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#1 Re: Help Me ! » Max and Min Problem » 2008-08-04 09:36:04

wooooooow

good job,, JaneFairfax

,, thank you very much

your solution is perfectly true
,,

thanks again

#2 Help Me ! » Max and Min Problem » 2008-08-03 21:08:11

saudi_boy
Replies: 5

Hey everybody,,,

can u help me with this question??

Find three positive numbers whose sum is 27 and such that the
sum of their squares is as small as possible.??

I think we need to use Partial Derivatives..

I really need the solution..

thanks..

#3 Re: Help Me ! » integration question » 2008-06-12 12:36:48

hiiiiii JaneFairfax ,,,

thank thank thank you very much

you are really intelligent

but can you explain to me the first part    I understand the second part of integration

but why you take the limit   
we know that the integration at zero is undefined..

maybe the question is strange  but I would like to know the exact meaning...

thanks again

#4 Help Me ! » integration question » 2008-06-12 11:10:55

saudi_boy
Replies: 2

hii all,,,

how r u??

My friend gave me a integral to solve

and I try but I couldn't,, I think that the question is wrong !!!

I think this integral is improper

see it
and try to help me to solve it

#5 Re: Help Me ! » How to do this integral? » 2008-03-30 06:31:19

yes the final solution , in my opinion, 1/4

you need to use integration by parts

mikau wrote:

lol! has everyone gone nuts? Double integrals?

Let u = ln(x + 1)

let dv = x dx

its easy as pi! smile

then use substitution with (x+1)=u   in second term of integral   ( du*v)


if you don't understand yet

i can help you more

tell me my brother if you get

#6 Re: Help Me ! » trigonometric identities..help!! » 2008-01-26 11:50:40

1.  [sin(x-y)] / [cos(x+y)] = [tan(x)-tan(y)] / [1-tan(x)tan(y)]

Left side


now divide {numerator}{denominator} by cos(y)cos(x)

#7 Re: Help Me ! » how to solve this » 2008-01-26 11:39:17

first simplify this expression



by usining division by -1 you will find f(x) = 0
thats means -1 is the first zero

you will get


divide by -1 again

and you can complete the answer

#8 Re: Help Me ! » Help with some calc » 2007-12-15 07:35:30

this limit is one of famous limits

you can solve this limit by l'Hopital's rule(if you don't know this rule tell me to explain it).

The limit of X --> 0 degrees   of (Sin x)/X = 1

The limit of X --> 0 degrees   of (tan x)/X = 1

The limit of X --> 0 degrees   of (1-cosx)/X = 0

#10 Re: Help Me ! » Trig Limit » 2007-10-09 21:42:27

Examples

1.



apply L’Hòpital’s rule.

2. note:Applying the rule a single time still results in an indeterminate form. In this case, the limit may be evaluated by applying l'Hôpital's rule three times:


#11 Re: Help Me ! » simplifying expressions » 2007-10-05 23:01:35

apply

we get


then simplify

apply

we'll get



finally

#12 Re: Help Me ! » Another quadratic prob. » 2007-05-10 06:08:18

This is another example

it help you
to understan


iBD20406.jpg

#13 Re: Help Me ! » Quartic!? » 2007-05-10 05:55:42

First we use synthetic division..

9OX19029.jpg


then we also use the same mathod with another number  -2
to solve

4QW19118.jpg


now we can factor

   or use the quadratic equation to solve


the solution set (3.-2.-1.4)

#14 Re: Help Me ! » Help with Math » 2007-03-05 06:25:00

hiii    Joey

It's easy

and I'll try to explain my solution

firstly

subtract 5 from each side

3x + 5 - 5 = 8 - 5

3x= 3

then
divide each side by 3

x = 1

finally

thank you for visit

#15 Re: Help Me ! » yes no » 2006-11-07 05:49:51

Toast

yes ,
I drew that's in paint,, big_smile
it was hard work,,:|

because I didn't knew anything about topic
smile:)

thank you very much,, upup

#17 Re: Help Me ! » yes no » 2006-11-06 05:25:00

hiii
unique

good work..

but there is a mistake..

see to my solution

9b95ee07c1.jpg

#18 Re: Help Me ! » help me please » 2006-11-02 19:41:05

thank you ALL
big_smile:D:cool::cool:

#19 Re: Help Me ! » help me please » 2006-11-02 18:00:41

hmmmmmmmmmm,,

What multipied by 2^x-1 equals 2^3x-2 ?.

(2^x-1)(2^2x-1)=2^3x-2

2(2^x-1)^3=2^3x-2

= 2y^3

Is that a correct answer ?

#20 Re: Help Me ! » factor » 2006-11-02 06:25:48

pi man,,
smile
I forgot that.

I'm sorry.

thank you for advice


RauLiTo

big_smile that's good...


nice to meet you here ..

big_smile:D
your brother
saudi_boy

#21 Re: Exercises » Simplifying Simple i Terms (Intermediate) » 2006-11-02 05:47:18

Devanté wrote:

Simplifying Simple i Terms - Intermediate Level

Express in simplest terms, in terms of i.

1. i[sup]1[/sup]
2. i[sup]3[/sup]
3. i[sup]0[/sup]
4. i[sup]19[/sup]
5. (i[sup]12[/sup])[sup]15[/sup]
6. i[sup]16[/sup] + i[sup]11[/sup]
7. i[sup]18[/sup] + i[sup]13[/sup]
8. (i[sup]6[/sup])[sup]30[/sup]
9. i[sup]25[/sup] - i[sup]20[/sup]
10. i[sup]29[/sup] × i[sup]14[/sup]
11. i[sup]23[/sup]
12. i[sup]36[/sup] ÷ i[sup]32[/sup]
13. i[sup]37[/sup] - i[sup]2[/sup]
14. (i[sup]28[/sup])[sup]33[/sup]
15. i[sup]8[/sup]
16. i[sup]34[/sup] × i[sup]44[/sup]
17. i[sup]50[/sup] + i[sup]40[/sup]
18. i[sup]49[/sup] ÷ i[sup]17[/sup]

-------------------

Answer,,

1.       i

2.       -i

3.       i^o=  1

4.       i^19 = i^3    =    -i

5.       1

6.       1 - i

7.        i^18 + i^13  =  i^2 + i^1    =   -1 +i

8.        1

9.         i - 1

10.        i^29 + i^14 =    i^1 + i^2 =      -i

11.         i^23 = i^3     = -i

12.         1

13.          i + 1

14.           1

15.            1

16.            -1

17.             0

18.            1


smile:)

#24 Re: Help Me ! » factor » 2006-11-01 22:08:47

77f11901e7.jpg


Is that a correct answer?

#25 Re: Help Me ! » help me please » 2006-11-01 20:19:35

pi man

thank U very much,,,

polylog

yessssss,,

that's good work.

thank you veeeery much..

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