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Yakvenalex [24]
3 years ago
13

Classify each pair of numbered angles.

Mathematics
1 answer:
Volgvan3 years ago
6 0

Answer:

1 and 2 adjacent

5 and 6 are linear pair

Step-by-step explanation:

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Can someone help for my math test? its 20 points and ill be giving brainiest!
weqwewe [10]

Answer:

the length in fraction form would be 16 1/2 rounded to the nearest 10 would be 20

Step-by-step explanation:

because 16.5=16 1/2

16.5=17

17=20

4 0
3 years ago
Read 2 more answers
If 30,000 cm2 of material is available to make a box with a square base and an open top, what is the largest possible volume (in
Cloud [144]

Answer:

The largest possible volume of the box is 2000000 cubic meters.

Step-by-step explanation:

The volume (V), in cubic centimeters, and surface area (A_{s}), in square centimeters, of the box with a square base are described below:

A_{s} = l^{2}+h\cdot l (1)

V = l^{2}\cdot h (2)

Where:

l - Side length of the base, in centimeters.

h - Height of the box, in centimeters.

By (2), we clear h within the formula:

h = \frac{V}{l^{2}}

And we apply in (1) and simplify the resulting expression:

A_{s} = l^{2}+ \frac{V}{l}

A_{s}\cdot l = l^{3}+V

V = A_{s}\cdot l -l^{3} (3)

Then, we find the first and second derivatives of this expression:

V' = A_{s}-3\cdot l^{2} (4)

V'' = -6\cdot l (5)

If V' = 0 and A_{s} = 30000\,cm^{2}, then we find the critical value of the side length of the base is:

30000-3\cdot l^{2} = 0

3\cdot l^{2} = 30000

l = 100\,cm

Then, we evaluate this result in the expression of the second derivative:

V'' = -600

By Second Derivative Test, we conclude that critical value leads to an absolute maximum. The maximum possible volume of the box is:

V = 30000\cdot l - l^{3}

V = 2000000\,cm^{3}

The largest possible volume of the box is 2000000 cubic meters.

4 0
3 years ago
PLZ HELP 10 POINTS
vaieri [72.5K]

Congruent means there's exact shape and angle magnitude since transformation requires relocation of an image there's no change to it's shape therefore it's congruent

5 0
3 years ago
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Find the area of the shape shown below pls
Afina-wow [57]

Step-by-step explanation:

area of missed triangle =1/2×2×4=4 cm2

area of whole rectangle =4×8=32 cm2

area of shape =32_4 =28cm2

6 0
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
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