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-="Math Clinic"=-

May mga chemical engineering student po ba dito?

meron...
present..:D
sa tambayan ng chem eng meron din..


@puge
madali lang yun..:thumbsup:



:thanks: sa mga sumasagot dito..
balik na ulit ako dito.. :yipee:
 
sna gumaling ako sa math...dti ko pa kseng prob to eh..mg papasukan nanamn ito nanaman ang prob ko..basa basa muna ako bka my mtutunan...guys mdli lng ba yung college algebra? yan kse mging subject nmn ng math IT nga pla ko slamat

Nadagdagan lang naman ng konting formulas sa college dre. Basta dapat alam mo pa din ang basics.

meron...
present..:D
sa tambayan ng chem eng meron din..


@puge
madali lang yun..:thumbsup:



:thanks: sa mga sumasagot dito..
balik na ulit ako dito.. :yipee:

Dati ka pang volunteer na tutor di ba? Naging busy ka ba this past few days? :think:

Anyway, good to have you back!
 
ok lang po na magpost kayo pero sana naman po hindi niyo palagi ipapagawa assignment nyo dito, try nyo muna isolve tapos tanong nyo dito yung hindi niyo maintindihan kung sakali..:salute:



bro may formula jan,

abscissa = x - ({y’[1 + (y’)^2]} / y’’)

ordinate = y + {[1 + (y’)^2] / y’’}

yung x at y jan, yan yung given points,

ito ang sagot, abscissa=3
ordinate=-2

wow galing my mga teknik ba kau para makuha agad gagamitin na formula?gagaling nyu professor ka ba sir?
anu magandang book sa mathematics nagrereview kasi ako ngaun EE course ko
 
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Question about Sets.

Pakicheck naman po kung tama pagkakaintindi ko
Given:

B = {1,3,5}
B' = {2,4,6,7,8,9,10}
C = {1,5,7,9}
C' = {2,3,4,6,8,10}
U = {1,2,3,4,5,6,7,8,9,10}

Problem:
(B' [union] C)' [intersection] (U [intersection] B')

Magbase tayo sa unang term. Using De Morgan's Law, tama ba ang pagkakaaply ko?

Eto ang sagot ko
(B [intersection] C')

Tama ba na nagdouble complement ako sa B'?


pasensya na, mukhang nakalimutan ko na din magsimplify neto..:slap:

pero nagtry ako isubstitute yung values pero parang may mali dun sa problem na binigay mo kasi walang sagot na lalabas..

ewan ko lang kung ako ang mali, pacheck na lang sa mga nakakaalam..:salute:


wow galing my mga teknik ba kau para makuha agad gagamitin na formula?gagaling nyu professor ka ba sir?
anu magandang book sa mathematics nagrereview kasi ako ngaun EE course ko


may derivation yan formula pero mahaba at nakakasakit ng ulo..:D

meron yang formula na yan dun sa engineering mathematics na libro ni besavilla, yung makapal na parang PEC na may part 1 at 2..:salute:
 
ah tnx po sa mga sumagot..blik po ako dto pg nhirapan ako slamat.gaya ng snbi nyo.. i memorize ko mga formulas at aralin mbuti...pra kseng nkkpg taka eh..yung mga clasmte ko kuhang kuha nla yung math pro ako hrap na hrp hahaha ako na yta my prob hehe
 
help

Patul0ng naman po kung paanu isolve ito at paanu nakuha.

Prove that "x=1minus squreroot of 5" is a soluti0n of the equati0n x^3-3x^2-2x+4=0

thanks sa tutul0ng,

ito nga po pala yung lumalabas na sagot using Pindotati0n hehe.

X1= 1+squarer00t of 5
x2= 1-squareroot of 5
x3=1

ang kailangan ko po mga sir ay soluti0n,thanks!
 
Since proving yan, you can substitute the value of X dun sa equation.


Kung mag-equal, edi solve na. :lol:


Ang kaso nga lang, mahabang substitution dahil na din sa [ X = 1 - squareroot of 5 ]. Pwede namang i-solve 'yung [ squareroot of 5 ] pero magkaka-decimal points. Baka hindi maging equal sa zero kung puro round-off.


Try ko solve mamaya. May gagawin lang ako.
 
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help

Patul0ng naman po kung paanu isolve ito at paanu nakuha.

Prove that "x=1minus squreroot of 5" is a soluti0n of the equati0n x^3-3x^2-2x+4=0

thanks sa tutul0ng,

ito nga po pala yung lumalabas na sagot using Pindotati0n hehe.

X1= 1+squarer00t of 5
x2= 1-squareroot of 5
x3=1

ang kailangan ko po mga sir ay soluti0n,thanks!

Here's my solution:



Proving x = [ 1 - ( sq.rt of 5 ) ] is a solution to this equation: ( x^3 ) - 3 ( x^2 ) - ( 2x ) + 4 = 0


( x^3 ) - 3 ( x^2 ) - ( 2x ) + 4 = 0

x = [ 1 - ( sq.rt of 5 ) ]; Let y = ( sq.rt of 5 )

Substituting the value of x to the given equation:

[ ( 1 - y ) ^ 3 ] - 3 [ ( 1 - y ) ^ 2 ] - 2 ( 1 - y ) + 4 = 0

expanding...**

( 16 - 8y ) - 3 ( 6 - 2y ) - 2 + 2y + 4 = 0

16 - 8y - 18 + 6y - 2 + 2y + 4 = 0

Rearranging:

16 - 18 - 2 + 4 - 8y + 6y + 2y = 0

0 = 0



**Dun sa pag-expand, short-cut na yan. Nilagyan ko ng colors para alam mo kung saan 'yun nanggaling.

Hindi ko po siya na-double check. Wala na akong time e. Dito ko lang siya rekta na-solve. Pakisabi kung may mali ako sa pag-solve ko.





 
Last edited:


Here's my solution:



Proving x = [ 1 - ( sq.rt of 5 ) ] is a solution to this equation: ( x^3 ) - 3 ( x^2 ) - ( 2x ) + 4 = 0


( x^3 ) - 3 ( x^2 ) - ( 2x ) + 4 = 0

x = [ 1 - ( sq.rt of 5 ) ]; Let y = ( sq.rt of 5 )

Substituting the value of x to the given equation:

[ ( 1 - y ) ^ 3 ] - 3 [ ( 1 - y ) ^ 2 ] - 2 ( 1 - y ) + 4 = 0

expanding...**

( 16 - 8y ) - 3 ( 6 - 2y ) - 2 + 2y + 4 = 0

16 - 8y - 18 + 6y - 2 + 2y + 4 = 0

Rearranging:

16 - 18 - 2 + 4 - 8y + 6y + 2y = 0

0 = 0



**Dun sa pag-expand, short-cut na yan. Nilagyan ko ng colors para alam mo kung saan 'yun nanggaling.

Hindi ko po siya na-double check. Wala na akong time e. Dito ko lang siya rekta na-solve. Pakisabi kung may mali ako sa pag-solve ko.









thanks po sir for the effort, ang gusto ko po kasi na makta ay yung sol. Kung paano lumabas yung answer na x=1-squareroot of 5, yung parang find the r0ot of the given pr0blem.
 
Proving po yan e. More on substituting the given value to a certain equation.


Edit:

Kung pipilitin mo kasing palabasin ang value ng X using the equation, 1 po ang lalabas dyan. Pwera na lang kung sa Calculator.

PM mo na lang si Sir Devilbat. Baka matulungan ka nya.


Edit ulit:

Here's another solution:

Given: ( x^3 ) - 3 ( x^2 ) - ( 2x ) + 4 = 0

where: a = -3 ; b = -2 ; c = +4

We can use these formulas:

Q = [ 3b - ( a^2 ) ] / 9 ; R = [ 9ab - 27c - 2 ( a^3 ) ] / 54

D = ( Q^3 ) + ( R^2 )

Substituting the given to the formulas:

Q = - ( 5 / 3 ) ; R = 0 ; D = - ( 125 / 27 )



We cannot solve the equation using Cardan's Method because D < 0

Hence, we will then use these formulas:

attachment.php


attachment.php


attachment.php


where θ can be solved using this equation:

cos θ = R / sq.rt. of ( - Q ^ 3 )

X1 = 0
X2 = 1 - ( sq.rt. of 5 )
X3 = 1 + ( sq.rt. of 5 )


Ikaw na bahalang mag-substitute ng values. I checked my answer twice! :D

Hindi kasi ako mapakali na hindi ko masagot e. Ayan na solution ko. :laugh:

 

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Last edited:
Proving po yan e. More on substituting the given value to a certain equation.


Edit:

Kung pipilitin mo kasing palabasin ang value ng X using the equation, 1 po ang lalabas dyan. Pwera na lang kung sa Calculator.

PM mo na lang si Sir Devilbat. Baka matulungan ka nya.


Edit ulit:

Here's another solution:

Given: ( x^3 ) - 3 ( x^2 ) - ( 2x ) + 4 = 0

where: a = -3 ; b = -2 ; c = +4

We can use these formulas:

Q = [ 3b - ( a^2 ) ] / 9 ; R = [ 9ab - 27c - 2 ( a^3 ) ] / 54

D = ( Q^3 ) + ( R^2 )

Substituting the given to the formulas:

Q = - ( 5 / 3 ) ; R = 0 ; D = - ( 125 / 27 )



We cannot solve the equation using Cardan's Method because D < 0

Hence, we will then use these formulas:

attachment.php


attachment.php


attachment.php


where θ can be solved using this equation:

cos θ = R / sq.rt. of ( - Q ^ 3 )

X1 = 0
X2 = 1 - ( sq.rt. of 5 )
X3 = 1 + ( sq.rt. of 5 )


Ikaw na bahalang mag-substitute ng values. I checked my answer twice! :D

Hindi kasi ako mapakali na hindi ko masagot e. Ayan na solution ko. :laugh:


thanks! :D
 
@mirandojesson: Post mo dito kung tama ang pagkakaintindi natin sa problem at kung tama ang pagkaka-solve. 2 ang solutions ko di ba?
 
Diff. Calculus : Solve the following using Squeeze Theorem:

#1. lim x--->0 1-cos 4x/1-cos2x = 4

#2. lim y--->1/2pie (y-1/2pie)^2 / 1-sin y = 2


...pa help naman po paano gagawin zero si #2 sa problem. I'm thinking using multiplying it to cos 90?:noidea:
 
Diff. Calculus : Solve the following using Squeeze Theorem:

#1. lim x--->0 1-cos 4x/1-cos2x = 4

#2. lim y--->1/2pie (y-1/2pie)^2 / 1-sin y = 2


...pa help naman po paano gagawin zero si #2 sa problem. I'm thinking using multiplying it to cos 90?:noidea:

Hindi ko alam 'yung mahabang solution nyan e. 'Yung shortcut sa Calculator parang alam ko pa.

i love math :D

Good for you! :hat:

can anyone help me in math? im struggling in high school level.

Can you speak tagalog? Anyway, we are willing to help. Just post your question and we'll try to help you.




--Galing ako sa bakasyon! Natambakan na masyado ang thread. Wala na bang bubuhay dito? :think:
 
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