JF Mathematics clinic Special thread

[Leak]

kama swali la pipe umelitoa mahala ama limekosewa au umekopi vibaya, only 2 pipes are emptying the tank and not 3, this is maths, not a language test!

ndio lilivyo! why only two pipes emptying the tank?

 

[Odili]

na mimi naongezea math problems

1. (logx)square=log((x)square) find the value of x
2. underoot(logx)=log(x underroot) find the value of x

Mkuu hebu jaribu kupitia hizi solution kwa case zote mbili kama kuna makosa utanirekebisha:-
(1)solution:

(logx)^2=log(x^2)
(logx)^2-log(x^2)=0
logx.logx-2logx=0
logx(logx-2)=0
logx=0 or logx-2=0
logx=0 or logx=2
x=10^0 or x=10^2
x=1 or x=100

(2)solution:
sqrt(logx)=log(sqrt(x))
(logx)^(1/2)=log((x)^(1/2))
squaring both sides we get
logx=log((x)^(1/2)).log((x)^(1/2))
logx=(1/2)logx.(1/2)logx
logx=(1/2).(1/2).logx.logx
logx=(1/4)logx.logx
multiplying by 4 both sides
4logx=logx.logx
4logx-logx.logx=0
logx(4-logx)=0
logx=0 or 4-logx=0
logx=0 or 4=logx
x=10^0 or x=10^4
x=1 or x=10000

~End~

 

[CHAMVIGA]

na mimi naongezea math problems

1. (logx)square=log((x)square) find the value of x
2. underoot(logx)=log(x underroot) find the value of x

1. (LOGX)^2 ni sawa na log(x)^2, hapo kwenye (logx)2=logx times logx. Na log(x)^2=2logx.
Then 2logx=logxlogx gawanya kwa logx bothside itabakia 2=logx. Hiyo 2=logx ipeleke kwenye exponential form itakuwa 10^2=x. Therefore x=100.

 

[CHAMVIGA]

Mkuu hebu jaribu kupitia hizi solution kwa case zote mbili kama kuna makosa utanirekebisha:-
(1)solution:

(logx)^2=log(x^2)
(logx)^2-log(x^2)=0
logx.logx-2logx=0
logx(logx-2)=0
logx=0 or logx-2=0
logx=0 or logx=2
x=10^0 or x=10^2
x=1 or x=100

(2)solution:
sqrt(logx)=log(sqrt(x))
(logx)^(1/2)=log((x)^(1/2))
squaring both sides we get
logx=log((x)^(1/2)).log((x)^(1/2))
logx=(1/2)logx.(1/2)logx
logx=(1/2).(1/2).logx.logx
logx=(1/4)logx.logx
multiplying by 4 both sides
4logx=logx.logx
4logx-logx.logx=0
logx(4-logx)=0
logx=0 or 4-logx=0
logx=0 or 4=logx
x=10^0 or x=10^4
x=1 or x=10000

~End~

Sawa mkuu.

 

[Bandido-2]

If 2×2=4
2+2=4
Why
1×1=1
1+1=2

 

[Leak]

Your about to play a dice game where your paid 5,000 each time you roll an odd number .You are allowed to roll the dice(kete) five times. A friend tells you that the dice used is biased such that even numbers are twice as likely to appear as odd numbers.

Based on this information what is the maximum amount you would pay to play this game?

 

[sembo]

X + Y + Z = 5
3X + 5Y - Z = 21
X - 7Y + 4Z = -19

Solve for X, Y and Z

 

[Odili]

X + Y + Z = 5
3X + 5Y - Z = 21
X - 7Y + 4Z = -19

Solve for X, Y and Z

Hebu nifuatilie taratibu na kwa makini na popote utakapokwama usisite kuniuliza:

x+y+z=5_____(1)
3x+5y-z=21_____(2)
x-7y+4z=-19_____(3)

add equation (1) into (2) to eliminate the variable 'z':
(x+y+z=5) + (3x+5y-z=21)
==> 4x+6y=26
divide by 2 throughout to simplify the equation:
(4x+6y=26)/2
===> 2x+3y=13_____(4)

multiply equation (2) by 4 and equation (3) by 1, then add them to eliminate the variable 'z':
4(3x+5y-z=21)+1(x-7y+4z=-19
==>13x+13y=65
divide by 2 throughout to simplify the equation:
===>x+y=5_____(5)

solve equation (4)&(5) by first multiplying equation (5) by 2 and subtracting (equation 5) from equation (4) to eliminate variable 'x':
(2x+3y=13) - 2(x+y=5)
==>(2x+3y=13)-(2x+2y=10)
==>y=3_____(6)

go back to equation (5):
x+y=5 (but y=3, from eqn(6)),
so
x+3=5
==>x=5-3
==>x=2_____(7)

go back to equation (1):
x+y+z=5
(but x=2, y=3 from eqn(7)&(6))
==>2+3+z=5
==>5+z=5
==>z=5-5
==>z=0_____(8)

So, going from equation (1) throught (8) we got the solution as:
x=2, y=3 and z=0.

Karibu tena.

 

[kuku wa kienyeji]

If 2×2=4
2+2=4
Why
1×1=1
1+1=2

Because

 

[Kibishi]

Prove 1=2

Viswali kama hivi huwa vinaitwa viswali mbuzi,maana yake ni rahisi mno.
Njia hii hapa.
From/let x=y
2x-x=2y-y
y-x=2y-2x
y-x=2(y-x)
Divide by y-x both sides.
=>1=2 hence proved.

 

[Edwin mkw]

Viswali kama hivi huwa vinaitwa viswali mbuzi,maana yake ni rahisi mno.
Njia hii hapa.
From/let x=y
2x-x=2y-y
y-x=2y-2x
y-x=2(y-x)
Divide by y-x both sides.
=>1=2 hence proved.

Yeah! Kwel w mbshi nmekuelew man

 

[medisonmuta]

-(-(-)(-)(-10x))=-5 solve for x

 

[Charles kikoti]

Itabidi nihamishie wanafunzi humu ndani.

 

[Kibishi]

Twendeni kazi,wale ambao hawajala,makande hayo hapo.
Solve for x.
2^x=4x
N.B:Jibu lake ni x=4,onyesha njia ya kueleweka.Ukileta ujanja ujanja katika steps zako nitajua,kwa sababu nazijua sheria zote za exponents,logarithms,radicals na algebra.

 

[Kibishi]

Mbona kimya kwenye hili jukwaa?
Hamna wanahisabati?

 

[sembo]

Mbona kimya kwenye hili jukwaa?
Hamna wanahisabati?

Mkuu, tupo una swali lolote?

 

[anthony silas]

Twendeni kazi,wale ambao hawajala,makande hayo hapo.
Solve for x.
2^x=4x
N.B:Jibu lake ni x=4,onyesha njia ya kueleweka.Ukileta ujanja ujanja katika steps zako nitajua,kwa sababu nazijua sheria zote za exponents,logarithms,radicals na algebra.

Let 2^x=16
But 16 =2^4

Therefore 2^x =2^4

2 and 2 are canceling

Remain x=4

So x=4

 

[Goodvision]

Hebu nifuatilie taratibu na kwa makini na popote utakapokwama usisite kuniuliza:

x+y+z=5_____(1)
3x+5y-z=21_____(2)
x-7y+4z=-19_____(3)

add equation (1) into (2) to eliminate the variable 'z':
(x+y+z=5) + (3x+5y-z=21)
==> 4x+6y=26
divide by 2 throughout to simplify the equation:
(4x+6y=26)/2
===> 2x+3y=13_____(4)

multiply equation (2) by 4 and equation (3) by 1, then add them to eliminate the variable 'z':
4(3x+5y-z=21)+1(x-7y+4z=-19
==>13x+13y=65
divide by 2 throughout to simplify the equation:
===>x+y=5_____(5)

solve equation (4)&(5) by first multiplying equation (5) by 2 and subtracting (equation 5) from equation (4) to eliminate variable 'x':
(2x+3y=13) - 2(x+y=5)
==>(2x+3y=13)-(2x+2y=10)
==>y=3_____(6)

go back to equation (5):
x+y=5 (but y=3, from eqn(6)),
so
x+3=5
==>x=5-3
==>x=2_____(7)

go back to equation (1):
x+y+z=5
(but x=2, y=3 from eqn(7)&(6))
==>2+3+z=5
==>5+z=5
==>z=5-5
==>z=0_____(8)

So, going from equation (1) throught (8) we got the solution as:
x=2, y=3 and z=0.

Karibu tena.

nahisi kizunguzungu walah!

 

[Shark]

Let 2^x=16
But 16 =2^4

Therefore 2^x =2^4

2 and 2 are canceling

Remain x=4

So x=4

Hili jibu umefoji. Ukiulizwa 16 imetoka wapi utabaki unatoa macho.

 

[Kibishi]

Let 2^x=16
But 16 =2^4

Therefore 2^x =2^4

2 and 2 are canceling

Remain x=4

So x=4

Umejaribu.
Katika algebra mara nyingi huwa inatakiwa u let variable na siyo namba.
Hiyo 16 umeitoa wapi katika 2^=16?,umetupiga changa la macho,jaribu tena mkuu.