2 L = 2M - 2 - 8 L = 2M - 10 : how old is each boy? Substitute (L-10) for M in the above equation, solve for L: L = 2(L-10) - 10 L = 2L - 20 - 10 L = 2L - 30 L - 2L = -30 -L = -30 L = -30/-1 L = +30 is Larry's age Then M = 30 - 10 = 20 is Marks age : : Check our solution using the statememt: "larrys age8 years from now will exceed twice marks age 3 years ago by 4 years". (30 + 8) = 2(20-3) + 4 38 = 2(17) + 4; confirms our solution : Did this method makes sense to you? Any questions?