I have a take-home test due tomorrow, and I need sum halp.
"Simplify the compound fraction. SHOW YOR WORK! (4 pts)
((1/x+1)+(1/2))/(3/2X^2+4X+2))
kthx ^_^
8th grade :-x
Seriously, help me.
Can't you just bump the topic without posting?
???
......
/me looks at the problem, then you, the problem, and you again, confused.
Start with the numbers.
I dun liek these problems...but this is seriously pre-calc? I'm in pre-calc and we're solving higher-degree polynomials now O_o
Quote from: Zovistograt on December 06, 2007, 05:51:23 PM
I dun liek these problems...but this is seriously pre-calc? I'm in pre-calc and we're solving higher-degree polynomials now O_o
You never know, schools go at different paces...
And ur nawt on D:
WE LEARN COMPOUND FRACTIONS NEXT SEMESTER! :(
well i'm not here to do YOUR work...your just here to do my work ^_^
Quote from: LightShock on December 06, 2007, 06:04:53 PM
WE LEARN COMPOUND FRACTIONS NEXT SEMESTER! :(
NOOOOOOOO :(
I'd help, but I gotta do my own homework. Soon. :(
Don't you just need to come up with a greatest common factor for all the denominators to cancel and simplify?
I've done similar problems and I remember having to do that.
Quote from: TheKoopaBros. on December 06, 2007, 06:33:32 PM
Don't you just need to come up with a greatest common factor for all the denominators to cancel and simplify?
I've done similar problems and I remember having to do that.
Yeah, but then I forget which things to cancel out. :(
Dude, that's hard as heck to read, but I'll help.
Quote from: Silverhawk79 on December 06, 2007, 05:10:49 PM
I have a take-home test due tomorrow, and I need sum halp.
"Simplify the compound fraction. SHOW YOR WORK! (4 pts)
[(1/x + 1) + (1/2)] / (3/2X2 + 4X + 2)
kthx ^_^
Much better, now to solve...
Teh answer:
x3 + 3x2
----------------------------
8x4 + 12x3 + 4x2 + 3x + 3
Multiply top and bottom by both 2x2 and (x +1). That's as simple as you can get without going into the realm of imaginary numbers.
Quote from: Termanian_Knight on December 06, 2007, 06:48:55 PM
Teh answer:
x3 + 3x2
----------------------------
8x4 + 12x3 + 4x2 + 3x + 3
Multiply top and bottom by both 2x2 and (x +1). That's as simple as you can get without going into the realm of imaginary numbers.
??? I got:
x(3x + 2)
--------------------------------------------
8x
3 + 4x
2 + 3
Quote from: NINNIN6 on December 06, 2007, 06:55:36 PM
Quote from: Termanian_Knight on December 06, 2007, 06:48:55 PM
Teh answer:
x3 + 3x2
----------------------------
8x4 + 12x3 + 4x2 + 3x + 3
Multiply top and bottom by both 2x2 and (x +1). That's as simple as you can get without going into the realm of imaginary numbers.
??? I got:
x(3x + 2)
--------------------------------------------
8x3 + 4x2 + 3
From what I see, I'm right if you solve it the way Silver wrote it, but Termanian_Knight is right if there's a parenthesis in this part...
1/(x + 1) + 1/2
Quote from: NINNIN6 on December 06, 2007, 06:55:36 PM
Quote from: Termanian_Knight on December 06, 2007, 06:48:55 PM
Teh answer:
x3 + 3x2
----------------------------
8x4 + 12x3 + 4x2 + 3x + 3
Multiply top and bottom by both 2x2 and (x +1). That's as simple as you can get without going into the realm of imaginary numbers.
??? I got:
x(3x + 2)
--------------------------------------------
8x3 + 4x2 + 3
Aaaaaaaa, make up your mind. D:
repeat 9th grader: wuts a compund number/fraction O_O
Quote from: Silverhawk79 on December 06, 2007, 07:01:33 PM
Quote from: NINNIN6 on December 06, 2007, 06:55:36 PM
Quote from: Termanian_Knight on December 06, 2007, 06:48:55 PM
Teh answer:
x3 + 3x2
----------------------------
8x4 + 12x3 + 4x2 + 3x + 3
Multiply top and bottom by both 2x2 and (x +1). That's as simple as you can get without going into the realm of imaginary numbers.
??? I got:
x(3x + 2)
--------------------------------------------
8x3 + 4x2 + 3
Aaaaaaaa, make up your mind. D:
Look at the post above yours. ::)
Quote from: Silverhawk79 on December 06, 2007, 07:01:33 PM
Aaaaaaaa, make up your mind. D:
So what is it at the beginning of the problem? (1/x) + 1 or 1/(x+1)?
Quote from: Ludwig3 on December 06, 2007, 07:15:50 PM
Quote from: Termanian_Knight on December 06, 2007, 07:13:26 PM
Quote from: Silverhawk79 on December 06, 2007, 07:10:35 PM
Quote from: Termanian_Knight on December 06, 2007, 07:08:13 PM
Quote from: Silverhawk79 on December 06, 2007, 07:01:33 PM
Aaaaaaaa, make up your mind. D:
So what is it at the beginning of the problem? (1/x) + 1 or 1/(x+1)?
1/(x+1)
I'M DA WINRAR!!!1!1
The answer I posted is the 'correct' one.
are you a senior? O_O
I'mma a magic man. :^D
Quote from: Termanian_Knight on December 06, 2007, 07:19:06 PM
Quote from: Ludwig3 on December 06, 2007, 07:15:50 PM
Quote from: Termanian_Knight on December 06, 2007, 07:13:26 PM
Quote from: Silverhawk79 on December 06, 2007, 07:10:35 PM
Quote from: Termanian_Knight on December 06, 2007, 07:08:13 PM
Quote from: Silverhawk79 on December 06, 2007, 07:01:33 PM
Aaaaaaaa, make up your mind. D:
So what is it at the beginning of the problem? (1/x) + 1 or 1/(x+1)?
1/(x+1)
I'M DA WINRAR!!!1!1
The answer I posted is the 'correct' one.
are you a senior? O_O
I'mma a magic man. :^D
i wish i was magickal at math :(
Also, I forget. Does f o g(x) mean I put f in for all the x values in g, or vice versa?
Quote from: Silverhawk79 on December 06, 2007, 07:27:10 PM
Also, I forget. Does f o g(x) mean I put f in for all the x values in g, or vice versa?
do inner terms first. calculate g(x) and put that in f(x).
Quote from: Zovistograt on December 06, 2007, 07:28:25 PM
Quote from: Silverhawk79 on December 06, 2007, 07:27:10 PM
Also, I forget. Does f o g(x) mean I put f in for all the x values in g, or vice versa?
do inner terms first. calculate g(x) and put that in f(x).
So I put g(x) in for all x values of f(x)?
Quote from: Silverhawk79 on December 06, 2007, 07:29:32 PM
Quote from: Zovistograt on December 06, 2007, 07:28:25 PM
Quote from: Silverhawk79 on December 06, 2007, 07:27:10 PM
Also, I forget. Does f o g(x) mean I put f in for all the x values in g, or vice versa?
do inner terms first. calculate g(x) and put that in f(x).
So I put g(x) in for all x values of f(x)?
f o g(x) = f(g(x))
Quote from: Zovistograt on December 06, 2007, 07:30:03 PM
Quote from: Silverhawk79 on December 06, 2007, 07:29:32 PM
Quote from: Zovistograt on December 06, 2007, 07:28:25 PM
Quote from: Silverhawk79 on December 06, 2007, 07:27:10 PM
Also, I forget. Does f o g(x) mean I put f in for all the x values in g, or vice versa?
do inner terms first. calculate g(x) and put that in f(x).
So I put g(x) in for all x values of f(x)?
f o g(x) = f(g(x))
I'm dense tonight. :(
ALSO ONE LAST QUESTION.
How does I find f-1(x)? :(
Quote from: Silverhawk79 on December 06, 2007, 07:41:35 PM
ALSO ONE LAST QUESTION.
How does I find f-1(x)? :(
It's the inverse of the function, exempli gratia f(x) = x, f
-1(x) = 1/x; f(x) = e
x, f
-1(x) = ln x.
Quote from: NINNIN6 on December 06, 2007, 07:38:45 PM
Quote from: Termanian_Knight on December 06, 2007, 07:13:26 PM
Quote from: Silverhawk79 on December 06, 2007, 07:10:35 PM
Quote from: Termanian_Knight on December 06, 2007, 07:08:13 PM
Quote from: Silverhawk79 on December 06, 2007, 07:01:33 PM
Aaaaaaaa, make up your mind. D:
So what is it at the beginning of the problem? (1/x) + 1 or 1/(x+1)?
1/(x+1)
I'M DA WINRAR!!!1!1
The answer I posted is the 'correct' one.
XP I'm too technical. >_>
No, I wondered the same thing, but the format of the problem seemed odd on paper.
Quote from: Silverhawk79 on December 06, 2007, 07:41:35 PM
ALSO ONE LAST QUESTION.
How does I find f-1(x)? :(
The inverse is found by substituting 'x' with 'y' in the equation and solving for 'y'. Note that 'y' really is f(x).
Quote from: NINNIN6 on December 06, 2007, 07:46:12 PM
Quote from: Silverhawk79 on December 06, 2007, 07:41:35 PM
ALSO ONE LAST QUESTION.
How does I find f-1(x)? :(
It's the inverse of the function, exempli gratia f(x) = x, f-1(x) = 1/x; f(x) = ex, f-1(x) = ln x.
O_o
reversing x and y is less complicated
Quote from: NINNIN6 on December 06, 2007, 07:46:12 PM
Quote from: Silverhawk79 on December 06, 2007, 07:41:35 PM
ALSO ONE LAST QUESTION.
How does I find f-1(x)? :(
It's the inverse of the function, exempli gratia f(x) = x, f-1(x) = 1/x; f(x) = ex, f-1(x) = ln x.
I know, but somehow...my answer for this one problem ends up weird.
f(x) = 6/(3x+2)
f
-1(x) = x(3y+2) - 2/3
That doesn't seem right...
...this thread confuses me.
Quote from: Silverhawk79 on December 06, 2007, 07:51:13 PM
Quote from: NINNIN6 on December 06, 2007, 07:46:12 PM
Quote from: Silverhawk79 on December 06, 2007, 07:41:35 PM
ALSO ONE LAST QUESTION.
How does I find f-1(x)? :(
It's the inverse of the function, exempli gratia f(x) = x, f-1(x) = 1/x; f(x) = ex, f-1(x) = ln x.
I know, but somehow...my answer for this one problem ends up weird.
f(x) = 6/(3x+2)
f-1(x) = x(3y+2) - 2/3
That doesn't seem right...
f(x) = 6/(3x+2)
y = 6/(3x+2)
x = 6/(3y+2)
and solve that for x.
Quote from: Zovistograt on December 06, 2007, 07:53:27 PM
Quote from: Silverhawk79 on December 06, 2007, 07:51:13 PM
Quote from: NINNIN6 on December 06, 2007, 07:46:12 PM
Quote from: Silverhawk79 on December 06, 2007, 07:41:35 PM
ALSO ONE LAST QUESTION.
How does I find f-1(x)? :(
It's the inverse of the function, exempli gratia f(x) = x, f-1(x) = 1/x; f(x) = ex, f-1(x) = ln x.
I know, but somehow...my answer for this one problem ends up weird.
f(x) = 6/(3x+2)
f-1(x) = x(3y+2) - 2/3
That doesn't seem right...
f(x) = 6/(3x+2)
y = 6/(3x+2)
x = 6/(3y+2)
and solve that for x.
That's what I just did. psyduck;
Quote from: Silverhawk79 on December 06, 2007, 07:54:27 PM
Quote from: Zovistograt on December 06, 2007, 07:53:27 PM
Quote from: Silverhawk79 on December 06, 2007, 07:51:13 PM
Quote from: NINNIN6 on December 06, 2007, 07:46:12 PM
Quote from: Silverhawk79 on December 06, 2007, 07:41:35 PM
ALSO ONE LAST QUESTION.
How does I find f-1(x)? :(
It's the inverse of the function, exempli gratia f(x) = x, f-1(x) = 1/x; f(x) = ex, f-1(x) = ln x.
I know, but somehow...my answer for this one problem ends up weird.
f(x) = 6/(3x+2)
f-1(x) = x(3y+2) - 2/3
That doesn't seem right...
f(x) = 6/(3x+2)
y = 6/(3x+2)
x = 6/(3y+2)
and solve that for x.
That's what I just did. psyduck;
Then you should have gotten f
-1(x) = 2x - 2/3;