(a) 
since 13 is prime.
(b) 
, and there are 81/3 = 27 multiples of 3 between 1 and 81, which leaves 81 - 27 = 54 numbers between 1 and 81 that are coprime to 81, so 
.
(c) 
; there are 50 multiples of 2, and 20 multiples of 5, between 1 and 100; 10 of these are counted twice (the multiples of 2*5=10), so a total of 50 + 20 - 10 = 60 distinct numbers not coprime to 100, leaving us with 
.
(d) 
; there are 51 multiples of 2, 34 multiples of 3, and 6 multiples of 17, between 1 and 102. Among these, we double-count 17 multiples of 2*3=6, 3 multiples of 2*17=34, and 2 multiples of 3*17=51; we also triple-count 1 number, 2*3*17=102. There are then 51 + 34 + 6 - (17 + 3 + 2) + 1 = 70 numbers between 1 and 102 that are not coprime to 102, and so 
.
30,576 because 30 fills the thousands and 5,7,6 fill the rest
Answer:
a = 36
Step-by-step explanation:
We use pythagoras theorem here.
h = 60
p = 48
b = a
h² = p² + b²
60² = 48² + b²
3600 = 2304 + b²
b² = 3600 - 2304
b² = 1296
b = 36
Answer 2.85
On rounding to nearest hundredths, we get than 2.85 as at thousandths place, there is 9 which is greater than 5, so 1 is added to the hundredth value.
Answer:
The answer is the second graph, Option 2
