I need u guys to help me out.solving a simultaneous equation using matrices.

i cn wrk on d 2*2 matrix buh wen it cmes to 3*3 dere is trouble.

dis an example

4x-3y+2z=-7
6x+2y-3z=33
2x-4y-z=-3


pls wen solve show ur workings,if possible u cn take a picture of it on were u solved it n upload so i wil understand ur steps.

fanks in advance.

VEVEDIHNO:
I need u guys to help me out.solving a simultaneous equation using matrices.

i cn wrk on d 2*2 matrix buh wen it cmes to 3*3 dere is trouble.

dis an example

4x-3y+2z=-7
6x+2y-3z=33
2x-4y-z=-3


pls wen solve show ur workings,if possible u cn take a picture of it on were u solved it n upload so i wil understand ur steps.

fanks in advance.

u cn easily use crammer's method to solve this if it is for UTME..

dejt4u:

u cn easily use crammer's method to solve this if it is for UTME..

nice job...boss.

its me. benbuks ...

#shalom.

agentofchange1:


nice job...boss.

its me. benbuks ...

#shalom.

bros bros..longest tym..how u doing?? Happy new year!

dejt4u:

u cn easily use crammer's method to solve this if it is for UTME..

OK
I JUST NEED TO SEE UR WORKINGS

Plz dnt be annoyed am posting physics question on maths forum its because dey aiint active dere plz help me out my Guruz


A alloy of mass 588g and a volume 100cm^3 is made of density 8.0g/cm^3 and alluminium of density 2.7g/cm^3.
Calculate the proportion;

(i)By Volume
(ii)By Mass of the constituents of the alloy.

Fankz in advance.

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Sample questions below:

VEVEDIHNO:
Plz dnt be annoyed am posting physics question on maths forum its because dey aiint active dere plz help me out my Guruz


A alloy of mass 588g and a volume 100cm^3 is made of density 8.0g/cm^3 and alluminium of density 2.7g/cm^3.
Calculate the proportion;

(i)By Volume
(ii)By Mass of the constituents of the alloy.

Fankz in advance.

I know my method might be long, but I do it so you can understand how i got there.

This is a long question but let's go!! You didnt specify the first metal used in the alloy, so let's just call it Metal
And also lets remember that Mass= Density * Volume

Let Volume of Metal be Equal to A

Therefore mass of Metal will be equal to 8.0g/cm^3 * A

Volume of Aluminium will be equal to 100cm^3 - A (the symbol - indicates minus)

Therefore the mass of Aluminium will be 2.7g/cm^3*(100cm^3-A).

We know the mass of both metals and we know the total mass. We use this to form an equation

Total mass= Mass of Metal+ Mass of Aluminium

588g = 8.0*A + 2.7*(100-A) ---- I removed the units to make it clearer

588 =8A+270-2.7A.

588-270=8A-2.7A

318 =5.3A

A=318/5.3=60

Solving for A will give A=60 cm^3 (Remember A is the volume of Metal)

Volume of Aluminium is 100cm^3-60cm^3 =40cm^3

Mass of Metal is= 8.0 * 60 = 480grams

Mass of Aluminium is = 588-480 = 108 grams

VEVEDIHNO:
I need u guys to help me out.solving a simultaneous equation using matrices.

i cn wrk on d 2*2 matrix buh wen it cmes to 3*3 dere is trouble.

dis an example

4x-3y+2z=-7
6x+2y-3z=33
2x-4y-z=-3


pls wen solve show ur workings,if possible u cn take a picture of it on were u solved it n upload so i wil understand ur steps.

fanks in advance.

I hope this helps

goofyone:


I hope this helps

Goofy pls am having a Lil problem.i hve followed your step in solving some problems faster and it has helped me a lot but am having a lil problem in sub.1 in factorisation.for eg
1)factorise: a2-b2+ap-bq+bp+aq
Ans=(a+b)(a-b+p-q)
if i sub. a=1,b=1,p=1,q=1,i get
1-1+1-1+1+1=2....but sub it in the ans (1+1)(1-1+1-1)=0 and nt 2..pls help me

plz some should help me solve 1. x^2 + x^1/4 = 20
2. solve 9^2x+1= 8^x-2 /3x

Adiwana:

Goofy pls am having a Lil problem.i hve followed your step in solving some problems faster and it has helped me a lot but am having a lil problem in sub.1 in factorisation.for eg
1)factorise: a2-b2+ap-bq+bp+aq
Ans=(a+b)(a-b+p-q)
if i sub. a=1,b=1,p=1,q=1,i get
1-1+1-1+1+1=2....but sub it in the ans (1+1)(1-1+1-1)=0 and nt 2..pls help me

If you look at the answer, you'll see it suggests that you should have the term -aq and not +aq.
(a+b)(a-b+p-q) = a(a-b+p-q) +b(a-b+p-q)
a x -q among the terms will give you -aq.

Using that will make the question 1-1+1-1+1-1=0

[quote author=goofyone post=30891573]
thanks man..been using this ya method even in factorising polymas.
Pls see another
x2(x2-1)^1/2 - (x2-1)^1/2
a)(x2-1)^1/2
b)(x2-1)
c)(x2-1)^-1
d)(x2-1)^-1/2.


1-(a-b)2
a) (1-a-b)(1-a+b)
b) (1-a+b)(1+a-b)
c) (1-a+b)(1-a+b)
d) (1-a-b)(1+a-b)
Thanks

goofyone:


If you look at the answer, you'll see it suggests that you should have the term -aq and not +aq.
(a+b)(a-b+p-q) = a(a-b+p-q) +b(a-b+p-q)
a x -q among the terms will give you -aq.

Using that will make the question 1-1+1-1+1-1=0

thanks man..been using this ya method even in factorising polymas.
Pls see another
x2(x2-1)^1/2 - (x2-1)^1/2
a)(x2-1)^1/2
b)(x2-1)
c)(x2-1)^-1
d)(x2-1)^-1/2.


1-(a-b)2
a) (1-a-b)(1-a+b)
b) (1-a+b)(1+a-b)
c) (1-a+b)(1-a+b)
d) (1-a-b)(1+a-b)
Thanks

@goofy,you go talk say i dey disturb you..but pls i need full explanation on that subtution method cos at times wen i subt. I get same answers in the option..for eg from the question,the ans may be 1 and i get 1 in two places.using this method will save me a lot of troubles of factorisation...thanx Big time

Adiwana:
@goofy,you go talk say i dey disturb you..but pls i need full explanation on that subtution method cos at times wen i subt. I get same answers in the option..for eg from the question,the ans may be 1 and i get 1 in two places.using this method will save me a lot of troubles of factorisation...thanx Big time

It's possible to get that kind of result. If you get 1 in two places or three places, then you should eliminate those that didn't give 1 as answer. Let's say you have four options and are able to eliminate two options out of four. You left with two more. Then you can test something more exclusive on these two. Say -1. If -1 still gives similar results, then move up to 2. Or -2. Just a number simple enough to manipulate. You get it?

goofyone:

It's possible to get that kind of result. If you get 1 in two places or three places, then you should eliminate those that didn't give 1 as answer. Let's say you have four options and are able to eliminate two options out of four. You left with two more. Then you can test something more exclusive on these two. Say -1. If -1 still gives similar results, then move up to 2. Or -2. Just a number simple enough to manipulate. You get it?

sure sir..you no sabi you don help save me a lot stress and time esp in simultaneous equation.i now solve it faster...thanx very much
still waiting for soln of those qquestions

Adiwana:

sure sir..you no sabi you don help save me a lot stress and time esp in simultaneous equation.i now solve it faster...thanx very much
still waiting for soln of those qquestions

You always welcome

Adiwana:


1-(a-b)2
a) (1-a-b)(1-a+b)
b) (1-a+b)(1+a-b)
c) (1-a+b)(1-a+b)
d) (1-a-b)(1+a-b)
Thanks

Let's look at this first:
Make a=1 and b =1
First the question: 1-(1-1)² = 1-0=1
Option a: (1-1-1)(1-1+1)=-1 x 1 = -1
Option b: (1-1+1)(1+1-1)=1 x 1 = 1
Option c: (1-1+1)(1-1+1)=1 x 1 = 1
Option d: (1-1-1)(1+1-1)= -1 x 1 = -1

So you have between option b and option c to decide. At this moment you want to immediately determine your answer. Though the next step should be to substitute -1 (which I'll do), but another thing you could immediately do is to look at the question and look for something you can quickly use to eliminate b or c as an answer.

Look at the b² in the question and determine what it should look like in the answer. In the question, if you simplify (which you SHOULDN'T do), you'll find that the b² term is going to end up as -b² for the simple reason that expanding (a-b)² will give b² and multiplying by the negative outside the bracket will give -b². You can do this without simplifying. In which case, if you look at option b and c, you'll see that only option b can end up with a -b² term and that makes it the answer. You get it?

But if you want to go the route of substituting -1, then attend to just option b and c.

Option b: (1+1-1)(1-1+1) = 1
Option c: (1-1+1)(1-1+1) = 1

Hmm... not working.
If you still haven't figured the answer, then you should vary the constants. Make a=1 and b=2

Option b: (1-1+2)(1+1-2)=2x0=0
Option c: (1-1+2)(1-1+2)=2x2=4

Then into the question: 1-(1-2)² = 1-1=0

So, b is indeed our answer. But you really wanna be very quick with this substitutions. In that respect, practice makes perfect.

I think you should try the upper question yourself and let's see your hand.

goofyone:


Let's look at this first:
Make a=1 and b =1
First the question: 1-(1-1)² = 1-0=1
Option a: (1-1-1)(1-1+1)=-1 x 1 = -1
Option b: (1-1+1)(1+1-1)=1 x 1 = 1
Option c: (1-1+1)(1-1+1)=1 x 1 = 1
Option d: (1-1-1)(1+1-1)= -1 x 1 = -1

So you have between option b and option c to decide. At this moment you want to immediately determine your answer. Though the next step should be to substitute -1 (which I'll do), but another thing you could immediately do is to look at the question and look for something you can quickly use to eliminate b or c as an answer.

Look at the b² in the question and determine what it should look like in the answer. In the question, if you simplify (which you SHOULDN'T do), you'll find that the b² term is going to end up as -b² for the simple reason that expanding (a-b)² will give b² and multiplying by the negative outside the bracket will give -b². You can do this without simplifying. In which case, if you look at option b and c, you'll see that only option b can end up with a -b² term and that makes it the answer. You get it?

But if you want to go the route of substituting -1, then attend to just option b and c.

Option b: (1+1-1)(1-1+1) = 1
Option c: (1-1+1)(1-1+1) = 1

Hmm... not working.
If you still haven't figured the answer, then you should vary the constants. Make a=1 and b=2

Option b: (1-1+2)(1+1-2)=2x0=0
Option c: (1-1+2)(1-1+2)=2x2=4

Then into the question: 1-(1-2)² = 1-1=0

So, b is indeed our answer. But you really wanna be very quick with this substitutions. In that respect, practice makes perfect.

I think you should try the upper question yourself and let's see your hand.

ok.
x2(x2-1)^1/2 - (x2-1)^1/2
let x=0
0^2(0^2-1)^1/2 - (0^2-1)^1/2
=0-(-1)^1/2
=1^1/2
=1
a) (x2-1)1/2=(0-1)=-1^1/2
b) (x2-1)=(0-1)=-1
c) (x2-1)^-1=(0-1)^-1=-1^-1=1
d) (x2-1)-1/2=(0-1)^-1/2=-1^-1/2
=-1^1/2
so is it C?

*someone should help with this*.
a rectangular plot measures 12m by 5m,a path of constant width run along one side and one end. if the total area of the plot and the path is 120msq.find the width of the path.
(hint:let the width of the path be xm)

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Emmanuelclean:
*someone should help with this*.
a rectangular plot measures 12m by 5m,a path of constant width run along one side and one end. if the total area of the plot and the path is 120msq.find the width of the path.
(hint:let the width of the path be xm)

(12+x)(5+x)=120
x^2 + 17x + 60-120=0
x^2 + 17x - 60=0
x=(-17+/-sqrt(17^2 - 4*1*-60))/(2*1)
x=(-17+/-23)/2
x = (-17+23)/2=6/2=3 or (-17-23)/2=-20

x looking more likely to be 3, as it wouldn't necessarily be negative.

If you test your answer you have 15*8=120msq

goofyone:

(12+x)(5+x)=120
x^2 + 17x + 60-120=0
x^2 + 17x - 60=0
x=(-17+/-sqrt(17^2 - 4*1*-60))/(2*1)
x=(-17+/-23)/2
x = (-17+23)/2=6/2=3 or (-17-23)/2=-20

x looking more likely to be 3, as it wouldn't necessarily be negative.

If you test your answer you have 15*8=120msq

brother could you please be helping me in Maths

so

nelsoelomi:
somebody pls help me with this equation.
1.Add 256 base 7 and 443 base 7

2.Subtract 311 base 5 and 1240 base 5

3.if 34n = 10011 base 2 find n

4. 14*25 = 374, find the number base used.


Thank in advance.

1.Add 256 base 7 and 443 base 7

532 base 7
!Convert 256 and 443 to base 10 and do the addition..convert your final answer back to base 7

2.Subtract 311 base 5 and 1240 base 5

424 base 5
!Convert 311 and 1240 to base 10 and do the subtraction..convert your final answer back to base 5

3.if 34n = 10011 base 2 find n

n = 3
!converting both sides to base 10, it will give you; 3n + 4 = 19, n = 3

4. 14*25 = 374, find the number base used.

base 8

cheers!

nelsoelomi:
somebody pls help me with this equation.

3.if 34n = 10011 base 2 find n


Thank in advance.

solution. convert the both sides to base 10. we have 3n+4=19 3n=19-4 3n=15 divide all through by 3 n=5

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