(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
.
Answer: C . 147π / 4 mi²
Concept:
The sector is the part of a circle is enclosed by two radii of a circle and their intercepted arc.
A = (θ / 360) πr²
θ = angle of the sector
π = constant
r = radius
Solve:
<u>Given variable</u>
θ = 270°
r = 7 mi
<u>Given formula</u>
A = (θ / 360) πr²
<u>Substitute values into the formula</u>
A = (270 / 360) π (7)²
<u>Simplify exponents</u>
A = (270 / 360) π 49
<u>Simplify by multiplication</u>
A = (147 / 4) π
A = 147π / 4
Hope this helps!! :)
Please let me know if you have any questions
14/32 in simplest form would be 7/16
Write each line to make the sum 7 it say I can't tell y'all the answer dang alot of people need help I got 18 seconds to tell yall