Report, edit, etc...Posted by WoAHorde2 on 2006-10-10 at 22:47:25
This is how this "1337" Algebra thread works. I post an Equation and ask for an answer. Whoever gets it right posts another Equation, so on and so forth.
[-4-5(y+2x)-23]+[4+5(3x+4y)]
Note: The -4 is NEGITIVE 4. It is the only negative in the equation.
SIMPLIFY
[-4-5(y+2x)-23]+[4+5(3x+4y)]
Note: The -4 is NEGITIVE 4. It is the only negative in the equation.
SIMPLIFY
Report, edit, etc...Posted by FatalException on 2006-10-10 at 22:52:16
ACKKKKKK!!! *Melts*
Report, edit, etc...Posted by Doodan on 2006-10-10 at 22:53:37
I wonder how long this "game" will last...
Report, edit, etc...Posted by WoAHorde2 on 2006-10-10 at 22:57:02
So do I Doodan 
. Ok I jsut solved the problem, although I' not 100% sure on my answer. And oh a hint, DISTRIBUTIVE PROPERTY
. Ok I jsut solved the problem, although I' not 100% sure on my answer. And oh a hint, DISTRIBUTIVE PROPERTYReport, edit, etc...Posted by Night on 2006-10-10 at 22:58:16
QUOTE(WoAHorde2 @ Oct 10 2006, 08:47 PM)
[-4-5(y+2x)-23]+[4+5(3x+4y)]
15y + 5x - 23 (fixed it)
What is the vertex of: 3x²+96x+3073 ?
ADDITION:
The extended formula (without brackets):
-4-5y-10x-23+4+15x+20y
Combine like terms:
-4+4-23 = -23
-10x+15x = 5x
-5y+20y = 15y
Report, edit, etc...Posted by WoAHorde2 on 2006-10-10 at 23:05:48
I believe your answer is right, althought I got 25x + 15y -23. We don't agree on x.
Add:
Well heres how I did it.
[-4-5(y+2x)-23]+[4+5(3x+4y)]
Use Dist Property
(-4-5y+10x-23)+4+15x+20y
Combine Like Terms On One Side
(-27-5y+10x)+(4+15x+20y)
Switch Numbers and Variables Over
-27 + 4 = -23
-5y + 20y = 15y
10x = 15x = 25 x
Final ANswer
25x + 15y - 23
Add:
Well heres how I did it.
[-4-5(y+2x)-23]+[4+5(3x+4y)]
Use Dist Property
(-4-5y+10x-23)+4+15x+20y
Combine Like Terms On One Side
(-27-5y+10x)+(4+15x+20y)
Switch Numbers and Variables Over
-27 + 4 = -23
-5y + 20y = 15y
10x = 15x = 25 x
Final ANswer
25x + 15y - 23
Report, edit, etc...Posted by Night on 2006-10-10 at 23:13:43
QUOTE(WoAHorde2 @ Oct 10 2006, 08:47 PM)
[-4-5(y+2x)-23]+[4+5(3x+4y)]
[right][snapback]574738[/snapback][/right]
[right][snapback]574738[/snapback][/right]
that part comes out to -5y - 10x. a NEGETIVE 10 x.
Now solve mine:
QUOTE(Night @ Oct 10 2006, 08:57 PM)
What is the vertex of: 3x²+96x+3073 ?
[right][snapback]574748[/snapback][/right]
[right][snapback]574748[/snapback][/right]
Report, edit, etc...Posted by WoAHorde2 on 2006-10-10 at 23:15:22
Oh I forgot to carry over the engitive, silly me
Report, edit, etc...Posted by Syphon on 2006-10-10 at 23:18:59
[-4-5(y+2x)-23]+[4+5(3x+4y)]
= -4 - 5y - 10x - 23 + 4 + 15x + 20y
= 5x + 15y -23
...
DUR DUR DUR.
= -4 - 5y - 10x - 23 + 4 + 15x + 20y
= 5x + 15y -23
...
DUR DUR DUR.
Report, edit, etc...Posted by Zeratul_101 on 2006-10-10 at 23:21:09
damnit night, why did you have to use such a large number. i think its -16, 2305
EDIT: doh, got my signs mixed up
EDIT: doh, got my signs mixed up again! i'm hopeless
EDIT: doh, got my signs mixed up
EDIT: doh, got my signs mixed up again! i'm hopeless
Report, edit, etc...Posted by Night on 2006-10-10 at 23:25:52
No the vertex is two numbers (the point where the axis of symatry intersects the parabola) comes in the form of (x,y) or aka: (p,q)
ADDITION:
Oh you did put two numbers, but no thats not the answer.
Heres the first step:
3(x²+32x+1024)+2305
Correct Zera! -16, 2305
... oops my mistake.
ADDITION:
Oh you did put two numbers, but no thats not the answer.
Heres the first step:
3(x²+32x+1024)+2305
Correct Zera! -16, 2305
... oops my mistake.
Report, edit, etc...Posted by Rantent on 2006-10-10 at 23:28:56
I hope you know that subtracting a numer is like adding a negative...
[-4-5(y+2x)-23]+[4+5(3x+4y)] = -4-5y-10x-23+4+15x+20y = 15y+5x-23 Simple as it goes.
[-4-5(y+2x)-23]+[4+5(3x+4y)] = -4-5y-10x-23+4+15x+20y = 15y+5x-23 Simple as it goes.
Report, edit, etc...Posted by Syphon on 2006-10-10 at 23:30:03
3x²+96x+3073
{Inster quadratic formula here}
x[sub]1[/sub] = -16 + 67.89698i and x[sub]2[/sub] = -16 - 67.89698i
Vertex is x[sub]1[/sub] + x[sub]2[/sub] / 2
I'm too lazy to simplify it.
EDIT - I need to note to you, that I am failing math class.
{Inster quadratic formula here}
x[sub]1[/sub] = -16 + 67.89698i and x[sub]2[/sub] = -16 - 67.89698i
Vertex is x[sub]1[/sub] + x[sub]2[/sub] / 2
I'm too lazy to simplify it.
EDIT - I need to note to you, that I am failing math class.
Report, edit, etc...Posted by Zeratul_101 on 2006-10-10 at 23:31:49
why the freak is there a one over there...
anyhow, this is what i got
0 = 3x^2 + 96x + 3073
-3073 = 3(x^2 + 32x)
768 - 3073 = 3(x^2 + 32x + 256)
-2305 = 3(x + 16)^2
y = 3(x + 16)^2 + 2305
anyhow, this is what i got
0 = 3x^2 + 96x + 3073
-3073 = 3(x^2 + 32x)
768 - 3073 = 3(x^2 + 32x + 256)
-2305 = 3(x + 16)^2
y = 3(x + 16)^2 + 2305
Report, edit, etc...Posted by Night on 2006-10-10 at 23:35:52
Here's a simple one.
If nm=0 does that prove that n+m=0?
FYI: Zera, I edited the post above.
If nm=0 does that prove that n+m=0?
FYI: Zera, I edited the post above.
Report, edit, etc...Posted by Zeratul_101 on 2006-10-10 at 23:36:43
errr, i can't think up any eqquations 
Report, edit, etc...Posted by Syphon on 2006-10-10 at 23:37:04
QUOTE(Zeratul_101 @ Oct 10 2006, 10:31 PM)
why the freak is there a one over there...
anyhow, this is what i got
0 = 3x^2 + 96x + 3073
-3073 = 3(x^2 + 32x)
768 - 3073 = 3(x^2 + 32x + 256)
-2305 = 3(x + 16)^2
y = 3(x + 16)^2 + 2305
[right][snapback]574772[/snapback][/right]
anyhow, this is what i got
0 = 3x^2 + 96x + 3073
-3073 = 3(x^2 + 32x)
768 - 3073 = 3(x^2 + 32x + 256)
-2305 = 3(x + 16)^2
y = 3(x + 16)^2 + 2305
[right][snapback]574772[/snapback][/right]
We're trying to find the vertex. (x = -16)
Report, edit, etc...Posted by Zeratul_101 on 2006-10-10 at 23:37:21
IRT to night
no
IRT to syphon
thats the axis of symmetry, not vertex
no
IRT to syphon
thats the axis of symmetry, not vertex
Report, edit, etc...Posted by CheeZe on 2006-10-10 at 23:38:51
Boring.
Report, edit, etc...Posted by Syphon on 2006-10-10 at 23:39:09
QUOTE(Night @ Oct 10 2006, 10:35 PM)
Here's a simple one.
If nm=0 does that prove that n+m=0?
FYI: Zera, I edited the post above.
[right][snapback]574775[/snapback][/right]
If nm=0 does that prove that n+m=0?
FYI: Zera, I edited the post above.
[right][snapback]574775[/snapback][/right]
No. 0 + R>0 ≠ 0, 0 x R>0 = 0
Report, edit, etc...Posted by Night on 2006-10-10 at 23:39:12
QUOTE(Syphon(MM) @ Oct 10 2006, 09:36 PM)
We're trying to find the vertex. (x = -16)
[right][snapback]574777[/snapback][/right]
[right][snapback]574777[/snapback][/right]
The proper way to show the vertex is: ( p , q ) = ( -16 , 2305 )
EDIT: What you have there is the equation for the axis of symatry (how u spell this?)
Report, edit, etc...Posted by Syphon on 2006-10-10 at 23:40:11
QUOTE(Zeratul_101 @ Oct 10 2006, 10:36 PM)
IRT to night
no
IRT to syphon
thats the axis of symmetry, not vertex
[right][snapback]574778[/snapback][/right]
no
IRT to syphon
thats the axis of symmetry, not vertex
[right][snapback]574778[/snapback][/right]
The vertex is the axis of symmetry. (Atleast in parabolas it is.)
Report, edit, etc...Posted by Zeratul_101 on 2006-10-10 at 23:41:57
axis of symettry only has an x coordinate, a vertex has both x and y
Report, edit, etc...Posted by HolySin on 2006-10-10 at 23:42:06
Let me screw this up for you guys with the older sibling of algerbra, Very Basic Calculus:
An airplane at an altitude of 10,000 ft is flying at a constant speed on a line that will take it directly over an observer on the ground. If, at a given instant, the observer notes that the angle of elevation of the airplane is 60º and is increasing at a rate of 1º per second, find the speed of the airplane.
An airplane at an altitude of 10,000 ft is flying at a constant speed on a line that will take it directly over an observer on the ground. If, at a given instant, the observer notes that the angle of elevation of the airplane is 60º and is increasing at a rate of 1º per second, find the speed of the airplane.
Report, edit, etc...Posted by Zeratul_101 on 2006-10-10 at 23:45:28
omg, don't bring back trig