- I’m assuming the expression implies the following:

(1-n)/(n^2 - 1)

The steps goes as follows:

= -(n-1)/(n^2-1) (factored a negative out)
= -(n-1)/[(n+1)(n-1)] (Factored denominator into a difference of squares, but don’t just blindly accept the fact, expand it and you should get the original denominator)

At this point, the (n-1) terms cancel, as you should know: a real number, besides 0, over itself is 1.

= -1/(n+1) , which is the answer

Note: You could also see it as 1/-(n+1) or 1/(-n-1)

8 0

2 years ago

(02.02 MC)

hoa [83]

So for the rhombus to carry onto itself, it needs to start in one position and end up in that same position. In this case, (x,y) will need to end up on (x,y) after the reflections are done. Here what the options do:

A: y-axis, x-axis, y-axis

(x,y) -> (-x,y) -> (-x,-y) -> (x,-y)

B: x-axis, y-axis, y-axis

(x,y) -> (x,-y) -> (-x,-y) -> (x,-y)

C: y=x, x-axis, y=x, y-axis

(x,y) -> (y,x) -> (y,-x) -> (-x,y) -> (x,y)

D: x-axis, y=x, x-axis, y=x

(x,y) -> (x,-y) -> (-y,x) -> (-y,-x) -> (-x,-y)

The only option that starts as (x,y) and ends as (x,y) is C: y=x, x-axis, y=x, y-axis, therefore making that the correct answer.

3 0

3 years ago