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3 years ago

The graph shows the system y=2x+1 y=2x^2+1 Which ordered pairs are solutions of the system?

Mathematics

1 answer:

zhenek [66]3 years ago

3 0

Hey!

To solve this problem, we would really have to graph both equations.

OPEN THE FIRST IMAGE

The first image I provided was the image of this equation graphed: y=2x+1

OPEN THE SECOND IMAGE

The second image I provided was the image of this equation graphed: y=2x^2+1

PLEASE DO NOT LOOK AT THE THIRD IMAGE YET!

The first image has two points. One at ( -0.5, 0 ) and the other at ( 0, 1 ).

The second image has only one point, which is at ( 0, 1 ).

OPEN THE THIRD IMAGE

Now, when both are combined we see that they intersect at two points, which are ( 0, 1 ) and ( 1, 3 ).

So, our answer is...

The ordered pairs that are the solution to the system are ( 0,1 ) and ( 1,3 ).

Hope this helped!

- Lindsey Frazier ♥

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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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Answer:

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and the degree is also 18

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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