Answer:
7238 rounded to the nearest whole number.
Step-by-step explanation:
To find the volume of the sphere, we use the formula.

The radius will be 24 units ÷ 2 = 12 units.
Hence, the volume is

When entered in a calculator you'll get a result of 7238.23 cubic units in or 7238 rounded to the nearest whole number.
Answer:
Check the explanation
Step-by-step explanation:
(a)Let p be the smallest prime divisor of (n!)^2+1 if p<=n then p|n! Hence p can not divide (n!)^2+1. Hence p>n
(b) (n!)^2=-1 mod p now by format theorem (n!)^(p-1)= 1 mod p ( as p doesn't divide (n!)^2)
Hence (-1)^(p-1)/2= 1 mod p hence [ as p-1/2 is an integer] and hence( p-1)/2 is even number hence p is of the form 4k+1
(C) now let p be the largest prime of the form 4k+1 consider x= (p!)^2+1 . Let q be the smallest prime dividing x . By the previous exercises q> p and q is also of the form 4k+1 hence contradiction. Hence P_1 is infinite
Now, the cosecant of θ is -6, or namely -6/1.
however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.
we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

recall that

therefore, let's just plug that on the remaining ones,

now, let's rationalize the denominator on tangent and secant,
4000 (4x10^3) times 400 (4x10^2) is equal to 16x10^5 ( 1,600,000)
135 because
15x1=15 15x7=105
15x2=30 15x8=120
15x3=45 15x9=135
15x4=60 15x10=150
15x5=75 15x11=165
15x6=90 15x12=190