Find the points on the cone z2 = x2 + y2 that are closest to the point (6, 2, 0). (x, y, z) = (smaller z-value)
(x, y, z) = (larger z-value)

The closest points on the cone are;

(6, 2, -√10) and (6, 2, √10)

Step-by-step explanation:

Let B(x, y, z) denote a point on the cone.

Therefore, the distance between the points (6, 2, 0) and B(x, y, z) is;

d = √[(x - 6)² + (y - 2)² + (z - 0)²]

d = √[(x - 6)² + (y - 2)² + z²]

Since we are given that z² = x² + y², we now have;

d = √[(x - 6)² + (y - 2)² + x² + y²]

Taking the square of both sides gives;

d² = [(x - 6)² + (y - 2)² + x² + y²]

x² is an increasing function. Thus, minimizing d is also the same as to minimize f (x, y) = d²

Thus, f' = 0. So;

df/dx = 2(x - 6) + 2x = 0

2x - 12 + 2x = 0

4x = 12

x = 12/4

x = 3

Similarly,

df/dy = 2(y - 2) + 2y = 0

2y - 4 + 2y = 0

4y - 4 = 0

4y = 4

y = 4/4

y = 1

Now,from earlier;

z² = x² + y²

Thus;

z = ±√(3² + 1²)

z = ±√10

Thus, the closest points on the cone are;

(6, 2, -√10) and (6, 2, √10)

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

10 - 75

step-by-step explanation:

there's nothing attached so it's most likely anything in that range.