₦airaland Forum

lavosier: Instincts bro....can u help xplain dah my question 4get abt ƌ no2 is done,,is jst ƌ no1.

...using the general formula for circle,x^2+y^2+2gx+2fy+c=0...it passes through the points {2,3},{3,2} and {-4,3}...which are in form of x and y...substitute them one by one.let's start from 2,3...substitute it to the general equation...2^2+3^2+2g.2+2f.3+c=0...4+9+4g+6f+c=0...13+4g+6f+c=0...4g+6f+c=-13...(*1)...substitute 3,2 in the general equation to give 3^2+2^2+2g.3+2f.2+c=0...9+4+6g+4f+c=0...13+6g+4f+c=0...6g+4f+c=-13...(*2)...substitute the points -4,3 in the equation to give -8g+6f+c=-25...(*3)....the 3 equations are 4g+6f+c=-13...(*1)
6g+4f+c=-13...(*2)
-8g+6f+c=-25...(*3)
use any means of solving simultaneous equation..subtract *3 from *1...4g+6f+c-(-8g+6f+c)=-13-(-25)...4g+6f+c+8g-6f-c=-13+25...12g=12...therefore [g=1]...subtract *1 from *2...6g+4f+c-(-8g+6f+c)=-13-(-25)...6g+4f+c+8g-6f-c=-13+25...14g-2f=12...remember that g=1...14.1-2f=12...2f=2 and [f=1]...substitute g=1,f=1 to any of the equation to get c...using equation *1...4g+6f+c=-13...4.1+6.1+c=-13...4+6+c=-13...[c=-23]...substitute the values of g,f and c to the general equation of a circle...x^2+y^2+2gx+2fy+c=0...x^2+y^2+2.1x+2.1y+(-23)=0...x^2+y^2+2x+2y-23=0...

lavosier: Yh.. sat @ ur bak,,dah guy dah had 268..

...that's great my bro...but,how did you know i was the one...

lavosier: Were u ƌ guy dah had Ă small passport on ur print out during ƌ putme 4 uniben??

...yeah...do you know me my general?

How can i solve polynomial questions of no simple root...

Hello my maths gurus...pls,let's keep the thread moving...a tree can never make a forest

Librate: Solve d equation 3tanQ = 2secQ..

If a sequence x, 7, y, 15 is an AP find d values of X and Y..

Simplify d followin log45-log9+log20, log4+log25

...for the second question...x,7,y,15 is an a.p...x is the first term...7 is the second term...y is the third term and 15 is the fourth term...for the second term...we have already known that a=x...using the formula for a.p that a+(n-1)d...for second term.n=2... substitute that...a+d=second term...for the third term...it will be a+2d and for the third term...try yours for the fourth term...guess your answer is a+3d(great)...therefore a+d=7(second term)...a+3d=15(fourth term)...subtract the second term from the fourth term to get a+3d-(a+d)=15-7....a+3d-a-d=8...2d=8...divide through by 2...d=4...since a+d=7 then substitute for d=4...a+4=7...a=7-4=3...since x=first term(a)...therefrod x=3 and y(third term)=a+2d...y=3+2*4=3+8...y is equal to 11...therefore x=3,y=11

Librate: Solve d equation 3tanQ = 2secQ..

If a sequence x, 7, y, 15 is an AP find d values of X and Y..

Simplify d followin log45-log9+log20, log4+log25

....for the first question 3tanQ=2secQ...remember that tanQ=sinQ/cosQ and secQ=1/cosQ....substitue those...3sinQ/cosQ=2/cosQ...multiply through by cosQ...3sinQ=2...sinQ=2/3...sinQ=0.6667...Q=arcsin(0.6667)...Q=41.8(to 1dp)...

Librate: Solve d simultaeous equation
2 raise to power -x, 4 raise to power -y =2.......(eqn..i)
3 raise to power -x, 9 raise to power 2y =3 .........(eqn...ii).,

...guess the question is this...2^(-x).4^(-y)=2...(eqn1) and 3^(-x).9^(2y)=3...(eqn2)...make the equation as the powers of their products...for equation 1...2^(-x).(2^2)^-y=2^1....(2^-x).2^(-2y)=2^1...using the law of indices 2^(-x-2y)=2^1...since they have the same base...-x-2y=1...multiply through by -1... x+2y=-1....(eqn3)..from equation 2...3^(-x).9^(2y)=3...3^(-x).(3^2)^(2y)=3^1...3^(-x).3^(4y)=3^1...from indices...3^(-x+4y)=3^1...since they have the same base...-x+4y=1...multiply through by -1 to get x-4y=-1...(eqn4)...consider equation 3 and 4...x+2y=-1 and x-4y=-1...subtract equation 4 from 3...x+2y-(x-4y)=-1-(-1)...x+2y-x+4y=-1+1...6y=0...therefore y=0...substitute y=0 to either of equation 3 or 4...let's use 3.x+2y=-1...x+2(0)=-1...therefore x=-1...Y=0,X=-1

Solve the simultaneous equation y'=y-2x...(i) and x'=x-2y...(ii)

Bolaji 16: Sum1 shud pls help me with this proof??

..where 'C' represents the combination sign...n'C'r+n'C'(r+1)=(n+1)'C'(r+1)...using the formula from combination...n'C'r=n!/(n-r)!r!....n'C'(r+1)=n!/{n-(r+1)}!(r+1)!=n!/(n-r-1)!(r+1)!....and (n+1)'C'(r+1)=(n+1)!/{n+1-(r+1)}!(r+1)!=(n+1)!/(n-r)!(r+1)!...(i)..note all those...substitute them to the question...we want to prove that.n!/(n-r)!r!+n!/(n-r-1)!(r+1)!=(n+1)!/(n-r)!(r+1)!...consider the one at the left...n!/(n-r)!r!+n!/(n-r-1)!(r+1)!....which gives n!/(n-r)(n-r-1)!+n!/(n-r-1)!(r+1)(r+1)!....factorize the ones that are factorizable...n!/(n-r-1)!r!{1/(n-r)+1/(r+1)}...solve the one the the nested bracket to get (n+1)/(n-r)(r+1)....take it to where the bracket is...n!/(n-r-1)!r!*(n+1)/(n-r)(r+1)...multiply the numerator by eachother,do the same to the denominator...(n+1)n!/(n-r)(n-r-1)!(r+1)r!...[remember (n+1)!=(n+1)n!,(n-r)!=(n-r)(n-r-1)! and (r+1)!=(r+1)r!]...substitute those...to give (n+1)!/(n-r)!(r+1)!...which is the same as equation 1 above...hence equal to (n+1)'C'(r+1)...

Hello pals,let's keep the thread moving

Any engineer here.MALE OR FEMALE.add me with the username MATHEMAGICIAN on 2go...

Allen kee: I have been trying to recheck mine but no such option on the school website.

.i'm grateful my bro...pls,contact me if you find solution to it.my mail is omoyeleolalekanjacob@gmail.com

honey: BEFORE I PROCEED, I WANT TO PROVE THAT (X^X^X^X)=X^{x^3}~~ (X^X^X^X) CAN ALSO WRITTEN AS (X^x)^x)^x) APPLYING BASIC LAW OF INDICES, THEN (X)^x*x*x =(X)^{x^3}, 2^2^2^2=256=2^{2^3}, after proof, we can restructure the question as d(X^{x^3})/dy~~~then i think power rule can do this, replacing x³ as u, then dy/dx= dy/du*du/dx : du/dx=3x²: dy/du= u*x^u-1, multiplying and replacing the values (3X²)*(x³*x^{x^3-1}), applying the law of indices while multiplying::: the solution now becomes {3X^{(x^3+4)}} , i hope i am right.

...you tried my bro,but where you differentiate x^(x^3)...it's not done that way

benbuks: waw u try bro but dia z a simpla method 2 dat..anyway weldon. still nt satisfied with ur proof

...i love that bro,solve it so that others can learn

Pls,i did fed.poly ado postutme and didn't wait to check my score...is there any way to check it?...

benbuks: prove that. 0!= 1
prove that 0.9999...=1
without approximation
..

...to proof that 0!=1...consider A,B and C...let's arrange them picking three at a time.i.e.permutation[arrangement of n object picking r at a time...nPr=n!/(n-r)!]...ABC,ACB,BAC,BCA,CAB,CBA...=6...it means arrangement of 3 object picking three at a time gives 6...i.e.3P3=6...3!/(3-3)!=6...3!/0!=6...since 3!=3*2*1=6...6/0!=6...0!=6/6...0!=1...to prove that 0.9999=1...consider the equation 2x-1...as x tends to 1,2x-1 also tends to 1...when x is 0.9,2x-1 is 0.8...when x is 0.99,2x-1 is 0.98...when x is 0.999,2x-1 is 0.998...it means as x tends to 1 2x-1 tends to 1...

benbuks: ok...take this....if y = X^x^x^x.... compute. dy/dx.

...y=X^x^x^x...take natural log of both sides...lny=ln(x^x)^(x^x)...from rule of log...that loga^b=bloga...therefore lny=X^xlnX^x...consider lnx^x,it also gives xlnx...therefore lny=x^x.xlnx...from indices.that a^b.a^c=a^(b+c)..which implies that x^x.x=x^(x+1)...therefore lny=x^(x+1)lnx...take d/dx of both sides...i.e.d/dx of lny=d/dx of x^(x+1)lnx...for the lhs...it's not possible to differentiate lny w.r.t to x...then,we make a manipulation that d/dx=d/dy*dy/dx...(i) and from rhs...we differentiate using product rule that d/dx(uv)=udv+vdu...so,d/dy*dy/dx of lny=d/dx of x^(x+1)lnx...d/dy of lny*dy/dx=d/dx of x^(x+1)lnx...1/y.dy/dx=d/dx of x^(x+1)lnx..1/y.dy/dx=x^(x+1)/x+lnx.d/dx of x^(x+1)..1/ydy/dx=x^x+lnx.d/dx of x^(x+1)[to solve d/dx of x^(x+1)..,let z be x^(x+1)...lnz=(x+1)lnx...1/zdy/dx=(x+1)/x+lnx(using product rule)...dy/dx=z{(x+1)/x+lnx}...dy/dx=x^(x+1).{x+1)/x+lnx}...expand to give...dz/dx=x^x(x+1)+x^(x+1)lnx]...substitute that to...1/ydy/dx=x^x+lnx.d/dx of x^(x+1)...1/y.dy/dx=x^x+lnx{x^x(x+1)+x^(x+1)lnx}...1/y.dy/dx=x^x+x^x(x+1)lnx+x^(x+1)(lnx)^2...dy/dx=y{x^x+x^x(x+1)lnx+x^(x+1)(lnx)^2)...dy/dx=x^x^x^x{x^x+x^x(x+1)lnx+x^(x+1)(lnx)^2}...

Asakel: Hmmmm! Diz wan iz strooong.

.how bro,pls explain better

Pls pals,i really need your help.....is it possible for me to rectify the picture i used in postutme coz the one i used was small....they complained about it when i wanted to do the postutme

For a=1.......let's chose equation i and ii...a+b+c=6...therefore b+c=5...(x) since a=1...a^2+b^2+c^2=14...therefore b^2+c^2=13...(y)...consider equation y...b^2+c^2=13...from the expansion of (b+c)^2=b^2+2bc+c^2...b^2+c^2=(b+c)^2-2bc...substitute that...(b+c)^2-2bc=13...from equation (x)...b+c=5...therefore 5^2-2bc=13...25-13=2bc...2bc=12,therefore bc=6...(z)...from there...b=6/c...substitute that to equation (x)...b+c=5...since b=6/c...6/c+c=5...multiply through by c to cancel the fraction...6+c^2=5c...c^2-5c+6=0...using any method of solution to quadratic equation...c=2 or 3...from (x)...b+c=5...b=5-c...substitute c=2,b will be equal to 3,substitute c=3,b will be equal to 2..............................................the solution when a=1 will be a=1,b=2,c=3 or a=1,b=3,c=2...TRY AND FINE THE VALUES OF B AND C WHEN a=2 and a=3

I tried to explain better but my text of more than 3900 words was hidden...i'm sorry,no be my fault o

[quote author=Calculusf(x)]...hmmmm...for the simultaneous equation...i tried using that numbers but complex...so let's try this one,a+b+c=6...(i).a^2+b^2+c^2=14...(ii) and a^3+b^3+c^3=36...[/quote]

Mikebis: Cn any1 pls help me wt dis questn.1.fnd dy/dx of (x^2+1)^x^2..(2)solv simultaneously x+y+z=8..eqn1
x^2+y^2+z^2=72..eqn2
x^3+y^3+z^3=216..eqn3.(3)prv dat x^3+y^3+z^3=(x+y+z)^2(x^2+y^2+z^2-xy-yz-zx)..(4)when x=3^2/3+3^-2/3 prv dat 9x^3-27x=82..pls help me,nd as do it may GOD increase ur knowledge.AMEN

...hmmmm...for the simultaneous equation...i tried using that numbers but complex...so let's try this one,a+b+c=6...(i).a^2+b^2+c^2=14...(ii) and a^3+b^3+c^3=36...let's consider equation ii,(a+b+c)^2={(a+b)+c}^2=(a+b)^2+2(a+b)c+c^2=a^2+2ab+b^2+2ac+2bc+c^2=a^2+b^2+c^2+2(ab+ac+bc)...that means (a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)...don't forget from the question that a+b+c=6 and a^2+b^2+c^2=14...therefore 6^2=14+2(ab+ac+bc)...22=2(ab+ac+bc)...divide both sides by 2...11=ab+ac+bc...(iv)...let's consider equation iii.(a+b+c)^3=a^3+b^3+c^3+3a^2b+3a^2c+3ab^2+3b^2c+3ac^2+3bc^2+6abc...(using pascal's triangle or newton's method for expansion)...don't forget that a+b+c=6 and a^3+b^3+c^3=36...substitute that to it...therefore 6^3=36+3a^2b+3a^2c+3ab^2+3b^2c+3ac^2+3bc^2+6abc...180=3a^2b+3a^2c+3ab^2+3b^2c+3ac^2+3bc^2+6abc...divide through by 3...60=a^2b+a^2c+ab^2+b^2c+ac^2+bc^2+2abc.then let's make some factorization...60=a^2(b+c)+b^2(a+c)+c^2(a+c)+2abc...[remember from equation i...a+b+c=6,therefore a+b=6-c,a+c=6-b and b+c=6-a...]substitute that...60=a^2(6-a)+b^2(6-b)+c^2(6-c)+2abc...expand,60=6a^2-a^3+6b^2-b^3+6c^2-c^3+2abc...60=6(a^2+b^2+c^2)-(a^3+b^3+c^3)+2abc...60=6*14-36+2abc...60=84-36+2abc...2abc=12...divide through by 2 to give abc=6...(v)...from equation (iv)...ab+ac+bc=11...factorize a...a(b+c)+bc=11...don't forget from equation i that b+c=6-a...therefore a(6-a)+bc=11...from equation (v)...abc=6...bc=6/a...a(6-a)+bc=11.substitute for bc...a(6-a)+6/a=11...multiply through by a...a^2(6-a)+6=11a...6a^2-a^3-11a+6=0...multiply through by -1...therefore a^3-6a^2+11a+6=0...from there...a=1,2 and 3...solutions for b and c still loading,me afraid of my text being hidden...

Pls,let's help each other in this room,the room is not the place to be testing people,but to bring questions and let people try it,if they do and after some days,nobody gets it,then solve it...that's the only way we can gain from one another...for me o,i don't bring or solve questions to be called guru but to help others...let everyone of us have that in mind pls.so,for all that have posted question and it has not been solved and you know the solution,pls.kindly provide the solution........THANK YOU

johnpaul1101: goodday everyone. Plz how do you find the third side of a triangle when two sides and the area of the triangle is given?

...using hero's formula is a complicated way...you can do it using this method...take one length as the base...from there,calculate the altitude using the area given and the base,through the use of the formula 1/2*base*height...after calculating the height...then use pythagora's theorem to calculate part of the base...i.e from where the altitude touches the base to the edge of the side given...after getting the side,you subtract it from the total length to get the other side of the base,then use pythagora's theorem again for the other side got and the altitude to calculate the unknown side.........THAT'S THE BETTER WAY OF SOLVING IT...ONE CAN MAKE MISTAKES WHEN USING HERO'S FORMULA COZ OF THE UNKNOWN...

Hidorano: Greetings to all math guru in the aose. Pls help me solve this. 1.Express 3x^2-6x+10 in the form a(x-b)^2+c where a, -b and c are integers. Hence state the minimum value of 3x^2-6x+10 and the value of x for when it occurs.

...from 3x^2-6x...make it a perfect square using the formula b^2=4ac...a=3 b=-6...make c the subject of formula...c=b^2/4a...(-6)^2/4*3=36/12...c=3...therefore the equation becomes...3x^2-6x+3+7...therefore,3(x^2-2x+1)+7...NOTE:x^2-2x+1=(x-1)^2...therefore it becomes...3(x-1)^2+7....................to find the minimum point of 3x^2-6x+10...find d/dx of the equation(which is the turning point of the equation)...6x-6=0...therefore x=1...and the formula for minimum point is d^2y/dx^2>0...then d^2y/dx^2=6 which shows it's a minimum point...the minimum point is 1...therefore to get the minimum value,substitute x=1 to the equation 3x^2-6x+10...when x=1...3*(1)^2-6*1+10...3-6+10=7.....therefore the minimum value is 7 and occurs when x=1

Hidorano: 3. How many whole number from 100 to 999 are divisible by (a). 4 (b). both 3 and 4. God bless u all.

...for(a)numbers which are divisible by 4 from 100-999...therefore it starts with 100 and has a common difference of 4 and last term will be 996(the closest to 999 which is divisible by 4) which is a.p...using the formula for last term of an a.p...nth=a+(n-1)d...where nth is the last term,a is the first term,n is the number of terms and d is the common difference...substitute those to the formula...996=100+(n-1)4...therefore 996-100=(n-1)4...896=(n-1)4...divide both sides by 4...224=n-1...therefore,make n the subject of formula...therefore n=225........(b)...the numbers divisible by 3 and 4 must start with 12 and multiple of 12...the first term closest to 100 which is a multiple of 12 is 108...and the last term is 996...the common difference is 12...then,substitute that to the formula for the last term of a.p....996=108+(n-1)12...996-108=(n-1)12...888= (n-1)12...divide both sides by 12...74=n-1...therefore n=75...

mathematician: your question is a bit bent, the correct equality is x² + e^2x, meaning that this is a linear differential equation, to get the solution, we say d/dx=d, d²/dx²=d², the equation now becomes {d²+2d+1}=x²+e^2x , the equation d²+2d+1 can be rewritten as (d+1)², the root of the quadratic equation is -1 ,-1, therefore the complementary equation can be given as c₁e^-x, now to get the auxilliary equation we say y=ax²+bx+c+de^-2x, now replacing y in the original equation and applying proper differentiation we have ax²+(4a+b)x+ (2a+2b+c) +de^-2x, a=1, b=-4, c=-6, d=1, therefore the complete solution for y equals c[sub]1[/sup]e^-x + x² - 4x -6 + e^-2x

...thank you very much general.i know it if it was that,but it's product...x^2e^2x

Pls my generals...solve y''+2y'+y=x^2e^2x...where y''=d^2y/dx^2 and y'=dy/dx,i have been on this question since but if the rhs is x^2 or e^2x...i can solve that but the combination of the two...hmmmmm...pls,help ooo...

On your questions(2&3)now @Mikebis...i pray i give positive reply

Mikebis,mathematician,general double dx,mr calculus and other gurus in the house...well done ooo,me back again ooo...the room is very interesting now...let's keep it moving