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: 15731 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

Numeric_Fractions/483463: I need a tutor to please help me?
Divide and simplify
r^2+8r/r^2+5r-24śr/r+3
1 solutions

Answer 330854 by ankor@dixie-net.com(15746) About Me  on 2011-08-19 19:57:22 (Show Source):
You can put this solution on YOUR website!
Divide and simplify
rewrite it:
%28r%5E2%2B8r%29%2F%28r%5E2%2B5r-24%29
-----------
r%2F%28r%2B3%29
invert the dividing fraction, and multiply
%28r%5E2%2B8r%29%2F%28r%5E2%2B5r-24%29 * %28r%2B3%29%2Fr
Factor
r%28r%2B8%29%2F%28%28r%2B8%29%28r-3%29%29 * %28r%2B3%29%2Fr
look closely, we can cancel r and (r+8), leaving a simple single fraction
1%2F%28r-3%29 * (r+3) = %28r%2B3%29%2F%28r-3%29


Linear_Algebra/483452: A rectangular table can be be set against one wall of a room, as shown, so that it is 5 feet from the two side walls and 8 feet from the fourth wall. If the perimeter of the room is 58 feet and the combined length of the three exposed sides of the table is 17.5 feet, then what are the dimensions of the table?
1 solutions

Answer 330852 by ankor@dixie-net.com(15746) About Me  on 2011-08-19 19:34:13 (Show Source):
You can put this solution on YOUR website!
A rectangular table can be be set against one wall of a room, as shown, so that it is 5 feet from the two side walls and 8 feet from the fourth wall.
If the perimeter of the room is 58 feet
:
let x = length of the table against the wall
let y = the width of the table
the room perimeter equation
2(x+10) + 2(y+8) = 58
2x + 20 + 2y + 16 = 58
2x + 2y + 36 = 58
2x + 2y = 58 - 36
2x + 2y = 22
simplify, divide by 2
x + y = 11
:
"and the combined length of the three exposed sides of the table is 17.5 feet,"
x + 2y = 17.5
:
Use elimination with these two equations
x + 2y = 17.5
x + 2 = 11
---------------subtraction eliminates x, find y
y = 6.5 ft is the width of the table
:
Find x:
x + y = 11
x + 6.5 = 11
x = 11 - 6.5
x = 4.5 ft is the length of the table against the wall
:
then what are the dimensions of the table? 6.5' by 4.5'
;
:
Check solution in the room perimeter equation
2(4.5+10) + 2(6.5+8) =


Travel_Word_Problems/483316: It takes Omar 10 minutes to bicycle 1 mile to school and 20 minutes to return the same way. What is his average speed in
miles per hour for the round trip?
1 solutions

Answer 330844 by ankor@dixie-net.com(15746) About Me  on 2011-08-19 18:54:26 (Show Source):
You can put this solution on YOUR website!
It takes Omar 10 minutes to bicycle 1 mile to school and 20 minutes to return the same way.
What is his average speed in miles per hour for the round trip?
:
10 + 20 = 30 min which is .5 hrs
Round trip of 2 mi takes .5 hrs
2%2F.5 = 4 mph is the average for the round trip


Polynomials-and-rational-expressions/483410: Could someone help me with this please, I'm having a hard time finding my answer..
For jumps that exceed 13 sec, the polynomial 173t − 369 can be used to approximate the distance, in feet, that a skydiver has fallen in t seconds. Approximately how far has a skydiver fallen 20 sec after having jumped from a plane?

1 solutions

Answer 330816 by ankor@dixie-net.com(15746) About Me  on 2011-08-19 15:50:00 (Show Source):
You can put this solution on YOUR website!

how about
173(20) − 369 = 3031 ft


Age_Word_Problems/483223: whats five times the sum of a number j and four

1 solutions

Answer 330756 by ankor@dixie-net.com(15746) About Me  on 2011-08-19 09:07:48 (Show Source):


Age_Word_Problems/483009: A says to B "my present age is five times of your age when i was as old as you are now." The sum of their ages is 48 years. Find their age.
1 solutions

Answer 330657 by ankor@dixie-net.com(15746) About Me  on 2011-08-18 21:23:15 (Show Source):
You can put this solution on YOUR website!
A says to B "my present age is five times of your age when i was as old as you are now."
A = 5B
:
The sum of their ages is 48 years.
A + B = 48
B = (48-A)
Replace B with (48-A) in the 1st equation
A = 5(48-A)
A = 240 - 5A
A + 5A = 240
6A = 240
A = 240%2F6
A = 40 yrs old
and
B = 48-40
B = 8 yrs old
:


Rate-of-work-word-problems/482971: A swimming pool is 50 meters long. A boy can swim across the length and then return to his starting position in 5 meters. What is his swimming speed in km/hr? (Ans. 1.2 km/hr)
1 solutions

Answer 330559 by ankor@dixie-net.com(15746) About Me  on 2011-08-18 15:58:04 (Show Source):
You can put this solution on YOUR website!
Assuming you mean:
A swimming pool is 50 meters long. A boy can swim across the length and then return to his starting position in 5 MINUTES.
What is his swimming speed in km/hr?
:
Find the meter/min, multiply by 60 to get meter/hr, divide by 1000 to get km/hr
100%2F5 * 60 * 1%2F1000 = 1.2 km/hr


Square-cubic-other-roots/482954: What is 4 more than the square root of 2x equals x? Please include all work.
1 solutions

Answer 330558 by ankor@dixie-net.com(15746) About Me  on 2011-08-18 15:48:16 (Show Source):
You can put this solution on YOUR website!
What is 4 more than the square root of 2x equals x?
sqrt%282x%29 + 4 = x
:
subtract 4 from both sides
sqrt%282x%29 = x - 4
:
square both sides
2x = (x-4)^2
:
FOIL (x-4(x-4)
2x = x^2 - 8x + 16
:
Subtract 2x from both sides
0 = x^2 - 8x - 2x + 16
:
A quadratic equation
x^2 - 10x + 16 = 0
:
Factor to
(x-8)(x-2) = 0
:
Two solutions
x = 8
x = 2
:
Check both solutions in original equation
x=8
sqrt%282%288%29%29 + 4 = 8
sqrt%2816%29 + 4 = 8; a good solution
and
x=2
sqrt%282%282%29%29 + 4 = 2
sqrt%284%29 + 4 = 2
2 + 4 does not = 2, therefore not a good solutions


Permutations/482825: 5 digits
100 digit is 2 more then my tens digit, my ones digit is 4 more then my thousands, the sum of my number is 8
1 solutions

Answer 330499 by ankor@dixie-net.com(15746) About Me  on 2011-08-17 21:22:50 (Show Source):
You can put this solution on YOUR website!
5 digits: a, b, c, d, e
:
100 digit is 2 more then my tens digit,
c = d+2
:
my ones digit is 4 more then my thousands,
e = b+4
:
the sum of my number is 18
a + b + c + d + e = 8
Substitute:
a + b + (d+2) + d + (b+4) = 18 (clearly a sum of 8 is impossible)
a + 2b + 2d + 6 = 18
a + 2b + 2d = 18 - 6
a + 2b + 2d = 12
How about a=2, b=3, d=2
then
a=2
b=3
c=4
d=2
e=7
the number: 23427 is one possibility


Travel_Word_Problems/482917: The elevator in the washington monument takes 60 seconds to rise 500ft. What is the average speed in miles per hour?
1 solutions

Answer 330493 by ankor@dixie-net.com(15746) About Me  on 2011-08-17 21:03:10 (Show Source):
You can put this solution on YOUR website!
The elevator in the washington monument takes 60 seconds to rise 500ft. What is the average speed in miles per hour?
:
Find how many feet in 1 hr, change that to mph, by dividing by 5280
%2860%2A500%29%2F5280 = 5.68 mph


Mixture_Word_Problems/482792: Every minute an bacteria splits into two halves, each one growing to the same size as parent. In 60 minute, the multitude of a germ completely fill the container they are in. to the nearest minute , how much time was needed to half fill the container?
1 solutions

Answer 330488 by ankor@dixie-net.com(15746) About Me  on 2011-08-17 20:47:43 (Show Source):
You can put this solution on YOUR website!
Every minute an bacteria splits into two halves, each one growing to the same size as parent.
In 60 minute, the multitude of a germ completely fill the container they are in.
to the nearest minute , how much time was needed to half fill the container?
:
wouldn't it be 59 min, since is doubles every minute?


Distributive-associative-commutative-properties/482901: My son and I have been working on his homework for a few hours now. I can solve the problem, but I can't show the work on how to solve it how it is 'suppose to be' done. The problem is 7-2n/7+2n=1/18. I know that I can use a 'reverse system to solve for the right side of the = sign, but I've been out of school too long to explain to him the correct way to solve it nor explain how to show the work. Please help me explain this to him
1 solutions

Answer 330486 by ankor@dixie-net.com(15746) About Me  on 2011-08-17 20:41:48 (Show Source):
You can put this solution on YOUR website!
Assume this is The problem:
%287-2n%29%2F%287%2B2n%29 = 1%2F18
Cross multiply and you have
7 + 2n = 18(7-2n)
7 + 2n = 126 - 36n
2n + 36n = 126 - 7
38n = 119
n = 119%2F38


Volume/482509: a trough is formed by nailing together, edge to edge, two boards 11 ft. in length, so that the right section is a right triangle. If 15 gal. of water are poured into the trough ad if the trough is held level so that a right section of the water is an isosceles right triangle, how deep is the water? (231 cu. in. = 1 gal.)
1 solutions

Answer 330434 by ankor@dixie-net.com(15746) About Me  on 2011-08-17 17:13:09 (Show Source):
You can put this solution on YOUR website!
a trough is formed by nailing together, edge to edge, two boards 11 ft. in length, so that the right section is a right triangle.
If 15 gal. of water are poured into the trough ad if the trough is held level so that a right section of the water is an isosceles right triangle, how deep is the water? (231 cu. in. = 1 gal.)
:
Find the vol of the water in cu/in
15 * 231 = 3465 cu/in
:
Change 11 ft: 11*12 = 132 inches
:
Find the triangular area of the end of the trough; 3465/132 = 26.25 sq/inches
:
Let x = width of the trough; A right triangle area =
1%2F2x*x = 26.25
x^2 = 2(26.25)
x = sqrt%2852.5%29
x = 7.2457 inches the width of each board
;
Find the width of the trough which is the the hypotenuse
h = sqrt%28x%5E2+%2B+x%5E2%29
h = sqrt%2852.5+%2B+52.5%29
h = sqrt%28105%29
h = 10.247 inches wide, also the base of the triangle
:
Find the altitude of the triangle, which is the depth of the water; we know it's area
1%2F2*10.247*d = 26.25
10.247d = 2(26.25)
10.247d = 52.5
d = 52.5%2F10.247
d = 5.12 inches is the depth of the water
:
:
Check this by finding the vol of water with these measurements
1%2F2*10.247*5.12*132 = 3462.366 cu/inches which is about 15 gal


Problems-with-consecutive-odd-even-integers/482494: find three consecutive even integers with the property that the second integer plus the third integer is 18 more than the first integer. how do you show work for this?
1 solutions

Answer 330425 by ankor@dixie-net.com(15746) About Me  on 2011-08-17 16:18:46 (Show Source):
You can put this solution on YOUR website!
let the three consecutive even integers be; x, (x+2), (x+4)
:
write an equation for the phrase:
"the second integer plus the third integer is 18 more than the first integer."
(x+2) + (x+4) = x + 18
2x + 6 = x + 18
2x - x = 18 - 6
x = 12, then 14 and 16 are next even integers
:
Check that in the statement;
the second integer plus the third integer is 18 more than the first integer."
14 + 16 = 12 + 18
:
Not that hard, right?


Expressions-with-variables/482653: rationalize the denominator
sqrt(u)-sqrt(v)/sqrt(u)+sqrt(v)
1 solutions

Answer 330422 by ankor@dixie-net.com(15746) About Me  on 2011-08-17 16:09:12 (Show Source):
You can put this solution on YOUR website!
Multiply by the conjugate of the denominator over itself
:
%28sqrt%28u%29-sqrt%28v%29%29%2F%28sqrt%28u%29%2Bsqrt%28v%29%29 * %28sqrt%28u%29-sqrt%28v%29%29%2F%28sqrt%28u%29-sqrt%28v%29%29 FOIL%2FFOIL: %28u-sqrt%28uv%29-sqrt%28uv%29%2Bv%29%2F%28u-+sqrt%28uv%29%2Bsqrt%28uv%29-v%29 = %28u-2sqrt%28uv%29%2Bv%29%2F%28u-v%29


Rectangles/482608: The width of a rectangle is 2x + 3 and its length is x + 7. If the area is 40 in^2, what are the dimensions of the Rectangle?
1 solutions

Answer 330419 by ankor@dixie-net.com(15746) About Me  on 2011-08-17 15:43:56 (Show Source):
You can put this solution on YOUR website!
The width of a rectangle is 2x + 3 and its length is x + 7.
If the area is 40 in^2, what are the dimensions of the Rectangle?
:
We know the area = width * length, therefore
(2x+3)*(x+7) = 40
FOIL
2x^2 + 14x + 3x + 21 = 40
2x^2 + 17x + 21 - 40 = 0
2x^2 + 17x - 19 = 0
You can use the quadratic formula, but this will factor to
(x-1)(2x+19) = 0
the positive solution is what we want here
x = 1
therefore
the width: 2(1) + 3 = 5
the length: 1 + 7 = 8
:
Check by finding the area: 5 * 8 = 40


Numbers_Word_Problems/482452: Translate into an equation and solve. (Use n to represent the smaller number.)
The sum of two numbers is eighteen. Four times the smaller is equal to two times the larger.
The sum of two numbers is eleven. Five less than three times the smaller is equal to the larger.
1 solutions

Answer 330418 by ankor@dixie-net.com(15746) About Me  on 2011-08-17 15:34:46 (Show Source):
You can put this solution on YOUR website!
n = the smaller number
m = the larger
:
Write an equation for each statement:
:
The sum of two numbers is eighteen.
m + n = 18
:
Four times the smaller is equal to two times the larger.
4n = 2m
We can simplify, divide both sides by 2
2n = m
replace m with 2n in the 1st equation
2n + n = 18
3n = 18
n = 18%2F3
n = 6
I'll let you figure out what m is, check your solutions in both equations
:
:
The sum of two numbers is eleven.
m + n = 11
:
Five less than three times the smaller is equal to the larger.
3n - 5 = m
In the 1st equation replace m with (3n-5)
3n - 5 + n = 11
3n + n = 11 + 5
4n = 16
n = 16%2F4
n = 4
I'll let you find m, check your solutions in both equations


Average/482606: fred scored an average of 32 points in the first 3 games of the series.However his average increased after the fourth game to 35. What was his score in game 4
1 solutions

Answer 330414 by ankor@dixie-net.com(15746) About Me  on 2011-08-17 15:19:33 (Show Source):
You can put this solution on YOUR website!
fred scored an average of 32 points in the first 3 games of the series.
However his average increased after the fourth game to 35.
What was his score in game 4
:
Let x = his score in the 4th game
%283%2832%29+%2B+x%29%2F4 = 35
:
%2896+%2B+x%29%2F4 = 35
Multiply both sides by 4
96 + x = 4(35)
96 + x = 140
x = 140 - 96
x = 44 is the score in the 4th game
:
Check this in the original equation


Travel_Word_Problems/482641: If you are going 50 kilometers per hour, how many meters per second are you traveling?
1 solutions

Answer 330388 by ankor@dixie-net.com(15746) About Me  on 2011-08-17 14:05:43 (Show Source):
You can put this solution on YOUR website!
If you are going 50 kilometers per hour, how many meters per second are you traveling?
:
%2850%2A1000%29%2F%2860%2A60%29 = 13.89 meters/sec


Linear-equations/482540: Jerod ran from the 2-mi marker to the finish line of a 5-mi race in 25 min. At this rate, how long would it take Jerod to run a 10-km race?

1 solutions

Answer 330380 by ankor@dixie-net.com(15746) About Me  on 2011-08-17 12:53:09 (Show Source):
You can put this solution on YOUR website!
Jerod ran from the 2-mi marker to the finish line of a 5-mi race in 25 min.
At this rate, how long would it take Jerod to run a 10-km race?
:
J ran 3 mi in 25 min
Change 3 mi to km; 1.609 * 3 = 4.827 km
:
Write a ratio equation to find the minutes to run a 10 km race
let t = time to do this
t%2F25 = 10%2F4.827
Cross multiply
4.827t = 10 * 25
t = 250%2F4.827
t = 51.8 min to run 10 km


Travel_Word_Problems/482503: If you could walk a mile in 30 minutes, how long would it take you to walk 2 kilometers?
1 solutions

Answer 330349 by ankor@dixie-net.com(15746) About Me  on 2011-08-17 09:50:11 (Show Source):
You can put this solution on YOUR website!
If you could walk a mile in 30 minutes, how long would it take you to walk 2 kilometers?
:
If you can walk a mi in 30 min, you are walking 2 mph
:
Change 2 km to miles
.6214 * 2 = 1.2428 mi
:
Find how long this would take at 2 mph; (Time = dist/speed)
t = 1.2428%2F2
t = .6214 hrs or .6214(60) = 37.28 min


Probability-and-statistics/482352: two bi cyclists, 52 miles apart, begin riding toward each other on a long straight avenue. one cyclist travels 16 miles per housr and the other 10 miles per hour. at the same time, spot ( a greyhound),starting at one cyclist, runs back and forth between the two cyclists as they approach each other. if spot runs 32 miles per hour and turns around instantly at each cyclist, how far has he run when the cyclists meet?
1 solutions

Answer 330305 by ankor@dixie-net.com(15746) About Me  on 2011-08-16 21:57:16 (Show Source):
You can put this solution on YOUR website!
two bi cyclists, 52 miles apart, begin riding toward each other on a long straight avenue.
one cyclist travels 16 miles per housr and the other 10 miles per hour.
at the same time, spot ( a greyhound),starting at one cyclist, runs back and forth between the two cyclists as they approach each other.
if spot runs 32 miles per hour and turns around instantly at each cyclist, how far has he run when the cyclists meet?
:
Let t = time for the two bikes to meet, write a dist equation; dist = speed * time
16t + 10t = 52
26t = 52
t = 52%2F26
t = 2 hrs for them to meet
:
The dog runs at 32 mph for two hours then.
32 * 2 = 64 mi dist run by the dog


Unit_Conversion_Word_Problems/482470: A jogger runs 94 yd in 10.00 seconds. What
would be his time for a 490 m run at the same
rate?
Answer in units of s.
1 solutions

Answer 330303 by ankor@dixie-net.com(15746) About Me  on 2011-08-16 21:50:45 (Show Source):
You can put this solution on YOUR website!
A jogger runs 94 yd in 10.00 seconds.
What would be his time for a 490 m run at the same rate?
:
We can use a ratio equation here: let t = time to run 490 m
t%2F10 = 490%2F94
cross multiply
94t = 10 * 490
94t = 4900
t = 4900%2F94
t = 52.13 sec to go 490 m


Quadratic_Equations/482438: Solve Navigation. The current in a typical Mississippi River shipping route flows at a rate of 4 mph. In order for a barge to travel 24 mi upriver and then return in a total of 5 hr, approximately how fast must the barge be able to travel in still water?
1 solutions

Answer 330297 by ankor@dixie-net.com(15746) About Me  on 2011-08-16 21:39:15 (Show Source):
You can put this solution on YOUR website!
The current in a typical Mississippi River shipping route flows at a rate of 4 mph.
In order for a barge to travel 24 mi upriver and then return in a total of 5 hr,
approximately how fast must the barge be able to travel in still water?
:
Let s = speed in still water
then
(s-4) = effective speed up-river
and
(s+4) = effective speed down-river
:
Write a time equation; time = dist/speed
:
Time up + time down = 5 hrs
24%2F%28%28s-4%29%29 + 24%2F%28%28s%2B4%29%29 = 5
:
Multiply by (s-4)(x+4), results
24(s+4) + 24(s-4) = 5(s+4)(s-4)
:
24s + 96 + 24s - 96 = 5(s^2-16)
:
48s = 5s^2 - 80
A quadratic equation
5s^2 - 48s - 80 = 0
Use the quadratic formula to find s; a=5; b=-48; c=-80
You should get a positive solution: s = 11.05 mph in still water


Miscellaneous_Word_Problems/482434: Jodie bikes 8 km/h faster than robert in the same time it takes robert to bike 30km, Jodie can bike 54km. How fast does each bicyclist travel?
I have done plenty of problems like this , however, now I just cant think right now. Can anyone help me?
1 solutions

Answer 330279 by ankor@dixie-net.com(15746) About Me  on 2011-08-16 20:46:28 (Show Source):
You can put this solution on YOUR website!
Jodie bikes 8 km/h faster than robert in the same time it takes robert to bike 30km, Jodie can bike 54km.
How fast does each bicyclist travel?
:
Let s = R's speed
then
(s+8) = J's speed
:
Write a time equation; time = dist/speed
:
J's time = R's time
54%2F%28%28s%2B8%29%29 = 30%2Fs
Cross multiply
54s = 30(s+8)
54s = 30s + 240
54s - 30s = 240
24s = 240
s = 240%2F24
s = 10 mph is R's speed
then
10 + 8 = 18 mph is J's speed
:
:
Confirm our solutions by finding the time of each, they should be equal
54/18 = 3 hrs
30/10 = 3 hrs


Quadratic_Equations/482315: Can a quadratic equation solution be determined by looking at a graph, and how?
1 solutions

Answer 330276 by ankor@dixie-net.com(15746) About Me  on 2011-08-16 20:25:29 (Show Source):
You can put this solution on YOUR website!
Can a quadratic equation solution be determined by looking at a graph, and how?
:
+graph%28+300%2C+200%2C+-3%2C+10%2C+-4%2C+15%2C+x%5E2+-+8x+%2B+12%29+
:
You can get all the information from the above graph to write an equation
The graph opens upward means the coefficient of x^2 is positive
The x intercepts, x=2; x=6 would be from the factors
(x-2)(x-6) = 0
FOIL
y = x^2 - 6x - 2x + 12
y = x^2 - 8x + 12; is the equation of the above


Equations/482121: how do you evaluate the power 1.2 to the third power and 12 to the sencond power
1 solutions

Answer 330250 by ankor@dixie-net.com(15746) About Me  on 2011-08-16 17:59:18 (Show Source):
You can put this solution on YOUR website!
how do you evaluate the power 1.2 to the third power and 12 to the second power
How about: 1.2 * 1.2 * 1.2
and
12 * 12


Travel_Word_Problems/482313: Donald and Eric swam as a team in a relay race. Donald swam at a steady 3 km/h while Eric returned at an average rate of 4 km/h. Their total time was posted as 5 h 15 min. How long did Eric swim?
1 solutions

Answer 330246 by ankor@dixie-net.com(15746) About Me  on 2011-08-16 17:50:19 (Show Source):
You can put this solution on YOUR website!
Donald and Eric swam as a team in a relay race.
Donald swam at a steady 3 km/h while Eric returned at an average rate of 4 km/h.
Their total time was posted as 5 h 15 min. How long did Eric swim?
:
Change 5 hrs 15 min to 5.25 hrs
:
Find the distance first
Let d = the one-way distance
:
Write a time equation; time = dist/speed
:
D's time + E's time = 5 hr 15 min
d%2F3 = d%2F4 = 5.25
multiply by 12 to clear the denominators
4d + 3d = 12(5.25)
7d = 63
d = 63/7
d = 9 km is the one way distance
:
Find Eric's time
t = 9/4
t = 2.25 hrs is E's swim time
:
:
Confirm the solution by finding the total time
9/4 = 2.25
9/3 = 3.00
------------
total 5.25 hrs


absolute-value/482379: As I have said on many occassions Algebra is very confusing to me to say the least...if someone could help me I would be very grateful
Solve by elimination x-7y=62, -x +6y=42
1 solutions

Answer 330237 by ankor@dixie-net.com(15746) About Me  on 2011-08-16 17:20:50 (Show Source):
You can put this solution on YOUR website!
Solve by elimination
x - 7y = 62,
-x + 6y = 42
-------------------adding eliminates x, find y
-y = 104
y has to be positive, multiply both sides by -1
y = -104
:
Find x using the 1st equation, replace y with -104
x - 7(-104) = 62
x + 728 = 62
x = 62 - 728
x = -666
:
:
Check solution in the 2nd equation, replace x and y
-(-666) + 6(-104) = 42
666 - 624 = 42, confirms our solutions of x=-666 and y=-104


Age_Word_Problems/481949: If stronium-90 has a half-life of fourteen days, and you have a sample that contains 30 kg of strontium-90, how much would remain after thirty-five days?
1 solutions

Answer 330236 by ankor@dixie-net.com(15746) About Me  on 2011-08-16 17:13:54 (Show Source):
You can put this solution on YOUR website!
If stronium-90 has a half-life of fourteen days, and you have a sample that contains 30 kg of strontium-90, how much would remain after thirty-five days?
:
A = Ao*2^(-t/h)
:
A = 30*2^(-35/14)
A = 30*2^-2.5
A = 30*.17678
A = 5.3 kg after 35 days


Age_Word_Problems/481943: Suppose you have 50 kg of a substance that has a half-life of five years. How much would remain after twenty years?
1 solutions

Answer 330234 by ankor@dixie-net.com(15746) About Me  on 2011-08-16 17:08:19 (Show Source):
You can put this solution on YOUR website!
Suppose you have 50 kg of a substance that has a half-life of five years.
How much would remain after twenty years?
:
The radioactive decay formula: A = Ao*2^(-t/h), where
A = amt after t time
Ao = initial amt
t = time of decay
h = half-life of substance
:
A = 50*2^(-20/5)
A = 50*2^-4
A = 50*.0625
A = 3.125 kg after 20 yrs