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zhuklara [117]
3 years ago
8

HELP ME i do not get it

Mathematics
1 answer:
Elena-2011 [213]3 years ago
8 0
Domain (x values) are all real numbers due to the fact that the graph extends in both directions forever.
Range (y values) is when y is greater than -4, because -4 is where it ends
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If an event of probability p is observed repeatedly during independent repetitions, the ratio of the observed frequency of that
Arlecino [84]

Answer:

b

Step-by-step explanation:

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More on fraction <br><br>21×5​
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Answer:

105.

Step-by-step explanation:

21+21+21+21+21=105.

4 0
3 years ago
PLZZZ HELPP anyonee plzzzz
mihalych1998 [28]

Answer:

The correct answer is B

Step-by-step explanation:

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4 years ago
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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
PLease help me with this!
astraxan [27]

Answer:

1. x2 - 9 > 0

x^2-3^2>0

(x+3)(x-3)>0

(x+3)>0 and (x-3)>0

x>-3       and  x>3

2. x2 - 8x + 12 > 0

   x^2 - 8x  +12>0

   x^2 -2x -6x +12 >0   (-8x is replaced by (-2x) + (-6x) )

   x(x-2) -6(x-2) >0

    (x-6)(x-2)>0

(x-6)>0    and (x-2)>0

    x>6     and     x>2

3. -x2 - 12x - 32 > 0

    -x^2 -12x -32 >0

     x^2 +12x +32 <0

      x^2 +4x +8x +32<0

      x(x+4) +8(x+4)<0

      (x+8)(x+4)<0

(x+8)<0   and  (x+4)<0

x<-8      and    x<-4

4. x2 + 3x - 20 >= 3x + 5

   x^2 +3x -20 >= 3x +5

   x^2 +3x -20 -3x >= 3x +5 -3x

     x^2 -20  >= 5

     x^2 -20 +20  >= 5  +20

     x^2 >=25

     x^2-25 >=0

      (x-5)(x+5)>=0

(x-5)>=0  and (x+5)>=0

  x>=5    and x>=-5

6 0
3 years ago
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