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3 years ago

Given:<2=<3 Prove: m<1+<3=180

Mathematics

1 answer:

s2008m [1.1K]3 years ago

7 0

Statement Reason

m<1 + m<2 = 180 linear pair

*(m<1 and "<2 form a straight line together, a straight line has a degree measure of 180)

m<2 = m<3 given

m<1 + m<3 = 180 transitivity

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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

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)

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3 years ago

Machine A produces bolts at a uniform rate of 120 every 40 seconds, and Machine B produces bolts at a uniform rate of 100 every

Nutka1998 [239]

Answer:

25 seconds

Step-by-step explanation:

Hi there!

In order to answer this question, first we need to know how many bolts per second are produced by each machine, this can be known by dividing the number of bolts by the time it takes.

For machine A:

$A = \frac{120 bolts}{40 s}= 3 \frac{bolts}{s}$