Adol16: Solve
1. 1/X + 1/y + 1/z = 1 find x,y and z


2. 5Loga base 4 + 48log4 base a = a/8.

1. Z=xy/(xy-x-y) by manipulating
Therefore (xy-x-y) is not equal to 0

2. That question is unnecessarily hard and stinks of wickedness, but I am going to give u a clue though

Log(a) to base 4= Log(a) to base a/Log(4) to base a

If you do this and substitute you getting a quadratic equation, a wicked one, smh

@richiez, i still dey wait

ositadima1:

Mr Spoiler alert, it would enlighten us some if you gave us the correct answer.

Like I mentioned earlier,this is a nonlinear equation which can only be solved numerically or by graphical method. In other word, neither of the two variables are linearly defined.

I've solved couple of question here.

Adol16: Solve
1. 1/X + 1/y + 1/z = 1 find x,y and z


2. 5Loga base 4 + 48log4 base a = a/8.

Question 1,

x can only be solved in terms of y and z, same way y can only be solved in terms of x and z; and z in terms of x and y.

1/x + 1/y + 1/z = 1

Let's solve for x in terms of y and z>

(yz + xz + xy)/xyz = 1
yz + xz + xy = xyz....equation (1)
xyz - xy - xz = yz
x(yz - y - z) = yz
x = yz/[yz - y - z]
x = yz/[y(z - 1) - z]

Now, we can solve for y in terms of x and z from equation (1)

yz + xz + xy = xyz
xyz - yz - xy = xz
y(yz - z - x) = xz
y = xz/(yz - x - z)
y = xz/[z(y - 1) - x]

And finally z in terms of x and y from equation (1)

yz + xz + xy = xyz
xyz - yz - xz = xy
z(xy - y - x) = xy
z = xy/[xy - x - y]
z = xy/[x(y - 1) - y]

donedy:

Like I mentioned earlier,this is a nonlinear equation which can only be solved numerically or by graphical method. In other word, neither of the two variables are linearly defined.

I've solved couple of question here.

I thought you said he should reach the solutiong analytically, dat was what I wanted to see.

please help with this:
"reduce 60 by 30%"
this is my working
60*30/100
=18
but 18 is not in the options

johnpaul1101: please help with this: "reduce 60 by 30%" this is my working 60*30/100 =18 but 18 is not in the options

nice try but the final solution is 60-18 = 42

good job ghetoguru!

^^^

Thanks sir!

Richiez:
nice try but the final solution is 60-18 = 42

yea the final answer is 42, but why substract it again

ositadima1: I thought you said he should reach the solutiong analytically, dat was what I wanted to see.

yea donedy i'd also like to see the solution. i appreciate ur efforts though. once again, welcome to the maths family

johnpaul1101:
yea the final answer is 42, but why substract it again

It is because 30%(i.e. 18) is the amount you're told to reduce 60. After reducing or removing 30% (18) you have 42 left.

In order words, how much of your salary is left after spending 30% of it on Ashawo every month.

Richiez:
yea donedy i'd also like to see the solution. i appreciate ur efforts though. once again, welcome to the maths family

thnx, you guys are cool.

Richiez:
yea donedy i'd also like to see the solution. i appreciate ur efforts though. once again, welcome to the maths family

I've said it earlier. it cannot be solved analytically because neither of the variables are linearly defined.

ositadima1:

1. Z=xy/(xy-x-y) by manipulating
Therefore (xy-x-y) is not equal to 0

2. That question is unnecessarily hard and stinks of wickedness, but I am going to give u a clue though

Log(a) to base 4= Log(a) to base a/Log(4) to base a

If you do this and substitute you getting a quadratic equation, a wicked one, smh

So true bro, it's a cruel question

Richiez: @double dy/dx, i dey see ur handwork, u'r a great man, i salute you!

Thank you bruv!

I salute all maths gurus in da house,
o.p. Check dis out.

Find the equation of the tangent to the parabola y^2=6x, which is parallel to the line y+2x=0...

Now, normal equation and formula for the tangent is yy1=2a(x+x1), Which means its tangent is at the point(x1,y1).
Now, this question doesn't doesnt give me the co-ordinates, so how do i tackle it?

Shed more light pls.

topmostg: I salute all maths gurus in da house,
o.p. Check dis out.

Find the equation of the tangent to the parabola y^2=6x, which is parallel to the line y+2x=0...

Now, normal equation and formula for the tangent is yy1=2a(x+x1), Which means its tangent is at the point(x1,y1).
Now, this question doesn't doesnt give me the co-ordinates, so how do i tackle it?

Shed more light pls.

This is basically simple and straight forward. Since the equation of the tangent has to be parallel to the line y + 2x = 0, It means that
the equation of the tangent has to have a slope of -2 (i.e. dy/dx = -2)

So, find the derivative of the given equation, y^2 = 6x, and set it equal to -2.

dy/dy: 2y*(dy/dx) = 6 ------------- Eq.i

Eq.i ==> dy/dx = 3/y ------------- Eq.ii

Thus, dy/dx = 3/y = -2 ------------ Eq.iii

Solving Eq.iii yields y = -3/2

We find x from substituting -3/2 for y in y^2 = 6x, which (-3/2)^2 = 6x

Therefore, x = (-3/2)^2/(6) = (9/4)/6 = 3/8

Now we know that the tangent has a slope of -2 and passes through the point (3/8, -3/2)

Basic Equation of line: y = y0 + m(x-x0)
y = -3/2 - 2( x - 3/8 )

y = -3/2 -2x + 3/4

which ==> 4y = -8x - 3

or

The equation of tangent: 4y + 8x + 3 = 0

donedy: I've said it earlier. it cannot be solved analytically because neither of the variables are linearly defined.

okay now i understand you. but a logical and not an analytical approach was used. anyways, graphical method remains the most acceptable

[quote author=Ghettoguru]

Question 1,

x can only be solved in terms of y and z, same way y can only be solved in terms of x and z; and z in terms of x and y.

1/x + 1/y + 1/z = 1

Let's solve for x in terms of y and z>

(yz + xz + xy)/xyz = 1
yz + xz + xy = xyz....equation (1)
xyz - xy - xz = yz
x(yz - y - z) = yz
x = yz/[yz - y - z]
x = yz/[y(z - 1) - z]

Now, we can solve for y in terms of x and z from equation (1)

yz + xz + xy = xyz
xyz - yz - xy = xz
y(yz - z - x) = xz
y = xz/(yz - x - z)
y = xz/

thanks i solved it that way b4 but the guy that gave me the question said it has real value solutions. Just wanted to b sure.

i dnt seem to get the second question the guy said the ans is a=256 which is true when i substituted it into the question.
can it b solved?
i got something like this
5(loga base 4)square + 48=a/8 loga base 4.

^ Write out the question properly, if possible write it out on a paper, then scan and attach. That way, we'll understand your question properly!

doubleDx: ^ Write out the question properly, if possible write it out on a paper, then scan and attach. That way, we'll understand your question properly!

whats up man, was looking for ur pix on the other thread.

ositadima1:

whats up man, was looking for ur pix on the other thread.

I'm okay, thanks man. Lol, I'll post one tomorrow Have you posted yours?

donedy: This is basically simple and straight forward. Since the equation of the tangent has to be parallel to the line y + 2x = 0, It means that the equation of the tangent has to have a slope of -2 (i.e. dy/dx = -2) So, find the derivative of the given equation, y^2 = 6x, and set it equal to -2. dy/dy: 2y*(dy/dx) = 6 ------------- Eq.i Eq.i ==> dy/dx = 3/y ------------- Eq.ii Thus, dy/dx = 3/y = -2 ------------ Eq.iii Solving Eq.iii yields y = -3/2 We find x from substituting -3/2 for y in y^2 = 6x, which (-3/2)^2 = 6x Therefore, x = (-3/2)^2/(6) = (9/4)/6 = 3/8 Now we know that the tangent has a slope of -2 and passes through the point (3/8, -3/2) Basic Equation of line: y = y0 + m(x-x0) y = -3/2 - 2( x - 3/8 ) y = -3/2 -2x + 3/4 which ==> 4y = -8x - 3 or The equation of tangent: 4y + 8x + 3 = 0

nice work, u'r on ur way to earning the title 'GENERAL'

Richiez:
nice work, u'r on ur way to earning the title 'GENERAL'

Bros don't forget to invite me at the inaugural ceremony, long live the generals, long live all maths luvers, you all rock!

ositadima1: Bros don't forget to invite me at the inaugural ceremony, long live the generals, long live all maths luvers, you all rock!

lol...nice idea General Osita, i foresee an inaugural ceremony 4 all Generals

It seems like this thread has been dormant for a while, but let me see if I can warm it up. The first to get this questions right within three responses will have 1K credited to his/her bank account by me.

1. Define a Circle using mathematical terminology.

2. Define a Straight Line in terms of an arc.

3. Calculate the area of the biggest rectangle that be inscribed in a circle of form x² + y² = a², where a is a constant.

donedy:

This is basically simple and straight forward. Since the equation of the tangent has to be parallel to the line y + 2x = 0, It means that
the equation of the tangent has to have a slope of -2 (i.e. dy/dx = -2)

So, find the derivative of the given equation, y^2 = 6x, and set it equal to -2.

dy/dy: 2y*(dy/dx) = 6 ------------- Eq.i

Eq.i ==> dy/dx = 3/y ------------- Eq.ii

Thus, dy/dx = 3/y = -2 ------------ Eq.iii

Solving Eq.iii yields y = -3/2

We find x from substituting -3/2 for y in y^2 = 6x, which (-3/2)^2 = 6x

Therefore, x = (-3/2)^2/(6) = (9/4)/6 = 3/8

Now we know that the tangent has a slope of -2 and passes through the point (3/8, -3/2)

Basic Equation of line: y = y0 + m(x-x0)
y = -3/2 - 2( x - 3/8 )

y = -3/2 -2x + 3/4

which ==> 4y = -8x - 3

or

The equation of tangent: 4y + 8x + 3 = 0

nice job dude,
that was awesome......

Kudos to all

maths fieldmarshals....

topmostg: nice job dude, that was awesome...... Kudos to all maths fieldmarshals....

lol welcome to the family

@donedy, i was waiting to see if someone else will attempt your question but i guess it's time i attempted it

Richiez: @donedy, i was waiting to see if someone else will attempt your question but i guess it's time i attempted it

The price is still up. I want to keep the fire because the thread seems to be fading slowly.

hmmmmmm......mathematics used to be my best but i'm sure i'm as rusty as rusty can be now. So, i've decided to follow this thread and c if i can still wake up d maths guru in me. I rily hope so. @poster, good thread!