Ortarico:

Hmmm, the answer will be an infinite number. You can put 1 million of the 4's in line, eg, 444444444444.... 44444 and it would be even. Or a billion. there is no limit.

Laplacian: ...d sum of two non-zero integers is equal to their product...find d numbers....

smurfy: @factorial1

We not only ask questions on this thread. We ATTEMPT to answer posted questions, too.
I'm yet to see you attempt any.

Sometimes, my solution can be completely wrong, but I trust Masters Laplacian, Maximus, Richiez, Jackpot, Ortarico, DoubleDx, Benbuks, etc. to put me through.

I must confess that I've learnt a lot on this thread. I now visit this thread more than I do Yahoo News, Punchng.com, BBC, CNN, etc.

I'd like to see you attempt a question or two soonest.

Lest I forget, thanks for your last set of questions. I'm eagerly following the solution provided by the masters.

factorial1: Oh! Sorry 4 dat, that's wat I actually wanted to du now(to ans. ur qt)
Note: I'm nt dah less busy, I'll try to be contributing too...

Laplacian: ...d sum of two non-zero integers is equal to their product...find d numbers....

smurfy: How many even numbers greater than 4000 can be formed using the digits 0, 1, 2, 3, 4, 5? [Note: A digit can be used only once.]

factorial1: The numbers are 2 and 2.

factorial1: Ok, here is the solution

If all six digits are used and the number is to be even, the last(right hand) digits must be either 0, 2 or 4 which gives 3 choices. When this's is done, there are 4 choices for the first digit, 3 for the 2nd, 2 for the third and 1 for the forth digit. So, we av 4 x 3 x 2 x 1 x3 = 62 even numbers using all five digits. Now, if all five digits are used and the number is to be even, the last digit cannot be 1, and either 0, 2 or 4 has already been used: therefore, der are 2 choices for the first, for the 2nd digits der are now 2 choices, 1 for d third n' 1 4 the forth= 2 x 2 x 1 x 1=4. Therefore the total num. of even numbers is 4+ 62= 66. Hope I'm correct sha...? coz the qt is somhw confusing to me seeing 0(zero) der

factorial1: Ok, here is the solution

If all six digits are used and the number is to be even, the last(right hand) digits must be either 0, 2 or 4 which gives 3 choices. When this's is done, there are 4 choices for the first digit, 3 for the 2nd, 2 for the third and 1 for the forth digit. So, we av 4 x 3 x 2 x 1 x3 = 62 even numbers using all five digits. Now, if all five digits are used and the number is to be even, the last digit cannot be 1, and either 0, 2 or 4 has already been used: therefore, der are 2 choices for the first, for the 2nd digits der are now 2 choices, 1 for d third n' 1 4 the forth= 2 x 2 x 1 x 1=4. Therefore the total num. of even numbers is 4+ 62= 66. Hope I'm correct sha...? coz the qt is somhw confusing to me seeing 0(zero) der

factorial1: Play with the questions below:
1. If the radius of a sphere is increased from 10cm to 10.1cm, what is the approximate increase in the surface area?

2. The height of a cylinder is 10cm and its radius is 4cm. Find the approximate increase in volume when the radius increases to 4.02cm

3. An error of 3% is made in measuring the radius of a sphere. Find the percentage error in volume?

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

5. If x=tany; find dy/dx?

Laplacian:
dats d correct answer but i want 2 c ur proof...

Calculusf(x):
hmmm ,for numbers greater than 4000(it can be 4,5 or 6 digits)if 4digits,we can see 4 and 5,and for the number to end with even,any of 0,2 or 4 must end the number,also note that the number must not be repeated,then we do it like this,make hole,4 for four digits,5 for five digits etc
4 DIGITS STARTING WITH 5
it can end with 0,2 or 4,then the fourth hole will have 3,it starts with 5,the first hole with have 1,don't forget there are six numbers in all,the last and the first are known therefore it gives 1*4*3*3=36
4 DIGITS STARTING WITH 4
since repeatition is not allowed,then it can end with 0 or 2,then the holes will have 1*4*3*2=24
total=36+24=60
5 DIGITS
it can end with 0,2 or 4.but note that 0 cannot be in the first hole so as not to make the number 4digits again(then only 4 digits can occupy the first hole)...then the holes are 4*4*3*2*3=288...
6 DIGITS
Same as 5 digits...it can end with 0,2 or 4,then 3 can occupy the last hole,and can start with 4 in the first hole coz 0 cannot start the digit...then 4*4*3*2*1*3=288...
TOTAL=288+288+60=636ways

Laplacian: ...d sum of two non-zero integers is equal to their product...find d numbers....

factorial1: *clears throat*
let's go,
firstly, let the two numbers be 'x' and 'y'. From the qt, it can be infered dhat the sum of this non-zero integers is equal to their product: the mathematical representation is x + y = xy.....(i)
Now, from equation (i), make x the subject of the relation/formulae, the result is x = xy - y.....(ii), also, make y the subject of the relation from eqn (i), the result is y = xy - x....(iii). Therefore, add eqn (ii) and (iii)...(i.e x + y) together and equate it with their product(i.e xy), x + y = xy, xy - x + xy - xy = xy, which is equal to 2xy - 2x = xy, collecting the like terms, we have 2xy - xy = 2x, xy(2 - 1) = 2x, therefore xy = 2x, divide both side by x, we have xy/x = 2x=x, from there....y = 2. So, subitutute y=2 into any of the above equation. I use eqn (i) which is x + y = xy, x + 2 = 2x, collecting the like terms x - 2x = -2, -x = -2. Note: minus cancelled out, therefor x = 2. With this...I think u are satisfied sir? The two non-zero integers in which their sum is equal to their product is/are ''2'' and ''2''. Thanks

Dane17: . x+y=xy ; x=xy-y ; x=y(x-1) ; y=x/x-1 ; therefore if x=2 ; y= 2/1 ; if x=3 ;y=3/2 if x=4 ; y=4/3 and so on.

smurfy: How many even numbers greater than 4000 can be formed using the digits 0, 1, 2, 3, 4, 5? [Note: A digit can be used only once.]

echibuzor:
**Clears throat**
Your head is there, but ehm... that bolded part should be 2xy-x-y

factorial1: Brov, u aint told to do try and error, laplacian asked for a proof, and I've done that. #Cheers

echibuzor:
All these questions are from one problem. Linear Approximation.. I ll tackle them later, I am on my way to work..
*****************************************************
1. The Surface Area of the sphere at correct measurement is 4*^r² = 4*3.142*100 = 8800/7 = 1257.14
Surface Area with error input 4*3.142*10.1² = 1281.90
Approximate increase = 1281.90 - 1257.14 = 24.76.
Now, if we are dealing with a complex function. You go by the linear approximation technique, the derivative of the surface area(without error) multiplied by the error in measurement. i.e d/dx (4^r²) = 8^r = 8 * 3.142 * 10 => 251.43 multiplied by (10.1 - 10);
25.143
(2, 3) Apply the second approach and you should get your answers faster...
Its pretty straight-forward though...

Dane17: . In solving ur question do I assume that 023514 is different from 23514

factorial1: oops, could'nt notice that though, guess I've to think of another method to use. thanks brov

Dane17: . u manipulated . I don't get dis part (Therefore, add eqn (ii) and
(iii)...(i.e x + y) together and equate it
with their product(i.e xy), x + y = xy, xy - x
+ xy - x y = xy, which is equal to 2xy - 2x
= xy,). how did u arrive at 2xy-2x=xy. Finally d answer to his question is not just 2 and 2 . they are many values like 3 and 1.5 which gives 4.5 . look at my previous solving carefully and u will understand

factorial1: oops, could'nt notice that though, guess I've to think of another method to use. thanks brov

factorial1: *clears throat*
let's go,
firstly, let the two numbers be 'x' and 'y'. From the qt, it can be infered dhat the sum of this non-zero integers is equal to their product: the mathematical representation is x + y = xy.....(i)
Now, from equation (i), make x the subject of the relation/formulae, the result is x = xy - y.....(ii), also, make y the subject of the relation from eqn (i), the result is y = xy - x....(iii). Therefore, add eqn (ii) and (iii)...(i.e x + y) together and equate it with their product(i.e xy), x + y = xy, xy - x + xy - x = xy, which is equal to 2xy - 2x = xy, collecting the like terms, we have 2xy - xy = 2x, xy(2 - 1) = 2x, therefore xy = 2x, divide both side by x, we have xy/x = 2x=x, from there....y = 2. So, subitutute y=2 into any of the above equation. I use eqn (i) which is x + y = xy, x + 2 = 2x, collecting the like terms x - 2x = -2, -x = -2. Note: minus cancelled out, therefor x = 2. With this...I think u are satisfied sir? The two non-zero integers in which their sum is equal to their product is/are ''2'' and ''2''. Thanks

Dane17: . x+y=xy ; x=xy-y ; x=y(x-1) ; y=x/x-1 ; therefore if x=2 ; y= 2/1 ; if x=3 ;y=3/2 if x=4 ; y=4/3 and so on.

Dane17: . x+y=xy ; x=xy-y ; x=y(x-1) ; y=x/x-1 ; therefore if x=2 ; y= 2/1 ; if x=3 ;y=3/2 if x=4 ; y=4/3 and so on.

echibuzor: A population of Ikamudu (yoruba : an insect that smells horribly when you step on it) exhibits logistic growth. If the carrying
capacity is 500 butterflies and r = 0.1 individuals/(individuals x month), what is the
maximum population growth rate for the population? (Hint: maximum population growth
rate occurs when N = K/2).

smurfy:

The numbers are integers, so only 2 and 2 qualify as answers.