godofphysics: hmmn! Daniel17 has actually killed d questn in a simple but logical way!

smurfy:

Solve for all odd numbers? Okay.

Here goes...

4-DIGIT NUMBERS

The last digit must be 1, 3 or 5. The first digit must be 4 or 5.
Six scenarios!

(a) first digit is 4, last digit is 1
(b) first digit is 4, last digit is 3
(c) first digit is 4, last digit is 5
(d) first digit 5, last digit 1
(e) first digit 5, last digit 3
(f) first digit 5, last digit...

Ha! We can't use 5 twice, so scenario (e) doesn't count. That means we'll only consider scenarios (a) to (e) only.

(a) gives 1*4*3*1 = 12
(b) gives 1*4*3*1 = 12
(c) gives 1*4*3*1 = 12
(d) gives 1*4*3*1 = 12
(e) gives 1*4*3*1 = 12.

Adding the scenarios gives 5*12 = 60 ways of forming odd numbers greater than 4000 from the digits 0 through 5.

5-DIGIT NUMBERS

We have three scenarios here since we don't have to bother too much about the first digit. We only ensure the digit zero doesn't come first.

(a) ends with digit 1 gives 4*4*3*2*1=96
(b) ends with digit 3 gives 4*4*3*2*1=96
(c) ends with digit 5 gives 4*4*3*2*1=96.

So there are 3*96 = 288 ways of forming odd 5-digit numbers from the digits 0 through 5.

6-DIGIT NUMBERS

Three scenarios. Let me skip a step or two...

(a) 4*4*3*2*1*1 = 96
(b) 4*4*3*2*1*1 = 96
(c) 4*4*3*2*1*1 = 96

So, there are 3*96 = 288 ways of getting 6-digit odd numbers from the digits 0 through 5.

Final answer...

The number of odd numbers greater than 4000 that can be formed from the digits 0, 1, 2, 3, 4, 5 is 60 + 288 + 288 = 636.

Hey, we obtained 684 for even and 636 for odd numbers. Why do we have less ways of forming odd numbers than even numbers?

It's partly because of the silly restriction placed on 4-digit numbers. It has cost the ODD NUMBERS team a whooping 48 points!

echibuzor:
The man said non-zero

echibuzor:

2. Volume of cylinder = ^r²h, d/dx (^r²h) = 2^rh
put in the correct measurement into this to get 80^
Multiply the result above with (4.02 - 4.00) => 0.02
Approximate increase = 0.02 * 80^ = 5.02655 cm

smurfy:

Solve for all odd numbers? Okay.

Here goes...

4-DIGIT NUMBERS

The last digit must be 1, 3 or 5. The first digit must be 4 or 5.
Six scenarios!

(a) first digit is 4, last digit is 1
(b) first digit is 4, last digit is 3
(c) first digit is 4, last digit is 5
(d) first digit 5, last digit 1
(e) first digit 5, last digit 3
(f) first digit 5, last digit...

Ha! We can't use 5 twice, so scenario (e) doesn't count. That means we'll only consider scenarios (a) to (e) only.

(a) gives 1*4*3*1 = 12
(b) gives 1*4*3*1 = 12
(c) gives 1*4*3*1 = 12
(d) gives 1*4*3*1 = 12
(e) gives 1*4*3*1 = 12.

Adding the scenarios gives 5*12 = 60 ways of forming odd numbers greater than 4000 from the digits 0 through 5.

5-DIGIT NUMBERS

We have three scenarios here since we don't have to bother too much about the first digit. We only ensure the digit zero doesn't come first.

(a) ends with digit 1 gives 4*4*3*2*1=96
(b) ends with digit 3 gives 4*4*3*2*1=96
(c) ends with digit 5 gives 4*4*3*2*1=96.

So there are 3*96 = 288 ways of forming odd 5-digit numbers from the digits 0 through 5.

6-DIGIT NUMBERS

Three scenarios. Let me skip a step or two...

(a) 4*4*3*2*1*1 = 96
(b) 4*4*3*2*1*1 = 96
(c) 4*4*3*2*1*1 = 96

So, there are 3*96 = 288 ways of getting 6-digit odd numbers from the digits 0 through 5.

Final answer...

The number of odd numbers greater than 4000 that can be formed from the digits 0, 1, 2, 3, 4, 5 is 60 + 288 + 288 = 636.

Hey, we obtained 684 for even and 636 for odd numbers. Why do we have less ways of forming odd numbers than even numbers?

It's partly because of the silly restriction placed on 4-digit numbers. It has cost the ODD NUMBERS team a whooping 48 points!

godofphysics: pls who solved d questn so dat i can view d person's posts. It wil save me d tyme of going 2ru all d pages. Thanks....
(I'm stil sweating wit d last few pages coz it's like i've not rly understood some of d solutions)

factorial1: Yup!

rhydex 247:
the soln is done by me. U can also view Master Richiez post.

godofphysics: tanx boss! Wen i grow up, i want 2 b lk u and smurfy...................next questn pls! Na me wan solve am!

rhydex 247: @ lanrexlan questn.
x+y=5... (1) nd x^x+y^x=13...(2)
Solutn.
frm eqn 1. x=5-y. Put x=5-y in eqn 2. We av
(5-y)^(5-y)+y^(5-y)=13... eqn * Nw we av reduced d problem to what value of y that must be put in eqn * to obtain 13. Hmmm. A thorough look at eqn 1 nd eqn 2 show dat d values of x nd y must b positive integers within the range 0<x<5 nd 0<y<5.
clearly, y=3. i.e
(5-3)^(5-3)+3^(5-3)=13. Nw dat we av y=3. It is more convenient to get x.
recal dat x=5-y. x=5-3. x=2.
hence d soln is x=2 nd y=3. All is well. @ richiez u knw wat i mean.

mastermynd:
So na here u dey hide abiiiii
#drags him by d hand#
Oya lets go home,,,,

godofphysics: #shrugs him off#dis is my new found home, BTW, hw do u knw i'm here?

godofphysics: #shrugs him off#dis is my new found home, BTW, hw do u knw i'm here?

factorial1: 4. The side of a square is 5cm. How much will the area of the square increase when the side expands by 0.01cm?

godofphysics: tanx boss! Wen i grow up, i want 2 b lk u and smurfy...................next questn pls! Na me wan solve am!

mastermynd:
#dragging him#
U dey stiff muscle abi
U knw whu to call

rhydex 247: Question 1.
Show that the four points whose position vectors are 3i-2j+4k, 6i+3j+k, 5i+7j+3k and 2i+2j+6k are coplanar.
Question 2.
If a ^ r=b+¥a and a.r=3 where a=2i+j-k and b=-i-2j+k. Then find r and ¥.
Question 3.
Prove the identity. (a ^ b).(c ^d)=(a.c)(b.d)-(a.d)(b.c). Where ^ means cap or cross

Ezechinwa: guys pls i need a mathematician to help me integrate this function comprehensively....

2x Sin (1/x) - Cos (1/ x)........


pls guys help me check that out !!!!!

Ezechinwa: guys pls i need a mathematician to help me integrate this function comprehensively....

2x Sin (1/x) - Cos (1/ x)........


pls guys help me check that out !!!!!

lavosier:
$2xsin(1/x)-$cos(1/x).,
Let's do d frst integral..
$2xsin(1/x)..let dv=2x,v=x^2,u=sin(1/x),du=cos(1/x).-1/x^2..
By part we av
X^2sin(1/x)-$x^2.Cos(1/x)-1/x^2
=X^2sin(1/x)+$cos(1/x)..
Remember ou first equ

$2xsin(1/x)-$cos(1/x)
We now av
X^2sin(1/x)+$cos(1/x)-$cos(1/x)
=x^2sin(1/x)...op em rite sha.....its so Gud 2 b back on nairaland...I grit all d profs in d ouse..

Ezechinwa:



guy..... may my God bless you....
may he give you strength to study more....
you have save me alot by solving this problem....

i have been trying to solve this problem since Oct lol....now i realise my mistake....
thankyou......

rhydex 247: una dey decline my question @ my maths general.

Laplacian:
2^2x=8x, or 2^(2x-3)=x or

2^(2x-3)/4^(2x-3)=x/4^(2x-3)

(1/2)^(2x-3)=x*(1/4)^(2x-3)

(0.5)^(2x-3)=x*(0.25)^(2x-3)

(1-0.5)^(2x-3)=x*(1-0.75)^(2x-3)

by binomial approx.

1-(2x-3)*0.5...=

x*[1-(2x-3)*0.75+...]

re-arrangin gives
1-x+3/2=x-3x/4*(2x-3)
or
4-4x+6=4x-6x^2+9x
or
6x^2-17x+6=0

for d second question;
(4/x)^x=2x or 4^x=2x^(x+1) or

2^(2x-1)=x^(x+1)...i hope u can go on from there?

benbuks: I salute my oga..@DX ..UR boy dey greet o.

doubleDx:

Greetings Chief benbuks! It's been a while, hope you are doing great? Lemme warm up with one of your questions ...may post later as I'm on mobile now! 1luv