those smileys are -8 respectivelyi.e (18,- (-7,-

Youngsage:

x + y =5 -------- eqn i.
x^x + y^y=31 ----- eqn ii.
From equatns i & ii, take the log of both sides
(xlog^x + ylog^y) = log 31------ eqn iii.
log(x+y) = log 5 ------------ eqn iv.
xylogxy= log 31 -------- eqn v.
log xy = log 5 --------- eqn vi.
(Using elimination method);
xylog5 = log31
xy = log 31/log 5 =6.2 approx. 6.

Recall, from eqn i, x + y = 5.
So x = 6/y or y = 6/x.
Therefore x=2 when y=3.
ans: x=2, and y=3.

Explain further on how you applied the elimination method

Richiez:


nice try bro, although you made some some mistakes especially at the point where you took log of both sides. e.g it ought to be log(x+y)= log5 and not logx + logy = log5. but i must admit, you did a very good job.

here's another approach;

x + y = 5.......(1)
x^x + y^y =31.......(2)

from eqn(1), x=5-y, hence we substitute this value for x in eqn(2).
(5-y)^(5-y) + y^y = 31.......(3)
now, we have reduced the problem to what value of y that must be inserted in eqn (3) to obtain 31.

a thorough look at eqns (1) & (2) shows that the values of x & y must be positive integers within the range 0<x<5 and 0<y<5
clearly, y=3
i.e (5-3)^(5-3) + 3^3
2^2 + 27
4+27=31
now that we have established that y=3, we shall substitute the value of y in eqn(1) to get x
recall that x+y=5
therefore; x+3=5 or x=5-3=2
hence, x=2 when y=3

Yours in more complex for leaners

Who can prove the almighty formula using this magic formula?

F'(x)=±sqrt(f'(x)^2 -4f(x))

This formula derives the quadratic formula.

Do you know? @ Richez

Limit n tend to infinity square root of n^2 +n _n =1/2
note _n is not in d square root

Find d supremum and infimum of n-1/2n

Profmaojo:
Limit n tend to infinity square root of n^2 +n _n =1/2
note _n is not in d square root

snap & post ..not clear

makasimatics:
Who can prove the almighty formula using this magic formula?

F'(x)=±sqrt(f'(x)^2 -4f(x))

This formula derives the quadratic formula.

Do you know? @ Richez

derive it let's see asap....

Which textbook can it be of help with this courses
1. Fluid mechanics
2. Mathematical modelling.

Arithmetic:
Evaluate ${(1/lnt)}dt.

∫(1/lnt)dt
=∫(1/lnt)(t.dt/t)
=∫(t/lnt)dlnt
=∫(e^lnt/lnt)dlnt
Change of variable! Put lnt=x
=∫(e^x/x)dx
=∫(1/x+1+x/2!+x^2/3!+x^3/4!...) dx
=lnx+x+x^2/2.2!+x^3/3.3!+x^4/4.4!+...+x^n/n.n!+...
now put x=lnt
=ln(lnt)+lnt+(lnt)^2/2.2!+(lnt)^3/3.3!+(lnt)^4/4.4!+...+(lnt)^n/n.n!+...

I dey laff o! Shey i nevr jonezz

Arithmetic:
Evaluate ${(1/t).e^t}dt.

∫(e^x/x)dx
=∫(1/x+1+x/2!+x^2/3!+x^3/4!...) dx
=lnx+x+x^2/2.2!+x^3/3.3!+x^4/4.4!+...+x^n/
n.n!+...

efficiencie:


∫(e^x/x)dx
=∫(1/x+1+x/2!+x^2/3!+x^3/4!...) dx
=lnx+x+x^2/2.2!+x^3/3.3!+x^4/4.4!+...+x^n/
n.n!+...

seems you're on track.

series's cool.

Find d area of d region between d curves; y=x^4 & y=2x-x^2.

First we find their point of intersection and ensure that none of d two curves crosses d x-axis bewteen dis pionts of intersection.
x^4=2x-x^2 or
(x^3+x-2)x=0.
Hence, x=0, x=1 are d only real solutions. Equatin each of d given eqns to zero, we obtain; x^4=0 i.e x=0. Then, 2x-x^2=0, x=0 or 2. Hence, none of d two equations crosses d x-axis between x=0 and x=1. So,
area bounded by y=x^4 is;
A1=§x^4dx=x^5/5 limit is from 0 to 1. So, A1=1/5.
Area bounded by y=2x-x^2 is,
A2=§(2x-x^2)dx=x^2-x^3/3 limit is from 0 to 1. Hence, A2=2/3.
So area between d two curves; A=A2-A1=2/3-1/5=7/15 units

Find d eqn of d cicle which passes thru d point (1,1), half a radius of sqr(10)/2 and whose centre lies on d line y=3x-7.
If d centre passes thru d given line, then d co-ordinate of d centre must satisfy d line.
Let d coordinate of d centre b (a,b), then; b=3a-7, the radius r=sqr(10) and d eqn of d circle is;
(a-1)^2+(b-1)^2=10
or substitutin for b gives
(a-1)^2+(3a - 8 ) ^2=10
put a-1=z, then,
z^2+(3z-5)^2=10
or
10z^2-30z+15=0
or
2z^2-6z+3=0
z=[3(+ or -)sqr(3)]/2.
From z=a-1, get a & from b=3a-7, get b.

Substitute for a and b in d eqn of d circle above

What must be d value of k in order that; 2x+y-3=0, kx+3y+1=0, x+y+7=0 may meet in a point. Discuss d cases when k=3 and k=6

d point of intersection of d first and last lines are obtained first. Subtractin their eqns give; x=10 and y=-17. For the second line to satisfy this point, we must have;
10k-3(17)+1=0, or k=52/10.
(i)caseI; if k=3, we have, 3x+3y+1=0 or by dividing thru by 3; x+y+1/3=0. Hence the line is parallel to the third line (since their slope is the same) and is higher the third line (and also the point of intersection) since the intercept is greater.
(ii) caseII; if k=6, then, 6x+3y+1=0 or by dividing by 3, 2x+y+1/3=0. Again, this line is parallel to the first line and is lower, below it.

Find the eqns of d sides of d triangle ABC where A, B, C are d points (5,7), (3,3), (7,1) respectively. Hence, show that the triangles ABC has angles of 90, 45 and 45. Verify this results by finding the lengths of d triangle.

Slope of line AB is (3-7)/(3-5)=2
eqn of line AB is y=2x-3

slope of line BC is (1-3)/(7-3)=-1/2
eqn of line BC is y=-x/2+9/2

slope of line AC is (1-7)/(7-5)=-3
eqn of line AC is y=-3x+22

now, two lines are perpendicular if the product of their slope is -1.
Multiplyin slope of AB and BC we get; 2*-1/2=-1.
Now, d angle between two lines is found from;
tan@=(m1-m2)/(1+m1*m2), where m1 and m2 are the slope of d lines. I dont have calculator, complete the rest.

D length of d tangent from d point (3,2) to d circle x^2 +y^2-2x-3y+k=0 is 9 units, find k.

Centre of d circle, C=(1,3/2).
Length of of given point from center L^2=(3-1)^2+(2-3/2)^2=17/4.
Since d the system forms a pythagorean triangle, the radius is found from;
r^2=17/4-9^2.
This shows that the circle does not have a REAL radius, and so, k cannot be real

Find the equation and radius of d second circle of the triangle formed by d three lines; 2y-9x+26=0, 9y+2x+32=0, 11y-7x-27=0.
I dont understand this questio. Is it the circumcircle we are lookin 4? or the inscribed circle?

Laplacian:
Find the equation and radius of d second circle of the triangle formed by d three lines; 2y-9x+26=0, 9y+2x+32=0, 11y-7x-27=0.
I dont understand this questio. Is it the circumcircle we are lookin 4? or the inscribed circle?

it is not second circle oooo......it is circumcircle i meant

emmyeuler1:
it is not second circle oooo......it is circumcircle i meant

k. Solve three equations choosing two at a time and u'll get the points of intersections. Substitute each points of intersection in the general equation of the circle; x^2+y^2+2ax+2by+c=0, and solve the three simultaneous equations for a, b and c.....its too bulky 2 solve without calculator...

hmmmm see mathematicians a beg....


@sir laplacian & prof. emmyeuler.


I greet thee in CAPITAL letters...

here's a lit2 problem that is becoming a problem to the problem solvers ...


try out

1. Find the value of x,y and z if x^2-yz=1,
y^2-xz=4, and z^2-xy=9.



anticipating.

agentofchange1:


OK..let's see if I can ...

or others can help out ..

Here

1. Integral 7x-4/2x^2-3x-2 dx

2. Integral x^3/x+1 dx

3. Integral cosx/7+sinx dx = In (7+sinx)+C

4. Integrate with respect to X
A. Sinxcosx
B. Sin6x
C. X^2 e^3

5. Y= sin^-1 (2x-1)
b. Y=tan^-1 (1+x/1-x)

yemstok:


Here

1. Integral 7x-4/2x^2-3x-2 dx

2. Integral x^3/x+1 dx

3. Integral cosx/7+sinx dx = In (7+sinx)+C

4. Integrate with respect to X
A. Sinxcosx
B. Sin6x
C. X^2 e^3

5. Y= sin^-1 (2x-1)
b. Y=tan^-1 (1+x/1-x)

am not with pen/paper now.
Q3 put. u=7+sinx

du=cosxdx
dx= du/sinx

thus => $cosx/u *du/cosx

= $du/u = ln (u) +C

hence we have ln(7+sinx) +C . ...(as expected)
Q4.
A $sinxcosxdx

set u= sinx
dx=du/cosx

=> $u*sinx*du/sinx

= >u^2 /2 + C
= 0.5sin^2 x + C .

B. similarly

let t= 6x
dt= 6dx
dx= dt/6

=>1/6 $sint .dt
= -1/6 cost + C
= -1/6 cos6x + C .

C) guess the question is like this

x^2 e^(3x)

let's use integration by parts (I.B.P)

put u=x^2 ; du =2xdx

dv=e^(3x) dx ; v=e^(3x) /3

by $udv =u.v -$vdu. , we have

x^2 e^(3x) /3 -2/3 $ x e^(3x) dx..............(*)

again let's compute $xe^(3x)dx from. (*) above

set p= x. ; dp = dx

dq=e^(3x)dx ; q=e^(3x) /3

by $pdq = p.q -$qdp ,

we have xe^(3x) /3 - 1/3 $ e^(3x) dx

=>xe^(3x) /3. -e^(3x) /9 ..........(**)

putting (**) in (*) to get the final ans..

=>x^2 e^(3x) /3 -2/3 [ xe^(3x) /3 -e^(3x) /9 ]. + C

simply. further..
rest solution coming later... m not with pen/paper ...will use partial fraction for 1 &2 .

check Q5. are we to integrate or differentiate .?

yemstok:


Here

1. Integral 7x-4/2x^2-3x-2 dx

2. Integral x^3/x+1 dx

3. Integral cosx/7+sinx dx = In (7+sinx)+C

4. Integrate with respect to X
A. Sinxcosx
B. Sin6x
C. X^2 e^3

5. Y= sin^-1 (2x-1)
b. Y=tan^-1 (1+x/1-x)

no 1::: 3ln(2x+1)+2ln(x-2)+K

agentofchange1:


am not with pen/paper now.
Q3 put. u=7+sinx

du=cosxdx
dx= du/sinx

thus => $cosx/u *du/cosx

= $du/u = ln (u) +C

hence we have ln(7+sinx) +C . ...(as expected)
Q4.
A $sinxcosxdx

set u= sinx
dx=du/cosx

=> $u*sinx*du/sinx

= >u^2 /2 + C
= 0.5sin^2 x + C .

B. similarly

let t= 6x
dt= 6dx
dx= dt/6

=>1/6 $sint .dt
= -1/6 cost + C
= -1/6 cos6x + C .

C) guess the question is like this

x^2 e^(3x)

let's use integration by parts (I.B.P)

put u=x^2 ; du =2xdx

dv=e^(3x) dx ; v=e^(3x) /3

by $udv =u.v -$vdu. , we have

x^2 e^(3x) /3 -2/3 $ x e^(3x) dx..............(*)

again let's compute $xe^(3x)dx from. (*) above

set p= x. ; dp = dx

dq=e^(3x)dx ; q=e^(3x) /3

by $pdq = p.q -$qdp ,

we have xe^(3x) /3 - 1/3 $ e^(3x) dx

=>xe^(3x) /3. -e^(3x) /9 ..........(**)

putting (**) in (*) to get the final ans..

=>x^2 e^(3x) /3 -2/3 [ xe^(3x) /3 -e^(3x) /9 ]. + C

simply. further..
rest solution coming later... m not with pen/paper ...will use partial fraction for 1 &2 .

check Q5. are we to integrate or differentiate .?

Thanks bro Q5 is integration

bolkay47:
no 1::: [b]3/2[/b]ln(2x+1)+2ln(x-2)+K

yea man you try


just a li2 correction (the bolded its 3/2. not 3 )

nice 1 try post full solutions next time

1luv ..happy solving.

yemstok:


Thanks bro Q5 is integration

OK bro...

will post solutions asap..


but wait oo..how much you go pay me for. burning my MB....? lolz. ( just kidding ..never mind )

agentofchange1:


OK bro...

will post solutions asap..


but wait oo..how much you go pay me for. burning my MB....? lolz. ( just kidding ..never mind )

Oya mail me your number make I give you small change.

yemstok:


Thanks bro Q5 is integration

Q5
A)

...let p=sin^-1 (2x-1)

Sinp=(2x-1)
Cosp dp=2dx

dx=(cosp dp)/2
∫sin^-1 (2x-1)dx= ∫p(cosp dp)/2= 1/2∫pcosp dp

Using integration by parts
U=p
du=dp

dv=cosp

v=sinp

1/2(p sinp- ∫sinp dp)
1/2(p sinp+cosp)+k

But p=sin^-1 (2x-1), sinp=(2x-1), cosp= √(1-sin^2p)= √(1-(2x-1)^2)= 2√(x-x^2)

1/2((2x-1)sin^-1 (2x-1)+2√(x-x^2))
1/2(2x-1)sin^-1 (2x-1)+√(x-x^2)+k

B)

ans = xarctan[(1+x)/(1-x)] - 1/2 ln | 1+x² | + C

Given that CosZ =L, where Z is an acute angle, find an
expression for (CotZ - CosecZ )/(SecZ +TanZ )
2.Find there numbers in G.P whose sum is 28 and whose
product is 512.
3. Find the derivative of Y with respect to X If
Y=√{(1+ x )/(1- x )
Where √=square root