P = 300 $; r = 0.06; t = 20; n = 1
A = P ( 1 + r/n)^n*t
= 300 ( 1 + 0.06) ²⁰
= 300 ( 1.06) ²⁰
= 962.14
Answer:3
If x=0 or x=1 it is trivial, so 0<x<1. Define a=1−x2, then 0<a<1.
Step-by-step explanation:
D...........................
Three numbers between 3.65 and 3.66 could be:
1) 3.651
2) 3.655
3) 3.657
4) 3.659
Hope this helps! :)
