Kwa wale wakali wa Algebra, Solve this

[r20]

hii ni chuo co advance wala o level

so kweli asee

 

[Mtili wandu]

Wakati yupo form 6, Ticha akawa anafundisha calculus, tukamuuliza sasa hizi tutatumia wapi? Na kweli mpk Leo sina napotumia

 

[Makanyaga]

Kuna kitabu kiliindwa jipime chekundu kilikuwa na maswali hayo

Jibu nimelishapata, nasubiri nipate muda wa kuli-publish. Hili ndo swali la kwenye level ya University na si lile la mabomba a,b,c!

 

[Kijana ushe2]

[(V/30)+(V/45)+(V/90)]t=V

6/90=1/t ==> t=15

Itachukua 15 minutes.

This is correct solution among of all in this paper

 

[Makanyaga]

Maswali ya rate huwa yanasumbua wakati mwingine.
Hebu tusaidie hili:

Tank la lita 50 limewekewa mchanganyiko wa maji na chumvi(soluble, kipimo/concentration ya 30 gramu/lita) kiasi cha lita 25 zimejazwa.
Kuna bomba limefunguliwa linaweka maji kwenye tank kiasi cha lita 3 kwa dakika na hapo hapo kuna mrija/tobo kwenye tank linachuruzisha maji kutoka kwenye tank kiasi cha lita 1¼ kwa dakika.
a) Tanki litajaa baada ya muda gani?
b) mchanganyiko wa kwenye tanki utasomaje wakati likijaa?

Kama nimeelewa swali lako, just from normal logic na si katika utaalamu wa mahesabu


Kuna lita 25 kwa kuanzia, na concentration ya 30g/l.


Kwa hiyo SOLUTE iliyo ndani ya mchanganyiko huo ni 25x30=750g


INFLOW rate ni 3 litres per minute na

OUTFLAW RATE ni 1¼=1.25 litres


Kwa hiyo TOTAL NET INFLOW ni 3-1¼=1.75 litres per minute


At a net inflow of 1.75 litres, (to make a total of 50 litres for the tank to be full) it will take 14.3 minutes for the tank to be full (25litres/1.75)


THEN:

How many grams of the solute will be lost during this time interval of 14.3 minutes (i.e. The time taken for the Tank to be full at 50 litres)


Net inflow=1.75litres per minute


Original concentration, C= 30g/litre


Time taken for the tank to be full at 50 litres=14.3 minutes


Note that at time t=0, C=30. Now we need to find C when t=14.3


The guiding differential equation (DE), relating the rate of change of the Solute S with respect to time t will be

dS/dt=S/(25+1.75t) or

dS/dt=S/(25+7/4t)

i.e. amount of solute divided by: the initial voulme of 25 litres plus its NET rate INCREASE WITH TIME of 1.75 liters per minute, but this second quantity being a function of time t

The above is a separable differential equation which can be solved by using the technique of using the INTEGRATING FACTOR or by SUBSTITUION with a certain variable such as THETA.


Solving the above DE will give a solution


S=[C (25+ (5/4) t)] ^ (-7/5) ………………………… (1)

(note: to the power of -7 divided by 5)

where C is a constant of integration


Now, with t=0 and S=30 in the above equation, C can be found to be 0.0352 so that equation (1) above becomes:


S=[0.0352 (25+ (5/4) t)] ^ (-7/5) ………………………… (2)


The amount of Solute S when t=14.3 can be found by simply substituting this value of t in equation (2) above, i.e. put t=14.3 to get the value of S.


I was in hurry, sorry for any errors/mistakes, if any! However I stand to be correct, just in case!


Nimeshindwa kuyatafsiri baadhi ya maneno ya kimahesabu kwa Kiswahili ndiyo maana nimechanganya lugha

 

[Nematic]

Kama nimeelewa swali lako, just from normal logic na si katika utaalamu wa mahesabu


Kuna lita 25 kwa kuanzia, na concentration ya 30g/l.


Kwa hiyo SOLUTE iliyo ndani ya mchanganyiko huo ni 25x30=750g


INFLOW rate ni 3 litres per minute na

OUTFLAW RATE ni 1¼=1.25 litres


Kwa hiyo TOTAL NET INFLOW ni 3-1¼=1.75 litres per minute


At a net inflow of 1.75 litres, (to make a total of 50 litres for the tank to be full) it will take 14.3 minutes for the tank to be full (25litres/1.75)


THEN:

How many grams of the solute will be lost during this time interval of 14.3 minutes (i.e. The time taken for the Tank to be full at 50 litres)


Net inflow=1.75litres per minute


Original concentration, C= 30g/litre


Time taken for the tank to be full at 50 litres=14.3 minutes


Note that at time t=0, C=30. Now we need to find C when t=14.3


The guiding differential equation (DE), relating the rate of change of the Solute S with respect to time t will be

dS/dt=S/(25+1.75t) or

dS/dt=S/(25+7/4t)

i.e. amount of solute divided by: the initial voulme of 25 litres plus its NET rate INCREASE WITH TIME of 1.75 liters per minute, but this second quantity being a function of time t

The above is a separable differential equation which can be solved by using the technique of using the INTEGRATING FACTOR or by SUBSTITUION with a certain variable such as THETA.


Solving the above DE will give a solution


S=[C (25+ (5/4) t)] ^ (-7/5) ………………………… (1)

(note: to the power of -7 divided by 5)

where C is a constant of integration


Now, with t=0 and S=30 in the above equation, C can be found to be 0.0352 so that equation (1) above becomes:


S=[0.0352 (25+ (5/4) t)] ^ (-7/5) ………………………… (2)


The amount of Solute S when t=14.3 can be found by simply substituting this value of t in equation (2) above, i.e. put t=14.3 to get the value of S.


I was in hurry, sorry for any errors/mistakes, if any! However I stand to be correct, just in case!


Nimeshindwa kuyatafsiri baadhi ya maneno ya kimahesabu kwa Kiswahili ndiyo maana nimechanganya lugha

Kwa ujumla umezingatia vigezo vyote, ilikuwa inalazimika derivatives zitumike, rate of change of solution to time.
Tukubaliane:
Sijaona alama ya hesabu iliyoshughulikia maji na chumvi yaliyokuwa yanatoka, kumbuka chumvi ilikuwa inamwagika, ndio ugumu wa hii nondo.
Hebu weka uzingativu wa ziada hapa.

Muda wa kujaza ulikuwa ni wa lita 25, umezingatia.
Sorry kwa kutokuwa hewani.

 

[Myahudi Jr II]

Hesabu ni kipaji, kuna jamaa nilisoma nae Hesabu atapata A lakini njoo GEO, HISTO, KISW ENGL ni D mtindo mmoja, Phy na Chem atapata C

 

[radicals]

Hesabu ni kipaji, kuna jamaa nilisoma nae Hesabu atapata A lakini njoo GEO, HISTO, KISW ENGL ni D mtindo mmoja, Phy na Chem atapata C

huyo ni mimi mkuu

Sawa kabisa! Hongera

 

[Myahudi Jr II]

huyo ni mimi mkuu

Inakuaje hivi asee?? nashindwa kuelewa

 

[radicals]

Inakuaje hivi asee?? nashindwa kuelewa

yani mi nikijikuta tu nipo addicted na maths, but masomo mengine kawaida sana nikachaguliwa egm tosa nilienda na A ya maths, D ya geog... six kumalza nkapata A maths, D geog na D uchum

 

[Makanyaga]

Kwa ujumla umezingatia vigezo vyote, ilikuwa inalazimika derivatives zitumike, rate of change of solution to time.
Tukubaliane:
Sijaona alama ya hesabu iliyoshughulikia maji na chumvi yaliyokuwa yanatoka, kumbuka chumvi ilikuwa inamwagika, ndio ugumu wa hii nondo.
Hebu weka uzingativu wa ziada hapa.

Muda wa kujaza ulikuwa ni wa lita 25, umezingatia.
Sorry kwa kutokuwa hewani.

Uko sahihi. Nilisahau kuweka negative sign sabbabu ya uharaka

Inatakiwa kuwa hivi:

dS/dt= - S/(25+7/4t).

Hii equation ndiyo final kwa ajili ya kutafuta S, hapa ilipo imeshafanyiwa kazi ndiyo maana inaonekana hivyo Kuna negative (-ve) sign inatakiwa kuwa hapo kma nilivyoonyesha. Otherwise kwa mtu anayejua hesabu haiwi shida kuendelea kutokea hapo, sina haja ya kuonyyesha kila kitu, mda hauruhusu!

 

[Nematic]

Uko sahihi. Nilisahau kuweka negative sign sabbabu ya uharaka

Inatakiwa kuwa hivi:

dS/dt= - S/(25+7/4t).

Hii equation ndiyo final kwa ajili ya kutafuta S, hapa ilipo imeshafanyiwa kazi ndiyo maana inaonekana hivyo Kuna negative (-ve) sign inatakiwa kuwa hapo kma nilivyoonyesha. Otherwise kwa mtu anayejua hesabu haiwi shida kuendelea kutokea hapo, sina haja ya kuonyyesha kila kitu, mda hauruhusu!

Napenda unavyozingatia masuala ya kihesabu kwenye uhalisia wa tukio.
Kuna sababu mbili za concentration kubalika :
Moja ni kuongezeka wingi wa maji.
Pili ni kumwagika kwa chumvi.
At every minute, kuna kupungua kwa kiasi cha chumvi kinachopotea, depreciation of amount of salt lost, hapo ndo ile depreciation sijaona...
Nina idea bila ukamili wa hoja, ndo maana inabidi tusaidiane.

 

[Makanyaga]

Napenda unavyozingatia masuala ya kihesabu kwenye uhalisia wa tukio.
Kuna sababu mbili za concentration kubalika :
Moja ni kuongezeka wingi wa maji.
Pili ni kumwagika kwa chumvi.
At every minute, kuna kupungua kwa kiasi cha chumvi kinachopotea, depreciation of amount of salt lost, hapo ndo ile depreciation sijaona...
Nina idea bila ukamili wa hoja, ndo maana inabidi tusaidiane.

Unless kama unaona kuwa equation yangu ina parameters ambazo haziko sahihi, ila kama ziko sahihi, the negative sign takes care of what you are trying to say. Hiyo negative sign inawakilisha kupungua kwa chumvi ndani ya mchanganyiko wa chumvi na maji, ukizingatia kuwa maji yenyewe yalikuwa hayapungui isipokuwa yalikuwa yanaongezeka na hayo ndiyo yaliyosababisha concentration ya chumvi kupungua!

 

[Myomb_0]

V imebeba umuhimu upi hapa, iko virtual kabisa

Sikuweka wazi ila ni ujazo wa tank(V=volume) na t ni mda ambao mabomba yote matatu yatatumia kujaza tank.

 

[Teknocrat]

If a = –1/b, b = –1/c, c = –1/d, d = –1/4, what is the value of a?

 

[Teknocrat]

 

[CARDLESS]

4829

 

[TANZANIAN NINJA]

hii ni chuo co advance wala o level

lasaba hiyo kwa akina junior

 

[pilato93]

Hii kitu nakumbuka tulifundishwa darasa la 5,ila ndo hivyo sikuzingatia maana hadi leo sijaonaga tenki yenye bomba 3 za kuingiza maji, ni kuchoshana tu.

😆 😆 😆 😆 😆 😆 😆

 

[Msimamoo]

If a = –1/b, b = –1/c, c = –1/d, d = –1/4, what is the value of a?

Ivi algebra ya A-level unaijua mkuu, hili watafanya darasa la tano kabisa