ONM
OMN
NMO
Hope that helped!
Answer:
I think that it is...
Step-by-step explanation:
A. Elastic?
Answer:
a)
a1 = log(1) = 0 (2⁰ = 1)
a2 = log(2) = 1 (2¹ = 2)
a3 = log(3) = ln(3)/ln(2) = 1.098/0.693 = 1.5849
a4 = log(4) = 2 (2² = 4)
a5 = log(5) = ln(5)/ln(2) = 1.610/0.693 = 2.322
a6 = log(6) = log(3*2) = log(3)+log(2) = 1.5849+1 = 2.5849 (here I use the property log(a*b) = log(a)+log(b)
a7 = log(7) = ln(7)/ln(2) = 1.9459/0.6932 = 2.807
a8 = log(8) = 3 (2³ = 8)
a9 = log(9) = log(3²) = 2*log(3) = 2*1.5849 = 3.1699 (I use the property log(a^k) = k*log(a) )
a10 = log(10) = log(2*5) = log(2)+log(5) = 1+ 2.322= 3.322
b) I can take the results of log n we previously computed above to calculate 2^log(n), however the idea of this exercise is to learn about the definition of log_2:
log(x) is the number L such that 2^L = x. Therefore 2^log(n) = n if we take the log in base 2. This means that
a1 = 1
a2 = 2
a3 = 3
a4 = 4
a5 = 5
a6 = 6
a7 = 7
a8 = 8
a9 = 9
a10 = 10
I hope this works for you!!
Answer:
Because 0 counts as the first possible number. 0-9 gives you 10 possible numbers.
Step-by-step explanation:
1st number = x
2nd number = 3x
3rd number = 2x-10
x +3x + 2x-10 =92
6x - 10 = 92
6x =102
x = 102/6
x = 17
1st number = 17
2nd number = 17*3 = 51
3rd number = 17 *2 = 34-10 = 24
17 + 51 + 24 = 92
3 numbers are 17, 51 and 24
