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Alexxandr [17]
3 years ago
12

What is the solution to the equation​

Mathematics
2 answers:
Eduardwww [97]3 years ago
3 0
Umm sir there isno equation i think u forgot to put a picture
timofeeve [1]3 years ago
3 0

Answer:

cute cat xifncijfifjcig

Step-by-step explanation:

please mark me brainliest

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3 years ago
The plane x+y+2z=8 intersects the paraboloid z=x2+y2 in an ellipse. Find the points on this ellipse that are nearest to and fart
DiKsa [7]

Answer:

The minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

Step-by-step explanation:

Here, the two constraints are

g (x, y, z) = x + y + 2z − 8  

and  

h (x, y, z) = x ² + y² − z.

Any critical  point that we find during the Lagrange multiplier process will satisfy both of these constraints, so we  actually don’t need to find an explicit equation for the ellipse that is their intersection.

Suppose that (x, y, z) is any point that satisfies both of the constraints (and hence is on the ellipse.)

Then the distance from (x, y, z) to the origin is given by

√((x − 0)² + (y − 0)² + (z − 0)² ).

This expression (and its partial derivatives) would be cumbersome to work with, so we will find the the extrema  of the square of the distance. Thus, our objective function is

f(x, y, z) = x ² + y ² + z ²

and

∇f = (2x, 2y, 2z )

λ∇g = (λ, λ, 2λ)

µ∇h = (2µx, 2µy, −µ)

Thus the system we need to solve for (x, y, z) is

                           2x = λ + 2µx                         (1)

                           2y = λ + 2µy                       (2)

                           2z = 2λ − µ                          (3)

                           x + y + 2z = 8                      (4)

                           x ² + y ² − z = 0                     (5)

Subtracting (2) from (1) and factoring gives

                     2 (x − y) = 2µ (x − y)

so µ = 1  whenever x ≠ y. Substituting µ = 1 into (1) gives us λ = 0 and substituting µ = 1 and λ = 0  into (3) gives us  2z = −1  and thus z = − 1 /2 . Subtituting z = − 1 /2  into (4) and (5) gives us

                            x + y − 9 = 0

                         x ² + y ² +  1 /2  = 0

however, x ² + y ² +  1 /2  = 0  has no solution. Thus we must have x = y.

Since we now know x = y, (4) and (5) become

2x + 2z = 8

2x  ² − z = 0

so

z = 4 − x

z = 2x²

Combining these together gives us  2x²  = 4 − x , so

2x²  + x − 4 = 0 which has solutions

x =  (-1+√33)/4

and

x = -(1+√33)/4.

Further substitution yeilds the critical points  

((-1+√33)/4; (-1+√33)/4; (17-√33)/4)   and

(-(1+√33)/4; - (1+√33)/4; (17+√33)/4).

Substituting these into our  objective function gives us

f((-1+√33)/4; (-1+√33)/4; (17-√33)/4) = (195-19√33)/8

f(-(1+√33)/4; - (1+√33)/4; (17+√33)/4) = (195+19√33)/8

Thus minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

4 0
3 years ago
Consider the sets below. u = {x | x is a real number} a = {x | x is an odd integer} r = {x | x = 3, 7, 11, 27} is r ⊂ a?
-BARSIC- [3]

The correct option is (B) yes because all the elements of set R are in set A.

<h3>What is an element?</h3>
  • In mathematics, an element (or member) of a set is any of the distinct things that belong to that set.

Given sets:

  1. U = {x | x is a real number}
  2. A = {x | x is an odd integer}
  3. R = {x | x = 3, 7, 11, 27}

So,

  • A = 1, 3, 5, 7, 9, 11... are the elements of set A.
  • R ⊂ A can be understood as R being a subset of A, i.e. all of R's elements can be found in A.
  • Because all of the elements of R are odd integers and can be found in A, R ⊂ A is TRUE.

Therefore, the correct option is (B) yes because all the elements of set R are in set A.

Know more about sets here:

brainly.com/question/2166579

#SPJ4

The complete question is given below:
Consider the sets below. U = {x | x is a real number} A = {x | x is an odd integer} R = {x | x = 3, 7, 11, 27} Is R ⊂ A?

(A) yes, because all the elements of set A are in set R

(B) yes, because all the elements of set R are in set A

(C) no because each element in set A is not represented in set R

(D) no, because each element in set R is not represented in set A

8 0
2 years ago
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