Which shapes can the composite figure be divided into to find the area?

Rhombus JKLM is dilated with center (-4, 7) and scale factor to produce rhombus J'K'L'M'. Fill in the table below to determine t

bija089 [108]

Transforming the rhombus JKLM by dilation with a scale factor of 1/4 and center at D(0, 3), gives;

J'(0, 2.5)

L(-16, 19)

<h3>What formula can be used to find the coordinates of the image following a dilation?</h3>

From the question, we have;

Center of dilation = D(0, 3)

Scale factor of dilation = 1/4

The given points are;

J(0, 1), L'(-4, 7)

The required points are J' and L

Given that the x-coordinate of J is the same as the center of dilation, we have;

The x-coordinate of J' = 0

The formula for finding the coordinates following a dilation is presented as follows;

Dilation;

(x, y) → (k(x - a) + a, k(y - b) + b)

Where;

Center of dilation = (a, b)

Scale factor = k

Therefore;

J(0, 1)

J'(0.25•(0 - 0) + 0, 0.25•(1 - 3) + 3)

The coordinates of J' is therefore;

J'(0, 2.5)

L'(-4, 7) = L(0.25(x - 0) + 0, 0.25(y - 3) + 3)

0.25(x - 0) + 0 = -4

Therefore;

x = -16

0.25•(y - 3) + 3 = 7

y - 3 = (7 - 3)/0.25 = 16

y = 19

The coordinates of L is therefore (-16, 19)

Learn more about dilation transformation here:

brainly.com/question/14263066

#SPJ1

3 0

2 years ago

Which of the following numbers is irrational

aivan3 [116]

Answer:

A) -7.888

Step-by-step explanation:

The answer is A because it is an never ending number, which means it is irrational

4 0

3 years ago

Read 2 more answers

According to the graph, what is the value of the constant in the equation below?

viva [34]

Answer:

Choice A: 0.4

Step-by-step explanation:

We are given the the equation that relates the height and width by a constant:

$height=\frac{constant}{width}$

Therefore

$constant=height*width$

From the points given on the graph we find

$constant = 0.5*0.8=0.4$

or

$constant =1.6*0.25=0.4$

So the constant is 0.4.

3 0

4 years ago

Find the next two numbers in the pattern. -800, 400, -200, 100, ...

frutty [35]

Answer:

-50,25 is the answer

Step-by-step explanation:

........................

5 0

2 years ago

Read 2 more answers

Let <img src="https://tex.z-dn.net/?f=f%28x%2Cy%29%20%3D%20%5Cfrac%7Bx%5E%7B2%7D%20y%7D%7Bx%5E%7B4%7D%2By%5E%7B2%7D%20%7D" id="T

stiv31 [10]

The correct answer is C.

The limit in A does exist:

$\displaystyle \lim_{x\to0} f(x,0) = \lim_{x\to0} \frac0{x^4} = 0$

The limit in B also exists: for any $k\in\Bbb R$ ,

$\displaystyle \lim_{x\to0} f(x,kx) = \lim_{x\to0} \frac{kx^3}{x^4+k^2x^2} = \lim_{x\to0}\frac{kx}{x^2+k^2} = 0$

But this alone does not prove the 2D limit exists. $y=kx$ only captures all the paths through the origin that are straight lines.

The limit in C also exists, but it's not the same as either of the limits along the paths used in A and B.

$\displaystyle \lim_{x\to0} f(x,x^2) = \lim_{x\to0} \frac{x^4}{2x^4} = \frac12$

That this value is non-zero tells us the original limit does not exist.

The claim in D is generally not correct. That $f(0,0)$ is undefined does not automatically mean the limit doesn't exist. A simpler example:

$\displaystyle \lim_{x\to0} \frac{x}{x} = \lim_{x\to0} 1 = 1$

yet $\frac00$ is undefined.

3 0

1 year ago