1.) <span>(n+1)n! = (n+1)!
</span>
<span>2.) (n+8)!/n+8 = (n+8)!/(n+8)
= (n+8)*(n+8-1)! / (n+8) = ((n+8)*(n+7)!) / (n+8)
</span><span>Cancel out the (n+8)
= (n +7)!
</span>
3.) 100 x 99 x 98 = (100 x 99 x 98 x 97!) / (97!) = (100!) / (97!).
4. (n-1)(n-2)(n-3) = <span>((n-1)(n-2)(n-3)</span>(n-4)!) /(n-4)!
= (n-1)! / (n-4)!.
Cheers. That's it.
Answer:
5/2
Step-by-step explanation:
use formula



Which transformations can be used to map a triangle with vertices A(2, 2), B(4, 1), C(4, 5) to A’(–2, –2), B’(–1, –4), C’(–5, –4
Romashka [77]
The triangles ABC and A'B'C' are shown in the diagram below. The transformation is a reflection in the line

. This is proved by the fact that the distance between each corner ABC to the mirror line equals to the distance between the mirror line to A'B'C'.