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#1 Help Me ! » Expandiing » 2007-02-20 06:13:30

Kiran
Replies: 1

ab/m ( -2m^(-2) / ba  +  3m / a^(-1) b  )

than

ab/m (-2/abm^(2) +  3am/b) < handle negative exponents
ab/m * -2/abm^2 + ab/m * 3am/b < distribute
now i have to do......

#2 Help Me ! » Construct » 2007-01-17 03:39:33

Kiran
Replies: 2

construct the perpendicular bisector to segment OK.


__|_________________________________|__
   O                                                         K

#3 Help Me ! » Point » 2007-01-12 06:28:57

Kiran
Replies: 2

find the coordinates of the point that lies two thirds f the way between (1,3) and (7,6)

#4 Help Me ! » evaluating » 2007-01-10 07:32:17

Kiran
Replies: 1

csc^2  405degrees - tan45degrees

#5 Help Me ! » X » 2007-01-09 04:32:59

Kiran
Replies: 1

find x:

|\
|  \
|    \
|      \
|        \
|          \  7
|            \
|              \
|                \
|                  \
|                    \
|                      \
--------------------- 40 degrees
           x

#6 Help Me ! » following equation » 2007-01-08 03:42:50

Kiran
Replies: 1

given that 0 degrees <_ x < 360 degrees , solve the following equation :

cos x - sqrt(1 - cos^2 x) = 0

#7 Re: Help Me ! » Equation of the line » 2007-01-04 10:03:22

i get it ....
m = (-2 - 3) / (0 - 4) = 1.25
y - y1 = m (x - x1)
y - 3 = 1.25 ( x -(-2))
y - 3 = 1.25 ( x + 2)
y = 1.25x + 2.5 + 3
y = 1.25x + 5.5 this is the equation!

#8 Help Me ! » Equation of the line » 2007-01-04 09:48:08

Kiran
Replies: 3

write the equation of the line that is equidistant from the points (4,3) and (0,-2).

#9 Re: Help Me ! » Terms » 2007-01-04 09:44:40

ok so look at what i did:

(x+y)^7
1st term > 1(x)^7
2nd term > 7(x)^6 . (y)^1
3rd term > 14(x)^5 . (y)^2

thats all i need right?

#10 Re: Help Me ! » Terms » 2007-01-04 05:01:28

http://www.algebralab.org/lessons/lesson.aspx?file=Algebra_BinomialExpansion.xml

look at the practice at the bottom of the page.
i have to do it like that?

#11 Re: Help Me ! » Correct>? » 2007-01-03 10:57:09

i have to graph it....can you help me on the graphing

#12 Help Me ! » Terms » 2007-01-03 10:49:48

Kiran
Replies: 6

write the third term of (x + z)^7 .

#13 Re: Jokes » ha ha ha » 2007-01-03 06:39:02

Joke is great.... lol vvvvvv good smile

#15 Re: Dark Discussions at Cafe Infinity » My avi » 2007-01-03 06:35:40

yeah i sooooooo like pics of kittens smile

#16 Help Me ! » Correct>? » 2007-01-02 10:46:36

Kiran
Replies: 1

if a certain parabola has its vertex at (2,-1) and its focus at (2,1), write its equation and graph.

(h,k) = (2,-1)
h = 2  k = -1
focus (h,k + p) is (2,1) so k + p = 1
k = -1  p = 2

y - k = 1/4p( x - h)^2  < equation
y - (-1) = 1/4(2)  (x - (2))^2  < substituted
y = 1/8 ( x - 2)^2 - 1 < solved

#17 Help Me ! » conic section » 2007-01-02 09:35:18

Kiran
Replies: 2

tell what conic section the following equation represents:

x^2 - 4y^2 = 4

#19 Re: Dark Discussions at Cafe Infinity » My avi » 2007-01-02 03:40:58

no problem suha smile

devante last day of christmas on january 5th..i didnt hear that before?

#20 Re: Dark Discussions at Cafe Infinity » Happy New Year! » 2007-01-02 03:39:37

HAPPY NEW YEAR TOO ALLL
May this year bring you all happiness and joy and may all your wishes and dreams come true inshallah smile All the best
how was your new years day? what you all did? i saw the ball drop in times square on tv....not much called my family members/relatives.....

#21 Re: Help Me ! » Equation solving wht next? » 2007-01-02 03:34:39

soooooooo by using the reference i did:
9x^2 + 3x - 8 = 0
x^2 + 3/9x - 8/9 = 0
x^2 + 3/9x = 8/9
x^2 + 3/9x + (3/81)^2 = 8/9 + (3/81)^2
(x + 3/81)^2 = 8/9 + (3/81)^2
(x + 1/27)^2= 8/9 + 9/6561 = 649/729
(x + 1/27)^2 - 649/729 = 0
27(x + 1/27)^2 - 649/729 = 0
27(x + 1/27)^2 = 649/27
(x + 1/27)^2 = 649/729
x + 1/27 = +- sqrt(649/729)
x = 1/27 +- sqrt(649/729)
x = -703.5 ... and -0.906....

correct

#22 Re: Help Me ! » Equation solving wht next? » 2006-12-28 07:58:28

they need us to complete the square

#23 Re: Guestbook » how im i going to cheat on my sats » 2006-12-28 05:11:27

guys how can u cheat on ur sats ? is that even possible keya?
i dont think so

#24 Re: Dark Discussions at Cafe Infinity » My avi » 2006-12-28 05:10:16

http://www.tafreeh.net/picgalleries.php?cat=Cute&subcat=Kitties&pg=1

or this one...u can click for other pictures besides kitties..and just save them on ur desktop..and load them at this site

#25 Re: Dark Discussions at Cafe Infinity » My avi » 2006-12-28 05:09:06

http://www.avatarmaker.in/free-avatar/

check this site out suha

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