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#101 Help Me ! » completed equation now what » 2006-11-24 06:20:54
- Kiran
- Replies: 11
find the distance between the point (5,2) and the line 3x - 4y = 2
y = 3x + b
2 = 3(5) + b
2 = 15 + b
b = -13
y = 3x - 13 < completed equation
now what?
#102 Re: Dark Discussions at Cafe Infinity » Who is the Fan Of Cricket?? » 2006-11-24 03:28:41
hmm
i so dont understand circket at all.
confusing game
sixer.....and other words such as that seems to go from one ear and out the other ear lol
#103 Re: Help Me ! » ok another important question » 2006-11-24 03:22:45
ok now as i have it...
y = 1/3 - 5/6
i did what you said.
1/3x = y + 5/6
1/3= x(y + 5/6)
x = [1/3]/(y + 5/6)
now i think maybe you meant by this:
x = 0.3 / (y + 0.83)
correct?
#104 Re: Introductions » hello » 2006-11-24 03:05:55
Thanks Coolcat and Wizard 
#105 Help Me ! » correct the way i done it » 2006-11-22 04:22:38
- Kiran
- Replies: 2
factor x^2 - 3x + 4 over the set of complex numbers.
i did
the quadratic way
by getting
x = -3 +- sqrt(-7) / 2
and then
x = -1.5 +- sqrt(-3.5)
now what?
#106 Re: Help Me ! » important question » 2006-11-22 03:58:30
ok i got up tp
x = -3 +- sqrt-103 / 8
now what
#107 Re: Help Me ! » important question » 2006-11-22 03:46:05
ok thanks
#108 Re: Introductions » hello » 2006-11-22 03:42:32
hi thanks Haruka for the welcome
#109 Re: Help Me ! » important question » 2006-11-22 03:38:47
ok so in my problem
4x^2 - 3x - 7 = 0
i have ot put it like so >
x = -3 +- sqrt(3^2 - 4(4)(7)) / 2(4)
#110 Help Me ! » ok another important question » 2006-11-22 03:35:11
- Kiran
- Replies: 3
if they ask to find teh inverse function
and the eqaution looks like
y = number/numberX - number/number
now in my book..tehy are saying
a original function looks like this > y = 2x - 1
solved for x looks like this > x = 1/2y + 1/2
inverse function looks liek this > y = 1/2x + 1/2
now the problem that we have looks like an inverse function....
so now what?
so turn that problem around for it to look like :
x = number/numberY - number/number
right?
#111 Help Me ! » important question » 2006-11-22 03:28:43
- Kiran
- Replies: 7
ok
if teh question asks to complete the sqaure right
and the eqaution looks like
numberX^2 - numberX - number = number or you can say zero
right
so in this eqaution do you use
quadratic formula?
it seems like you do!
#112 Re: Help Me ! » anti log????? » 2006-11-22 03:24:32
ok thank you .... but as i check around in google.... they showed me a way to right anti log on the calculator how you do that?
my calculator is a bit different but anyways do tell me how you put that antilog in it
thanks
#113 Help Me ! » anti log????? » 2006-11-21 13:19:31
- Kiran
- Replies: 5
anti log3 4
ok anti log how you put that in calculator?
#114 Help Me ! » same as unique » 2006-11-21 04:37:54
- Kiran
- Replies: 3
ok like the question unique had i am almost the same question did i do it right ?
solve :
4sqrt (x + 2) = 3
so
x + 2 = 81
subtract 2 from 81
x = 79
#115 Help Me ! » volume » 2006-11-21 04:36:02
- Kiran
- Replies: 1
find the volume of a right circular cone with a base radius of 3 meters and a slant height of 5 meters.
V = 1/3 ( area of base) (h)
radius = 3m
diameter = 9m = base
v = 1/3 (9)(5)
v = 1/3 ( 45)
v = 15
correct
#116 Help Me ! » Next » 2006-11-20 04:37:56
- Kiran
- Replies: 2
solve for x
log4 (x - 1) + log4 (x + 2) = 1
log4 (x - 1) (x + 2) = 1
(x - 1)(x + 2) = 4^1
now what?
#117 Re: Help Me ! » messed up do help me out » 2006-11-20 04:18:39
thanks ALOT for your help i so understood now
#118 Help Me ! » messed up do help me out » 2006-11-20 03:56:34
- Kiran
- Replies: 3
The vertices of triangle JKl are J(0,0) , K(3,-1) , L(-8,-7)
find the coordinates of the vertices after a reflection over the given axis : yaxis
i got wrong answer ..half answer
i wrote :
J = (0,0)
K = (-3,1)
L = (8,7)
HELP
please
#119 Help Me ! » messed up do help me out » 2006-11-20 03:53:42
- Kiran
- Replies: 2
Triangle LMN has vertices L(-6,5) , M(-4,-3), N(-2,4).
Find the coordinates of the vertices of its image after it is reflected over the x - axis and then translated by (-3,0)
ok my answers came out ot be which came out to be half wrong so help 
thanks
L = (-9,5)
M = (-7,-3)
N = (-5,4)
#120 Help Me ! » hi » 2006-11-16 06:39:40
- Kiran
- Replies: 1
find the volume , to the nearest tenth , of a pyramid that is 6 feet tall and whose base is an equilateral triangle with sides each 10 feet long.
i did
v = 1/3bh
i got
60 feet...help
i got this answer wrong
#121 Re: Help Me ! » simplify » 2006-11-16 06:28:06
what do you mean take teh second power away form the first one?
i got
for top one > y^5/4
for bottom one > y^-6
now i have to divide i got -5/24
#122 Re: Help Me ! » hi » 2006-11-15 10:33:14
hey cool thanks
that wasnt that hard
#123 Re: Help Me ! » simplify » 2006-11-15 10:08:06
ok thanks for the help now you are asking me to do teh y terms... is this what i did look right?
y * y^3/4 y^3/4
-------------- = -------- divide = y^-1/8 correct?
y^3 * y^-2 y^-6
#124 Help Me ! » simplify » 2006-11-15 07:02:47
- Kiran
- Replies: 9
#125 Re: Introductions » hello » 2006-11-15 06:32:26
Welcome! Just, don't steal Coolcat23's avatar without first asking.
I am doing particularly fine. You will find a lot of mathematical help here.
hello devante and Neha thank you for the welcome
oh i am sorry i didnt ask. i didn't know that i was suppost to ask anyways my bad
ok thats great . i need alot of help thanks