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:
Step-by-step explanation: x - 6
The given equation can be re-written as y = ---------
-3
Arbitrarily choose x = 0. Then:
x - 6 0-6
y = --------- = ----------- = 2, so (0, 2) is a point on the graph which is also the
-3 -3 y-intercept
Arbitrarily choose x = 6. Then y = 0, and (6, 0) is another point on the graph
which happens to be the x-intercept
arbitrarily choose x = 12. Then y = (12 - 6) / (-3) = -2. Then (12, -2) is
another point on the
graph.
Plot (12, -2), (6, 0) and (0, 2). Draw a line through these three points.
For the answer to the question above, we will use this formula to solve this problem
<span>d = kv^2 </span>
<span>plugging in, </span>
<span>4.2 = k*10^2 </span>
<span>k = 0.042 </span>
<span>d = 0.042v^2 </span>
<span>the revoised equation will be </span>
<span>d = 0.9*0.042v^2 , i.e. </span>
<span>d = 0.0378v^2
I hope my answer helped you</span>
Answer:
I think "−13.4°C<−13.2°C because −13.2°C is warmer than −13.4°C" is the right answer
Lcm of 60 and n = 420
Stated: n <90 and a multiple of 6
420/60 = 7 thus n must be a multiple of 7
Therefore n =42 or 84