Compare 1/7 to consecutive multiples of 1/9. This is easily done by converting the fractions to a common denominator of LCM(7, 9) = 63:
1/9 = 7/63
2/9 = 14/63
while
1/7 = 9/63
Then 1/7 falls between 1/9 and 2/9, so 1/7 = 1/9 plus some remainder. In particular,
1/7 = 1/9¹ + 2/63.
We do the same sort of comparison with the remainder 2/63 and multiples of 1/9² = 1/81. We have LCM(63, 9²) = 567, and
1/9² = 7/567
2/9² = 14/567
3/9² = 21/567
while
2/63 = 18/567
Then
2/63 = 2/9² + 4/567
so
1/7 = 1/9¹ + 2/9² + 4/567
Compare 4/567 with multiples of 1/9³ = 1/729. LCM(567, 9³) = 5103, and
1/9³ = 7/5103
2/9³ = 14/5103
3/9³ = 21/5103
4/9³ = 28/5103
5/9³ = 35/5103
6/9³ = 42/5103
while
4/567 = 36/5103
so that
4/567 = 5/9³ + 1/5103
and so
1/7 = 1/9¹ + 2/9² + 5/9³ + 1/5103
Next, LCM(5103, 9⁴) = 45927, and
1/9⁴ = 7/45927
2/9⁴ = 14/45927
while
1/5103 = 9/45927
Then
1/5103 = 1/9⁴ + 2/45927
so
1/7 = 1/9¹ + 2/9² + 5/9³ + 1/9⁴ + 2/45927
One last time: LCM(45927, 9⁵) = 413343, and
1/9⁵ = 7/413343
2/9⁵ = 14/413343
3/9⁵ = 21/413343
while
2/45927 = 18/413343
Then
2/45927 = 2/9⁵ + remainder
so
1/7 = 1/9¹ + 2/9² + 5/9³ + 1/9⁴ + 2/9⁵ + remainder
Then the base 9 expansion of 1/7 is
0.12512..._9
Answer:
9
Step-by-step explanation:
-7 and -7 are the same and -5 and 4 are 9 spaces apary.
Hope this helps plz hit the crown <3
Answer:
c
Step-by-step explanation:
Answer:
390
Step-by-step explanation:
The formula of sum of arithmetic sequence is 
The number of this series are 2, 4, 6, 8, 10, 12, . . . . , 60.
The first term ( a ) = 2
The 30th even number is 60
The Tn is the term of the sequence
So the equation is
30/2 x ( 2 + 
)
= 60
30/2 x ( 2 + 60 )
30/2 x (62)
1860/2
390
Answer:
there is no way of solving this because the expression does not equal anything.
Step-by-step explanation:
