Answer: 17 easy how could you not get it but yea thats that
Step-by-step explanation:
Answer
Step-by-step explanation:
A=2πrh+2πr2=2·π·3·6+2·π·32≈169.646
<h2>
Answer:</h2>
The last answer - 9x squared y squared
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</h2>
Since

, we know that

follows a Poisson distribution with parameter

.
Now assuming

denote the mean and standard deviation of

, respectively, then we know right away that

and

.
So,
If

represent a family of surfaces for different values of the constant

. The gradient of the function

defined as

is a vector normal to the surface

.
Given <span>the paraboloid

.
We can rewrite it as a scalar value function f as follows:

The normal to the </span><span>paraboloid at any point is given by:

Also, the normal to the given plane

is given by:

Equating the two normal vectors, we have:
</span>

Since, -1 = 2 is not possible, therefore
there exist no such point <span>
on the paraboloid
such that the tangent plane is parallel to the plane 3x + 2y + 7z = 2</span>
.