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: 15623 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, >>NextWord_Problems_With_Coins/214173: I am a parent I want to help my child but i don't know how to do this. I'm not asking for the answer,(but i'm not turning it down either:>), but the necessary steps to take to solve this problem, so I may be able to help her. The question is Joshua had 142 coins, all nickels and quarters. The total amount was $15.30. HOw many coins of each type did he have? Solve this problem with Algebra using variables to stand for the number of nickels and the number of quarters. Thank you so much for whatever help you can provide me with. This question was not in a textebook. 1 solutions
Answer 161768 by ankor@dixie-net.com(15638) on 2009-09-16 15:36:44 (Show Source):
You can put this solution on YOUR website!Joshua had 142 coins, all nickels and quarters.
The total amount was $15.30.
HOw many coins of each type did he have?
:
Let n = no. of nickels; let q = no. of quarters
:
write an equation for each statement:
:
"Joshua had 142 coins,"
n + q = 142
:
"The total amount was $15.30."
.05n + .25q = 15.30
:
HOw many coins of each type did he have?
:
Go back to the 1st equation, and arrange it for substitution in the 2nd equation
n + q = 142
q = (142-n)
:
Replace q with (42-n) in the 2nd equation, find n
.05n + .25(142-n) = 15.30
:
Multiply what's inside the brackets
.05n + 35.5 - .25n = 15.30
;
.05n - .25n = 15.30 - 35.50
:
-.20n = -20.20
n =  ; minus into a minus is plus
n = +101 nickels
:
Find q:
q = 142 - 101
q = 41 quarters
:
:
See if this is right, substitute for n and q
.05(101) + .25(41) = 15.30
5.05 + 10.25 = 15.30; confirms our solution
:
Make sense to you?
|
Age_Word_Problems/214165: 40 years ago i was 1/3 the age i will be in 4 years. How old am I? 1 solutions
Answer 161755 by ankor@dixie-net.com(15638) on 2009-09-16 15:23:24 (Show Source):
You can put this solution on YOUR website!a = your age now
;
40 years ago i was 1/3 the age i will be in 4 years.
a - 40 =  (a+4)
Multiply both sides by 3, to get rid of the annoying fraction
3a - 120 = a + 4
3a - a = 4 + 120
2a = 124
a = 62 yr is your age now
;
:
Check in the statement:
40 years ago i was 1/3 the age i will be in 4 years.
62 - 40 =  (62 + 4)
22 =  (66)
|
Linear-systems/213787: sqrt(x+1)-sqrt(x-1)=2 1 solutions
Answer 161748 by ankor@dixie-net.com(15638) on 2009-09-16 15:10:31 (Show Source):
You can put this solution on YOUR website! = 2
:
Add  to both sides
 = 
:
Square both sides, FOIL the right side
x + 1 = (x-1) +  + 4
:
x + 1 = x - 1 + 4 + 
:
x + 1 = x + 3 + 
:
x - x + 1 - 3 = 
:
-2 = 
:
Square both sides
4 = 16(x - 1)
:
4 = 16x - 16
:
4 + 16 = 16x
:
20 = 16x
x = 
x = 1.25; however substitution in the original equation reveals this is not a solution. There is no solution
|
Numbers_Word_Problems/214070: Find the six-digit number in which the first digit is one less than the second digit and half the third; the fourth digit is one more than the third and is also the sum of the first and second digits; and the fifth and sixth digits make a number that is the sum of the first four digits. The sum of all the digits is 19. 1 solutions
Answer 161746 by ankor@dixie-net.com(15638) on 2009-09-16 14:45:14 (Show Source):
You can put this solution on YOUR website!Write an equation for each statement
:
Find the six-digit number
u, v, w, x, y, z
:
in which the first digit is one less than the second digit
u = v-1
:
and half the third;
u = .5w
:
the fourth digit is one more than the third
x = w+1
:
and is also the sum of the first and second digits;
x = u + v
:
the fifth and sixth digits make a number that is the sum of the first four digits.
10y + z = u + v + w + x
:
The sum of all the digits is 19.
u + v + w + x + y + z = 19
:
We can replace the 1st 4 digits with 10y + z
10y + z + y + z = 19
11y + 2z = 19
:
From this we can write an equation that has only one positive integer solution
2z = 19-11y
z = 
y = 1
z = 
z = 
z = 4
:
u + v + w + x + y + z = 19
Replace y and z
u + v + w + x + 1 + 4 = 19
u + v + w + x = 19 - 5
u + v + w + x = 14
:
Replace u with .5w, replace x with w+1
.5w + v + w + (w+1) = 14
2.5w + v = 14 - 1
2.5w + v = 13
:
We know that u = v-1 and we know that u = .5w, therefore
.5w = v - 1
or
v = .5w + 1
Replace v with (.5w+1)
2.5w + .5w + 1 = 13
3w = 13 - 1
3w = 12
w = 4
:
Find v
v = .5(4) + 1
v = 3
:
So far we have:
v=3, w=4, y=1, z=4
find x
x = w+1
x = 4+1
x = 5
:
find u
u = .5w
u = .5(4)
u = 2
:
The number: 234514
:
Check this:
2 + 3 + 4 + 5 + 1 + 4 = 19
|
Travel_Word_Problems/214124: It's 56 miles to work, it takes 10 minutes longer to get to work than coming home, and you travel 8 mph faster coming home. What is your avg. speed coming home? 1 solutions
Answer 161737 by ankor@dixie-net.com(15638) on 2009-09-16 13:59:02 (Show Source):
You can put this solution on YOUR website!It's 56 miles to work, it takes 10 minutes longer to get to work than coming home,
and you travel 8 mph faster coming home.
What is your avg. speed coming home?
:
Let s = speed coming home
then
(s-8) = speed going to work
:
Convert 10 min to  hr
:
Write a time equation: Time = dist/speed
:
Time from work + 10 min = time to work
 +  = 
Multiply equation by 6s(s-8), results:
56*6*(s-8) + s(s-8) = 6s*56
;
336(s-8) + s^2 - 8s = 336s
:
336s - 2688 + s^2 - 8s - 336s = 0
:
Arrange as a quadratic equation
s^2 - 8s - 2688 = 0
Factor to
(s-56)(s+48) = 0
positive solution
s = 56 mph on the return home
:
:
Check solution (speed to work = 48 mph)
56/56 + 1/6 = 56/48
1 + .167 = 1.167
|
Graphs/214069: Hi, I am having a hard time figuring what formula or equation needs to be applied to this problem. There's a slope graph. (0,96) (2.5,196)
Not sure if this is considered quadratic or just a simple word problem that need to be converted into an equation.
The graph below shows the height in feet, of a ball thrown straight up with an initial speed of 80 feet per second from an initial height of 96 feet after t seconds. Use the graph to answer the following questions.
a) What is the height of the ball at 1.5 seconds?
b) What is the maximum height of the ball and at what time does it occur?
c) Find the H intercept of the graph and explain what is happening to the ball at that point.
d) Find the T intercept of the graph and explain what is happening to the ball at that point. 1 solutions
Answer 161705 by ankor@dixie-net.com(15638) on 2009-09-16 10:32:13 (Show Source):
You can put this solution on YOUR website!The graph below shows the height in feet, of a ball thrown straight up with an initial speed of 80 feet per second from an initial height of 96 feet after t seconds. Use the graph to answer the following questions.
:
Perhaps we can construct the graph from the equation: h = -16t^2 + 80t + 96
-16t^2 is the downward pull of gravity
80t is the upward velocity of the ball
96 is initial height
:

:
a) What is the height of the ball at 1.5 seconds?
The graph indicates it to be about 180 ft
:
b) What is the maximum height of the ball and at what time does it occur?
Looks like this occurs at 2.5 sec when the ball is at 196 ft
:
c) Find the H intercept of the graph and explain what is happening to the ball at that point.
H intercept is on the y axis, occurs when x = 0, that would be 96 ft
:
d) Find the T intercept of the graph and explain what is happening to the ball at that point.
T is on the x axis, intercept when y = 0, looks like at 6 secs when it hits the ground
:
it is a quadratic equation
|
Travel_Word_Problems/213913: Two planes leave simultaneously from the same airport, one flying due east and the other due south. The eastbound plane is flying 100 miles per hour faster than the southbound plane. After 4 hours the planes are 3080 miles apart. Find the speed of each plane. (Use 1 decimal place.) 1 solutions
Answer 161656 by ankor@dixie-net.com(15638) on 2009-09-15 22:00:55 (Show Source):
You can put this solution on YOUR website!Two planes leave simultaneously from the same airport, one flying due east and the other due south.
The eastbound plane is flying 100 miles per hour faster than the southbound plane.
After 4 hours the planes are 3080 miles apart.
Find the speed of each plane. (Use 1 decimal place.)
:
The distance between the planes is the hypotenuse of a right triangle
a^2 + b^2 = c^2
:
Let s = speed of the southbound plane
then
(s+100) = speed of the eastbound
:
Dist = speed * time
:
a = 4s; south distance
b = 4(s+100); east distance
c = 3080; distance between the planes after 4 hrs
:
(4s)^2 + (4(s+100))^2 = 3080^2
:
16s^2 + (4s+400)^2 = 9486400
:
16s^2 + (4s+400)^2 = 9486400
:
16s^2 + 16s^2 + 3200s + 160000 = 948640
;
16s^2 + 16s^2 + 3200s + 160000 - 9486400 = 0
;
32s^2 + 3200s - 9326400 = 0
:
Simplify divide by 32:
s^2 + 100s - 291450 = 0
:
Using the quadratic formula; a=1; b=100; c=291450
;
I got a positive solution of 492.2 mph the south bound plane
:
then 592.2 mph the east bound plane
;
:
Check solution on a calc; enter  = 3080
|
Proportions/213924: I am taking GED classes. We do not have a textbook just work sheets.
1. if 3 cans of asparagus cost $7.00, How much do 12 cans cost?
2. if it takes a basket maker 20 days to make 30 baskets, how many baskets can be made in 60 days? 1 solutions
Answer 161590 by ankor@dixie-net.com(15638) on 2009-09-15 19:27:18 (Show Source):
You can put this solution on YOUR website!1. if 3 cans of asparagus cost $7.00, How much do 12 cans cost?
A ratio equation is one way:
 = 
Cross multiply
3x = 7*12
3x = 84
x = 
x = $28 for 12 cans
:
:
2. if it takes a basket maker 20 days to make 30 baskets,
how many baskets can be made in 60 days?
 = 
20x = 60*30
20x = 1800
x = 
x = 90 baskets in 60 days
|
Travel_Word_Problems/213850: Bryan took a 4-mile walk. The second half of his walk he averaged 1 mile per hour less than the first half, and it took him 1/3 hour longer. What rate did he average on the first half of the walk?
This question has confused me and i do not know how to set it up.
I know the total distance is 4 miles.
He averaged 1/3 hour more on the second half.
He walked 1 mile per hour less
1 solutions
Answer 161578 by ankor@dixie-net.com(15638) on 2009-09-15 19:06:59 (Show Source):
You can put this solution on YOUR website!Bryan took a 4-mile walk.
The second half of his walk he averaged 1 mile per hour less than the first
half, and it took him 1/3 hour longer.
What rate did he average on the first half of the walk?
:
Let s = speed on the 1st 2 mi
then
(s-1) = speed on the last 2 mi
:
Write a time equation: Time = 
1st half time + 1/3 hr = 2nd half time
 +  = 
Multiply equation by 3s(s-1)
3(s-1)*2 + s(s-1) = 3s(2)
:
6s - 6 + s^2 - s = 6s
:
s^2 - s + 6s - 6s - 6 = 0
:
s^2 - s - 6 = 0
Factor
(s-3)(s+2) = 0
Positive solution
s = 3 mph speed on the 1st half
:
:
Check by finding the times
2/3 hr
2/2 - 1 hr; 1/3 hr difference
|
Matrices-and-determiminant/213861: This question is from textbook Algebra 2
A jeweler plans to combine two silver alloys to make 50 grams of alloy that is 75% silver and contains 37.5 grams of pure silver (75% of 50 grams). If some silver alloy is 80% silver and the other is 60% silver, how many grams of each are needed? 1 solutions
Answer 161552 by ankor@dixie-net.com(15638) on 2009-09-15 16:47:47 (Show Source):
You can put this solution on YOUR website!A jeweler plans to combine two silver alloys to make 50 grams of alloy that is
75% silver and contains 37.5 grams of pure silver (75% of 50 grams).
If some silver alloy is 80% silver and the other is 60% silver, how many grams of each are needed?
:
Let x = amt of 80% silver required
Resulting amt is to be 50, therefore:
(50-x) = amt of 60% silver
:
.8x + .6(50-x) = .75(50)
:
.8x + 30 - .6x = 37.5
:
.8x - .6x = 37.5 - 30
:
.2x = 7.5
:
x = 
x = 37.5 gr of 80% silver
and
50 - 37.5 = 12.5 gr of 60% silver
:
Check solutions in the amt of silver equation
.8(37.5) + .6(12.5) = .75(50)
30 + 7.5 = 37.5
|
Travel_Word_Problems/213855: The captain of the Titanic was warned of ice ahead, but he thought his ship was unsinkable, and continued his pace of 60 feet every 1.5 seconds. How fast was he traveling in miles per hour? How fast was he traveling in nauts per hour? 1 solutions
Answer 161543 by ankor@dixie-net.com(15638) on 2009-09-15 16:08:02 (Show Source):
You can put this solution on YOUR website! pace of 60 feet every 1.5 seconds. How fast was he traveling in miles per hour?
:
 = 27.27 mph
:
How fast was he traveling in nauts per hour?
 = 24 knots
|
Mixture_Word_Problems/213844: Ellen takes 1/4 of the oranges in a bowl Jeff takes 1/3 of the remaining oranges in the bowl. Ellen and Jeff later decide that they don't need all their oranges so they each put 1 back in the bowl. Carlos takes 1/2 of the oranges that remain in the bowl There are now 4 oranges left in the bowl. How many oranges were originally in the bowl? 1 solutions
Answer 161537 by ankor@dixie-net.com(15638) on 2009-09-15 15:55:28 (Show Source):
You can put this solution on YOUR website!Ellen takes 1/4 of the oranges in a bowl Jeff takes 1/3 of the remaining oranges in the bowl.
Ellen and Jeff later decide that they don't need all their oranges so they each put 1 back in the bowl.
Carlos takes 1/2 of the oranges that remain in the bowl.
There are now 4 oranges left in the bowl.
How many oranges were originally in the bowl?
;
Let x = amt originally in the bowl
:
"Ellen takes 1/4 of the oranges in a bowl"
x -  x =  x remaining
;
Jeff takes 1/3 of the remaining oranges in the bowl.
that means he leaves 2/3
 *  x =  x
 x reduces to  x remaining
:
Ellen and Jeff later decide that they don't need all their oranges so they each put 1 back in the bowl.
 x + 2
:
Carlos takes 1/2 of the oranges that remain in the bowl There are now 4 oranges left in the bowl.
 (  x + 2) = 4
multiply both sides by 2
 x + 2 = 8
 x = 8 - 2
 x = 6
multiply both sides by 2
x = 2(6)
x = 12 oranges originally
:
:
Check solution:
Ellen takes 1/4 of the oranges in a bowl
1/4 of 12 = 3, leaving 9
Jeff takes 1/3 of the remaining oranges in the bowl.
1/3 of 9 = 3. leaving 6
Ellen and Jeff later decide that they don't need all their oranges so they each put 1 back in the bowl.
6 + 2 = 8
Carlos takes 1/2 of the oranges that remain in the bowl.
half of 8 = 4
|
Proportions/213793: an adult sets out from home to pick up their child from school, and walks at a pace that will take 10 minutes to get to the school. At exactly the same time,the child leaves school to walk home at a pace that will take 20 minutes to get home. After how long will they meet? How would you justify your solution to the child? 1 solutions
Answer 161529 by ankor@dixie-net.com(15638) on 2009-09-15 15:22:42 (Show Source):
You can put this solution on YOUR website!an adult sets out from home to pick up their child from school,
and walks at a pace that will take 10 minutes to get to the school.
At exactly the same time,the child leaves school to walk home at a pace that
will take 20 minutes to get home.
After how long will they meet?
;
Let t = time interval when they walk towards each other
:
From the information given, we can say the adult walks twice as fast as the child
Let s = walking speed of the child
then
2s = walking speed of the adult
and
3s = their relative speed walking towards each other
:
Write a dist equation Dist = speed * time
:
Combined dist = child walking 20 min to home
3s(t) = 20s
divide both sides by s
3t = 20
t = 
t = 6  min
or
Combined dist = adult walking 10 min
3s(t) = 10(2s)
3s(t) = 20t
divide both sides by s
3t = 20
t = 
t = 6  min
:
How would you justify your solution to the child?
:
You can say since I walk twice as fast as you do, I will cover twice the
distance.
therefore
in 6  min, child covers  the dist
and
in 6  min, adult covers  the dist (twice as fast)
|
Rectangles/213483: Two rectangles have the same width. The length of one is 1 foot longer than the width. The length of the other is 2 feet longer than the width. The larger rectangle has 6 more square feet than the smaller. What is the width of the rectangles? 1 solutions
Answer 161436 by ankor@dixie-net.com(15638) on 2009-09-14 21:59:03 (Show Source):
You can put this solution on YOUR website!Two rectangles have the same width.
let w = width of both rectangles
:
The length of one is 1 foot longer than the width.
L1 = w + 1
The length of the other is 2 feet longer than the width.
L2 = w + 2
;
The larger rectangle has 6 more square feet than the smaller.
W*L2 - W*L1 = 6
:
What is the width of the rectangles?
Replace L2 in the above equation with (W+2);
Replace L1 in the above equation with ((W+1)
W*(W+2) - W*(W+1) = 6
W^2 + 2W - W^2 - W = 6
2W - W = 6
W = 6 is the width of both rectangles
;
:
Check solution
8*6 - 7*6 = 6
|
Quadratic_Equations/213547: A college student has finished 49 credit hours with a GPA of 2.5. To get into the program she wishes to enter, she must have a GPA of 3.3. How many additional credit hours of 4.0 work will raise her GPA to 3.3? 1 solutions
Answer 161380 by ankor@dixie-net.com(15638) on 2009-09-14 19:05:30 (Show Source):
You can put this solution on YOUR website!A college student has finished 49 credit hours with a GPA of 2.5.
To get into the program she wishes to enter, she must have a GPA of 3.3.
How many additional credit hours of 4.0 work will raise her GPA to 3.3?
:
Let x = no. of 4.0 credits required
:
 = 3.3
 = 3.3
Multiply both sides by (x+49)
4x + 122.5 = 3.3(x+49)
:
4x + 122.5 = 3.3x + 161.7
:
4x - 3.3x = 161.7 - 122.6
:
.7x = 39.1
x = 
x = 55.857 ~ 56 additional 4.0 credit hrs, to get a 3.3 gpa (sounds almost impossible!)
:
:
Check: final total credit hr 49 + 56 = 105
 =
 =
 = 3.3
|
Rectangles/213554: A farmer wants to make a rectangular enclosure along the side of a barn and then divide the enclosure into two pens with a fence constructed at a right angle to the barn. If 300 yards of fencing are available, what are the dimensions of the largest section that can be enclosed?
Firstly, the wording of this particular problem has thrown both myself, family, and friends for a loop. I have drawn, erased, and redrawn several rough sketches of this barn-and-fence problem and still feel as though it is wrong. I am at a loss of going about how to solve it. Any assistance would be greatly appreciated!
Thank you in advance
-B 1 solutions
Answer 161342 by ankor@dixie-net.com(15638) on 2009-09-14 17:20:48 (Show Source):
You can put this solution on YOUR website!A farmer wants to make a rectangular enclosure along the side of a barn and then divide the enclosure into two pens with a fence constructed at a right angle to the barn. If 300 yards of fencing are available, what are the dimensions of the largest section that can be enclosed?
:
_________________Barn
|_____|_____|
:
I think it looks something like this:
Then we can write an equation for the 300 yds of fencing
L + 3W = 300
:
L = (300-3W)
:
Area = L*W
Replace L with (300-3W)
A = (300-3W) * W
Arrange as a quadratic equation
A = -3W^2 + 300W
Find the axis of symmetry for Width that gives max area x = -b/(2a)
in this equation a=-3; b=300
W = -300/(2*-3)
W = -300/-6
W = +50 ft is the width for max area
:
Find the length
L = 300 - 3(50)
L = 150 ft is the length
:
Max area would be 150 * 50 = 7500 sq/ft
|
Money_Word_Problems/213507: This question is from textbook
The revenue for selling x units of the product is R=115.95x. The cost of producing x units is
C=95x+750.
To obtain a profit, the revenue must be greater than the cost. For what values of x will this product return a profit. 1 solutions
Answer 161298 by ankor@dixie-net.com(15638) on 2009-09-14 15:09:31 (Show Source):
You can put this solution on YOUR website!The revenue for selling x units of the product is R=115.95x.
The cost of producing x units is C=95x+750.
To obtain a profit, the revenue must be greater than the cost.
For what values of x will this product return a profit.
:
R > C
115.95x > 95x + 750
Subtract 95x from both sides
115.95 - 95x > 750
20.95x > 750
divide both sides by 20.95
x > 
x > 35.8 units
Have to sell at least 36 units to make a profit
;
:
Check in the original equation
115.95(36) > 95(36) + 750
4174.20 > 3420 + 750
4174.20 > 4170; a slight profit of $4.20
|
Linear-equations/213427: This question is from textbook PRECALCULUS
WHAT TYPE OF FORMULA AM I SUPPOSE TO USE IN ORDER TO FIND THE VALUE OF X 0R Y. THE DIRECTIONS SAY, "FIND THE VALUE OF X OR Y SO THAT THE LINE THROUGH THE PAIR OF POINTS HAS THE GIVEN SLOPE." I'M NOT SURE HOW TO DO THAT. THE GIVEN IS (-2,3) AND (4,y) with the slope of -3. 1 solutions
Answer 161206 by ankor@dixie-net.com(15638) on 2009-09-14 08:01:31 (Show Source):
You can put this solution on YOUR website!WHAT TYPE OF FORMULA AM I SUPPOSE TO USE IN ORDER TO FIND THE VALUE OF X 0R Y. THE DIRECTIONS SAY, "FIND THE VALUE OF X OR Y SO THAT THE LINE THROUGH THE PAIR OF POINTS HAS THE GIVEN SLOPE." I'M NOT SURE HOW TO DO THAT. THE GIVEN IS (-2,3) AND (4,y) with the slope of -3.
:
Use the slope formula to find y: that is  = m
Assign the given values as follows:
x1=-2; y1=3
x2 =4; y2= y
m = -3
:
Substitute the given values as follows and find y
 = -3
 = -3
 = -3
Multiply both sides by 6
y - 3 = 6(-3)
y = -18 + 3
y = -15; the pair is: 4,-15
;
Use the point/slope formula to find the equation
y - y1 = m(x - x1)
Substitute the given values
y - 3 = -3(x -(-2))
y - 3 = -3(x + 2)
y - 3 = -3x - 6
y = -3x - 6 + 3
y = -3x - 3 is the equation
:
Check solution by substituting the x value of 4
y = -3(4) - 3
y = = -12 - 3
y = -15 which is the value we got for y
|
Angles/213212: This is a question from a worksheet... One of two copmlemtentary angles added to one-half of the other yields 72degrees. Find the half the measure of the larger.
not quite sure how to work the problem........thank you for your time! i really appreciate it! 1 solutions
Answer 161091 by ankor@dixie-net.com(15638) on 2009-09-13 16:39:30 (Show Source):
You can put this solution on YOUR website! One of two complemtentary angles added to one-half of the other yields 72degrees.
Find the half the measure of the larger.
:
the two angles: x and (90-x)
:
x + .5(90-x) = 72
x + 45 - .5x = 72
x - .5x = 72 - 45
.5x = 27
Multi both sides by 2
x = 54 degrees, has to be the larger
:
.5(54) = 27 degrees
|
Linear_Equations_And_Systems_Word_Problems/213169: This question is from textbook Prentice Hall Algebra I Louisiana
A moving van leaves a house traveling at an average rate of 35mph. The family leaves the house in a car 3/4 hour later traveling the same route at an average rate of 50mph. How long will it take the family to catch up with the van? 1 solutions
Answer 161056 by ankor@dixie-net.com(15638) on 2009-09-13 12:29:15 (Show Source):
You can put this solution on YOUR website!A moving van leaves a house traveling at an average rate of 35mph.
The family leaves the house in a car 3/4 hour later traveling the same route at an average rate of 50mph.
How long will it take the family to catch up with the van?
:
Change 3/4 hr to .75 hr
:
Let t = time required by the family to catch up
then
(t+.75) = Van travel time when it is caught
:
When this happens, they will have traveled the same distance
Write a distance equation: dis = speed * time
:
Car dist = Van dist
50t = 35(t+.75)
50t = 35t + 26.25
50t - 35t = 26.25
15t = 26.25
t = 
t = 1.75 hrs for the family to catch the van
;
:
Check solution by ensuring that they traveled the same dist:
50 * 1.75 = 87.5 mi
35 * (1.75+.75) = 87.5
|
Rate-of-work-word-problems/213071: allan can finish a set of furniture in 8 days. ben can finish the same set of furniture in 6 days. if allan starts working for 2 days, and then ben joined him, in how many days will the set of furniture be finished by the two men? 1 solutions
Answer 160965 by ankor@dixie-net.com(15638) on 2009-09-12 22:02:08 (Show Source):
You can put this solution on YOUR website!allan can finish a set of furniture in 8 days.
ben can finish the same set of furniture in 6 days.
if allan starts working for 2 days, and then ben joined him,
in how many days will the set of furniture be finished by the two men?
:
let t = time working together
:
Let the completed job = 1
:
 +  = 1
Multiply equation by 24, results
3(t+2) + 4t = 24
3t + 6 + 4t = 24
3t + 4t = 24 - 6
7t = 18
t = 
t = 2.57 days working together
;
:
Check
 +  =
.57 + .43 = 1
|
Age_Word_Problems/213069: Ricky is 26 yrs old and Jenny is 5 yrs old. in how many yrs will ricky be twice as old as jenny? 1 solutions
Answer 160964 by ankor@dixie-net.com(15638) on 2009-09-12 21:52:22 (Show Source):
You can put this solution on YOUR website!: Ricky is 26 yrs old and Jenny is 5 yrs old.
in how many yrs will ricky be twice as old as jenny?
:
Let t = no. of yrs for this to be true
:
26 + t = 2(5+t)
26 + t = 10 + 2t
26 - 10 = 2t - t
16 = t
;
:
Check:
26 + 16 = 2(5 + 16)
42 = 2(21)
|
Angles/212961: This question is from textbook Geometry for Enjoyment and Challenge
I have a picture with segment XY is perpendicular to segment YW. Ray YZ forms two angles within the 90 degrees of angle XYW. these ar identified as angle x (angle XYZ) and angle y (angle ZYW). The next picture i have has segment AB is perpendicular to segment BC. A segment BD is within this 90 degrees showing angles (2x)degrees (angle ABD) and (y-50) degrees (angle DBC). I need to find the measure of angle DBC.
I used two equations and two unknowns and came up with x =50 degrees and y = 40 degrees. that doesn't work when i use y=40 in the second picture with the angle DBC = (y-50)degrees. This would make that angle a negative number. 1 solutions
Answer 160960 by ankor@dixie-net.com(15638) on 2009-09-12 21:09:43 (Show Source):
You can put this solution on YOUR website!I have a picture with segment XY is perpendicular to segment YW. Ray YZ forms two angles within the 90 degrees of angle XYW. these ar identified as angle x (angle XYZ) and angle y (angle ZYW). The next picture i have has segment AB is perpendicular to segment BC. A segment BD is within this 90 degrees showing angles (2x)degrees (angle ABD) and (y-50) degrees (angle DBC). I need to find the measure of angle DBC.
I used two equations and two unknowns and came up with x =50 degrees and y = 40 degrees. that doesn't work when i use y=40 in the second picture with the angle DBC = (y-50)degrees. This would make that angle a negative number.
:
That's true but that has to be the answer, if x=50 then ABD = 100 then DBC has to
be 40 - 50 = -10: 100 - 10 = 90 degrees. I think you did it right!
|
Surface-area/213062: Henry is ten more than two times as old as Cheryl. In 20 years the sum of their ages will be 92. How old is Cheryl now? 1 solutions
Answer 160953 by ankor@dixie-net.com(15638) on 2009-09-12 20:26:01 (Show Source):
You can put this solution on YOUR website!Let h = Henry's present age
Let c = Cheryl's present age
;
Write an equation for each statement:
:
Henry is ten more than two times as old as Cheryl.
h = 2c + 10
:
In 20 years the sum of their ages will be 92.
(h+20) + (c+20) = 92
h + c + 40 = 92
h + c = 92 - 40
h + c = 52
:
How old is Cheryl now?
:
Replace h with (2c+10)
2c + 10 + c = 52
2c + c = 52 - 10
3c = 42
c = 
c = 14 yrs is Cheryl's present age
;
:
Check solution:
Find h: h + 14 = 52; h = 38
Check solution in the statement:
"In 20 years the sum of their ages will be 92."
38 + 20 + 14 + 20 = 92
|
Rectangles/213061: The width of a rectangle is 9 less than twice its length. If the area of the rectangle is 100 centimeters squared, what is the length of the diagonal?
So far i have this:
W= 2L-9
L= ?
L(2L-9)=100
2L^2-9L-100=0
I tried to factor it and used the quadratic formula and I still couldn't get answer.
Please help. Thank you! 1 solutions
Answer 160951 by ankor@dixie-net.com(15638) on 2009-09-12 20:11:30 (Show Source):
|
Proportions/213023: by weight, the ratio of chlorine to sodium in table salt is 35.46 to 23.00. How much sodium is contained in 50.00kg or salt? Has to be set up as an appropriate proportion and then be solved. 1 solutions
Answer 160944 by ankor@dixie-net.com(15638) on 2009-09-12 19:30:07 (Show Source):
You can put this solution on YOUR website!by weight, the ratio of chlorine to sodium in table salt is 35.46 to 23.00. How much sodium is contained in 50.00kg or salt? Has to be set up as an appropriate proportion and then be solved.
:
Let x = amt of sodium in 50 kg
then
(50-x) = amt of chlorine in 50 kg
;
A ratio equation:
 = 
cross multiply
35.46x = 23(50-x)
35.46x = 1150 - 23x
35.46x + 23x = 1150
58.46x = 1150
x = 
x = 19.67 kg of sodium
;
:
check:
Chlorine: 50 - 19.67 = 30.33
:
23/35.46 = .6486
19.67/30.33 = .6485; close enough
|
Exponential-and-logarithmic-functions/213021: I'ts been a while since I've solved these kinds of questions (ok 20+ years!) so I need a refresher. I looked a similar problems on-line, but am struggling with how to set it up. Here's the problem:
Sam invested $5000 for one year, part at 9% annual interest and the rest at 12% annual interest. The interest from the investment at 9% was $198 more than the interest from the investment at 12%. How much money did she invest at 9%? 1 solutions
Answer 160918 by ankor@dixie-net.com(15638) on 2009-09-12 16:58:39 (Show Source):
You can put this solution on YOUR website!Sam invested $5000 for one year, part at 9% annual interest and the rest at 12% annual interest.
The interest from the investment at 9% was $198 more than the interest from the investment at 12%.
How much money did she invest at 9%?
:
Let x = amt invested at %9
Total invested given as 5000, therefore
(5000-x) = amt invested at 12%
:
.09x - 298 = .12(5000-x)
.09x - 298 = 600 - .12x
.09x + .12x = 600 + 298
.21x = 898
x = 
x = $4,276.19 invested at 9%
then
5000 - 4276.19 = $723.08 invested at %12
:
:
Is this true? find the interest of each
.09*4276.19 = 384.86
.12 * 723.08 = 86.86
-------------------
difference = 298; confirms these solutions
|
|