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, >>NextQuadratic_Equations/149896: This question is from textbook
A rectangular garden has dimensions of 18 ft by 13 ft. A gravel path of uniform width is to be built around the garden. How wide can the path be if there is enough gravel for 516 square feet? 1 solutions
Answer 110026 by checkley77(12569) on 2008-07-27 16:11:52 (Show Source):
You can put this solution on YOUR website!18*13=234 ft^2 for the garden.
234+516=750 is the area for the garden plus the path.
(18+2x)(13+2x)=750
234+36x+26x+4x^2=750
4x^2+62x+234-750=0
4x^2+62x-516=0
2(2x^2+31x-258)=0
2(2x+43)(x-6)=0
x-6=0
x=6 for the width of the path.
proof:
(18+2*6)(13+2*6)+750
(18+12)(13+12)=750
30*25=750
750=750
|
Percentage-and-ratio-word-problems/149677: Lucy has invested some money invested at 18% and three times as much invested at 12%. her total annual income from the 2 investment is 85,000 pesos. how much was invested at each rate? 1 solutions
Answer 109881 by checkley77(12569) on 2008-07-25 18:17:01 (Show Source):
You can put this solution on YOUR website!.18x+.12*3x=85,000
.18x+.36x=85,000
.54x=85,000
x=85,000/.54
x=157,407.40 invested @ 18%
3*157,407.40=472,222.20 invested @ 12%
proof:
.18*157,407.40+.12*472,222.20=85,000
28,333.33+56,666.67=85,000
85,000=85,000
|
Money_Word_Problems/149685: Walt made an extra $9000.00 lst year from his part time job. He invested part of the money at 9% and the rest at 8%. He made a total of $770.00 in interest. How much was invested at 8%? 1 solutions
Answer 109879 by checkley77(12569) on 2008-07-25 18:07:13 (Show Source):
You can put this solution on YOUR website!.09x+.08(9,000-x)=770
.o9x+720-.08x=770
.01x=770-720
.01x=50
x=50/.01
x=5,000 invested @ 9%
9,000-5,000=4,000 invested @ 8%
proof:
.09*5000+.08*4000=770
450+320=770
770=770
|
Money_Word_Problems/149770: Please help with this problem:Becoming a millionaire: How much would you have to invest now at 6.5% interest compounded semi-annually to have a million dollars in 40 years? 1 solutions
Answer 109871 by checkley77(12569) on 2008-07-25 17:46:36 (Show Source):
You can put this solution on YOUR website!1,000,000=x(1+.065/2)^40*2
1,000,000=x(1.0325)^80
1,000,000=12.9183x
x=1,000,000/12.9183
x=$77,409.57 would be the initial investment.
|
sets-and-operations/149452: how do you find three consecutive numbers who's product is 15, 600? 1 solutions
Answer 109649 by checkley77(12569) on 2008-07-23 17:49:25 (Show Source):
You can put this solution on YOUR website!LET X-1, X & X+1 BE THE THREE NUMBERS.
(X-1)X(X+1)=15,600
A SIMPLE TEST CAN DETERMINE THE VALUE OF THE MIDDLE NUMBER.
X^3-X=15,600
X^3=15,600+X
NOW FIND THE CUBE ROOT OF 15,600:
CUBERT15,600=24.98
NOW ADD THE MEXT WHOLE NUMBER ABOVE 24.98 (25) TO 15,600+25=15,625.
NOW FIND THE CUBERT15,625
CUBERT15,625=25 WHICH IS THE MIDDLE NUMBER.
24,25,26 ARE THE 3 NUMBERS.
PROOF:
24*25*26=15,600
15,600=15,600
|
Quadratic_Equations/149427: In a right triangle, one of the legs is 2 feet longer than the shorter leg. The hypotenuse is 10 feet. Find the lenghths of the missing 2 sides of the triangle. 1 solutions
Answer 109642 by checkley77(12569) on 2008-07-23 17:17:18 (Show Source):
You can put this solution on YOUR website!X^2+(X+2)^2=10^2
X^2+X^2+4X+4=100
2X^2+4X+4-100=0
2X^2+4X-96=0
2(X^2+2X-48)=0
2(X+8)(X-6)=0
X-6=0
X=6 FEET FOR THE SHORTER SIDE.
6+2=8 FEET FOR THE LONGER SIDE.
PROOF:
6^2+8^2=10^2
36+64=100
100=100
|
Rectangles/149450: This question is from textbook page 397
THE PERIMETER OF A RECTANGLE IS 96. THE LENGTH OF THE RECTANGLE IS SIX LESS THAN TWICE THE WIDTH. FIND THE DIMENSIONS OF THE RECTANGLE. 1 solutions
Answer 109640 by checkley77(12569) on 2008-07-23 17:11:39 (Show Source):
You can put this solution on YOUR website!P=2L+2W
96=2(2W-6)+2W
96=4W-12+2W
96=4W-12
4W=96+12
4W=108
W=108/4
W=18 FOR THE WIDTH.
2*18-6=36-6=30 FOR THE LENGTH.
PROOF:
96=2*30+2*18
96=60+36
96=96
|
Graphs/149239: 2x=y+1
2x-y=5
solve by graphing 1 solutions
Answer 109528 by checkley77(12569) on 2008-07-22 19:32:40 (Show Source):
You can put this solution on YOUR website!2x=y+1
Y=2X-1 (RED LINE)
2x-y=5
Y=2X-5 (GREEN LINE)
 (graph 300x200 pixels, x from -6 to 5, y from -10 to 10, of TWO functions 2x -1 and 2x -5).
THSE PARALLEL LINES INDICATE THAT THERE IS NO S0LUTION TO THESE 2 EQUATIONS.
|
Quadratic_Equations/149278: A rectangular garden has dimensions of 18 feet by 13 feet. A gravel path of equal width is to be built around the garden. How wide can the path be if there is enough gravel for 516 square feet? 1 solutions
Answer 109526 by checkley77(12569) on 2008-07-22 19:16:50 (Show Source):
You can put this solution on YOUR website!18*13=234 FT^2 FOR THE GARDEN.
234+516=750 FOR THE TOTAL AREA WITH THE PATH.
(18+2X)(13+2X)=750
234+26X+36X+4X^2=750
4X^2+62X+234-750=0
4X^2+62X-516=0
2(2X^2+31X-258)=0
2(2X-12)(X+21.5)=0
2X-12=0
2X=12
X=6 FT. IS THE ANSWER FOR THE WIDTH OF THE PATH.
PROOF:
(18+2*6)(13+2*6)=750
(18+12)(13+12)=750
30*25=750
750=750
|
Quadratic_Equations/149282: Please, I need help with the next step, something is not right.
.05(113 – 2y) + 8 + .10(2y) = $10.45 1 solutions
Answer 109520 by checkley77(12569) on 2008-07-22 18:49:31 (Show Source):
You can put this solution on YOUR website!.05(113 – 2y) + 8 + .10(2y) = $10.45
5.65-.1Y+8+.2Y=$10.45
.1Y=10.45-5.65-8
.1Y=-3.2
Y=-3.2/.1
Y=-32 ANSWER.
PROOF:
.05(113-2*-32)+8+.1(2*-32)=10.45
.05(113+64)+8+.1*-64=10.65
.05*177+8-6.4=10.65
8.85+8-6.4=10.65
10.45=10.45
|
Expressions-with-variables/149122: Solve each question by the by the addition method
1) 3x+2y=3
4x-3y=-13
2)3x-y=5
-6x+2y=1 1 solutions
Answer 109396 by checkley77(12569) on 2008-07-21 19:30:05 (Show Source):
You can put this solution on YOUR website!1) 3x+2y=3 MULTIPLY THIS EQUATION BY 3
4x-3y=-13 MULTIPLY THIS EQUATION BY 2 & THEN ADD THEM.
9X+6Y=9
8X-6Y=-26
--------------------------------
17X=-17
X=-17/17
X=-1 ANSWER.
3*-1+2Y=3
-3+2Y=3
2Y=3+3
2Y=6
Y=6/2
Y=3 ANSWER.
-----------------------------------------------------------------------
2)3x-y=5 MULTIPLY BY 2 & ADD.
-6x+2y=1
6X-2Y=10
----------------------------------
0+0=11 THERE ARE NO SOLUTIONS FOR THESE 2 EQUATIONS. THESE EQUATIONS REPRESENT PARALLEL LINES IN GEOMETRY.
|
Quadratic_Equations/149125: Problem: Translate the following into a quadratic equation, and solve it: The length of a rectangular garden is three times its width; if the area of the garden is 75 square meters, what are it's dimensions? 1 solutions
Answer 109394 by checkley77(12569) on 2008-07-21 19:19:55 (Show Source):
You can put this solution on YOUR website!area=L*W
75=3W*W
75=3W^2
W^2=75/3
W^2=25
W=SQRT25
W=5 WIDTH OF THE GARDEN.
3*5=15 FOR THE LENGTH OF THE GARDEN.
PROOF:
75=15*5
75=75
|
Miscellaneous_Word_Problems/149130: To determine the number of deer in a game preserve,a man catches 715 deer, tags them and lets them loose,later, 312 of them are caught, 156 of them are tagged. How many of them are in the preserve? 1 solutions
Answer 109393 by checkley77(12569) on 2008-07-21 19:16:34 (Show Source):
|
Numbers_Word_Problems/148935: Find three consecutive positive integers such that the product of the first and third, minus the second, is 1 more than 4 times the third. 1 solutions
Answer 109275 by checkley77(12569) on 2008-07-20 19:00:13 (Show Source):
You can put this solution on YOUR website!LET X BE THE FIRST NTEGER, (X+1) BE THE SECOND & 9X+2) BE THE THIRD.
X(X+2)-(X+1)=4(X+2)+1
X^2+2X-X-1=4X+8+1
X^2+2X-X-4X-1-8-1=0
X^2-3X-10=0
(X-5)(X+2)=0
X-5=0
X=5 ANSWER FOR THE FIRST INTEGER.
5+1=6 THE SECOND.
5+2=7 THE THIRD.
PROOF:
5*7-6=4*7+1
35-6=28+1
29=29
|
Expressions-with-variables/148939: x=y+6
y=-2-x
My daughter is trying to solve this equation by substitution. I am useless in attempting to help her. We would greatly appreciate you explaining the answer. 1 solutions
Answer 109268 by checkley77(12569) on 2008-07-20 18:46:51 (Show Source):
You can put this solution on YOUR website!x=y+6
NOW SUBSTITUTE (Y+6) FOR X IN THE OTHER EQUATION.
y=-2-x
Y=-2-(Y+6)
Y=-2-Y-6
Y+Y=-8
2Y=-8
Y=-8/2
Y=-4 ANSWER.
X=-4+6
X=2 ANSWER.
PROOF:
-4=-2-2
-4=-4
|
Linear_Equations_And_Systems_Word_Problems/148931: Please help me put the word problem below in linear equations. Thank you so much in advance. Please help me.
How do I go about this with the question?
An express and local train leave gray's Lake at 3 P.M. and head for Chicago 50 miles away. The express travels twice as fast as the local, and arrives 1 hour ahead of it. Find the speed of each train. 1 solutions
Answer 109266 by checkley77(12569) on 2008-07-20 18:43:39 (Show Source):
You can put this solution on YOUR website!LET THE TIME OF THE LOCAL TRAIN=50/X
LET THE TIME OF THE EXPRESS=50/2X
50/X=50/2X+1
50/X-50/2X=+1
(2*50-50)/2X=+1
(100-50)/2X=+1
50/2X=+1
2X=50
X=50/2
X=25 MPH FOR THE LOCAL
25*2=50 MPH FOR THE EXPRESS.
PROOF:
50/25=50/50+1
2=1+1
2=2
|
|