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3 years ago

(-8,-2) and (-4,6) write a equation

Mathematics

1 answer:

anastassius [24]3 years ago

5 0

The equation of line passing through (-8, -2) and (-4, 6) is y = 2x + 14

Solution:

Given that

Line is passing through point (− 8 ,− 2) and ( -4 , 6 )

Equation of line passing through point $\left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right)$ is given by:

$y-y_{1}=\frac{\left(y_{2}-y_{1}\right)}{\left(x_{2}-x_{1}\right)}\left(x-x_{1}\right)$ ----- eqn 1

$\text { In our case } x_{1}=-8, y_{1}=-2, x_{2}=-4, y_{2}=6$

Substituting given value in (1) we get

$\begin{array}{l}{y-(-2)=\frac{(6-(-2))}{(-4-(-8))}(x-(-8))} \\\\ {=>y+2=\frac{8}{4}(x+8)} \\\\ {=>y+2=2(x+8)} \\\\ {=>y+2=2 x+16} \\\\ {=>-2 x+y=14}\end{array}$

Thus the required equation of line is y = 2x + 14

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The plane x + y + z = 12 intersects paraboloid z = x^2 + y^2 in an ellipse.(a) Find the highest and the lowest points on the ell

emmasim [6.3K]

Answer:

Highest (-3,-3)

Lowest (2,2)

Farthest (-3,-3)

Closest (2,2)

Step-by-step explanation:

To solve this problem we will be using Lagrange multipliers.

Let us find out first the restriction, which is the projection of the intersection on the XY-plane.

From x+y+z=12 we get z=12-x-y and replace this in the equation of the paraboloid:

$\bf 12-x-y=x^2+y^2\Rightarrow x^2+y^2+x+y=12$

completing the squares:

$\bf x^2+y^2+x+y=12\Rightarrow (x+1/2)^2-1/4+(y+1/2)^2-1/4=12\Rightarrow\\\\\Rightarrow (x+1/2)^2+(y+1/2)^2=12+1/2\Rightarrow (x+1/2)^2+(y+1/2)^2=25/2$

and we want the maximum and minimum of the paraboloid when (x,y) varies on the circumference we just found. That is, we want the maximum and minimum of

$\bf f(x,y)=x^2+y^2$

subject to the constraint

$\bf g(x,y)=(x+1/2)^2+(y+1/2)^2-25/2=0$

Now we have

$\bf \nabla f=(\displaystyle\frac{\partial f}{\partial x},\displaystyle\frac{\partial f}{\partial y})=(2x,2y)\\\\\nabla g=(\displaystyle\frac{\partial g}{\partial x},\displaystyle\frac{\partial g}{\partial y})=(2x+1,2y+1)$

Let $\bf \lambda$ be the Lagrange multiplier.

The maximum and minimum must occur at points where

$\bf \nabla f=\lambda\nabla g$

that is,

$\bf (2x,2y)=\lambda(2x+1,2y+1)\Rightarrow 2x=\lambda (2x+1)\;,2y=\lambda (2y+1)$

we can assume (x,y)≠ (-1/2, -1/2) since that point is not in the restriction, so

$\bf \lambda=\displaystyle\frac{2x}{(2x+1)} \;,\lambda=\displaystyle\frac{2y}{(2y+1)}\Rightarrow \displaystyle\frac{2x}{(2x+1)}=\displaystyle\frac{2y}{(2y+1)}\Rightarrow\\\\\Rightarrow 2x(2y+1)=2y(2x+1)\Rightarrow 4xy+2x=4xy+2y\Rightarrow\\\\\Rightarrow x=y$

Replacing in the constraint

$\bf (x+1/2)^2+(x+1/2)^2-25/2=0\Rightarrow (x+1/2)^2=25/4\Rightarrow\\\\\Rightarrow |x+1/2|=5/2$

from this we get

x=-1/2 + 5/2 = 2 or x = -1/2 - 5/2 = -3

and the candidates for maximum and minimum are (2,2) and (-3,-3).

Replacing these values in f, we see that

f(-3,-3) = 9+9 = 18 is the maximum and

f(2,2) = 4+4 = 8 is the minimum

Since the square of the distance from any given point (x,y) on the paraboloid to (0,0) is f(x,y) itself, the maximum and minimum of the distance are reached at the points we just found.

We have then,

(-3,-3) is the farthest from the origin

(2,2) is the closest to the origin.

3 0

3 years ago

4. The reflecting dish of a parabolic microphone has a cross-section in the shape of a parabola. The microphone itself is placed

elixir [45]

Assume the parabola is placed on a graph where the x-axis is the top of the dish.
The vertex is then at (0,-30) The x-intercepts or zeros are at (-30,0) and (30,0)

The equation of such parabola would be:
$y = a(x+30)(x-30)$
Plug in vertex to find value of 'a'
$-30 = a(0+30)(0-30) \\ \\ a = \frac{-30}{(-30)(30)} = \frac{1}{30}$

Now find the focus given that $p = \frac{1}{4a}$
$p = \frac{1}{4(1/30)} = \frac{30}{4} = 7.5$

Answer: the microphone should be placed 7.5 inches from vertex.

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3 years ago

Angles 1 and 2 are complementary. If angle 1 = 35, then what is the measure of angle 2?

enot [183]

Answer:

doge

Step-by-step explanation:

EAnswers avatar

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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