We know that
volume of a cylinder=area of the base*height
area of the base=4x^5 in²
height=<span>24x^5y^8/ 4x^5 in
volume=</span>24x^5y^8 in³
height=24x^5y^8/ 4x^5 ------> (24/4)*(x^5/x^5)*(y^8)------> (6)*(1)*(y^8)-----> 6y^8
the answer is
B: 6y^8
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
Answer:
r =2.9985yd
Step-by-step explanation:
Given that the circumference of a circle is 2
r
where r is the radius and = 3.1415926....
You have
2r = 18.84 yd
2(3.1415926)r = 18.84 yd
6.2831r = 18.84yd
divide through by 6.2831
r = 2.9985yd = aprox 3yds
Answer:
x=6 . m<PQS=82 m<SQR=61 :)
Step-by-step explanation:
(13x+4) + (10x-1) = 141
combine like terms
23x+3=141
subtract 3 from both sides
23x=138
divide both sides by 23
x=6
substitute x into both original equations
m<PQS=13(6)+4
m<PQS=78+4
m<PQS= 82
m<SQR=10(6)+1
m<SQR=60+1
M<SQR=61
Answer:
-33
Step-by-step explanation:
· 
(-33)
