Algebra Cafe: Solve Math Problems Here

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Author Topic: Algebra Cafe: Solve Math Problems Here  (Read 2306 views)
oldie (m)
Re: Algebra Cafe: Solve Math Problems Here
« #64 on: April 02, 2008, 07:12 PM »

Some of these problems have mathematical solutions, but not practical solutions!
Like this one:

Quote
Rasheed can mow his mother's lawn in 6 minutes. His brother Peter can mow it in 9 minutes. How long will it take them to do it together, if each has his own lawnmower.

The answer depends on many variables
Are the mowers clones of each other? Same efficiency?
The time that will be used by the two brothers can never be determined in practice
Dem fit quarrel self and the job is never done Grin Grin

So always remember prototypes are always different from the real thing
But I like the zeal put in by contributors
donchichi
Re: Algebra Cafe: Solve Math Problems Here
« #65 on: April 03, 2008, 05:17 AM »

Quote from: dinozzo on April 02, 2008, 04:36 PM
ok, if I'm to go with that I'll have 0.5(6+x) + 1.25x = 12.75

1.75x=9.75
x=5.57, I don't know doesn't look like a good answer, maybe the analogy i'm using is all wrong then

@kobe whats the answer, I can't think of any other way to solve it Tongue

exactly 4Him

Quote from: 4Him on April 02, 2008, 04:32 PM
If you use this equation you don't get a round figure. He can't have mowed the lawn 5.6 times.

Guys, i agree that u'll not get an integer for an answer, but i didnt set the question Grin. i also tried multiplying by 6 but just looking at the question and statement in particular, you should actually add 6 to it. But its all good.
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #66 on: April 03, 2008, 07:39 AM »

Rasheed can mow his mother's lawn in 6 minutes. His brother Peter can mow it in 9 minutes. How long will it take them to do it together, if each has his own lawnmower.

t(rasheed) = 6 minutes
t(peter) = 9 minutes

Rasheed mows one lawn over 6 minutes
Peter mows one lawn over 9 minutes
Rate of work for Rasheed = L/6
Rate of work for Peter = L/9

Now we want to mow one lawn when both are doing it together.
Time it takes to mow one lawn
amount of work done = rate at which the work is done multiplied by the time they worked.
So we want time.
1L= (L/6 + L/9)t
t = L/(L/6+L/9) = 54/15mins
LearnBook
Re: Algebra Cafe: Solve Math Problems Here
« #67 on: April 03, 2008, 07:56 AM »

Dele shares out 15 sweets.   He gives Ifeoma 1 for every 4 he eats.   How many does Ifeoma get?
donchichi
Re: Algebra Cafe: Solve Math Problems Here
« #68 on: April 03, 2008, 07:58 AM »

Quote from: LearnBook on April 03, 2008, 07:56 AM
Dele shares out 15 sweets. He gives Ifeoma 1 for every 4 he eats. How many does Ifeoma get?

15?
LearnBook
Re: Algebra Cafe: Solve Math Problems Here
« #69 on: April 03, 2008, 07:59 AM »

Yes 15.
oldie (m)
Re: Algebra Cafe: Solve Math Problems Here
« #70 on: April 03, 2008, 08:34 AM »

Quote from: LearnBook on April 03, 2008, 07:56 AM
Dele shares out 15 sweets. He gives Ifeoma 1 for every 4 he eats. How many does Ifeoma get?

Should it not be 3?
4x + x = 15
5x = 15
x = 3

Correct me if I am wrong
PL+C
Re: Algebra Cafe: Solve Math Problems Here
« #71 on: April 03, 2008, 09:22 AM »

I can't see the ladies contributing anything here, if na Romance thread now, they will tell you how Solomon slept with over 1000 women closing his eyes, kissing them and doing legs like lizard.

The only two ladies that tried to show face only applauded the guys but no contributions from them.  Whether the guys are working jargons they don't know. Cheesy
donchichi
Re: Algebra Cafe: Solve Math Problems Here
« #72 on: April 03, 2008, 09:53 AM »

Quote from: oldie on April 03, 2008, 08:34 AM
Should it not be 3?
4x + x = 15
5x = 15
x = 3

Correct me if I am wrong

The question already said He shares out 15 sweets and the only recipient mentioned is ifeoma. So ifeoma gets all 15 sweets that was shared. If the question was rephrased as "Dele has 15 sweets", then the answer would be 3.
oldie (m)
Re: Algebra Cafe: Solve Math Problems Here
« #73 on: April 03, 2008, 10:03 AM »

@ donchichi
Thanks!
RichyBlacK (m)
Re: Algebra Cafe: Solve Math Problems Here
« #74 on: April 03, 2008, 11:35 AM »

A die is cast five times, what is the probability of getting at least a six?
LearnBook
Re: Algebra Cafe: Solve Math Problems Here
« #75 on: April 03, 2008, 01:24 PM »

donchichi:  shares out is the right phrase.  It's more like 4 for Dele, 1 for Ifeoma.    Dele is sharing but that does not exclude him from the process.

Anyway, enough of the semantics here.  Yes, the correct answer is 3.    Dele gets 12, Ifeoma gets 3 and the ratio 12:3 is equivalent to 4:1.

Have a good day.
bawomolo (m)
Re: Algebra Cafe: Solve Math Problems Here
« #76 on: April 03, 2008, 02:16 PM »

Quote
A die is cast five times, what is the probability of getting at least a six?

(1/6)*5=5/6
GeeCee (m)
Re: Algebra Cafe: Solve Math Problems Here
« #77 on: April 03, 2008, 03:09 PM »

I bow my head for the mathematicians in the house. A very big WELL DONE 2 u.

But wait, where are the ladies?
Olumide7 (m)
Re: Algebra Cafe: Solve Math Problems Here
« #78 on: April 03, 2008, 04:31 PM »

Integrate: xlnx
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #79 on: April 03, 2008, 05:15 PM »

integral of xlnx using int by parts = uv-integral (vdu)
u = lnx dv = x
du = 1/xdx
v = integral of dv = x2/2

applying formula.

uv = x2/2lnx
integral vdu = integral (x2/2 * 1/x dx) = x/2

integral xlnx = x2/2lnx - 1/4x2 + C
4Him (m)
Re: Algebra Cafe: Solve Math Problems Here
« #80 on: April 03, 2008, 05:29 PM »

integration was the main reason i withdrew from advanced maths class in high school.
Too confusing.
spikedcylinder (f)
Re: Algebra Cafe: Solve Math Problems Here
« #81 on: April 03, 2008, 05:52 PM »

Reading through this thread, all i can think to say is, Una dey craze,  Grin Grin
Imani (f)
Re: Algebra Cafe: Solve Math Problems Here
« #82 on: April 03, 2008, 10:28 PM »

Quote from: PL+C on April 03, 2008, 09:22 AM
I can't see the ladies contributing anything here, if na Romance thread now, they will tell you how Solomon slept with over 1000 women closing his eyes, kissing them and doing legs like lizard.

The only two ladies that tried to show face only applauded the guys but no contributions from them. Whether the guys are working jargons they don't know. Cheesy


Mr bigmouth PLC, how dare you claim the ladies are "olodo" Angry Grin. Of course Romance topics are more interesting.

Anyway, i have to admit, since this thread started, i have been trying to dust off all the "cobwebs" from my brain Grin. I did do A level Maths but that was like 8 years ago now. But no excuses, i will try to rise to some of the simplier chalenges Lips sealed

Quote from: LearnBook on April 03, 2008, 01:24 PM
donchichi: shares out is the right phrase. It's more like 4 for Dele, 1 for Ifeoma. Dele is sharing but that does not exclude him from the process.

Anyway, enough of the semantics here. Yes, the correct answer is 3. Dele gets 12, Ifeoma gets 3 and the ratio 12:3 is equivalent to 4:1.


Quote from: LearnBook on April 03, 2008, 07:59 AM
Yes 15.

So, which one is it?. I initially agreed with Oldie, using 4:1 ratio, however, after seeing donchichi's working, I can see why he arrived at 15. What is missing is the key word, is it Dele "shared out" 15 sweets thereby excluding him from the process or "has" 15 sweets, which includes him? Either answer is correct, depending on whether Dele is included or not.
Imani (f)
Re: Algebra Cafe: Solve Math Problems Here
« #83 on: April 03, 2008, 10:32 PM »

1. Expand (4+2x)6 in ascending powers of X up to the power of X3

2. Solve the following quadratic equations 3x2 + 5x -8 =0

3.  X2 + 2X -8 =0
huxley
Re: Algebra Cafe: Solve Math Problems Here
« #84 on: April 03, 2008, 10:59 PM »

What is the product of:

(x-a)(x-b)(x-c)(x-d) . . . .  (x-z)
Damest09 (f)
Re: Algebra Cafe: Solve Math Problems Here
« #85 on: April 04, 2008, 02:13 AM »

wow, this is good. I wonder what kind of job u people r doing with this level of brain u posses and how much money  make by it.
Damest09 (f)
Re: Algebra Cafe: Solve Math Problems Here
« #86 on: April 04, 2008, 02:15 AM »

Well done anyway
The Sly
Re: Algebra Cafe: Solve Math Problems Here
« #87 on: April 04, 2008, 02:21 AM »

Quote
I can't see the ladies contributing anything here, if na Romance thread now, they will tell you how Solomon slept with over 1000 women closing his eyes, kissing them and doing legs like lizard.

The only two ladies that tried to show face only applauded the guys but no contributions from them. Whether the guys are working jargons they don't know.
  Grin Grin Grin Grin Grin Grin Grin

You nko?   Cheesy why don't u contribute to the thread. . .but if na to photoshop people's pic. . that is what u know how to do. .
Shiorrrrrrrrr!!

donchichi
Re: Algebra Cafe: Solve Math Problems Here
« #88 on: April 04, 2008, 02:38 AM »

Quote from: Imani on April 03, 2008, 10:32 PM
1. Expand (4+2x)6 in ascending powers of X up to the power of X3

2. Solve the following quadratic equations 3x2 + 5x -8 =0

3.  X2 + 2X -8 =0

1. Can be solved using some theorem i can't remember Grin

2. (3x+8)(x-1) = 0 ; x = 1, x = -8/3

3. (x-2)(x+4) = 0; x = 2, x = -4
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #89 on: April 04, 2008, 06:49 AM »

I don't understand what you mean by in powers of X up to 3, but

Expansion of (2X+4)6 using the binomial theorem is:

Coefficients are : 1 6 15 20 15 6 1

1(2x)640+6(2x)541+15(2x)442+20(2x)343+15(2x)244+6(2x)145+1(2x)046

Simplifies to:
64x6+768x5+3840x4+10240x3+15360x2+12288x+4096
Imani (f)
Re: Algebra Cafe: Solve Math Problems Here
« #90 on: April 04, 2008, 06:51 AM »

Quote from: Imani on April 03, 2008, 10:32 PM
1. Expand (4+2x)6 in ascending powers of X up to the power of X3

2. Solve the following quadratic equations 3x2 + 5x -8 =0

3.  X2 + 2X -8 =0

1. Use binomial theorem. To find the answer subsititute 4 for a and 2x for b:

46 + (6 C1)(45)(2x) + (6C2)(44)(2x)2 + (6 C3)(43)(2x)3
4096 + (6 ×1024 ×2x) + (15 ×256 ×4x2) + (20 ×64 ×8x3)
4096 + 12288x + 15360x2 + 10240x3

2. Use the quadratic equation formulae

3. Solve by factorising

the answers are in the below post

Quote from: donchichi on April 04, 2008, 02:38 AM

2. (3x+8)(x-1) = 0 ; x = 1, x = -8/3

3. (x-2)(x+4) = 0; x = 2, x = -4


Correct answers to numbers 2 and 3. welldone, i will post the workings later.


kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #91 on: April 04, 2008, 07:02 AM »

Quote from: huxley on April 03, 2008, 10:59 PM
What is the product of:

(x-a)(x-b)(x-c)(x-d) . . . . (x-z)

Product of (x-a)(x-b)(x-c)(x-d) . . . . (x-z)
= (x-a)(x-b)(x-c)(x-d) . . . (x-x),  (x-z)
= (x-a)(x-b)(x-c)(x-d) . . . .(0),  . (x-z)
=0*(x-a)(x-b)(x-c)(x-d) . . . . (x-z)
= 0
Imani (f)
Re: Algebra Cafe: Solve Math Problems Here
« #92 on: April 04, 2008, 07:14 AM »

Quote from: kobe on April 04, 2008, 07:02 AM
Product of (x-a)(x-b)(x-c)(x-d) . . . . (x-z)
= (x-a)(x-b)(x-c)(x-d) . . . (x-x), (x-z)
= (x-a)(x-b)(x-c)(x-d) . . . .(0), . (x-z)
=0*(x-a)(x-b)(x-c)(x-d) . . . . (x-z)
= 0

Kobe, you are good. Wink Kiss. I didnt get as far as (x-x). but your working is spot on and makes sense. And of course anything multipled by zero is zero.
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #93 on: April 04, 2008, 07:28 AM »

good ke.  all of you guys are just too good. even you sister imani, i'm amazed at your superpower.

I gotta post more problems.
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #94 on: April 04, 2008, 07:40 AM »

A boot that costs x dollars is marked down by 20 percent off its original price on Monday, then an additional 10 percent on Tuesday. What is the price of the jacket, in terms of x?
kobe (m)
Re: Algebra Cafe: Solve Math Problems Here
« #95 on: April 04, 2008, 07:48 AM »

Given x2 - y2 = 77 and x + y = 11 find x?
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