Answer
C but then again i might be wrong
Answer:
Tn = 2/3^(n-1)
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 t common ratio
From the sequence
a = 2/9
r = (2/3)/(2/9) = 2/(2/3) =3
Substitute
Tn = 2/9(3)^(n-1)
Tn = 2/3^(n-1)
Hence the required equation is Tn = 2/3^(n-1)
Answer:
Yes
Step-by-step explanation:
a^2+b^2=c^2
7^2+11^2=c^2
49+121=c^2
170=c^2
then square it
c=13
<u>7</u><u>,</u><u>1</u><u>1</u><u>,</u><u>13</u><u> </u><u>is</u><u> </u><u>a</u><u> </u><u>pythagorean</u><u> </u><u>triple</u>
The answer is D the 2 vertical lines make all numbers positive hope this helps