(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 
.
Heya!!!
Answer to your question is
Option A
Kindly refer to attachment for solution
Hope it helps *_*
Answer:
Length of diagonal in the first top figure is 13 yd
Okay for this one, the formula of a diagonal in a cuboid is the root of sum of squares
length, breadth and height
so
D = root of 12^2 + 4^2 + 3^2
D = root of 144 + 16 + 9
D = root 169
D = 13yd
Length of diagonal in the bottom figure is 10.77 which is 10.8m
For this one, you have to find the diameter of base first
Since radius of the base is 5 diameter = 10
Since its a right angled triangle, Hypotenuse square = sum of sides square
Diagonal^2 = 10^2 + 4^2
D^2 = 100 + 16
D^2 = 116
D = root 116
D = 10.77
D = 10.8 m
Answer:
0
Step-by-step explanation:
you can subtract from point 1 to point 2
(9,-2) - (8,3)
9-8= 1
-2-3= -5
therefore horizontal leg is 1
vertical leg is -5
you can also sketch and count the units :)
