Write an equation with integer coefficients that has the given roots:<br><br> x= -4, 6

Determine whether the following scenario represents a direct variation and describe why or why not: It takes 4 painters 8 hours

marysya [2.9K]

Answer:

No

Step-by-step explanation:

It would take 1 painter 32 hours to paint a house (because 4*8=32). If we divide 32 by 12 we get 3, so no, it is not a direct variation because if it was it would only take 3 hours for the 12 painters to paint the house.

4 0

2 years ago

A company paid $48 for 2 cases of printer paper. each case contained 12 packages of paper. next month the company’s office manag

enyata [817]

Cost would be $4,320.

6 0

3 years ago

Given the following sets.

laila [671]

B = {a, b, c, d}

C = {0, a, 2, b}

B ∪ C = {a, b, c, d, 0, 2}

<h3>Answer: E)</h3>

Everything from the set B and everything from the set C give to one set. Duplicated elements are written only once.

8 0

2 years ago

Read 2 more answers

A fishing guide charges a $150 fee per fishing trip plus $60 per person in a group. The guide charged a group $630 for a fishing

harina [27]

Answer:

8 people

Step-by-step explanation:

$630-$150= 480

480 divided by 60 dollars= 8 people

8 0

2 years ago

Read 2 more answers

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:

a)

Highest (-3,-3)

Lowest (2,2)

b)

Farthest (-3,-3)

Closest (2,2)

Step-by-step explanation:

To solve this problem we will be using Lagrange multipliers.

a)

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

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

b)

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

Write an equation with integer coefficients that has the given roots: x= -4, 6