New!
Get regular updates about newly solved problems
via algebra.com's RSS system.
Recent problems solved by 'checkley77'
checkley77 answered: 12560 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, >>NextMoney_Word_Problems/224660: a total of $4,500 is invested in two funds that pay 4% and 5% simple intrest. the annunal intrest is $210.00. how much is invested in the fund paying 4% interest? 1 solutions
Answer 167803 by checkley77(12569) on 2009-10-10 19:20:58 (Show Source):
You can put this solution on YOUR website!.05x+.04(4,500-x)=210
.05x+180-.04x=210
.01x=210-180
.01x=30
x=30/.01
x=3,000 amount invested @ 5%.
4,500-3,000=1,500 amount invested @ 4%.
Proof:
.05*3,000+.04*1,500=210
150+60=210
210=210
|
Travel_Word_Problems/224661: an airplane flying into a headwind travels 3000 miles in 6 hours and 15 minutes. on the return flight, the same distance is traveled in 5 hours. find the speed of the airplane in still air. 1 solutions
Answer 167802 by checkley77(12569) on 2009-10-10 19:16:25 (Show Source):
You can put this solution on YOUR website!d=rt
r=d/t
r=3,000/6.25=480 mph. against the wind.
r=3000/5=600 mph. with the wind.
(600-480)2=120/2=60 mph is the speed of the wind.
Proof:
600-60=540 mph for the plane in still air.
480+60=540 ditto.
|
Travel_Word_Problems/224675: If I am going 75 miles per hour and my friend is going 55 miles per hour and he leaves 10 minutes before me, how long will it take me to catch him? 1 solutions
Answer 167800 by checkley77(12569) on 2009-10-10 19:04:57 (Show Source):
You can put this solution on YOUR website!d=rt or
d=75(x-1/6)
d=55x
The 2 distances are equal thus:
75(x-1/6)=55x
75x-75/6=55x
75x-55x=75/6
20x=75/6
20x=12.5
x=12.5/20
x=5/8 or .625 or 37.5 minutes after your friend leave you will catch up with him.
Proof:
75(5/8-1/6)=55*5/8
75(5*6-8)/48=275/8
75(30-8)/48=275/8
75*22/48=275/8
1,650/48=275/8
48*275=8*1,650
13,200=13,200
|
Money_Word_Problems/224572: Jody invested $5000 less in an account paying 4% simple interest than she did in an account paying 3% simple interest. At the end of the first year, the total interest from both accounts was $675. Find the amount invested in each amount invested in each account. 1 solutions
Answer 167759 by checkley77(12569) on 2009-10-10 13:05:57 (Show Source):
You can put this solution on YOUR website!.04(X-5,000)+.03X=675
.04X-200+.03X=675
.07X=675+200
.07X=875
X=875/.07
X=12,500 AMOUNT INVESTED @ 3%
12,500-5,000=7,500 INVESTED @ 3%
PROOF:
.04*7,500+.03*12,500=675
300+375=675
675=675
|
Linear_Equations_And_Systems_Word_Problems/224580: In 1993, the number of Americans dying from cancer was 6 times the number that died from accidents. If the number of deaths from these two causes totaled 630,000 how many americans died from each cause? 1 solutions
Answer 167757 by checkley77(12569) on 2009-10-10 12:58:39 (Show Source):
You can put this solution on YOUR website!X+6X=630,000
7X=630,000
X=630,000/7
X=90,000 DIED IN ACCIDENTS.
6*90,000=540,000 DIED OF CANCER.
PROOF:
90,000+540,000=630,000
630,000=630,000
|
Money_Word_Problems/224476: Dara invested some money at 3% simple interest and $6000 more than that at 4.5% simple interest. After 1 year her total interest from the two accounts was $870. How much did she invest at each rate? 1 solutions
Answer 167741 by checkley77(12569) on 2009-10-10 10:47:10 (Show Source):
You can put this solution on YOUR website!.03X+.045(X+6,000)=870
.03X+.045X+270=870
.075X=870-270
.075X=600
X=600/.075
X=8,000 AMOUNT INVESTED @ 3%.
8,000+6,000=14,000 AMOUNT INVESTED @ 4.5%.
PROOF:
.03*8,000+.045*14,000=870
240+630=870
870=870
|
Numbers_Word_Problems/224511: a man invested $25,000, part of it was invested at 3% and the rest at 4%. the total annual income from two investments is $950.00. how much was invested at each of these rates? 1 solutions
Answer 167738 by checkley77(12569) on 2009-10-10 10:38:17 (Show Source):
You can put this solution on YOUR website!04X+.03(25,000-X)=950
.04X+750-.03X=950
.01X=950-750
.01X=200
X=200/.01
X=20,000 AMOUNT INVESTED @ 4%
25,000-20,000=5,000 INVESTED @ 3%
PROOF:
.04*20,000+.03*5,000=950
800+150=950
950=950
|
Numbers_Word_Problems/224519: a man invested $25,000, part of it was invested at 3% and the rest at 4%. the total annual income from two investments is $950.00. how much was invested at each of these rates? 1 solutions
Answer 167734 by checkley77(12569) on 2009-10-10 10:29:03 (Show Source):
You can put this solution on YOUR website!.04X+.03(25,000-X)=950
.04X+750-.03X=950
.01X=950-750
.01X=200
X=200/.01
X=20,000 AMOUNT INVESTED @ 4%
25,000-20,000=5,000 INVESTED @ 3%
PROOF:
.04*20,000+.03*5,000=950
800+150=950
950=950
|
Travel_Word_Problems/224527: A train leaves Danville Union and travels north at a speed of 75 km/h. Two hours later, an express train leaves on a parallel track and travels north at 125 km/h. How far from the station will they meet? 1 solutions
Answer 167730 by checkley77(12569) on 2009-10-10 10:24:42 (Show Source):
You can put this solution on YOUR website!D=RT
D=75T
D=125(T-2)
75T=125(T-2)
75T=125T-250
75T-125T=-250
-50T=-250
T=-250/-50
T=5 HOURS IS THE TIME UNTIL THEY MEET.
PROOF:
75*5=125(5-2)
375=125*3
375=375
|
Linear-equations/224356: change the equation to slope-intercept form and determine the slope and y interceptof the line.
3x+y=8
a) slope is
b) (0,?) (y intercept) 1 solutions
Answer 167688 by checkley77(12569) on 2009-10-09 22:33:44 (Show Source):
You can put this solution on YOUR website!Y=mX+b m=slope & b=y intercept.
3x+y=8
y=-3x+8
a) slope is -3
b) (0,8) (y intercept)
 (graph 300x200 pixels, x from -6 to 5, y from -10 to 10, -3x +8).
|
Expressions-with-variables/224361: Solve the system by using the substitution method
x=1/4y-5/4
y=2/5x+14/5
(x,y)=( , ) 1 solutions
Answer 167687 by checkley77(12569) on 2009-10-09 22:30:01 (Show Source):
You can put this solution on YOUR website!x=y/4-5/4
y=2/5x+14/5
y=2/5(y/4-5/4)+14/5
y=2/5(y/4)+2/5(-5/4)+14/5
y=2y/20-10/20+14/5
y=2y/20-10/20+4*14/20
y=(2y-10+56)/20
y=(2y+46)/20
20y=2y+46
20y-2y=46
18y=46
y=46/18
y=23/9 ans.
x=(23/9)*1/4-5/4
x=23/36-5/4
x=23/36-(5*9)/36
x=23/36-45/36
x=-22/36
x=-11/18
|
Expressions-with-variables/224362: Solve by using either the addition method or the substitution method
1/40x-1/5y=-1/8
x+1/15y=46/15
The solution is ( , ) 1 solutions
Answer 167680 by checkley77(12569) on 2009-10-09 21:49:08 (Show Source):
You can put this solution on YOUR website!X/40-Y/5=-1/8 MULTIPLY BY 1/3 & ADD.
X+Y/15=46/15
X/120-Y/15=-1/24
----------------------
X+X/120=46/15-1/24
(120X+X)/120=(46*24-1*15)/360 MULTIPLY BOTH SIDES BY 120
121X=(1,104-15)/3
3*121X=1,089
363X=1,089
X=1,089/363
X=3 ANS.
3+Y/15=46/15
Y/15=46/15-3
Y/15=(46-3*15)/15
Y/15=(46-45)/15
Y/15=1/15
Y=1 ANS.
PROOF:
3/40-1/5=-1/8
3/40-8/40=-1/8
-5/40=-1/8
-1/8=-1/8
The solution is ( , )
|
Expressions-with-variables/224365: A chemistry student wants to mix an 13% acid solution with a 51% acid solution to g 13L of a 29% acid solution. How many liters of the 13% solution and how many liters of the 51% solution should be mixed
The mixture contains 7.52632 L of 13% acid solution and 5.47363 L of 51%acid solution
19 __ 1 solutions
Answer 167674 by checkley77(12569) on 2009-10-09 21:26:14 (Show Source):
You can put this solution on YOUR website!.51X+.13(13-X)=.29*13
.51X+1.69-.13X=3.77
.38X=3.77-1.69
.38X=2.08
X=2.08/.38
X=5.47368 LITERS OF 51% ACID IS USED.
13-5.47368=7.52632 LITERS OF 13% ACID IS USED.
PROOF:
.51*5.47368+.13*7.52632=3.77
2.79158+.97842=3.77
3.77=3.77
|
Equations/224366: I have read this question over and over and I just don't understand it, help please!
The sum of the page numbers on facing pages of a book is 157; what are the pages numbers? 1 solutions
Answer 167671 by checkley77(12569) on 2009-10-09 21:12:02 (Show Source):
|
Equations/224367: A girl buys three apples and seven oranges for $4.56. If an orange cost .28 more than an apple, how much does each fruit cost? 1 solutions
Answer 167670 by checkley77(12569) on 2009-10-09 21:09:32 (Show Source):
You can put this solution on YOUR website!3X+7(X+.28)=4.56
3X+7X+1.96=4.56
10X=4.56-1.96
10X=2.60
X=2.60/10
X=.26 COST OF AN APPLE.
.26+.28=.54 COST OF AN ORANGE.
PROOF:
3*.26+7*.54=4.56
.78+3.78=4.56
4.56=4.56
|
Equations/224369: Tom’s uncle owns a triangular piece of land. The perimeter fence that surrounds the land measures 378 yards. The shortest side is 30 yards longer than one-half of the longest side. The second longest side is 2 yards shorter than the longest side. What is the length of each side? 1 solutions
Answer 167668 by checkley77(12569) on 2009-10-09 21:02:05 (Show Source):
You can put this solution on YOUR website!.5X+30+X-2+X=378
2.5X=378-30+2
2.5X=350
X=350/2.5
X=140 IS THE LENGTH OF THE LONGEST SIDE.
.5*140+30=70+30=100 IS THE SHORTEST SIDE.
140-2=138 FOR THE MIDDLE LENGTH SIDE.
pROOF:
140+100+138=378
378=378
|
Travel_Word_Problems/224371: Two cyclists startat the same time from opposite ends of a cours that is 51 miles long. One cyclist is riding at a rate of 16mph, and the second cyclist is riding at a rate of 18mph. How long after they begin will they meet? 1 solutions
Answer 167667 by checkley77(12569) on 2009-10-09 20:56:56 (Show Source):
|
expressions/224384: Could you help me with this problem?
Dina and Masha start out on a 10km bikepath at the same time. When Dina reaches the end of the 10km, Masha still has 2km left to bike. Dina's biking speed is 5km per hour, find Masha's biking speed. 1 solutions
Answer 167666 by checkley77(12569) on 2009-10-09 20:53:03 (Show Source):
|
|