Answer:
The series 1/5, 2/15, 4/45, 8/135... converges and sums up to 3/5
Step-by-step explanation:
Consider the infinite geometric series
1/5, 2/15, 4/45, 8/135...
With first term, a=1/5
common ratio, r = ⅔
The series converge because the common ratio, |r|<1.
The sum to infinity of a geometric series, S= a/(1-r)
S= 1/5 ÷ (1-⅔) = 1/5 ÷ 1/3 = 3/5
Therefore, the geometric series 1/5, 2/15, 4/45, 8/135... sums up to 3/5.
Answer:
$111
Step-by-step explanation:
37x3 = 111
111/3 =37
Answer:9
Step-by-step explanation:
Consecutive integers are 1 apart
x,x+1,x+2
(x)(x+1)(x+2)=-120
x^3+3x^2+3x=-120
add 120 to both sides
x^3+3x^2+3x+120=0
factor
(x+6)(x^2-3x+20)=0
set each to zero
x+6=0
x=-6
x^2-3x+20=0
will yeild non-real result, discard
x=-6
x+1=-5
x+2=-4
the numbers are -4,-5,-6
use trial and error and logic
factor 120
120=2*2*2*3*5
how can we rearange these numbers in (x)(y)(z) format such that they multiply to 120?
obviously, the 5 has to stay since 2*5=10 which is out of range
so 2*2*2*3 has to arrange to get 3,4 or 4, 6 or 6,7
obviously, 7 cannot happen since it is prime
3 and 4 results in in 12, but 2*2*2*3=24
therfor answer is 4 and 6
they are all negative since negaive cancel except 1
the numbers are -4,-5,-6
Answer:
4.5 × 10^5 or 450000 :)))))