benbuks: ..ok. now ur talking ...


will get bk.to u much later ..

Ok

PatEinstEin:
Ok

Enter the principle amount, interest rate, and number of years in the textboxes. When you click Calculate, the program uses the following code to display your balance for the following years.
 
Private Sub cmdCalculate_Click()
Dim principle As Double
Dim interest_rate As Double
Dim num_years As Integer
Dim i As Integer
Dim current_value As Double

lstBalances.Clear

principle = Val(txtPrinciple.Text)
interest_rate = Val(txtInterestRate.Text)
num_years = Val(txtYears.Text)

For i = 1 To num_years
current_value = principle * (1 + interest_rate) ^ i
lstBalances.AddItem i & ": " & _
vbTab & FormatCurrency(current_value)
Next i
End Sub
 
The program simply loops through the years, evaluating the compound interest formula:
balance = principle * Math.Pow(1 + interestRate, i)
Interesting tidbit: To estimate how long it will take to double your money, you can use the "Rule of 72." Divide the interest rate into 72 and the result tells you roughly how many years it will take to double your money. For example, at 7.2% it'll take about 10 years. It's a pretty decent estimate.

you will make use of command bottom , text box , label box .

wish i could show you a diagram for d program design.

benbuks: Enter the principle amount, interest rate, and number of years in the textboxes. When you click Calculate, the program uses the following code to display your balance for the following years.
 
Private Sub cmdCalculate_Click()
Dim principle As Double
Dim interest_rate As Double
Dim num_years As Integer
Dim i As Integer
Dim current_value As Double

lstBalances.Clear

principle = Val(txtPrinciple.Text)
interest_rate = Val(txtInterestRate.Text)
num_years = Val(txtYears.Text)

For i = 1 To num_years
current_value = principle * (1 + interest_rate) ^ i
lstBalances.AddItem i & ": " & _
vbTab & FormatCurrency(current_value)
Next i
End Sub
 
The program simply loops through the years, evaluating the compound interest formula:
balance = principle * Math.Pow(1 + interestRate, i)
Interesting tidbit: To estimate how long it will take to double your money, you can use the "Rule of 72." Divide the interest rate into 72 and the result tells you roughly how many years it will take to double your money. For example, at 7.2% it'll take about 10 years. It's a pretty decent estimate.

Wow! *dancing shoki*
Thanks a lot boss...do u mind teaching me some Visual Basic Programming, no matter how small pls use this thread in order not to derail: https://www.nairaland.com/1832504/pls-write-program-visual-basic thanks

GOD BLESS YOU GUYS, YOU ARE REALLY GURUS. I DESPERATELY WANT TO BE LIKE YOU GUYS.

PatEinstEin:
Wow! *dancing shoki*
Thanks a lot boss...do u mind teaching me some Visual Basic Programming, no matter how small pls use this thread in order not to derail: https://www.nairaland.com/1832504/pls-write-program-visual-basic thanks

guy..2b honest with you , am not an expert on this , its long i did it ..though still have some idea on it...

go to programing section if you really want to learn
programing or follow this link for clarity on your assignment
www.functionx.com/vb/applications/compoundinterest.htm


or get a pc. den get VB . software & practice with it . that's how i could help on this....

1luv

A company won a contract to supply furniture to a leading Oil and Maritime company. They were out of stock and sub-contracted the job to another company who charged 100,000 naira for the supply.

Before sending their final quotes to the client, they added 5% VAT fee and 20% as their profit for the transaction. How much was their final quote?

jackpot: Sir, you called?

Try differentiating your result, lets see

will that pi go away eventually, to give us back the integrand?

Given the derivative: d(π(4n+1)x/4-1/(1+x^2))/dx

=π(4n+1)/4-d(1/(1+x^2))/dx

Bt ∫(arctanx)dx=1/(1+x^2)
hence arctanx=d(1/(1+x^2))/dx

=π(4n+1)/4-d(1/(1+x^2))/dx

=π(4n+1)/4-arctanx

=arctan[tan(π(4n+1)/4-arctanx)]

=arctan[(tan(π(4n+1)/4)-tan(arctanx))/(1+ tan(π(4n+1)/4)tan(arctanx))]

=arctan[(1-x)/(1+1.x)]

=arctan[(1-x)/(1+x)] which is the integrand!

So happy I discovered this thread....leggo!!...time to watch and learn

(x/2)^x = 16

solve..

x=4


so show workings..


asap

Freiburger: https://www.nairaland.com/1834634/need-mathematics-questions

Q1
. x + y + z= 1 .........................(B)

x² +y² + z²= 35...........(E)

x ³+ y³ + z³= 97............(N)


SOLUTION :

From (E)

we observe that
(x+y+z)²= x²+ y² + z² + 2 (xy + xz + yz)

=> (1)²=35 +2(xy +xz+yz)

=> xy +xz + yz = -17 ..... ........(J)

also ,

(x+y+z) (x²+ y² +z² -xy-xz - yz )=x³ +y³ +z³ -3xyz

=> (1)(35 +17)=97-3xyz

=>3xyz=45
=>xyz=15............ ................(A)

Thus lets have
x+y+z=1.................................(B)
xy+xz+ yz =-17.....................(J)
xyz=15...................................(N)


From (B) , x+y= 1-z .................(M)
From (N), xy =15/z....................(I)

put these in (J)

=> 15/z + z(1-z) = -17

=> 15+z²-z³ = -17z

=> z³-z²-17z-15 =0

guess you can finish up from there ....

Q2

SOLUTION
lets use binomials

3^x =(1+2)^x

=>
3^x=1+2x+(4x(x-1))/2 +…

=> 3^x =1+2x² +…

similarly ,
4^x=(2-3x+9x² ) /2 + …
&
5^x =1-4x+8x² + …

recall that
3^x +4^x=5^x

=> 1-4x+8x² +…=1+2x² +...+(2-3x+9x² )/2 +…

regrouping identical terms & factorizing

=> 3x²-5x-2=0

thus x=2 or -1/3

hence x=2

benbuks: (x/2)^x = 16

solve..

x=4


so show workings..


asap

given; (x/2)^x=16=2⁴, by the unique factoriztaion theorem, (x/2) must be a power of two since the right hand side of the given eqn is a power of two. Hence, (x/2)=2^k for some integer k. Hence, 2^kx=2⁴ or kx=4 and there are only three possibilities; k=1, x=4 Or k=x=2 Or k=4, x=1. Testing each option in the given equation shows that only the first one is valid, Hence x=4

Laplacian:
given; (x/2)^x=16=2⁴, by the unique factoriztaion theorem, (x/2) must be a power of two since the right hand side of the given eqn is a power of two. Hence, (x/2)=2^k for some integer k. Hence, 2^kx=2⁴ or kx=4 and there are only three possibilities; k=1, x=4 Or k=x=2 Or k=4, x=1. Testing each option in the given equation shows that only the first one is valid, Hence x=4

hmm...prof..prof...

nice1 .

gat be an alternative solution 4d sake of others who might nt understand what you did above ...
or what do you think ..?

@efficienci and benbuks,pls help integrate [1/x^3-1]

pls gurus in the house assist me in solving this:
prove that LOGaA•LOGbB=LOGbA•LOGaB
please note that the small letters are the bases of the individual logarithms.

Kendzyma: @efficienci and benbuks,pls help integrate [1/x^3-1]

[b]SOLUTION

we use partial fractions

1/(x³ -1 )

=> 1/(x-1)(x² +x +1] =p /(x-1) + (qx+r)/(x² +x +1)

=> 1= px²+px+p + qx² -qx+rx-r ............ (*)

putting x=-1 , 0 & 1

thus we have 1/3, -1/3, -2/3 as p , q & r respectively

=> $1/3(x-1) - $(x+2)/(x²+x1] dx

=> 1/3 ln(x-1) -1/3 $(x+2)/(x² +x+1 ) ............(**)


hmm. young man , @ this point we shall need great mathematical strength to do some manipulations

now lets consider $(x+2)/(x² +x+1 ] dx

=> 1/2 $(2x+4)/(x² +x +1 ] dx

=> 1/2$[(2x+1) +3 )/(x² +x+1 ) dx

= 1/2 $(2x+1)/(x^2 +x +1) dx + 3/2 $1/(x² +x + 1 ) dx

=> 1/2 ln(x² +x +1 ) + $ 1/[(x+1/2)² dx + (√3/2)² ) dx

recall your standard integral for arc.tan.
we thus have

=> 1/2 ln(x² +x + 1) + 3/2 * 2/√3 arc tan (2x+1) /√3 now put this in (**) to obtain your final answer .

hmm this your question looks innocent , but far from that [/b]

Soneh: pls gurus in the house assist me in solving this:
prove that LOGaA•LOGbB=LOGbA•LOGaB
please note that the small letters are the bases of the individual logarithms.

Given LOGaA•LOGbB=LOGbA•LOGaB

Starting frm d RHS:
LOGaA•LOGbB
=(LogA/Loga).(LogB/Logb)
=(LogA/Logb).(LogB/Loga)
=LOGbA•LOGaB

Which is the LHS

Nb LOGxX=LogX/Logx

benbuks:

[b]SOLUTION

we use partial fractions

1/(x³ -1 )

=> 1/(x-1)(x² +x +1] =p /(x-1) + (qx+r)/(x² +x +1)

=> 1= px²+px+p + qx² -qx+rx-r ............ (*)

putting x=1 , 0 & 1

thus we have 1/3, -1/3, -2/3 as p , q & r respectively

=> $1/3(x-1) - $(x+2)/(x²+x1] dx

=> 1/3 ln(x-1) -1/3 $(x+2)/(x² +x+1 ) ............(**)


hmm. young man , @ this point we shall need great mathematical strength to do some manipulations

now lets consider $(x+2)/(x² +x+1 ] dx

=> 1/2 $(2x+4)/(x² +x +1 ] dx

=> 1/2$[(2x+1) +3 )/(x² +x+1 ) dx

= 1/2 $(2x+1)/(x^2 +x +1) dx + 3/2 $1/(x² +x + 1 ) dx

=> 1/2 ln(x² +x +1 ) + $ 1/[(x+1/2)² + (√3/2)² ) dx

recall your standard integral for arc.tan.
we thus have

=> 1/2 ln(x² +x + 1) + 3/2 * 2/√3 arc tan (2x+1) /√3 now put this in (**) to obtain your final answer .

hmm this your question looks innocent , but far from that [/b]

,couldnt go beyond d partial fraction.tanx bro.

pls someone should help me with the solution and explanation to this question :
If. A and B are the roots of ax^2+bx+C=o form the equation whose roots are: A-2B, B-2A
PLS note that A and B are alpha and beta respectively.

efficiencie:

Given LOGaA•LOGbB=LOGbA•LOGaB

Starting frm d RHS:
LOGaA•LOGbB
=(LogA/Loga).(LogB/Logb)
=(LogA/Logb).(LogB/Loga)
=LOGbA•LOGaB

Which is the LHS

Nb LOGxX=LogX/Logx

a million thanks and God bless

Freiburger: https://www.nairaland.com/1834634/need-mathematics-questions

1.) Given 3^x-2^x=65, we use Modular arithmetic, working in Mod3 we get;
-2^x # -1 (Mod3)
Or
(-1)^x # 1 (Mod3) where # means "congruent to". The congruence holds iff x is even. Hence, x=2k, and we have
3^2k-2^2k=65
Or
(3^k-2^k)(3^k+2^k)=65*1=13*5. Now there are four possibilities;
(3^k+2^k)=65
(3^k-2^k)=1
Or
(3^k+2^k)=1
(3^k-2^k)=65
Or
(3^k+2^k)=5
(3^k-2^k)=13
Or
(3^k+2^k)=13
(3^k-2^k)=5

the first three eqns have no solution, adding or subtractin d last eqn gives k=2 and hence x=2k=2*2=4

Soneh: pls someone should help me with the solution and explanation to this question :
If. A and B are the roots of ax^2+bx+c=0 form the equation whose roots are: A-2B, B-2A
PLS note that A and B are alpha and beta respectively.

the roots of the given eqn are, A nd B,
then, sum of the root, A+B=-b/a,
Product, AB= c/a,

for the required eqn,
roots are (A-2B) and (B-2A),
sum = (A-2B)+(B-2A) = -(A+B) = -(-b/a) = b/a,
product = (A-2B)(B-2A) = AB-2(A²+B²)+4AB = 5AB - 2(A²+B²)
=9AB-2(A+B)²
=[s]9(c/a) -2(-b/a) = (9c+2b)/a..

But the quadratic equation is of the form,
x² - (sum of the roots)x + (product of the roots)= 0
using the values aboves,

the required eqn is therefore,
x² - b(x)/a + (9c+2b)/a = 0,
ax² - bx + (9c+2b) = 0[/s]

modifications.,
product = (A-2B)(B-2A) = AB-2(A²+B²)+4AB = 5AB - 2(A²+B²)
=9AB-2(A+B)²
=9(c/a) -2(-b/a)² = (9ac-2b²)/a²..

But the quadratic equation is of the form,
x² - (sum of the roots)x + (product of the roots)= 0
using the values aboves,

the required eqn is therefore,
x² - b(x)/a + (9ac-2b²)/a² = 0,
multiplying all by a²
(ax)² - abx + (9ac-2b²) = 0...QED
sorry for the previous mistake.. Tnx to oga benbuks

use first principle differentiate x^1/3
hence or otherwise compute f ' [ x^1/n ]

also by first principle

NB: use of binomials/series prohibited

asap.

thanks .

guys kindly give this a try plzzz..

Laplacian:
given; (x/2)^x=16=2⁴, by the unique factoriztaion theorem, (x/2) must be a power of two since the right hand side of the given eqn is a power of two. Hence, (x/2)=2^k for some integer k. Hence, 2^kx=2⁴ or kx=4 and there are only three possibilities; k=1, x=4 Or k=x=2 Or k=4, x=1. Testing each option in the given equation shows that only the first one is valid, Hence x=4

ok boss

alternatively

GIVEN (x/2)^x =16

=> x/2= (4²)<sup>1/x

put x/2 = t

= 1/t = 2/x (taking reciprocals )

=> t = (4)[sup]1/t</sup>

now we have

t^t = 2²

by definition,

=> t = 2

since x/2 = t

=> x/2 = 2

hence x = 4

dejt4u:
the roots of the given eqn are, A nd B,
then, sum of the root, A+B=-b/a,
Product, AB= c/a,

for the required eqn,
roots are (A-2B) and (B-2A),
sum = (A-2B)+(B-2A) = -(A+B) = -(-b/a) = b/a,
product = (A-2B)(B-2A) = AB-2(A²+B²)+4AB = 5AB - 2(A²+B²)
=9AB-2(A+B)²
=9(c/a) -2(-b/a) = (9c+2b)/a..

But the quadratic equation is of the form,
x² - (sum of the roots)x + (product of the roots)= 0
using the values aboves,

the required eqn is therefore,
x² - b(x)/a + (9c+2b)/a = 0,
ax² - bx + (9c+2b) = 0 ...QED

pls explain

Soneh:
pls explain

i tried as much as possible to it self explanatory.. Pls tell(quote the area or lines) where the problem is exactly nd i will explain better..

Soneh: pls someone should help me with the solution and explanation to this question :
If. A and B are the roots of ax^2+bx+C=o form the equation whose roots are: A-2B, B-2A
PLS note that A and B are alpha and beta respectively.

my boss @ dej4u has justice already...BUT....

given ax² +bx + c =0

by definition ,

the sum of roots , A+B = -b/a

the product of roots , AB =c/a

but here we are given new roots build a quadratic equation with ,

A-2B & B-2A

now for sum , we have (A-2B) +(B-2A) =A-2B+B-2A
= (A+B) -2(A+B)

recall that A+B =- b/a
=> -b/a -2(-b/a)
= (-b+2b)/a = b/a

& product , (A-2B)(B-2A) =AB-2A² -2B² +4AB

=>

5AB -2(A² + B²)

but A² +B² =(A+B)² -2AB

=> 5AB -2[ ( A+B)² -2AB ]

=> 5c/a -2[ (b/a)² -2c/a]

=> 5c/a -2b² /a² + 4c/a

=> 9c/a - 2(b/a)²

using the general model for building quadratic equations , given roots ,

x² - (sum of roots ) x + product of roots =0

=> x² -(b/a)x +[ 9c/a -2(b/a)² ] =0

simplify further …

hope you get .?

dejt4u:
the roots of the given eqn are, A nd B,
then, sum of the root, A+B=-b/a,
Product, AB= c/a,

for the required eqn,
roots are (A-2B) and (B-2A),
sum = (A-2B)+(B-2A) = -(A+B) = -(-b/a) = b/a,
product = (A-2B)(B-2A) = AB-2(A²+B²)+4AB = 5AB - 2(A²+B²)
=9AB-2(A+B)²
=[s]9(c/a) -2(-b/a) = (9c+2b)/a..

But the quadratic equation is of the form,
x² - (sum of the roots)x + (product of the roots)= 0
using the values aboves,

the required eqn is therefore,
x² - b(x)/a + (9c+2b)/a = 0,
ax² - bx + (9c+2b) = 0[/s]

modifications.,
product = (A-2B)(B-2A) = AB-2(A²+B²)+4AB = 5AB - 2(A²+B²)
=9AB-2(A+B)²
=9(c/a) -2(-b/a)² = (9ac-2b)/a²..

But the quadratic equation is of the form,
x² - (sum of the roots)x + (product of the roots)= 0
using the values aboves,

the required eqn is therefore,
x² - b(x)/a + (9ac-2b)/a² = 0,
multiplying all by a²
(ax)² - abx + (9ac-2b) = 0...QED
sorry for the previous mistake.. Tnx to oga benbuks

pls I thought the last part is supposed to be. (ax)^2-(ab)x+(9ac-2b^2)

benbuks:

my boss @ dej4u has justice already...BUT....

given ax² +bx + c =0

by definition ,

the sum of roots , A+B = -b/a

the product of roots , AB =c/a

but here we are given new roots build a quadratic equation with ,

A-2B & B-2A

now for sum , we have (A-2B) +(B-2A) =A-2B+B-2A
= (A+B) -2(A+B)

recall that A+B =- b/a
=> -b/a -2(-b/a)
= (-b+2b)/a = b/a

& product , (A-2B)(B-2A) =AB-2A² -2B² +4AB

=>

5AB -2(A² + B²)

but A² +B² =(A+B)² -2AB

=> 5AB -2[ ( A+B)² -2AB ]

=> 5c/a -2[ (b/a)² -2c/a]

=> 5c/a -2b² /a² + 4c/a

=> 9c/a - 2(b/a)²

using the general model for building quadratic equations , given roots ,

x² - (sum of roots ) x + product of roots =0

=> x² -(b/a)x +[ 9c/a -2(b/a)² ] =0

simplify further …

hope you get .?

(ax)^2-(ab)x+(9ac-2b^2)
hope I am right

Soneh:
pls I thought the last part is supposed to be. (ax)^2-(ab)x+(9ac-2b^2)

yea..you ar correct..

...Good morning to you all my lovely , wonderful , funny & ingenious mathematicians & other viewers/followers who learnt / i learnt lots from in the house , of which time & memory will fail me enumerating , however lets see how possible ..

Today being my 1 full year ( 12 months , 52 weeks ,365days ,8760hrs,560640mins ,33638400sec.) on our one & only NAIRALAND FORUM (1 of a thousand ) my NAIRALAND-birthday. (lol)

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great thanks to our ever innovative & patriotic
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Osewa & co .
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