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ENGINEERING KA BA? [See 1st page for USEFUL SITES]
- Thread starter vhinpotz
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Re: Tambayan ng mga Engineering Students at Graduates.[See 1st page for USEFUL SITES]
sir pa request namn po ng .Engineering Economy “the great problem solver”
by Jaime R. Tiong plzzz po
sir pa request namn po ng .Engineering Economy “the great problem solver”
by Jaime R. Tiong plzzz po
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Re: Tambayan ng mga Engineering Students at Graduates.[See 1st page for USEFUL SITES]
mga sir patulong po sa catalog designing ng crushing and mining plant.. bale ang raw material ko is ung copper ore.. den ung given capacity ng plant is 53tph.. tapos dinivide ko po ung capacity ng bulk density ng copper ore den i come up with 2.17 cubic ft/sec.. eto na po ba ung volume flowrate na magiging basis ko sa mga equipment na gagamitin ko? den ito po ung flowsheet ko sa crushing....
1.coarse ore bin
2.vibrating feeder
3.primary crusher (jaw crusher)
4.conveyor belt
5.secondary crusher
6.vibrating screen
problema ko po kng pano ko idedesign ung coarse ore bin.. un po bang volume flowrate na 2.17cu ft/sec dun ko ba ibabase ung flow ng bin ko? kung oo pano po?
mga sir patulong po sa catalog designing ng crushing and mining plant.. bale ang raw material ko is ung copper ore.. den ung given capacity ng plant is 53tph.. tapos dinivide ko po ung capacity ng bulk density ng copper ore den i come up with 2.17 cubic ft/sec.. eto na po ba ung volume flowrate na magiging basis ko sa mga equipment na gagamitin ko? den ito po ung flowsheet ko sa crushing....
1.coarse ore bin
2.vibrating feeder
3.primary crusher (jaw crusher)
4.conveyor belt
5.secondary crusher
6.vibrating screen
problema ko po kng pano ko idedesign ung coarse ore bin.. un po bang volume flowrate na 2.17cu ft/sec dun ko ba ibabase ung flow ng bin ko? kung oo pano po?
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Re: Tambayan ng mga Engineering Students at Graduates.[See 1st page for USEFUL SITES]
anung course mo??
anung course mo??
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Re: Tambayan ng mga Engineering Students at Graduates.[See 1st page for USEFUL SITES]
BSME po
BSME po
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Re: Tambayan ng mga Engineering Students at Graduates.[See 1st page for USEFUL SITES]
w0w! Thanks talaga! ~^w^~
s0brang helpful nito!
Na-save ko n ang thread n to sa ph0ne.
--------;<&.,
w0w! Thanks talaga! ~^w^~
s0brang helpful nito!
Na-save ko n ang thread n to sa ph0ne.
--------;<&.,
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Re: Tambayan ng mga Engineering Students at Graduates.[See 1st page for USEFUL SITES]
mga ka symbianize cno po d2 may sample copy ng engineering contract kelangan ko po kc requirements sa ME LAWS sna po my mag upload d2,.
mga ka symbianize cno po d2 may sample copy ng engineering contract kelangan ko po kc requirements sa ME LAWS sna po my mag upload d2,.
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Re: Tambayan ng mga Engineering Students at Graduates.[See 1st page for USEFUL SITES]
pa-help naman po dito sa exponential function na to.. medyo nahihirapan po kasi eh, yung equal kasi zero (0)..
eto po yung equation :
27(3^2x) - 12(3^x)+1=0
(kailangan ko lang po talaga, medyo bumabagsak po kase..)
malaking tulong po kung matutulungan nyo ko
pa-help naman po dito sa exponential function na to.. medyo nahihirapan po kasi eh, yung equal kasi zero (0)..
eto po yung equation :
27(3^2x) - 12(3^x)+1=0
(kailangan ko lang po talaga, medyo bumabagsak po kase..)
malaking tulong po kung matutulungan nyo ko
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Re: Tambayan ng mga Engineering Students at Graduates.[See 1st page for USEFUL SITES]
mga masters, pasolve naman po nito ∫x^2 dx/ (9-x^2)^3...trigo sub po...hindi ko kasi magets kung paano, finals week na namin ngayon...TIA
mga masters, pasolve naman po nito ∫x^2 dx/ (9-x^2)^3...trigo sub po...hindi ko kasi magets kung paano, finals week na namin ngayon...TIA
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Re: Tambayan ng mga Engineering Students at Graduates.[See 1st page for USEFUL SITES]
saang school ka po? magkano per sem sa inyo.?
saang school ka po? magkano per sem sa inyo.?
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Re: Tambayan ng mga Engineering Students at Graduates.[See 1st page for USEFUL SITES]
boss sabi kasi sa amin mechanical indemand ngayun pero kung ngayun ka pa lang mag aaral ehhh di sure kung anu na in demand in 6yrs, pero pag liscence engineer kana kahit anung clase makakuha ka ng trabaho tiyaga lang.
boss sabi kasi sa amin mechanical indemand ngayun pero kung ngayun ka pa lang mag aaral ehhh di sure kung anu na in demand in 6yrs, pero pag liscence engineer kana kahit anung clase makakuha ka ng trabaho tiyaga lang.
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Re: Tambayan ng mga Engineering Students at Graduates.[See 1st page for USEFUL SITES]
Kung simplification ang problema dito: (Completing the square)
27 (3^2x) - 12 (3^x) + 1 = 0
27 (3^x)^2 - 12 (3^x) + 1 = 0
let 3^x = y
27 (y^2) - 12 y + 1 = 0
Using Quadratic Formula:
where: a = 27 ; b = - 12 ; c = 1
y = (1/3) ; y = (1/9)
y - (1/3) = 0 ; y - (1/9) = 0
but y = 3^x
3^x - (1/3) = 0 ; 3^x - (1/9) = 0
[(3^x) - (1/3)] * [(3^x) - (1/9)] = 0
Note: I'm not 100% sure with this one, though.
Mas madali kasing mag-solve kapag ballpen at papel ang gamit.
Sana makatulong sa'yo ito dude!
pa-help naman po dito sa exponential function na to.. medyo nahihirapan po kasi eh, yung equal kasi zero (0)..
eto po yung equation :
27(3^2x) - 12(3^x)+1=0
(kailangan ko lang po talaga, medyo bumabagsak po kase..)
malaking tulong po kung matutulungan nyo ko
Kung simplification ang problema dito: (Completing the square)
27 (3^2x) - 12 (3^x) + 1 = 0
27 (3^x)^2 - 12 (3^x) + 1 = 0
let 3^x = y
27 (y^2) - 12 y + 1 = 0
Using Quadratic Formula:
where: a = 27 ; b = - 12 ; c = 1
y = (1/3) ; y = (1/9)
y - (1/3) = 0 ; y - (1/9) = 0
but y = 3^x
3^x - (1/3) = 0 ; 3^x - (1/9) = 0
[(3^x) - (1/3)] * [(3^x) - (1/9)] = 0
Note: I'm not 100% sure with this one, though.
Mas madali kasing mag-solve kapag ballpen at papel ang gamit.
Sana makatulong sa'yo ito dude!
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Re: Tambayan ng mga Engineering Students at Graduates.[See 1st page for USEFUL SITES]
ano poh kukunin diyan? integrate lan poh ba?
ano poh kukunin diyan? integrate lan poh ba?
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Re: Tambayan ng mga Engineering Students at Graduates.[See 1st page for USEFUL SITES]
SALAMAT dude!
last na lang :
5^4x+1 + 5 = 26•5^2x
(yung dot multiply po yun)
ganito po ba ang gagawin :
5^4x+1 + 5 = 130^2x
(minultiply ko po yung 26 at 5 kaya 130, pero di ko na alam kung
ano ang susunod kong gagawin..)
anu pong factor sa 5 na makukuha ko po ay 130?
patulong po.. salamat ng marami
Kung simplification ang problema dito: (Completing the square)
27 (3^2x) - 12 (3^x) + 1 = 0
27 (3^x)^2 - 12 (3^x) + 1 = 0
let 3^x = y
27 (y^2) - 12 y + 1 = 0
Using Quadratic Formula:
where: a = 27 ; b = - 12 ; c = 1
y = (1/3) ; y = (1/9)
y - (1/3) = 0 ; y - (1/9) = 0
but y = 3^x
3^x - (1/3) = 0 ; 3^x - (1/9) = 0
[(3^x) - (1/3)] * [(3^x) - (1/9)] = 0
Note: I'm not 100% sure with this one, though.
Mas madali kasing mag-solve kapag ballpen at papel ang gamit.
Sana makatulong sa'yo ito dude!
SALAMAT dude!
last na lang :
5^4x+1 + 5 = 26•5^2x
(yung dot multiply po yun)
ganito po ba ang gagawin :
5^4x+1 + 5 = 130^2x
(minultiply ko po yung 26 at 5 kaya 130, pero di ko na alam kung
ano ang susunod kong gagawin..)
anu pong factor sa 5 na makukuha ko po ay 130?
patulong po.. salamat ng marami
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