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Kazeer [188]
3 years ago
14

Help means should be well explained :) ​

Mathematics
2 answers:
Marysya12 [62]3 years ago
6 0

There are 7 numbers

Calculate sum of digits of all numbers

  • 331=3+3+1=7
  • 482=4+8+2=14
  • 551=5+5+1=11
  • 263=2+6+3=11
  • 383=3+8+3=14
  • 362=3+6+2=11
  • 284=2+8+4=14

We observe that

\\ \sf\longmapsto 482\sim383\sim284

\\ \sf\longmapsto 551\sim263\sim362

  • 331 is odd one

Option C

Slav-nsk [51]3 years ago
5 0
551, it is too big of a number so it is the outlier. :)
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In circle L with KNM= 53, find the angle measure of minor arc KM
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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
Negative x + 6y = 16 and 8x - 6y = -2 which statements about the system of equations are true check all that are apply the x var
MakcuM [25]

Answer:

y variable will be elimited when adding the system of equations

x = 2 or y = 3

Step-by-step explanation:

- x + 6y = 16

8x - 6y = -2

Add both equations

(-x + 8x) + {6y + (-6y)} = 16 + (-2)

7x + 6y - 6y = 16 - 2

7x + 0 = 14

7x = 14

x = 14/7

x = 2

- x + 6y = 16

- 2 + 6y = 16

6y = 16 + 2

6y = 18

y = 18/6

y = 3

the x variable will be eliminating when adding the system of equations and the

y variable will be elimited when adding the system of equations

or the sum of the system equations is negative 8X = 14 or

x = 2 or y = 3

or there's only one solution to the system equations

4 0
2 years ago
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