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#26 Help Me ! » Traingle » 2006-12-14 06:28:35
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- Replies: 4
Solve the triangle below for angle x and then compute the area of the triangle.
*
* *
* *
* *
3in * * 5in
* *
* *
* *
* *
* 75dg xdg *
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#27 Help Me ! » equation of line » 2006-12-14 06:09:02
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- Replies: 1
Find the equation of the line with a slope of -3/4 and a y-intercept of -3
i did:
slope intercept equation of the line whose slope is -3/4 and whose y intercept is -3 is
y = -3/4x +(-3)
#28 Help Me ! » sketch the graph » 2006-12-04 08:37:17
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- Replies: 2
sketch the graph of y = -1 + 4sin 1/2((theta) + 30degrees)
ummmmm help 
#29 Re: Help Me ! » equation » 2006-11-29 03:38:32
sorry guys i messed up i had ot write
following equation given that 0 <_theta < 2pi :
tan2(theta) - sqrt(3) = 0
#30 Re: Dark Discussions at Cafe Infinity » Who is the Fan Of Cricket?? » 2006-11-28 05:07:07
i think now mostly almost all games / sports women are joining in with men am i right?
like for boxing and such kinds of sports i think that is't a sport for women to join in. only for men. women just join in so they can show off men that they can do what ever that you men do . am i right. its not compition right...that's how life is. somethings...some sports men can do which women cant and some women can do which men cant thats it
am i right.
i want to learn cricket ..... that would be so awesome if i threw a sixer on my first lesson ...lol in my dreams right
#31 Help Me ! » equation » 2006-11-28 04:41:40
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- Replies: 4
following equation given that 0 <_? < 2pi :
tan2? - sqrt(3) = 0
hi guys help me on this thanks
#32 Help Me ! » central angle » 2006-11-22 04:40:44
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- Replies: 1
the central angle of a sector of a circle is 0.6 radian.
find the area of the sector if the radius is 4 meters.
i did:
0.6 = 0.6 / 360 * pi(4)^2 = 0.026666666
correct?
#33 Re: Help Me ! » can u help me solve this » 2006-11-21 03:15:20
yup....in steps it looks like:
?(2x+3) = 3
=> (?(2x+3))² = 3²
=> 2x+3 = 9
=> 2x = 9-3
=> 2x = 6
=> x= (6/2)
=> x=3that would be if you have to all your steps.....well im sure you understand it now
you know when you put
?(2x+3) = 3
there has to be a 3 on the top of the sqrt sign
like this is ? so teh three is on the top side of teh ? liek this
3?
i know how to do it that way that you just said but the 3 messed me up.
#34 Help Me ! » can u help me solve this » 2006-11-20 11:26:17
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- Replies: 7
solve
3
sqrt(2x + 3) = 3
#35 Re: Dark Discussions at Cafe Infinity » Given the opportunity... » 2006-11-17 10:22:01
yeah who doesn't
and to answer teh question....
i wouldnt want to live forever......too much to take in and too much challenges
right
#36 Re: Help Me ! » figure » 2006-11-16 08:32:40
ok thanks... for the help
so DF is 17.5
#37 Help Me ! » figure2 » 2006-11-16 07:54:37
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- Replies: 1
find value of x
i did
m angle A = 1/2 (m CEarc - m BDarc)
25 = 1/2 (100 - x)
25 = 50 - 1/2x
1/2x - 50 = 25
add 50 both sides
1/2x = 75
x = 150
which is wrong
HELP
#38 Help Me ! » figure » 2006-11-16 07:51:50
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- Replies: 2
l || m || n
find DF
i did
6/5 = 15/x
6x = 75
x = 12.5
which is wrong HELP
#39 Re: Help Me ! » radius & coordinates » 2006-11-15 04:39:37
hello help anyone?
#40 Re: Help Me ! » radius & coordinates » 2006-11-14 09:35:26
ok i know what you are saying...
you have given me the standard circle form which i can do
which i see in the book.
but in teh book we have numbers instead of letters..can i change teh letter into number and letters back again for me to do this?
liek in the book
the standard form is > (x + 3)^2 + (y - 3)^2 = 16
then they did
x^2 + 6x + 9 + y^2 - 6y + 9 = 16
and finished by collecting like terms
x^2 + y^2 + 6x - 6y + 2 = 0
so in our problem:
(x - a)^2 + (y - b)^2 = r^2 right
so we can do teh same way like the way they did in the book?
like
x^2 - a^2 + y^2 - b^2 + ay - ab = r^2 then what i think i have the wrong way in doing this...help
#41 Re: Maths Teaching Resources » Announcement » 2006-11-14 02:57:39
is this new site?
you want us to join?
#42 Help Me ! » radius & coordinates » 2006-11-14 02:55:15
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- Replies: 6
find the radius and coordinates of the center of the circle whose equation is
9x^2 + 9y^2 - 18x + 36y - 36 = 0
ok how do i do this...
#43 Re: Guestbook » I W A N T S O M E O R E O S!!! » 2006-11-10 05:34:30
OMG!
I want some OREOS!
And some milk!
Yummmmmy
I am sooo hungry!
<3
Meeeeeeeeeagan*MWAH*
iLy
umm why are you telling us that?????????
:
you want us to buy it for you get you some oreos.....?
#44 Re: Help Me ! » divide » 2006-11-10 03:25:33
ok
x+3 / x^3 + 2x - 1
x - 1
-------------------
x + 3 | x^3 + 2x - 1
x^3 + 3x
----------------
5x - 1
-1x - 3
----------------
+1x + 3
--------------
6x + (-2)
now what do i do? and please explain it well to me how to do it if i did this wrong!
#45 Re: Introductions » hi! (._"_.) » 2006-11-10 03:15:08
hey hello haruka welcome to teh forum hope you have a great time here. and offcourse you'll make friends..here people are very friendly
ok then takecare
#46 Re: Guestbook » Hi how are you guys doing? » 2006-11-09 08:01:48
hey welcome to the site
thats great you liike this site and math.very good
well math is very important. it is used every single day
it is all on us to learn math. if others can't explain it well so what can we do they arent very good teachers right. we have to force our selves to learn math. so we can succeed in the future. don't try to look if otehrs can help or not cuz sometimes they cant help or they dont know how to. you have to help yourself am i right guys?
ok anyways hope you enjoy it over here
takecare
#47 Help Me ! » correct » 2006-11-09 03:50:55
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- Replies: 6
find the domain of the function defined by the following equation :
f (x) = sqrt(x + 6)
x + 6 > 0
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subtract 6 both sides
x < -6
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(-00,-6) < is the domain of F
?
#48 Help Me ! » divide » 2006-11-09 02:37:30
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- Replies: 4
divide
x^3 + 2x - 1 by x + 3
so
x^3 + 2x - 1
---------------
x + 3
x^3 + 3x - 2
the way i did it ..it seems to me that i did it wrong
can you correct me on this?
#49 Re: Help Me ! » Correct » 2006-11-08 07:42:26
so am i right?
#50 Re: Help Me ! » evaluate » 2006-11-08 03:21:06
this is much hard....
the other log problems looked much eaiser than this.