Close off the hemisphere 
by attaching to it the disk 
of radius 3 centered at the origin in the plane 
. By the divergence theorem, we have
where 
is the interior of the joined surfaces 
.
Compute the divergence of 
:
Compute the integral of the divergence over . Easily done by converting to cylindrical or spherical coordinates. I'll do the latter:
So the volume integral is
From this we need to subtract the contribution of
that is, the integral of over the disk, oriented downward. Since in , we have
Parameterize by
where 
and 
. Take the normal vector to be
Then taking the dot product of with the normal vector gives
So the contribution of integrating over is
and the value of the integral we want is
(integral of divergence of <em>F</em>) - (integral over <em>D</em>) = integral over <em>S</em>
==> 486π/5 - (-81π/4) = 2349π/20
AE = AC = 4
m<CAB = 60 (equilateral triangle)
m<CAE = 90 (square)
m<BAE = 150 (= 60 + 90)
Triangle BAE is isosceles since AB = AE;
therefore, m<AEB = m<ABE.
m<AEB + m<ABE + m<BAE = 180
m<AEB + m< ABE + 150 = 180
m<AEB + m<AEB = 30
m<AEB = 15
In triangle ABE, we know AE = AB = 4;
we also know m<BAE = 150, and m<AEB = 15.
We can use the law of sines to find BE.
BE/(sin 150) = 4/(sin 15)
BE = (4 sin 150)/(sin 15)
BE = 7.727
Answer:
Fraction Equivalent Fractions
1/5 2/10 4/20
2/5 4/10 8/20
3/5 6/10 12/20
4/5 8/10 16
Step-by-step explanation:
THIS IS ANSWER PLS MARK BRAINLIST
Mel should use the least common multiple to solve the problem
<u>Solution:</u>
Given, Mel has to put the greatest number of bolts and nuts in each box so each box has the same number of bolts and the same number of nuts.
We have to find that should Mel use the greatest common factor or the least common multiple to solve the problem?
He should use least common multiple.
Let us see an example, suppose 12 bolts and nuts are to be fit in 6 boxes.
Then, if we took H.C.F of 12 and 6, it is 6, which means 6 bolts and nuts in each box, but, after filling 2 boxes with 6 bolts and nuts, there will be nothing left, which is wrong as remaining boxes are empty.
So the remaining method to choose is L.C.M.
Hence, he should use L.C.M method.
