See tutors' answers!

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


   
By Tutor
 | By Problem Number | 

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: 15732 problems
Jump to solutions: 0..29 , 30..59 , 60..89 , 90..119 , 120..149 , 150..179 , 180..209 , 210..239 , 240..269 , 270..299 , 300..329 , 330..359 , 360..389 , 390..419 , 420..449 , 450..479 , 480..509 , 510..539 , 540..569 , 570..599 , 600..629 , 630..659 , 660..689 , 690..719 , 720..749 , 750..779 , 780..809 , 810..839 , 840..869 , 870..899 , 900..929 , 930..959 , 960..989 , 990..1019 , 1020..1049 , 1050..1079 , 1080..1109 , 1110..1139 , 1140..1169 , 1170..1199 , 1200..1229 , 1230..1259 , 1260..1289 , 1290..1319 , 1320..1349 , 1350..1379 , 1380..1409 , 1410..1439 , 1440..1469 , 1470..1499 , 1500..1529 , 1530..1559 , 1560..1589 , 1590..1619 , 1620..1649 , 1650..1679 , 1680..1709 , 1710..1739 , 1740..1769 , 1770..1799 , 1800..1829 , 1830..1859 , 1860..1889 , 1890..1919 , 1920..1949 , 1950..1979 , 1980..2009 , 2010..2039 , 2040..2069 , 2070..2099 , 2100..2129 , 2130..2159 , 2160..2189 , 2190..2219 , 2220..2249 , 2250..2279 , 2280..2309 , 2310..2339 , 2340..2369 , 2370..2399 , 2400..2429 , 2430..2459 , 2460..2489 , 2490..2519 , 2520..2549 , 2550..2579 , 2580..2609 , 2610..2639 , 2640..2669 , 2670..2699 , 2700..2729 , 2730..2759 , 2760..2789 , 2790..2819 , 2820..2849 , 2850..2879 , 2880..2909 , 2910..2939 , 2940..2969 , 2970..2999 , 3000..3029 , 3030..3059 , 3060..3089 , 3090..3119 , 3120..3149 , 3150..3179 , 3180..3209 , 3210..3239 , 3240..3269 , 3270..3299 , 3300..3329 , 3330..3359 , 3360..3389 , 3390..3419 , 3420..3449 , 3450..3479 , 3480..3509 , 3510..3539 , 3540..3569 , 3570..3599 , 3600..3629 , 3630..3659 , 3660..3689 , 3690..3719 , 3720..3749 , 3750..3779 , 3780..3809 , 3810..3839 , 3840..3869 , 3870..3899 , 3900..3929 , 3930..3959 , 3960..3989 , 3990..4019 , 4020..4049 , 4050..4079 , 4080..4109 , 4110..4139 , 4140..4169 , 4170..4199 , 4200..4229 , 4230..4259 , 4260..4289 , 4290..4319 , 4320..4349 , 4350..4379 , 4380..4409 , 4410..4439 , 4440..4469 , 4470..4499 , 4500..4529 , 4530..4559 , 4560..4589 , 4590..4619 , 4620..4649 , 4650..4679 , 4680..4709 , 4710..4739 , 4740..4769 , 4770..4799 , 4800..4829 , 4830..4859 , 4860..4889 , 4890..4919 , 4920..4949 , 4950..4979 , 4980..5009 , 5010..5039 , 5040..5069 , 5070..5099 , 5100..5129 , 5130..5159 , 5160..5189 , 5190..5219 , 5220..5249 , 5250..5279 , 5280..5309 , 5310..5339 , 5340..5369 , 5370..5399 , 5400..5429 , 5430..5459 , 5460..5489 , 5490..5519 , 5520..5549 , 5550..5579 , 5580..5609 , 5610..5639 , 5640..5669 , 5670..5699 , 5700..5729 , 5730..5759 , 5760..5789 , 5790..5819 , 5820..5849 , 5850..5879 , 5880..5909 , 5910..5939 , 5940..5969 , 5970..5999 , 6000..6029 , 6030..6059 , 6060..6089 , 6090..6119 , 6120..6149 , 6150..6179 , 6180..6209 , 6210..6239 , 6240..6269 , 6270..6299 , 6300..6329 , 6330..6359 , 6360..6389 , 6390..6419 , 6420..6449 , 6450..6479 , 6480..6509 , 6510..6539 , 6540..6569 , 6570..6599 , 6600..6629 , 6630..6659 , 6660..6689 , 6690..6719 , 6720..6749 , 6750..6779 , 6780..6809 , 6810..6839 , 6840..6869 , 6870..6899 , 6900..6929 , 6930..6959 , 6960..6989 , 6990..7019 , 7020..7049 , 7050..7079 , 7080..7109 , 7110..7139 , 7140..7169 , 7170..7199 , 7200..7229 , 7230..7259 , 7260..7289 , 7290..7319 , 7320..7349 , 7350..7379 , 7380..7409 , 7410..7439 , 7440..7469 , 7470..7499 , 7500..7529 , 7530..7559 , 7560..7589 , 7590..7619 , 7620..7649 , 7650..7679 , 7680..7709 , 7710..7739 , 7740..7769 , 7770..7799 , 7800..7829 , 7830..7859 , 7860..7889 , 7890..7919 , 7920..7949 , 7950..7979 , 7980..8009 , 8010..8039 , 8040..8069 , 8070..8099 , 8100..8129 , 8130..8159 , 8160..8189 , 8190..8219 , 8220..8249 , 8250..8279 , 8280..8309 , 8310..8339 , 8340..8369 , 8370..8399 , 8400..8429 , 8430..8459 , 8460..8489 , 8490..8519 , 8520..8549 , 8550..8579 , 8580..8609 , 8610..8639 , 8640..8669 , 8670..8699 , 8700..8729 , 8730..8759 , 8760..8789 , 8790..8819 , 8820..8849 , 8850..8879 , 8880..8909 , 8910..8939 , 8940..8969 , 8970..8999 , 9000..9029 , 9030..9059 , 9060..9089 , 9090..9119 , 9120..9149 , 9150..9179 , 9180..9209 , 9210..9239 , 9240..9269 , 9270..9299 , 9300..9329 , 9330..9359 , 9360..9389 , 9390..9419 , 9420..9449 , 9450..9479 , 9480..9509 , 9510..9539 , 9540..9569 , 9570..9599 , 9600..9629 , 9630..9659 , 9660..9689 , 9690..9719 , 9720..9749 , 9750..9779 , 9780..9809 , 9810..9839 , 9840..9869 , 9870..9899 , 9900..9929 , 9930..9959 , 9960..9989 , 9990..10019 , 10020..10049 , 10050..10079 , 10080..10109 , 10110..10139 , 10140..10169 , 10170..10199 , 10200..10229 , 10230..10259 , 10260..10289 , 10290..10319 , 10320..10349 , 10350..10379 , 10380..10409 , 10410..10439 , 10440..10469 , 10470..10499 , 10500..10529 , 10530..10559 , 10560..10589 , 10590..10619 , 10620..10649 , 10650..10679 , 10680..10709 , 10710..10739 , 10740..10769 , 10770..10799 , 10800..10829 , 10830..10859 , 10860..10889 , 10890..10919 , 10920..10949 , 10950..10979 , 10980..11009 , 11010..11039 , 11040..11069 , 11070..11099 , 11100..11129 , 11130..11159 , 11160..11189 , 11190..11219 , 11220..11249 , 11250..11279 , 11280..11309 , 11310..11339 , 11340..11369 , 11370..11399 , 11400..11429 , 11430..11459 , 11460..11489 , 11490..11519 , 11520..11549 , 11550..11579 , 11580..11609 , 11610..11639 , 11640..11669 , 11670..11699 , 11700..11729 , 11730..11759 , 11760..11789 , 11790..11819 , 11820..11849 , 11850..11879 , 11880..11909 , 11910..11939 , 11940..11969 , 11970..11999 , 12000..12029 , 12030..12059 , 12060..12089 , 12090..12119 , 12120..12149 , 12150..12179 , 12180..12209 , 12210..12239 , 12240..12269 , 12270..12299 , 12300..12329 , 12330..12359 , 12360..12389 , 12390..12419 , 12420..12449 , 12450..12479 , 12480..12509 , 12510..12539 , 12540..12569 , 12570..12599 , 12600..12629 , 12630..12659 , 12660..12689 , 12690..12719 , 12720..12749 , 12750..12779 , 12780..12809 , 12810..12839 , 12840..12869 , 12870..12899 , 12900..12929 , 12930..12959 , 12960..12989 , 12990..13019 , 13020..13049 , 13050..13079 , 13080..13109 , 13110..13139 , 13140..13169 , 13170..13199 , 13200..13229 , 13230..13259 , 13260..13289 , 13290..13319 , 13320..13349 , 13350..13379 , 13380..13409 , 13410..13439 , 13440..13469 , 13470..13499 , 13500..13529 , 13530..13559 , 13560..13589 , 13590..13619 , 13620..13649 , 13650..13679 , 13680..13709 , 13710..13739 , 13740..13769 , 13770..13799 , 13800..13829 , 13830..13859 , 13860..13889 , 13890..13919 , 13920..13949 , 13950..13979 , 13980..14009 , 14010..14039 , 14040..14069 , 14070..14099 , 14100..14129 , 14130..14159 , 14160..14189 , 14190..14219 , 14220..14249 , 14250..14279 , 14280..14309 , 14310..14339 , 14340..14369 , 14370..14399 , 14400..14429 , 14430..14459 , 14460..14489 , 14490..14519 , 14520..14549 , 14550..14579 , 14580..14609 , 14610..14639 , 14640..14669 , 14670..14699 , 14700..14729 , 14730..14759 , 14760..14789 , 14790..14819 , 14820..14849 , 14850..14879 , 14880..14909 , 14910..14939 , 14940..14969 , 14970..14999 , 15000..15029 , 15030..15059 , 15060..15089 , 15090..15119 , 15120..15149 , 15150..15179 , 15180..15209 , 15210..15239 , 15240..15269 , 15270..15299 , 15300..15329 , 15330..15359 , 15360..15389 , 15390..15419 , 15420..15449 , 15450..15479 , 15480..15509 , 15510..15539 , 15540..15569 , 15570..15599 , 15600..15629 , 15630..15659 , 15660..15689 , 15690..15719 , 15720..15749, >>Next

Numbers_Word_Problems/511264: My number has 3 different digits, and all of them are even. This is divisibale by 5 and 8. The sum of its digits is greater than12.
1 solutions

Answer 342308 by ankor@dixie-net.com(15747) About Me  on 2011-10-08 22:05:53 (Show Source):
You can put this solution on YOUR website!
My number has 3 different digits, and all of them are even. This is divisibale by 5 and 8. The sum of its digits is greater than12.
:
How about 680?


Miscellaneous_Word_Problems/511387: a number consists of two digits the digits in the ten's place is 3 times the digit in the unit place if 54 is subtracted from the number the digits are reversed . The number is
1 solutions

Answer 342261 by ankor@dixie-net.com(15747) About Me  on 2011-10-08 16:31:31 (Show Source):
You can put this solution on YOUR website!
a number consists of two digits the digits in the ten's place is 3 times the digit in the unit place if 54 is subtracted from the number the digits are reversed.
The number is
:
Let the two digit number = 10x + y
:
"the ten's place is 3 times the digit in the unit place"
x = 3y
:
"if 54 is subtracted from the number the digits are reversed."
10x + y - 54 = 10y + x
10x - x = 10y - y + 54
9x = 9y + 54
simplify, divide by 9
x = y + 6
Replace x with 3y
3y = y + 6
3y - y = 6
2y = 6
y = 3
then
x = 3(3)
x = 9
:
93 is the number
:
:
Check solution in the statement:
"if 54 is subtracted from the number the digits are reversed."
93 - 54 = 39


Age_Word_Problems/511286: A man is 4 years older than his wife and 5 times as old as his son. When the son was born, the age of the wife was six-sevenths that of her husband's age. Find the age of each.
1 solutions

Answer 342204 by ankor@dixie-net.com(15747) About Me  on 2011-10-08 10:33:05 (Show Source):
You can put this solution on YOUR website!
A man is 4 years older than his wife and 5 times as old as his son.
When the son was born, the age of the wife was six-sevenths that of her husband's age.
Find the age of each.
:
Write an equation for each statement:
:
"A man is 4 years older than his wife"
m = w + 4
subtract 4 from both sides
w = (m-4)
:
"and 5 times as old as his son."
m = 5s
:
"When the son was born, the age of the wife was six-sevenths that of her husband's age."
w - s = 6%2F7(m-s)
multiply both sides by 7
7(w-s) = 6(m-s)
7w - 7s = 6m - 6s
7w = 6m - 6s + 7s
7w = 6m + s
replace w with (m-4)
7(m-4) = 6m + s
7m - 28 = 6m + s
7m - 6m = s + 28
m = s + 28
replace m with 5s
5s = s + 28
5s - s = 28
4s = 28
s = 7 yrs is the son's age
then
5(7) = 35 yrs is Dad's age
and
35-4 = 31 yrs is Mom's age
:
;
confirm these solutions in the statement:
"When the son was born, the age of the wife was six-sevenths that of her husband's age."
31 - 7 = 6%2F7(35 - 7)
24 = 6%2F7*28
24 = 24;


Rate-of-work-word-problems/511281: A man and a boy perform a task in 2 days. They completed one half of the task the first day with the man working 8 hours and the boy working 10 hours.
On the next day, they finished the task with the man working 4 hours and the boy working 15 hours. Find how many hours would take for each to do the task working alone.
1 solutions

Answer 342199 by ankor@dixie-net.com(15747) About Me  on 2011-10-08 09:48:35 (Show Source):
You can put this solution on YOUR website!
A man and a boy perform a task in 2 days.
They completed one half of the task the first day with the man working 8 hours and the boy working 10 hours.
On the next day, they finished the task with the man working 4 hours and the boy working 15 hours.
Find how many hours would take for each to do the task working alone.
:
Let m = time required by the man working alone
let b = time required by the boy working alone
Let the completed job = 1
:
1st day equation:
8%2Fm + 10%2Fb = 1%2F2
2nd day equation:
4%2Fm + 15%2Fb = 1%2F2
:
multiply the 2nd equation by 2, subtract the 1st equation
8%2Fm + 30%2Fb = 1
8%2Fm + 10%2Fb = 1%2F2
------------------- eliminates m, find b
20%2Fb = 1%2F2
b = 2(20)
b = 40 hrs, boy alone
:
Find m
8%2Fm + 10%2F40 = 1%2F2
reduce fraction
8%2Fm + 1%2F4 = 1%2F2
multiply by 4m
4(8) + m = 2m
32 = 2m - m
m = 32 hrs, man alone
:
:
Confirm this using the 2nd day equation
4%2F32 + 15%2F40 = 1%2F2
1%2F8 + 3%2F8 = 1%2F2


Travel_Word_Problems/511072: If you walked 7 1/4 miles at a rate of 2 1/2 miles per hour, how long in hours and minutes will it take to complete the walk?
1 solutions

Answer 342198 by ankor@dixie-net.com(15747) About Me  on 2011-10-08 09:19:17 (Show Source):
You can put this solution on YOUR website!
If you walked 7 1/4 miles at a rate of 2 1/2 miles per hour, how long in hours and minutes will it take to complete the walk?
:
7.25%2F2.5 = 2.9 hrs, 2 + .9(60) = 2 hrs 54 min


Proportions/511040: if a certain amount of food will last 50 days for 200 soldiers, how long will it last if 50 soldiers are added at the end of 20 days?
1 solutions

Answer 342130 by ankor@dixie-net.com(15747) About Me  on 2011-10-07 22:08:13 (Show Source):
You can put this solution on YOUR website!
if a certain amount of food will last 50 days for 200 soldiers, how long will it last if 50 soldiers are added at the end of 20 days?
:
50 * 200 = 10000 food days, total available food
:
Let d = no. of days for this scenario
200d + 50(d-20) = 10000
200d + 50d - 1000 = 10000
250d = 10000 + 1000
250d = 11000
d = 44 days


Linear-equations/511027: Will you please help me solve this equation? 2x+4=28
Thank you.:)
1 solutions

Answer 342121 by ankor@dixie-net.com(15747) About Me  on 2011-10-07 21:56:54 (Show Source):
You can put this solution on YOUR website!
2x + 4 = 28
Subtract 4 from both sides
2x = 28 - 4
2x = 24
divide both sides by 2
x = 24%2F2
x = 12
:
:
Confirm this by replacing x with 12 in the original equation
2(12) + 4 = 28


Numbers_Word_Problems/511030: I am a 4 digit whole number my digits decrease by 2 from the thousand place to the ones place each of my digits is an even number . what number am i ?

1 solutions

Answer 342117 by ankor@dixie-net.com(15747) About Me  on 2011-10-07 21:51:01 (Show Source):
You can put this solution on YOUR website!
I am a 4 digit whole number
a, b, c, d
:
my digits decrease by 2 from the thousand place to the ones place
a, (a-2), (a-4), (a-6), (a-6)
each of my digits is an even number . what number am i ?
a has to be 8, then the number is:
8642


Distributive-associative-commutative-properties/511029: 7=-2x+7
1 solutions

Answer 342116 by ankor@dixie-net.com(15747) About Me  on 2011-10-07 21:48:39 (Show Source):
You can put this solution on YOUR website!
7 = -2x + 7
subtract 7 from both sides
0 = -2x
therefore
x = 0


Triangles/511010: The largest angle of a triangle is 6 times the smallest angle and it is 3 times the other angle. Find the size of the smallest angle.
A. 40 degrees
B. 30 degrees
C. 20 degrees
D. 10 degrees
So far I have tried all answers but they don't come to 180 degrees. My first thought was 30 degrees because then the other angle would be 90 but then that would require the largest angle to be 180. Now there is a 200 degree triangle but I am not sure if I am correct.
All help would be greatly appreciated!
Thank you,
Ann
1 solutions

Answer 342114 by ankor@dixie-net.com(15747) About Me  on 2011-10-07 21:42:26 (Show Source):
You can put this solution on YOUR website!
Let A = the largest angle
Let B = the "other angle
let C = the smallest angle
:
Write an equation for each statement, get them all in terms of A:
:
"The largest angle of a triangle is 6 times the smallest angle"
A = 6C
Divide both sides by 6, and we have
C = 1%2F6A
:
and it is 3 times the other angle.
A = 3B
Divide both sides by 3
B = 1%2F3A
:
A + B + C = 180
therefore
A + 1%2F3A + 1%2F6A = 180
Get rid of those annoying fractions, multiply by 6 and we have
6A + 2A + A = 6(180)
9A = 1080
A = 1080%2F9
A = 120 degrees is the largest angle
:
Find the size of the smallest angle.
C = 1%2F6A
C = 1%2F6*120
C = 20 degrees
:
:
You can check this; find the other angle and ensure they add up to 180


Age_Word_Problems/511004: hi i need help with this kind of word problem because i dont understand it i would really appreciate it if you can also show the work so i can understand it more thank you
kristine, jake, and lexi are cousins. jake is 2 years older than kristine and lexi is 8 years younger than jake. the sum of their ages is 35. find lexi's age.
thanks again!
1 solutions

Answer 342113 by ankor@dixie-net.com(15747) About Me  on 2011-10-07 21:28:46 (Show Source):
You can put this solution on YOUR website!
kristine, jake, and lexi are cousins.
Write an equation for each statement
:
jake is 2 years older than kristine
j = k + 2
we want to get k in terms of j, like the next equation,
subtract 2 from both sides and we can write it
k = j - 2
:
lexi is 8 years younger than jake.
l = j - 8
:
the sum of their ages is 35.
k + j + l = 35
Replace k with (j-2) and l with (j-8)
(j-2) + j + (j-8) = 35
combine like terms
3j - 10 = 35
3j = 35 + 10
3j = 45
j = 45%2F3
j = 15 yr is Jake's age
:
Find L's age
l = 15 - 8
l = 7 yrs is Lexi's age
:
:
You should check this by finding k's age and see that the sum of all three ages add up 35




Linear_Equations_And_Systems_Word_Problems/510894: two girls each have a different number of candies. jasmine said if you give me 5 i will have as many you. Wendy said if you give me 5 i will have twice as many as you.how many candies did each girl have?

1 solutions

Answer 342103 by ankor@dixie-net.com(15747) About Me  on 2011-10-07 20:41:11 (Show Source):
You can put this solution on YOUR website!
Let j = original no. Jasmine had
let w = original no. Wendy had
:
two girls each have a different number of candies.
:
"jasmine said if you give me 5 i will have as many you."
j + 5 = w - 5
j = w - 5 - 5
j = w - 10
:
Wendy said if you give me 5 i will have twice as many as you
w + 5 = 2(j-5)
w + 5 = 2j - 10
w = 2j - 10 - 5
w = 2j - 15
replace j with (w-10)
w = 2(w-10) - 15
w = 2w - 20 - 15
w - 2w = -35
-w = -35
w = 35 candies had Wendy
then
j = 35 - 10
j = 25 candies had Jasmine
:
Check and see that this is true in both statements








Numbers_Word_Problems/510994: one half the sum of a givin number and six is the same as ten decreased by the difference obtained when five is subtracted from the number. find the number
1 solutions

Answer 342101 by ankor@dixie-net.com(15747) About Me  on 2011-10-07 20:27:44 (Show Source):
You can put this solution on YOUR website!
one half the sum of a given number and six is the same as ten decreased by the difference obtained when five is subtracted from the number. find the number
:
Solve this equation
1%2F2(n+6) = 10 - (n-5)
:
1%2F2(n+6) = 10 - n + 5
:
1%2F2(n+6) = 15 - n
multiply both sides by 2
n + 6 = 2(15-n)
you should be able to finish this, check your solution in the original equation


Percentage-and-ratio-word-problems/511287: The numerator and denominator of a fraction are together equal to 100. Increase the numerator by 18 and decrease the denominator by 16 and the fraction is doubled. What is the fraction?
1 solutions

Answer 342100 by ankor@dixie-net.com(15747) About Me  on 2011-10-07 20:11:48 (Show Source):
You can put this solution on YOUR website!
The numerator and denominator of a fraction are together equal to 100.
n + d = 100
d = (100-a)
:
Increase the numerator by 18 and decrease the denominator by 16 and the fraction is doubled.
%28%28n%2B18%29%29%2F%28%28d-16%29%29 = %282n%29%2Fd
Cross multiply
d(n+18) = 2n(d-16)
dn + 18d = 2dn - 32n
18d = 2dn - dn - 32n
18d = dn - 32n
:
replace d with (100-n)
18(100-n) = n(100-n) - 32n
1800 - 18n = 100n - n^2 - 32n
:
Arrange as a quadratic equation on the left
n^2 - 18n -100n + 32n + 1800 = 0
n^2 - 86n + 1800 = 0
:
You can use the quadratic formula here but it will factor to:
(n-50){n-36) = 0
Two solutions (remember n + d = 100)
n = 50, then d = 50, but of course 50/50 = 1, not really considered a fraction
and
n = 36, then d = 64
:
36%2F64 is the fraction we want, reduces to 9%2F16
:
:
Check this in the original equation:
%28%2836%2B18%29%29%2F%28%2854-16%29%29 = 54%2F48 reduces to: 18%2F16 twice the original fraction



Exponents/510926: My question is when doing a problem example, 10x-22=29-7x . I under stand that I need to get my X on one side of the equation and I do that by adding or subracting on both sides. How do you know witch side to pick. and wether to add or subract?
10x-22= 29-7x
-7x 29
3x -51 =
-10x -51 divid each by 10 x= 5.1
Thank you
1 solutions

Answer 342056 by ankor@dixie-net.com(15747) About Me  on 2011-10-07 15:58:27 (Show Source):
You can put this solution on YOUR website!
10x-22=29-7x
Generally we want it on the left, but it is not necessary, sometimes it's easier to select the side which will make x positive (the final solution, x has to be positive)
:
add 7x to both sides, that gets rid of x on the right side
10x + 7x - 22 = 29 -7x +7x
17x - 22 = 29
:
add 22 to both sides, that gets the numbers on the right side
17x - 22 + 22 = 29 + 22
17x = 51
;
Divide both sides by 17 to get 1x
x = 51%2F17
x = 3
:
:
See if this works out in the original equation, replace x with 3
10(3) - 22 = 29 - 7(3)
30 - 22 = 29 - 21
8 = 8; equality reigns, x=3 is correct!


Travel_Word_Problems/510833: plane A is 40 miles south and 100 miles east of plane B. plane A is flying 2 miles west for ever mile it flies north, while plane B is flying 3 miles east for every mile it flies south. where do there paths cross?

1 solutions

Answer 342053 by ankor@dixie-net.com(15747) About Me  on 2011-10-07 15:43:15 (Show Source):
You can put this solution on YOUR website!
plane A is 40 miles south and 100 miles east of plane B. plane A is flying 2 miles west for ever mile it flies north, while plane B is flying 3 miles east for every mile it flies south. where do there paths cross?
:
Write a linear equation for the line taken by each plane
With O origin,
Plane B: When x = 0, y = 40
"plane B is flying 3 miles east for every mile it flies south."
Using the rise/run rule the slope will be -1/3 (negative because it's moving southward)
y = -1%2F3x + 40
+graph%28+300%2C+200%2C+-20%2C+130%2C+-10%2C+60%2C+-.333x%2B40%29+
:
Plane A: x intercept = +100, find the y intercept
"plane A is flying 2 miles west for ever mile it flies north,"
Slope = -1/2; find the y intercept (b)
-1%2F2(100) + b = 0
-50 + b = 0
b = 50
the equation for A: y = -1%2F2x + 50, plot this with B
intercept x=60 mi east, y=20 mi north
+graph%28+300%2C+200%2C+-20%2C+130%2C+-10%2C+60%2C+-.333x%2B40%2C+-.5x%2B50%29+



Age_Word_Problems/510937: Matthew is 3 times as old as Jenny. In 7 years he will be twice as old as she will be then. How old is each now.
I have figured out the answer, but not how to work the problem. I tried:
x + 1/3X = 2 times X + 7 + 1/3 X + 7 which is completely wrong. I have not done a problen where I didn't have a specific number for an answer. Reid
1 solutions

Answer 341940 by ankor@dixie-net.com(15747) About Me  on 2011-10-06 22:04:56 (Show Source):
You can put this solution on YOUR website!
Matthew is 3 times as old as Jenny.
m = 3j
In 7 years he will be twice as old as she will be then.
m + 7 = 2(j+7)
m + 7 = 2j + 14
m = 2j + 14 - 7
m = 2j + 7
Replace m with 3j
3j = 2j + 7
3j - 2j = 7
j = 7 yrs
then obviously
m = 21 yrs




Rectangles/510709: THE POPULATION OF TWO TOWNS ARE CHANGING AT STEADY RATES. ONE TOWN HAS A POPULATION OF 25,500. ITS POPULATION IS INCREASING BY 200 PEOPLE EACH YEAR. THE OTHER TOWN HAS A POPULATION OF 47,900. ITS POPULATION IS DECREASING BY 800 PEOPLE EACH YEAR. IF THE RATE FOR EACH TOWN REMAINS THE SAME,IN HOW MANY YEARS WILL THE POPULATION BE THE SAME ?

1 solutions

Answer 341930 by ankor@dixie-net.com(15747) About Me  on 2011-10-06 21:46:35 (Show Source):
You can put this solution on YOUR website!
THE POPULATION OF TWO TOWNS ARE CHANGING AT STEADY RATES.
ONE TOWN HAS A POPULATION OF 25,500. ITS POPULATION IS INCREASING BY 200 PEOPLE EACH YEAR.
THE OTHER TOWN HAS A POPULATION OF 47,900. ITS POPULATION IS DECREASING BY 800 PEOPLE EACH YEAR.
IF THE RATE FOR EACH TOWN REMAINS THE SAME,IN HOW MANY YEARS WILL THE POPULATION BE THE SAME ?
:
let t = no. of years for this to happen
An equation for the 1st town
f(t) = 200t + 25500
An equation for the 2nd town
f(t) = -800t + 47900
:
Find when the populations are equal
1st town = 2nd town
200t + 25500 = -800t + 47900
200t + 800t = 47900 - 25500
1000t = 22400
t = 22400%2F1000
t = 22.4 yrs the will be the same
:
:
Check this by finding the population of each after 22.4 yrs
200(22.4) + 25500 = 29980
-800(22.4)+ 47900 = 29980


Proportions/510705: It took 2 hours for 1 person to sweep the yard. How long would it take 4 persons working at the same speed to do the same job?
My attempt:
Firstly, it has to take less than 2 hours since the number of persons has been increased.
The # of persons has been increased by 3 so:
Increase in persons/ Time = 3 / 2
1.5 hours.

1 solutions

Answer 341925 by ankor@dixie-net.com(15747) About Me  on 2011-10-06 21:34:43 (Show Source):
You can put this solution on YOUR website!
It took 2 hours for 1 person to sweep the yard.
How long would it take 4 persons working at the same speed to do the same job?
:
Assume they each had broom, and worked at the same rate (2hrs to complete the job)
:
Let t = time required by 4 men doing the job
Let the completed job = 1
:
Each will do a fraction of the job, and the fractions will add up to 1
t%2F2 + t%2F2 + t%2F2 + t%2F2 = 1
Multiply thru by 2
t + t + t + t = 2
4t = 2
t = 2%2F4
t = 1%2F2 hr with all 4 men sweeping


Travel_Word_Problems/510665: The receiver in a football game catches a pass 45 yards from the end zone. He runs 12 yards per second
toward the end zone. The defender is 10 yards behind the receiver. He runs 14 yards per second toward the end zone. Will the defender catch the receiver before he reaches the end zone? If not, by how much time will the receiver beat him?
1 solutions

Answer 341896 by ankor@dixie-net.com(15747) About Me  on 2011-10-06 20:47:55 (Show Source):
You can put this solution on YOUR website!
The receiver in a football game catches a pass 45 yards from the end zone.
He runs 12 yards per second toward the end zone.
The defender is 10 yards behind the receiver.
He runs 14 yards per second toward the end zone.
If not, by how much time will the receiver beat him?:
:
Find the time it takes the receiver to cover 45 yds (time = dist/speed)
45%2F12 = 3.75 sec
Find the time it takes the defender to cover 55 yds
55%2F14 ~ 3.9 sec
:
Receiver will reach the goal line: 3.9 - 3.75 = .15 sec before the defender


Points-lines-and-rays/510614: An ice cube that measures 3 centimeters on each side contains how many cubic centimeters of ice? Include correct units with your solution.
1 solutions

Answer 341889 by ankor@dixie-net.com(15747) About Me  on 2011-10-06 19:29:48 (Show Source):
You can put this solution on YOUR website!
3 * 3 * 3 = 27 cu/cm


Money_Word_Problems/510402: Natalie had some nickels, Dirk had some dimes, and Quincy had some quarters. Dirk has 5 more dimes than Quincy has quarters. If Natalie gives a nickle to Dirk, Dirk gives a dime to Quincy and Quincy gives a quarter to Natalie, they will all have the same amount of money. How many coins did each originally have?
1 solutions

Answer 341826 by ankor@dixie-net.com(15747) About Me  on 2011-10-06 09:48:16 (Show Source):
You can put this solution on YOUR website!
Natalie had some nickels, Dirk had some dimes, and Quincy had some quarters.
Dirk has 5 more dimes than Quincy has quarters.
If Natalie gives a nickle to Dirk, Dirk gives a dime to Quincy and Quincy gives a quarter to Natalie, they will all have the same amount of money.
How many coins did each originally have?
:
Let n = no. nickels
let d = no. of dimes
let q = no. 0f quarters
:
"Dirk has 5 more dimes than Quincy has quarters."
d = q + 5
:
"If Natalie gives a nickle to Dirk, Dirk gives a dime to Quincy and Quincy gives a quarter to Natalie, "
Results:
Natalie amt:
.05n - .05 + .25
.05n + .20
Dirk amt:
.10d + .05 - .10
.10d - .05
Quin amt:
.25q + .10 - .25
.25q - .15
:
"They all have the same amt of money"
.25q - .15 = .10d -.05
.25q = .10d - .05 + .15
.25q = .10d + .10
We know,"Dirk has 5 more dimes than Quincy has quarters." d = q+5, replace d
.25q = .10(q+5) + .10
.25q = .10q + .50 + .10
.25q - .10q = .60
.15q = .60
q = .60%2F.15
q = 4 quarters
then
d = 4 + 5
d = 9 dimes
Find n
.05n + .20 = .10(9) - .05
.05n = .90 - .05 - .20
.05n = .65
n = .65%2F.05
n = 13 nickels
:
Summarize, originally, 13 nickels, 9 dimes, 4 quarters
:
You can check this finding the amt each had after all these coin transfers, they each had 85 cents


Travel_Word_Problems/510217: Jason' s house is 5 2/3 miles from the airport. He drives to the airport at an average speed of 61 mph, then drives home at an average speed of 57 mph. Approximately how many minutes was his round trip?
1 solutions

Answer 341797 by ankor@dixie-net.com(15747) About Me  on 2011-10-05 22:08:53 (Show Source):
You can put this solution on YOUR website!
Jason' s house is 5 2/3 miles from the airport.
He drives to the airport at an average speed of 61 mph, then drives home at an average speed of 57 mph.
Approximately how many minutes was his round trip.
:
Convert 5 2/3 to 5.667 miles
:
Write time equation time = dist/speed
:
t = 5.667%2F61 + 5.667%2F57 = .0929 + .0994 = .1923 hr
Convert to minutes
t = .1923 * 60 = 11.5 minutes


Numeric_Fractions/509893: 6 7/8 divided by 21 3/5 = 4 3/4 divided x
I cross multiplied getting 6 7/8x = (7 3/4)(21 3/5)
divided by 6 7/8 on both sides
x=(7 3/4)(21 3/5)/6 7/8
i tried turned the fractions into improper fractions but from this point i need step by step instruction. I have not been able to resolve correctly from this point.
1 solutions

Answer 341756 by ankor@dixie-net.com(15747) About Me  on 2011-10-05 19:36:48 (Show Source):
You can put this solution on YOUR website!
6 7/8 divided by 21 3/5 = 4 3/4 divided x
Using improper fractions
%2855%2F8%29%2F%28108%2F5%29 = %2819%2F4%29%2Fx
cross mutiply
%2855%2F8%29x = %28108%2F5%29%2A%2819%2F4%29
%2855%2F8%29x = %282052%2F20%29
multiply both sides by 40, results:
5(55x) = 2(2052)
275x = 4104
x = 4104%2F275
which is
x = 14254%2F275


Mixture_Word_Problems/510000: A radiator contains 13 quarts of fluid, 25% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 50% antifreeze?
1 solutions

Answer 341699 by ankor@dixie-net.com(15747) About Me  on 2011-10-05 17:19:03 (Show Source):
You can put this solution on YOUR website!
A radiator contains 13 quarts of fluid, 25% of which is antifreeze.
How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 50% antifreeze?
:
Let x = amt of original antifreeze to be drained, and the amt of pure antifreeze to be added
:
a simple equation
.25(13-x) + x = .50(13)
3.25 - .25x + x = 6.5
-.25x + 1x = 6.5 - 3.25
.75x = 3.25
x = 3.25/.75
x = 41%2F3 quarts removed, then the same amt of pure antifreeze added


Age_Word_Problems/509824: Mother was 25 yeras old when I was born. Father was 2 years older than Mother. The sum of our present ages is 91. How old am I?
1 solutions

Answer 341697 by ankor@dixie-net.com(15747) About Me  on 2011-10-05 17:10:59 (Show Source):
You can put this solution on YOUR website!
Let x = your age now
then
(x+25) = Mom's age
and
(x+27) = Dad's age
:
The sum
x + (x+25) + (x+27) = 91
3x + 52 = 91
3x = 91 - 52
3x = 39
x = 39/3
x = 13 yrs is your present age
:
:
Check this; find the sum of all the ages:
13 + (13+25) + (13+27) =
13 + 38 + 40 = 91


Travel_Word_Problems/509874: a car travels at an average speed of 30.0 m/s for 0.8 h. find the total distance traveled in km.
1 solutions

Answer 341695 by ankor@dixie-net.com(15747) About Me  on 2011-10-05 17:01:40 (Show Source):
You can put this solution on YOUR website!
a car travels at an average speed of 30.0 m/s for 0.8 h. find the total distance traveled in km.
:
Find the total no. of meters traveled in .8*3600 sec, divide that by 1000 to km
:
%28%2830%2A.8%2A3600%29%29%2F%28%281000%29%29 = 86.4 km


Linear_Equations_And_Systems_Word_Problems/509792: mary began walking home from school, heading south at a rate of 4 mph. Sharon left school at the same time heading north at 6 mph. how long will it take for them to be 3 miles apart?

1 solutions

Answer 341690 by ankor@dixie-net.com(15747) About Me  on 2011-10-05 16:39:30 (Show Source):
You can put this solution on YOUR website!
mary began walking home from school, heading south at a rate of 4 mph.
Sharon left school at the same time heading north at 6 mph.
how long will it take for them to be 3 miles apart?
:
Let t = time (in hrs) for them to be 3 mi apart
:
Write a distance equation; dist = speed*time
4t + 6t = 3
10t = 3
t = 3%2F10
t = .3 hrs (.3*60 = 18 min)
:
:
Check by finding the distance each walked
.3(4) = 1.2 mi
.3(6) = 1.8 mi
---------------
total: 3 mi


Money_Word_Problems/509795: Three individuals form a partnership and agree to divide the profits equally. X invests 9,000, Y invests 7,000 and Z invests 4,000. If the profits are 4800, how much less does X receive than if the profits were divided in proportion to the amount invested? Answer is 560. How is this arrived at is my question. I need the mechanics of this solution. Please and thank you sincerely, Claudette
1 solutions

Answer 341638 by ankor@dixie-net.com(15747) About Me  on 2011-10-05 10:47:46 (Show Source):
You can put this solution on YOUR website!
Three individuals form a partnership and agree to divide the profits equally.
X invests 9,000, Y invests 7,000 and Z invests 4,000.
If the profits are 4800, how much less does X receive than if the profits were divided in proportion to the amount invested?
:
To divide the profits according to amt invested, we can use fractions with the denominator of 9+7+4 = 20, therefore:
:
x should get 9%2F20 of the profits
y should get 7%2F20
z should get 4%2F20
:
Assume the total profit was 4800 and divided equally
4800%2F3 = 1600 each
:
If the profits divide based on invested amts, then x: 9%2F20*4800 = $2160
:
to find the difference in the two methods of distribution:
2160 = 1600 = $560 difference


Travel_Word_Problems/509436: if a boat traveled 160 miles in 5 hours. what is its speed in feet per second
1 solutions

Answer 341527 by ankor@dixie-net.com(15747) About Me  on 2011-10-04 21:33:03 (Show Source):
You can put this solution on YOUR website!
if a boat traveled 160 miles in 5 hours. what is its speed in feet per second
:
Convert no. of miles to ft, divide by the no. of seconds in 5 hrs
:
%28%28160%2A5280%29%29%2F%28%285%2A3600%29%29 = 46.93 ft/sec


Reduction-of-unit-multipliers/509334: please help me on this word question. A car is traveling at a speed of 55 miles per hour, what is the cars speed in feet per secound?
1 solutions

Answer 341518 by ankor@dixie-net.com(15747) About Me  on 2011-10-04 21:19:22 (Show Source):
You can put this solution on YOUR website!
A car is traveling at a speed of 55 miles per hour,
what is the cars speed in feet per second?
:
Convert miles to feet then divide that by the no. of seconds in an hour
:
%28%2855%2A5280%29%29%2F3600 = 802%2F3 ft per second