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Alexxandr [17]
4 years ago
6

Find the circumference of the circle to the nearest tenth. Use 3.14 for pi.

Mathematics
1 answer:
Grace [21]4 years ago
6 0
The correct answer is C, or 157 m.
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2,6,12,16,32,36,72 complete the number pattern​
deff fn [24]

Answer:

Next is 76, 82, 86, and 92

Step-by-step explanation:

4 0
3 years ago
Item 10
vfiekz [6]

Answer:

21 Students

Step-by-step explanation:

140 x .15 = 21

5 0
3 years ago
The center of a circle is (4, 6), and an endpoint of a diameter is (2, 5). What is the other endpoint of the diameter?
valentinak56 [21]
Consider this option:
1. if the point (4;6) is the centre of the circle and the point (2;5) is the first endpoint of its diameter, then point (4;6) is the middle point of the diameter (it means that is the middle between the 1st and the 2d endpoints of diameter).
2. using the property described above:
for x of the 2d endpoint of the diameter: x=4*2-2=6;
for y of the 2d endpoint of the diameter: y=6*2-5=7.

answer: (6;7)
3 0
3 years ago
Read 2 more answers
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
ASAP PLEASE HELP I BEEN WORKING ON IT FOR 5 HOURS
Luba_88 [7]
So in this case, the solution is where the graphs intersect. Put the equations in y-intercept form.
-4x+y=-6
y=4x-6

8x-2y=14
-2y=14-8x
y=4x-7

These lines have the same slope, which is 4. Therefore, they are parallel. Parallel lines never intersect, so there is NO SOLUTION.
7 0
3 years ago
Read 2 more answers
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