Nairaland Mathematics Clinic - Education (161)

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Re: Nairaland Mathematics Clinic by AlphaMaximus(m): 6:39pm On Jan 22, 2015

Re: Nairaland Mathematics Clinic by bolkay47(m): 8:05pm On Jan 22, 2015

AlphaMaximus:

20-(60*(0.1/100)*20)=18.8 square feet of grass left.....dont really see what was complicated about the question. Open to correction though

the question is on GP....it should somehow form a sequence. That's what I don't understand.

Re: Nairaland Mathematics Clinic by AlphaMaximus(m): 9:26pm On Jan 22, 2015

bolkay47:
the question is on GP....it should somehow form a sequence. That's what I don't understand.

alright.....well in that case, this is how it goes:
We have info that field is depleted by 0.1% with each swipe of the cutlass, thus each swipe leaves the field devoid of 1/1000th its preceding area.
Therefore, mathematically, the first term, a=20, the ratio , r=1/1000, and the number of terms/swipes, n=60........in this case since we are interested in the amount left from whole, we subtract "r" from 1 and we have : 1-1/1000= 999/1000.....lets call 999/1000 "R".....then we use the formula S_remainder= a (R)ⁿ and we have:
S_remainder= 20*(999/1000)⁶⁰=18.83472524
To further corroborate the validity of the formula, lets examine a case where each flow of a pipe depletes a 1,000,000litre tank by 1/4.......the first time, 250,000(one-fourth of 1,000,000) will be removed leaving 750,000,the second time, 187,500 (one-fourth of 750,000), leaving 562,500, the third time,140,625(one-fourth of 562,500), will be removed leaving 421,875
.....instead of this rather cumbersome approach, the previously used formula can be used to attain the 421,875 litres left......a=1,000,000 and n=3, and R=1-r=1-1/4=3/4....therefore;
S_remainder=a(R)ⁿ=1,000,000*(3/4)³=421,875 litres. Thus, the method is justified.

Re: Nairaland Mathematics Clinic by bolkay47(m): 10:04pm On Jan 22, 2015

AlphaMaximus:

alright.....well in that case, this is how it goes:
We have info that field is depleted by 0.1% with each swipe of the cutlass, thus each swipe leaves the field devoid of 1/1000th its preceding area.
Therefore, mathematically, the first term, a=20, the ratio , r=1/1000, and the number of terms/swipes, n=60........in this case since we are interested in the amount left from whole, we subtract "r" from 1 and we have : 1-1/1000= 999/1000.....lets call 999/1000 "R".....then we use the formula S_remainder= a (R)ⁿ and we have:
S_remainder= 20*(999/1000)⁶⁰=18.83472524
To further corroborate the validity of the formula, lets examine a case where each flow of a pipe depletes a 1,000,000litre tank by 1/4.......the first time, 250,000(one-fourth of 1,000,000) will be removed leaving 750,000,the second time, 187,500 (one-fourth of 750,000), leaving 562,500, the third time,140,625(one-fourth of 562,500), will be removed leaving 461,875
.....instead of this rather cumbersome approach, the previously used formula can be used to attain the 461,875 litres left......a=1,000,000 and n=3, and R=1-r=1-1/4=3/4....therefore;
S_remainder=a(R)ⁿ=1,000,000*(3/4)³=421,875 litres. Thus, the method is justified.

this is just excellent.... Although I don't know the answer yet cos its an assignment but WAO.#respect..

Re: Nairaland Mathematics Clinic by AlphaMaximus(m): 10:33pm On Jan 22, 2015

bolkay47:
this is just excellent.... Although I don't know the answer yet cos its an assignment but WAO.#respect..

Im certain thats the answer. cheesy

Open to contrary opinions.

Re: Nairaland Mathematics Clinic by AlphaMaximus(m): 10:41pm On Jan 22, 2015

Re: Nairaland Mathematics Clinic by efficiencie(m): 2:39am On Jan 23, 2015

jaryeh:
Chai! Long time........

I need help on this pls..

tag: jackpot, Alphamaximus, dejt4u, doubledx, Richie and other generals. I don forget una names finish o.

Given that:
arcsina+arcsinb+arcsinc=π

arcsina=π-arcsinb-arcsinc

Take the cosine of both sides:

cos(arcsina)=cos(π-arcsinb-arcsinc)

cos(arcsina)=cos(π-(arcsinb+arcsinc))

cos(arcsina)=
-cos(arcsinb+arcsinc)

cos(arcsina)=
-cos(arcsinb)cos(arcsinc)+ sin(arcsinb)sin(arcsinc)

√(1-sin^2(arcsina))=
-(√(1-sin^2(arcsinb))
√(1-sin^2(arcsinc)))+
sin(arcsinb).sin(arcsinc)

a√(1-a^2)=-a.√(1-b^2).√(1-c^2)+abc...1

and similarly the following is derived:

b√(1-b^2)=-b√(1-a^2).√(1-c^2)+abc...2

c√(1-c^2)=-c√(1-b^2).√(1-a^2)+abc...3

On adding up:

a√(1-a^2)+b√(1-b^2)+c√(1-c^2) =-a.√(1-b^2).√(1-c^2)
-b.√(1-a^2).√(1-c^2)
-c.√(1-b^2).√(1-a^2)+3abc ...4

Going back to the original equatn:
arcsina+arcsinb+arcsinc=π

arcsina+arcsinb=π-arcsinc
On taking the sine of both sides

sin(arcsina+arcsinb)=
sin(π-arcsinc)

sin(arcsina)cos(arcsinb)+ sin(arcsinb)cos(arcsina) =c

a.√(1-b^2)+b.√(1-a^2) =c ...5

and similarly the followin is derivd
a.√(1-c^2)+c.√(1-a^2) =b ...6

c.√(1-b^2)+b.√(1-c^2) =a ...7

And hence multiply 5 by
√(1-c^2), 6 by √(1-b^2) and 7 by √(1-a^2) the followin results:

a.√(1-b^2).√(1-c^2)+b.√(1-a^2).√(1-c^2)=c.√(1-c^2) ...8

a.√(1-c^2).√(1-b^2) +c.√(1-a^2).√(1-b^2) =b√(1-b^2) ...9

c.√(1-b^2).√(1-a^2)+b.√(1-c^2). √(1-a^2)=a.√(1-a^2) ...10

On adding up

2a.√(1-b^2).√(1-c^2)+
2b.√(1-a^2).√(1-c^2)+
2c.√(1-b^2).√(1-a^2) =
a.√(1-a^2)+b.√(1-b^2)+
c.√(1-c^2) ...11

On multiplyn equatn 4 by 2 we av:

2a√(1-a^2)+2b√(1-b^2)+2c√(1-c^2) =-2a.√(1-b^2).√(1-c^2)
-2b.√(1-a^2).√(1-c^2)
-2c.√(1-b^2).√(1-a^2)+6abc

on substituting the value of:
2a.√(1-b^2).√(1-c^2)+
2b.√(1-a^2).√(1-c^2)+
2c.√(1-b^2).√(1-a^2) from equatn 11

The following results:
2a√(1-a^2)+2b√(1-b^2)+2c√(1-c^2) =-a.√(1-a^2)-b.√(1-b^2)-
c.√(1-c^2) +6abc

3a√(1-a^2)+3b√(1-b^2)+3c√(1-c^2) =6abc

a√(1-a^2)+b√(1-b^2)+c√(1-c^2) =2abc

QED!

Too long! Any shortcut solutions!?

Re: Nairaland Mathematics Clinic by bolkay47(m): 7:46am On Jan 23, 2015

AlphaMaximus:

NB: I accidentally typed 461,875 instead of 421,875 in my formula justification exercise.......take note

thanks so much sir...but why did you assume "a" to be 20?

Re: Nairaland Mathematics Clinic by agentofchange1(m): 8:14am On Jan 23, 2015

greet thee all

try this plz.

Q1)Use Lagrange multiplier to find the values of x,y,z that minimises the objective function f(x,y,z) =11xy+14yz+15xz, subject to the constrain xyz=147840

Q2) A rectangular box open at the top is to have a volume of 32cm-cube , what must be the dimension of the box so that the total surface is maximum .

Re: Nairaland Mathematics Clinic by efficiencie(m): 11:25am On Jan 23, 2015

agentofchange1:
greet thee all

try this plz.

Q1)Use Lagrange multiplier to find the values of x,y,z that minimises the objective function f(x,y,z) =11xy+14yz+15xz, subject to the constrain xyz=147840

Q2) A rectangular box open at the top is to have a volume of 32cm-cube , what must be the dimension of the box so that the total surface is maximum .

let the dimensions of the box be: l×b×h
Volume of cuboid:
v=lbh and v=32
Hence the constraint function is:
lbh=32
Total surface area (and objective function):
a=lb+2lh+2bh
The lagrange function is:
L=lb+2lh+2bh-λ(lbh-32)
Diff partially wrt l,b,h and λ and obtaining d FOCs
δL/δl=b+2h-λbh=0...1
δL/δb=l+2h-λlh=0...2
δL/δh=2l+2b-λlb=0...3
lbh-32=0...4
On solvin d 4 equatns we have that:
l=4,b=4,h=2

1 Like

Re: Nairaland Mathematics Clinic by AlphaMaximus(m): 12:56pm On Jan 23, 2015

bolkay47:
thanks so much sir...but why did you assume "a" to be 20?

It wasnt an assumption......"a" represents the initial term, which in this case is 20.

Re: Nairaland Mathematics Clinic by AlphaMaximus(m): 12:59pm On Jan 23, 2015

efficiencie:

Given that:
arcsina+arcsinb+arcsinc=π

arcsina=π-arcsinb-arcsinc

Take the cosine of both sides:

cos(arcsina)=cos(π-arcsinb-arcsinc)

cos(arcsina)=cos(π-(arcsinb+arcsinc))

cos(arcsina)=
-cos(arcsinb+arcsinc)

cos(arcsina)=
-cos(arcsinb)cos(arcsinc)+ sin(arcsinb)sin(arcsinc)

√(1-sin^2(arcsina))=
-(√(1-sin^2(arcsinb))
√(1-sin^2(arcsinc)))+
sin(arcsinb).sin(arcsinc)

a√(1-a^2)=-a.√(1-b^2).√(1-c^2)+abc...1

and similarly the following is derived:

b√(1-b^2)=-b√(1-a^2).√(1-c^2)+abc...2

c√(1-c^2)=-c√(1-b^2).√(1-a^2)+abc...3

On adding up:

a√(1-a^2)+b√(1-b^2)+c√(1-c^2) =-a.√(1-b^2).√(1-c^2)
-b.√(1-a^2).√(1-c^2)
-c.√(1-b^2).√(1-a^2)+3abc ...4

Going back to the original equatn:
arcsina+arcsinb+arcsinc=π

arcsina+arcsinb=π-arcsinc
On taking the sine of both sides

sin(arcsina+arcsinb)=
sin(π-arcsinc)

sin(arcsina)cos(arcsinb)+ sin(arcsinb)cos(arcsina) =c

a.√(1-b^2)+b.√(1-a^2) =c ...5

and similarly the followin is derivd
a.√(1-c^2)+c.√(1-a^2) =b ...6

c.√(1-b^2)+b.√(1-c^2) =a ...7

And hence multiply 5 by
√(1-c^2), 6 by √(1-b^2) and 7 by √(1-a^2) the followin results:

a.√(1-b^2).√(1-c^2)+b.√(1-a^2).√(1-c^2)=c.√(1-c^2) ...8

a.√(1-c^2).√(1-b^2) +c.√(1-a^2).√(1-b^2) =b√(1-b^2) ...9

c.√(1-b^2).√(1-a^2)+b.√(1-c^2). √(1-a^2)=a.√(1-a^2) ...10

On adding up

2a.√(1-b^2).√(1-c^2)+
2b.√(1-a^2).√(1-c^2)+
2c.√(1-b^2).√(1-a^2) =
a.√(1-a^2)+b.√(1-b^2)+
c.√(1-c^2) ...11

On multiplyn equatn 4 by 2 we av:

2a√(1-a^2)+2b√(1-b^2)+2c√(1-c^2) =-2a.√(1-b^2).√(1-c^2)
-2b.√(1-a^2).√(1-c^2)
-2c.√(1-b^2).√(1-a^2)+6abc

on substituting the value of:
2a.√(1-b^2).√(1-c^2)+
2b.√(1-a^2).√(1-c^2)+
2c.√(1-b^2).√(1-a^2) from equatn 11

The following results:
2a√(1-a^2)+2b√(1-b^2)+2c√(1-c^2) =-a.√(1-a^2)-b.√(1-b^2)-
c.√(1-c^2) +6abc

3a√(1-a^2)+3b√(1-b^2)+3c√(1-c^2) =6abc

a√(1-a^2)+b√(1-b^2)+c√(1-c^2) =2abc

QED!

Too long! Any shortcut solutions!?

Just right.

Re: Nairaland Mathematics Clinic by bolkay47(m): 1:26pm On Jan 23, 2015

AlphaMaximus:

It wasnt an assumption......"a" represents the initial term, which in this case is 20.

thumbs up .its very clear now...

Re: Nairaland Mathematics Clinic by efficiencie(m): 1:34pm On Jan 23, 2015

efficiencie: the partial derivative of (δx/δy) with respect to 'x' is equal to what?

Re: Nairaland Mathematics Clinic by Oyasub11: 3:28pm On Jan 23, 2015

agentofchange1:
greet thee all

try this plz.

Q1)Use Lagrange multiplier to find the values of x,y,z that minimises the objective function f(x,y,z) =11xy+14yz+15xz, subject to the constrain xyz=147840
.

X=56,Y=60,Z=44.The workings is too long don't know how to post it.

Re: Nairaland Mathematics Clinic by agentofchange1(m): 4:22pm On Jan 23, 2015

[quote author=Oyasub11 post=30060049]X=56,Y=60,Z=44.The workings is too long don't know how to post it.[/quot

try snap & post abeg

Re: Nairaland Mathematics Clinic by Oyasub11: 4:37pm On Jan 23, 2015

[quote author=agentofchange1 post=30061439][/quote]

Re: Nairaland Mathematics Clinic by Oyasub11: 4:58pm On Jan 23, 2015

[quote author=agentofchange1 post=30061439][/quote]How do I go about it,please put me through.

Re: Nairaland Mathematics Clinic by agentofchange1(m): 9:38am On Jan 24, 2015

Oyasub11:
How do I go about it,please put me through.

take a photo of the solution ...then save with a name u will easily remember.... now when posting , b4 u click 'summit'. scroll down a bit where you see three spaces . ('choose" wink

now click there then locate the directory you saved the image , then post

that's it bro...try it..

Re: Nairaland Mathematics Clinic by Oyasub11: 10:48am On Jan 24, 2015

Here is it.

Re: Nairaland Mathematics Clinic by Oyasub11: 10:59am On Jan 24, 2015

Another

Re: Nairaland Mathematics Clinic by Oyasub11: 11:07am On Jan 24, 2015

Part 2

Re: Nairaland Mathematics Clinic by Nobody: 12:47pm On Jan 24, 2015

richiez,solve this?if x=sin@,simplify x/(1-x^2)^1/2.i.e 1-x^2 in square root.

Re: Nairaland Mathematics Clinic by neighy(m): 3:12pm On Jan 24, 2015

It might seem simple but guys please do justice to this Questions and give me a mention after every solving... Am Preparing For an Exam Thou... Thanks...

1) If 12 root e = X root 7 Find X where e=12

2) What is the area enclosed by the curve Y=4x-x^2, X=1,X=2, and the X axis

3) the factor of X^99+1

4)If F(x)=x/x-1, what is f(x+1)...

5) Find Lim 4+h-2/h... H}2...

#More To Come...

Re: Nairaland Mathematics Clinic by Nobody: 3:41pm On Jan 24, 2015

solve these?if x=sin@,simplify x/(1-x^2).note that @ stands for tita.

Re: Nairaland Mathematics Clinic by efficiencie(m): 6:27pm On Jan 24, 2015

jaryeh:
Chai! Long time........
I need help on this pls..
tag: jackpot, Alphamaximus, dejt4u, doubledx,
Richie and other generals. I don forget una
names finish o.

This looks clean enough!

Given
arcsina+arcsinb+arcsinc=π

Let
arcsina=A
arcsinb=B
arcsinc=C
hence:
a=sinA
b=sinB
c=sinC
Starting from the LHS of the required equation:
a√(1-a^2)+b√(1-b^2)+c√(1-c^2)
=sinA.√(1-sinA^2)+sinB√(1-B^2)+sinC√(1-sinC^2)
=sinA.cosA+sinB.cosB+sinC.cosC
=(1/2)sin2A+(1/2)sin2B+
(1/2)sin2C
=(1/2)(sin2A+sin2B+sin2C)
=(1/2)(2sin(A+B)cos(A-B) +sin2C)
but A+B+C=π
=(1/2)(2sin(π-C).cos(A-B)
+sin2C)
=(1/2)(2sin(C)cos(A-B)+sin2C)
=(1/2)(2sin(C)cos(A-B)+2sinC.cosC)
=sinC.(cos(A-B)+cosC)
=sinC.(cos(A-B)+cos(π-A-B))
=sinC.(2cos((π/2)-B)
cos(A-(π/2)))
=sinC.(2cos((π/2)-B)
cos((π/2)-A))
=sinC.(2sinB.sinA)
=2sinA.sinB.sinC
=2abc
which is the RHS of the equation.

cheesy

^QED!

Re: Nairaland Mathematics Clinic by bolkay47(m): 9:12pm On Jan 24, 2015

jgenius:
solve these?if x=sin@,simplify x/(1-x^2).note that @ stands for tita.

sin@/1-sin^2@
But Sin^2@+cos^2@=1
So,cos^2@=1-sin^2@
: sin@/cos^2@
Sin@/cos@*1/cos@
Sin@/cos@=tan@
and 1/cos@=sec@
tan@*sec@
=tan@sec@...

Re: Nairaland Mathematics Clinic by Umartins1(m): 10:17pm On Jan 24, 2015

bolkay47:
sin@/1-sin^2@
But Sin^2@+cos^2@=1
So,cos^2@=1-sin^2@
: sin@/cos^2@
Sin@/cos@*1/cos@
Sin@/cos@=tan@
and 1/cos@=sec@
tan@*sec@
=tan@sec@...

Hi,,, well done.

Do you know me? This is Umartins on myschool. We do compete together in the classroom. I don't have time for that again cos of poor network.

Anyway, I think you should re-visit the guy's question. The denominator is inside a square root.

Re: Nairaland Mathematics Clinic by bolkay47(m): 10:35pm On Jan 24, 2015

Umartins1:

Hi,,, well done.

Do you know me? This is Umartins on myschool. We do compete together in the classroom. I don't have time for that again cos of poor network.

Anyway, I think you should re-visit the guy's question. The denominator is inside a square root.

I didn't notice that sir..thanks
Long time though..

Re: Nairaland Mathematics Clinic by Umartins1(m): 10:45pm On Jan 24, 2015

bolkay47:
I didn't notice that sir..thanks
Long time though..

07066224158...... Whats+

Re: Nairaland Mathematics Clinic by bolkay47(m): 10:56pm On Jan 24, 2015

neighy:
It might seem simple but guys please do justice to this Questions and give me a mention after every solving... Am Preparing For an Exam Thou... Thanks...

1) If 12 root e = X root 7 Find X where e=12

2) What is the area enclosed by the curve Y=4x-x^2, X=1,X=2, and the X axis

3) the factor of X^99+1

4)If F(x)=x/x-1, what is f(x+1)...

5) Find Lim 4+h-2/h... H}2...

#More To Come...

{1}---note: ¥ means root)
12¥e=x¥7::12*¥12=x*¥7
Divide both sides by ¥7
X=12¥12/¥7 using surd
You will arrive at x=24¥21/7..

{2}. Y=4x-x^2 at x=1 and x=2
(Note $ means integral)
$4x-x^2dx
$4x^2/2-x^3/3
Therefore$4x-x^2= 2x^2-x^3/3+C
[2(2^2)-2^3/3]-[2-1/3]
(8-8/3)-(2-1/3)
16/3-5/3=11/3
Ans=3.7square units.

{3}. NO IDEA

{4}. F(x+1)=(x+1)/(x+1-1)=(x+1)/x

{5}. I DON'T GET THE QUESTION.

Thanks. Mr neighy.

Re: Nairaland Mathematics Clinic by efguy1(m): 11:08pm On Jan 24, 2015

Youngsage:

x + y =5 -------- eqn i.
x^x + y^y=31 ----- eqn ii.
From equatns i & ii, take the log of both sides
(xlog^x + ylog^y) = log 31------ eqn iii.
log(x+y) = log 5 ------------ eqn iv.
xylogxy= log 31 -------- eqn v.
log xy = log 5 --------- eqn vi.
(Using elimination method);
xylog5 = log31
xy = log 31/log 5 =6.2 approx. 6.

Recall, from eqn i, x + y = 5.
So x = 6/y or y = 6/x.
Therefore x=2 when y=3.
ans: x=2, and y=3.

But 2^2 + 3^2 is not equal to 31

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