Find the distance between (-1, 4) and (3, 1).

2 The value of y is directly proportional

rusak2 [61]

Answer:

y = 17.5

Step-by-step explanation:

Given that y is directly proportional to x then the equation relating them is

y = kx ← k is the constant of proportion

To find k use the condition y = 35 when x = 140, then

35 = 140k ( divide both sides by 140 )

0.25 = k

y = 0.25x ← equation of proportion

When x = 70, then

y = 0.25 × 70 = 17.5

4 0

2 years ago

Triangle J is shown below. James drew a scaled version of Triangle J using a scale factor of 4 and labeled it

Novay_Z [31]

Answer:

The area of triangle K is 16 times greater than the area of triangle J

Step-by-step explanation:

we know that

If Triangle K is a scaled version of Triangle J

then

Triangle K and Triangle J are similar

If two triangles are similar, then the ratio of its areas is equal to the scale factor squared

Let

z -----> the scale factor

Ak ------> the area of triangle K

Aj -----> the area of triangle J

so

$z^{2}=\frac{Ak}{Aj}$

we have

$z=4$

substitute

$4^{2}=\frac{Ak}{Aj}$

$16=\frac{Ak}{Aj}$

$Ak=16Aj$

therefore

The area of triangle K is 16 times greater than the area of triangle J

4 0

2 years ago

Read 2 more answers

The radius of a circle r in centimeters can be found by solving r to the second power = 70. What is the radius of the circle r,

slavikrds [6]

Answer:

bhb jkabdjkea 7- 70 bhakbhkd

3 0

2 years ago

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

1 what is sum of the interior angles of ABCD what is the sum of the exterior angles of CHFE 3 which sides are parallel to CE 4 w

expeople1 [14]

Given that ABCD is a parallelogram, the sum of its interior angles is:
360°

<h3>What is a parallelogram?</h3>

A parallelogram is a rectilinear two-dimensional figure with four sides whose opposite sides are parallel. See the image attached.

<h3>What is the sum of the exterior angles of CHFE?</h3>

The sum of the exterior angles of any Triangle or any polygon is 360°. Hence, the sum of the exterior angles of CHFE (a polygon) is 360°.

<h3>Which sides are parallel to CE?</h3>

The sides that are parallel to CE are:

FH

AD and

<h3>IJ.

Which side is perpendicular to CE?</h3>

The side that are perpendicular to CE is side side HH, hence we can state that:

CE ⊥ EH, because both lines intersect each other at right angles.

Because the sum of angles in a triangle is 180° and we already have ∠CEH to be equal to ∠90° hence,

∠C + ∠H? = 180° - 90°

= 90°

<h3>What is the measure of the exterior angles of a Regular Pentagon?</h3>

The measure of the exterior angles of a regular pentagon is 72°

Learn more about Parallelograms at:
brainly.com/question/970600
#SPJ1

5 0

2 years ago