Greg Said:

Algebra 2 teachers, Please help?

We Answered:

Personally I prefer the McDougle-Littell text. I tutor in the evenings at a local Sylvan Learning Center and find that the text presents the data in an easily understood format and it is student friendly.

Good Luck with you schooling ...

Morris Said:

Help with a few Algebra 2 concepts, please?

We Answered:

1. use your LCM (least common multiple ) of 3 and 6
which is 6
so multiply both sides by 6
6( (x/3) - (y/6) ) = 1 (6)
2x - y = 6
then manipulate the variables
- y = 6 - 2x
- y = -2x + 6
y = 2x - 6

2. you mean A as in Ax + By = C?
or y = Ax + B?
it's unclear what you're trying to do here...but
-2/5 x + y = 3
add 2/5 x to both sides
2/5 x - 2/5 x + y = 3 + 2/5 x
y = 3 + (2/5)x
dophse forgot to apply the multiply 5 to all terms :(

3. x = 5 which means it gives us the point (5, 0)
using the slope-intercept form
y = mx + b
slope is 1 so m = 1
y = (1)x + b
y = x + b
now plug in (5, 0) for x and y
0 = 5 + b
b = -5
the y intercept, the "b", is -5


HOpe this helps :D

Steven Said:

classzone algebra 2 concepts and skills activation code?

We Answered:

use perhaps the Riemann's dzeta function

Andre Said:

PLease give an example of these two abstract algebra concepts?

We Answered:

U(15) is not cyclic.
The elements of U(15) are 1,2,4,7,8,11,13,14
and their orders are 1,4,2, 4,4,2,4,and 2.
Since there is no element of order 8, U(15) is not cyclic.
More generally, U(n) is cyclic if and only if
n = p, 2p, p² or 2p², where p is an odd prime.

2. S_3 fills the bill quite nicely. Any proper subgroup
has order 2 or 3 and is therefore cyclic, and so abelian.

Vincent Said:

Can anyone tell me some core concepts on algebra 2?

We Answered:

http://www.mathtutor.com/software/algebr…

That should help.