The sum of any geometric sequence (if it converges, r^2<1) is of the form:
s(n)=a(1-r^n)/(1-r), a=initial term, r=common ratio, n=number of terms...
s(n)=30(1-0.4^n)/(0.6)
s(n)=50(1-0.4^n)
Since r<1 the sum of the infinite series is just:
s=50
Answer:
10xy^3
Step-by-step explanation:
Rewrite then factor as a difference of cubes:
1 - 64<em>x</em>³ = 1³ - (4<em>x</em>)³
= (1 - 4<em>x</em>) (1² + 4<em>x</em> + (4<em>x</em>)²)
= (1 - 4<em>x</em>) (1 + 4<em>x</em> + 16<em>x</em>²)
Answer:
3.5
Step-by-step explanation:
edgenuity2020
Answer: 58in^2
Step-by-step explanation:
This is the most reasonable answer.
