The cone equation gives

which means that the intersection of the cone and sphere occurs at

i.e. along the vertical cylinder of radius

when

.
We can parameterize the spherical cap in spherical coordinates by

where

and

, which follows from the fact that the radius of the sphere is 3 and the height at which the sphere and cone intersect is

. So the angle between the vertical line through the origin and any line through the origin normal to the sphere along the cone's surface is

Now the surface area of the cap is given by the surface integral,




Answer:
The correct option is C.
Step-by-step explanation:
The least common multiple (LCM) of any two numbers is the smallest number that they both divide evenly into.
The given terms are
and
.
The factored form of each term is


To find the LCM of given numbers, multiply all factors of both terms and common factors of both terms are multiplied once.


The LCM of given terms is
. Therefore the correct option is C.
Answer:
Point d
Step-by-step explanation:
because the x is -2 and the y is 3 so you look for the x and then the y
So, since we have a cubic equation with 4 terms, the first thing we should try is factoring by grouping, so:

Now that we've factored our equation, we can use ZPP and break it up:

So, our solutions are:

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