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: 15638 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, >>NextDivisibility_and_Prime_Numbers/268718: a group of children walk to a corner store and buy cans of soda for a total of $3.36 and candy bars for a total of $2.45. Each child has one can of soda and one candy bar. How many children were in the group> What is the cost of one can of soda nd one cany bar?
1 solutions
Answer 196918 by ankor@dixie-net.com(15653)   on 2010-02-11 21:10:12 (Show Source):
You can put this solution on YOUR website!a group of children walk to a corner store and buy cans of soda for a total of $3.36 and candy bars for a total of $2.45.
Each child has one can of soda and one candy bar.
How many children were in the group> What is the cost of one can of soda and one candy bar?
:
Lets do this in cents, 336 and 245
Find the common prime factors
336: 2*2*2*2*3*7
245: 5*7*7
:
The common factor is 7, therefore: 7 children in the group
:
336/7 = 48 cents for a soda
245/7 = 35 cents for a candy bar
|
Travel_Word_Problems/268621: An air rescue plane averages 300 miles per hour in still air. It carries enough fuel for 5 hours of flying time. If, upon takeoff it encounters a head wind of 30 mi/hr and the wind remains constant, how far can the plane fly and then return safely?
Is the rate simply 300-30? How do I find the distance traveled?
1 solutions
Answer 196911 by ankor@dixie-net.com(15653) on 2010-02-11 20:50:32 (Show Source):
You can put this solution on YOUR website!An air rescue plane averages 300 miles per hour in still air.
It carries enough fuel for 5 hours of flying time.
If, upon takeoff it encounters a head wind of 30 mi/hr and the wind remains constant, how far can the plane fly and then return safely?
:
Effective ground speed
300 - 30 = 270 mph against the wind
300 + 30 = 330 mph with the wind
:
Let d = the one-way distance; write a time equation, time = dist/speed
:
outbound time + return time = 5 hrs
 +  = 5
Multiply by 81900 (270*330), eliminate the denominators, results:
330d + 270d = 5(81900)
600d = 445500
6d = 4455
d = 
d = 742.5 mi, the plane can fly out and return safely
:
:
Check this by finding the total time
742.5/270 + 742.5/330 =
2.75 + 2.25 = 5
|
Miscellaneous_Word_Problems/268260: The sum of 3 numbers is 42. The second number is 2 more than twice the first number and the third number is 3 less than the second number. Find the 3 numbers.
1 solutions
Answer 196800 by ankor@dixie-net.com(15653) on 2010-02-11 15:43:42 (Show Source):
You can put this solution on YOUR website!Three numbers: a, b, c
:
Write an equation for each sentence.
:
The sum of 3 numbers is 42.
a + b + c = 42
:
The second number is 2 more than twice the first number
b = 2a + 2
:
the third number is 3 less than the second number.
c = b - 3
from the 2nd equation, replace b with (2a+2)
c = (2a + 2) - 3
c = 2a - 1
:
Find the 3 numbers.
:
a + b + c = 42
Substitute for b and c
a + (2a+2) + (2a-1) = 42
5a + 1 = 42
5a = 42 - 1
5a = 41
a = 
a = 8.2 is the 1st number
then
b = 2(8.2) + 2
b = 16.4 + 2
b = 18.4
and
c = 2(8.2) - 1
c = 16.4 - 1
c = 15.4
:
Check:
8.2 + 18.4 + 15.4 = 42
|
Triangles/268259: in an isosceles triangle,each of the equal sides measures 29 inches and the altitude drawn to the base measure 21 inches.find the length of the base
1 solutions
Answer 196798 by ankor@dixie-net.com(15653) on 2010-02-11 15:40:18 (Show Source):
You can put this solution on YOUR website!I got the solution of one problem mixed up with another problem. Let's see if I can straighten this mess out
;
in an isosceles triangle, each of the equal sides measures 29 inches and the altitude drawn to the base measure 21 inches.find the length of the base
:
We can use pythag: a^2 + b^2 = c^2, it consist of two equal right triangles
where
a = half the base
b = 21
c = 29
:
a^2 + 21^2 = 29^2
a^2 + 441 = 841
a^2 = 841 - 441
a^2 = 400
a = 
a = 20 in is half the base
then 40 inches is the length of the base
|
Rate-of-work-word-problems/268391: .A and B can do a piece of work in 12 days,B and C can do it in 15 days and C and A can do
the same work in 20 days. How long would each take to complete the job?
1 solutions
Answer 196783 by ankor@dixie-net.com(15653) on 2010-02-11 14:48:43 (Show Source):
You can put this solution on YOUR website!Let the piece of work = 1
:
Write a shared work equation for each phrase:
:
A and B can do a piece of work in 12 days,
 +  = 1
:
B and C can do it in 15 days and
 +  = 1
;
C and A can do the same work in 20 days.
 +  = 1
:
Use elimination; Mult the 1st eq by 5, mult the 3nd eq by 3
 +  = 5
+  = 3
------------------------------Subtraction eliminates A, leaving us with
- = 2
:
Multiply the 2nd equation by 4, add to the above equation
- = 2
+ = 4
-------------------------------addition eliminates C, find B
 = 6
120 = 6B
B = 
B = 20 hrs working alone
:
Use the 2nd equation to find C, substitute 20 for B, find C
 + = 1
Multiply by 20C, results:
15C + 20(15) = 20C
15C + 300 = 20C
300 = 20C - 15C
5C = 300
C = 
C = 60 hrs working alone
:
Use the 3rd equation to find A
+  = 1
Multiply by 60A
60(20) + 20A = 60A
1200 = 60A - 20A
40A = 1200
A = 
A = 30 hrs working alone
:
Summarize: A=30; B=20; C=60
;
:
Check it in the 1st equation
+ = 1
 +  =
 +  = 1
:
YOu can confirm it in the other two equations.
|
expressions/268223: Which set of factors of the number 420 has the least possible sum?Which set of factors of the number 420 has the greatest possible sum?Be sure that the two sets of factors both have a product of 420?
1 solutions
Answer 196766 by ankor@dixie-net.com(15653) on 2010-02-11 12:57:12 (Show Source):
You can put this solution on YOUR website!Which set of factors of the number 420 has the least possible sum?
They would be the closest to the square root of 420: 20+21 = 41
:
Which set of factors of the number 420 has the greatest possible sum?
210 + 2 = 212
|
Evaluation_Word_Problems/268221: Marian's younger sister's age is 2 years greater than 1/2 Marian's age. If Marian is 12 years old, how old is her sister? Solve each word problem by setting up an equation.
1 solutions
Answer 196760 by ankor@dixie-net.com(15653) on 2010-02-11 12:42:55 (Show Source):
You can put this solution on YOUR website!Let m = Marian's age
Let s = sister's age
:
Write an equation for the statement:
"Marian's younger sister's age is 2 years greater than 1/2 Marian's age."
s = .5m + 2
:
"If Marian is 12 years old,"
m = 12
:
how old is her sister?
:
Replace m with 12 in the 1st equation
s = .5(12) + 2
s = 6 + 2
s = 8 yrs
|
Graphs/268130: I have Three problems on my work sheet that I do not understand how to do. The instructions on the paper says use Substitution to solve each system of equations. My three problems look like this 1. x+2y=13 3x-5y=6 2. x+5y=4 3x+15y=-1 3. x-5y=10 2x-10y=20..... If you could would you please help me figure these problems out. Thank you so much.
1 solutions
Answer 196756 by ankor@dixie-net.com(15653) on 2010-02-11 12:07:27 (Show Source):
You can put this solution on YOUR website! My three problems look like this
Will write a step-by-step solution on the 1st problem
1.
x + 2y = 13
3x -5y = 6
:
Choose one equation to be used for substitution, the 1st equation would be simpliest
x + 2y = 13
x = -2y + 13
Now we can substitute (-2y+13) for x in the 2nd equation
3x -5y = 6
Substitute
3(-2y+13) - 5y = 6
Multiply what's inside the brackets
-6y + 39 - 5y = 6
Combine the y's on the left, the numbers on the right:
-6y - 5y = 6 - 39
-11y = -33
The variable (y) has to be positive, multiply both sides by -1
11y = 33
y = 
y = 3
Find x using the substitution equation x = -2y+13, replace y with 3
x = -2(3) + 13
x = -6 + 13
x = +7
Our solution is x=7, y=3
Check this in the 2nd original equation: 3x -5y = 6, replace x & y
3(7) - 5(3) = 6
21 - 15 = 6; confirms our solution
:
:
2.
This system does not have a solution, will become apparent as we proceed:
x + 5y = 4
3x +15y= -1
The 1st equation for substitution
x = -5y + 4
Substitute for x in the 2nd equation
3(-5y+4) + 15y = -1
-15y + 12 + 15y = -1
-15y + 15y = -1 - 12
0y = -13, solution undefined, leave it at that
:
3.
This system is not solvable either, the two equation are the same
x - 5y = 10
2x- 10y = 20
simplify the 2nd equation, divide by 2
x - 5y = 10
|
Miscellaneous_Word_Problems/268139: The length of a rectangular swimming pool is 4 meters more than 4 times its width. The perimeter of the pool is 248 meters. What are the length and width of the pool?
Thank you!!
1 solutions
Answer 196741 by ankor@dixie-net.com(15653) on 2010-02-11 10:55:27 (Show Source):
You can put this solution on YOUR website!The length of a rectangular swimming pool is 4 meters more than 4 times its width.
The perimeter of the pool is 248 meters.
What are the length and width of the pool?
:
The perimeter
2L + 2W = 248
simplify, divide by 2
L + W = 124
:
It says,"length is 4 meters more than 4 times its width." so we can say
L = 4W + 4
:
Replace L with (4W+4) in the simplified perimeter equation, find W:
(4W+4) + W = 124
5W = 124 - 4
5W = 120
W = 
W = 24 m is the width
:
I'll let you find L by using the equation L = 4W + 4
:
Then check the solutions in the original perimeter equation
|
Travel_Word_Problems/268135: Alice borrowed a bicycle from her friend, Bob, and wanted to return it. She rode the bicycle from her house to Bob's house at 7 miles per hour. She jogged back home from Bob's house to her house at a pace of 5 miles per hour. It took Alice 36 minutes (i.e., 3/5 hour) total travel time to make her round trip.
(a) How long (in minutes) did each leg of Alice’s trip take? That is, how long did it take Alice to travel to Bob’s house? How long did it take Alice to travel home from Bob’s house?
(b) What is the distance between Alice's and Bob's houses?
1 solutions
Answer 196734 by ankor@dixie-net.com(15653) on 2010-02-11 10:40:26 (Show Source):
You can put this solution on YOUR website!She rode the bicycle from her house to Bob's house at 7 miles per hour.
She jogged back home from Bob's house to her house at a pace of 5 miles per hour.
It took Alice 36 minutes (i.e., 3/5 hour) total travel time to make her round trip.
:
36 min =.6 hr
(a) How long (in minutes) did each leg of Alice’s trip take?
That is, how long did it take Alice to travel to Bob’s house?
How long did it take Alice to travel home from Bob’s house?
:
Let t = time (in hrs) to travel to Bob's house
then
(.6-t) = time to travel from his house
:
Dist both ways is the same; write a dist equation: dist = speed * time
To dist = From dist
7t = 5(.6-t)
7t = 3 - 5t
7t + 5t = 3
12t = 3
t = 
t = .25 hrs; that is .25(60) = 15 min to
and
.6-.25 = .35; that is .35(60) = 21 min from
:
:
(b) What is the distance between Alice's and Bob's houses?
:
.25(7) = 1.75 mi
check it using the return time and speed
.35(5) = 1.75 mi
|
Money_Word_Problems/268140: A person invested a sum of money at 9% annual simple interest and invested twice that amount at 11% annual simple interest. If the total yearly income from both investments was $1,395 how much was invested at each rate?
Thank you for your help. It's greatly appreciated.
1 solutions
Answer 196731 by ankor@dixie-net.com(15653) on 2010-02-11 10:27:07 (Show Source):
You can put this solution on YOUR website!A person invested a sum of money at 9% annual simple interest and invested twice that amount at 11% annual simple interest.
If the total yearly income from both investments was $1,395 how much was invested at each rate?
:
Let x = amt invested at 9%
then
2x = amt invested at 11%
:
a total interest equation
.09x + .11(2x) = 1395
:
.09x + .22x = 1395
:
.31x = 1395
x = 
x = $4500 invested at 9%
and, obviously:
2(4500) = $9000 invested at 11%
:
:
Check this by finding the total interest with these values:
.09(4500) + .11(9000) =
405 + 990 = 1395
|
Miscellaneous_Word_Problems/268138: I would to know if I could have some help here on this problem.
A druggist mixes a 10% soulution of iodine with a 15% solution of iodine. How much of each solution should be used in order to obtain 20 cubic centimeters of a 12% solution?
Thank you for your help
1 solutions
Answer 196710 by ankor@dixie-net.com(15653) on 2010-02-11 09:13:28 (Show Source):
You can put this solution on YOUR website!A druggist mixes a 10% solution of iodine with a 15% solution of iodine.
How much of each solution should be used in order to obtain 20 cubic centimeters of a 12% solution?
:
Let x = amt of 15% solution
then
(20-x) = amt of 10% solution
:
A typical mixture equation
.15x + .10(20-x) = .12(20)
.15x + 2 - .10x = 2.4
.15x - .10x = 2.4 - 2
.05x = .4
x = 
x = 8 cc of 15% solution required
then
20 - 8 = 12 cc of 10% solution
;
:
Check answer in original mixture equation
.15(8) + .10(12) = .12(20)
1.2 + 1.2 = 2.4
|
Linear-systems/268214: how do i find the solution using substitution to:
15x-10y=-1
15y=-2+5x
1 solutions
Answer 196676 by ankor@dixie-net.com(15653) on 2010-02-10 21:35:41 (Show Source):
You can put this solution on YOUR website!15x - 10y =-1
15y = -2 + 5x
:
Use the 2nd equation for substitution
Rearrange to
5x = 15y + 2
divide by 5
x = (3y + .4)
:
Substitute (3y+.4) for x in the 1st equation
15(3y + .4) - 10y = -1
45y + 6 - 10y = -1
45y - 10y = -1 -6
35y = -7
y = 
y = -.2
;
Find x:
x = 3y + .4
x = 3(-.2) + .4
x = -.6 + .4
x = -.2
:
:
Check solution in the 1st equation
15(-.2) - 10(-.2) =-1
-3 + 2 = -1
|
Polynomials-and-rational-expressions/268289: Hi I need help with an example problem so I can do all the other problems. The problem is p to the power of negative 3, times q to the power of negative two, all over q to the power of negative 3, times r to the power of five. This is all to the power of 4. I get most of these problems, but I don't get the more complicated ones. Could you work it out step by step?
1 solutions
Answer 196650 by ankor@dixie-net.com(15653) on 2010-02-10 20:09:06 (Show Source):
You can put this solution on YOUR website!The problem is p to the power of negative 3, times q to the power of negative two, all over q to the power of negative 3, times r to the power of five. This is all to the power of 4.
:
Assume the problem is:
combine the q's
:
:
|
Signed-numbers/268153: Here is my problem
4^3-2+6*5/4*9/6^2=100
I have to make three necessary changes to make this equation true. I cannot change numbers but can change operations and add ( )
If you can help even just a little...I have been working on this for 3 days!
THANK YOU
GNS
1 solutions
Answer 196643 by ankor@dixie-net.com(15653) on 2010-02-10 19:51:39 (Show Source):
You can put this solution on YOUR website!Here is my problem
4^3-2+6*5/4*9/6^2=100
I have to make three necessary changes to make this equation true. I cannot change numbers but can change operations and add ( )
:
 =100
How about
 = 100
|
Miscellaneous_Word_Problems/267856: Mr. Kelly’s company manufactures a cylindrical soup can that has a diameter of 6 inches and a volume of 226 cubic inches. If the diameter stays the same and the height is doubled, what will happen to the can’s volume?
How do you find the original height?
1 solutions
Answer 196585 by ankor@dixie-net.com(15653) on 2010-02-10 15:55:11 (Show Source):
You can put this solution on YOUR website!Mr. Kelly’s company manufactures a cylindrical soup can that has a diameter of 6 inches and a volume of 226 cubic inches. If the diameter stays the same and the height is doubled, what will happen to the can’s volume?
:
The volume will be doubled, (same as two identical cans)
:
How do you find the original height?
:
6" diameter: 3" radius
:
 = 226
 = 226
h = 
h = 7.99 ~ 8 inches
:
:
You can prove it to yourself
 = 226
 = 452
|
Quadratic_Equations/268002: How do you know if a quadratic equation will have one,two, or no solutions? how do you find a quadratic equation if you are only given the solution? Is it possible to have different quadratic equations with the same solution? explain
1 solutions
Answer 196576 by ankor@dixie-net.com(15653) on 2010-02-10 14:26:08 (Show Source):
You can put this solution on YOUR website!How do you know if a quadratic equation will have one,two, or no solutions?
You use the discriminant rules
In the form ax^2 + bx + c, the discriminant: D = -b^2 - 4*a*c
If D > 0 we have two real number solutions that are not equal
If D = 0 we have one real solution (a double root)
If D < 0 we have two complex (non-real) number solutions
:
how do you find a quadratic equation if you are only given the solution?
An example would be easy to understand. Say we have two solutions (roots)
x = 7, x = -3
They can be derived from the factors
(x-7)(x+3) = 0
FOIL
x^2 + 3x - 7x - 21 = 0
y = x^2 - 4x - 21 would be the equation
:
Is it possible to have different quadratic equations with the same solution?
Yes, the equation y = -x^2 + 4x + 21 will have the same solutions as the one above altho the vertex value of y will be different
:
Illustrated by the graph
|
Exponential-and-logarithmic-functions/267905: I am having difficulty on the steps to solve this equation:
e^-x = (e^4)^x+3
Any help would be greatly appreciated!
1 solutions
Answer 196569 by ankor@dixie-net.com(15653) on 2010-02-10 14:02:21 (Show Source):
You can put this solution on YOUR website!Assume the problem is:
 = 
:
Use the nat logs
 = 
:
use the log equiv of exponents
 = 
:
we know the ln of e = 1, and ln(e^4) = 4, therefore it is greatly simplified
 = 
-x = 4(x+3)
-x = 4x + 12
-12 = 4x + x
-12 = 5x
x = 
:
:
Check this on your calc
enter e^(-(-12/5)) = 11.023
enter (e^4)^((-12/5)+3)= 11.023, equality reigns
|
Numeric_Fractions/267975: john spent a quarter of his life growing up as a boy, 1/6 of his life in college and 1/2 of his life as a teacher.He spent his last six years in retirement. How old was he when he died?
1 solutions
Answer 196558 by ankor@dixie-net.com(15653) on 2010-02-10 13:20:18 (Show Source):
You can put this solution on YOUR website!john spent a quarter of his life growing up as a boy,
1/6 of his life in college and 1/2 of his life as a teacher.
He spent his last six years in retirement.
How old was he when he died?
:
let a = his age when he died
:
Age = boy + student + teacher + retired
a =  a +  a +  a + 6
Multiply by 12 to get rid of the denominators, results
12a = 3a + 2a + 6a + 12(6)
12a = 11a + 72
12a - 11a = 72
a = 72 his age when he passed away, much missed and remembered, I am sure.
|
Expressions-with-variables/267731: a plane takes 3 hours to travel 1,503 miles with the wind. It can travel only 1,179 miles against the wind in the same time. Find the speed of the wind and the speed of the plane in still air.
1 solutions
Answer 196543 by ankor@dixie-net.com(15653) on 2010-02-10 11:59:03 (Show Source):
You can put this solution on YOUR website!a plane takes 3 hours to travel 1,503 miles with the wind.
It can travel only 1,179 miles against the wind in the same time.
Find the speed of the wind and the speed of the plane in still air.
:
Let s = still air speed of the plane
Let w = speed of the wind
then
(s-w) = effective speed against the wind
and
(s+w) = effective speed with the wind
;
Write a distance equation for each trip; Dist = time * speed
:
3(s + w) = 1503
3(s - w) = 1179
we can simplify both of these, divide by 3
s + w = 501
s - w = 393
-------------addition eliminates w
2s = 894
s = 
s = 447 mph in still air
:
Find w
447 + w = 501
w = 501 - 447
w = 54 mph is the wind
:
:
Check solution in the 2nd equation
3(447 - 54) =
3(393) = 1179 mi
|
Linear-equations/267902: I'm hopeless !
In a triangle, the largest angle is 100 degrees greater than the smallest angle. The largest angle is three times as large as the third angle. What is the measure of each angle?
1 solutions
Answer 196519 by ankor@dixie-net.com(15653) on 2010-02-10 10:14:20 (Show Source):
You can put this solution on YOUR website!I don't think you are hopeless at all, how about a step-by-step method here?
Let x = the largest angle
:
"the largest angle is 100 degrees greater than the smallest angle."
(x-100) = the smallest angle
:
"The largest angle is three times as large as the third angle."
 x = the 3rd angle
:
Now we have an expression for all 3 angles using x
:
What is the measure of each angle?
I'm sure you know that a triangle has a total of 180 degrees
:
Large + small + 3rd angle = 180
x + (x-100) + x = 180
2x + x - 100 = 180
2x + x = 180 + 100
2x + x = 280
Multiply by 3 to get rid of the denominator, results
3(2x) + x = 3(280)
:
6x + x = 840
;
7x = 840
x = 
x = 120 degrees is the largest angle
then
* 120 = 40 degrees is the 3rd angle
and
120 - 100 = 20 degrees is the smallest angle
:
;
Check our answers
120 + 40 + 20 = 180 degrees
:
:
Hope this relieves you of that hopeless feeling!!
If you have any questions about this, you can email me. ankor@att.net
|
Equations/267904: 3/4x=15 why would the answer be 20? I understand that you have to multiply the reciprocal 4/3 by both sides however I am lost after that. Could you please explain this to me?
1 solutions
Answer 196503 by ankor@dixie-net.com(15653) on 2010-02-10 09:42:28 (Show Source):
You can put this solution on YOUR website!3/4x=15 why would the answer be 20? I understand that you have to multiply the reciprocal 4/3 by both sides however I am lost after that. Could you please explain this to me?
:
Perhaps it will make sense if you do it in two steps
:
 x = 15
Multiply by 4 to get rid of the denominator, results:
3x = 4(15)
3x = 60
Divide both sides by 3
x = 
x = 20
|
|
