Answer:A
Step-by-step explanation:
Answer:
6, 2, 2/3, 2/9, 2/27, 2/81
Step-by-step explanation:
The nth term of a geometric progression is expressed as;
Tn = ar^n-1
a is the first term
n is the number of terms
r is the common ratio
Given
a = 6
r = 1/3
when n = 1
T1 = 6(1/3)^1-1
T1 = 6(1/3)^0
T1 = 6
when n = 2
T2= 6(1/3)^2-1
T2= 6(1/3)^1
T2 = 2
when n = 3
T3 = 6(1/3)^3-1
T3= 6(1/3)^2
T3= 6 * 1/9
T3 = 2/3
when n = 4
T4 = 6(1/3)^4-1
T4= 6(1/3)^3
T4= 6 * 1/27
T4 = 2/9
when n = 5
T5 = 6(1/3)^5-1
T5= 6(1/3)^4
T5= 6 * 1/81
T5 = 2/27
when n = 6
T6 = 6(1/3)^6-1
T6= 6(1/3)^5
T6= 6 * 1/243
T6 = 2/81
Hence the first six terms are 6, 2, 2/3, 2/9, 2/27, 2/81
Answer:
As x decreases without bound, f(x) increases without bound
As x increases without bound, f(x) approaches 0
Step-by-step explanation:
As x gets more and more negative f(x) gets bigger and bigger
f(x) = (7/10) ^ x but x is negative so flip it and then x is a larger number
(10/7) ^ large number so it will get larger
as → -∞ f(x) →∞
As x gets more and more positive f(x) gets smaller and smaller
f(x) = (7/10) ^ x closer and closer to 0. The denominator gets larger faster than the numerator = 1 / large number
as → ∞ f(x) →0
