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3 years ago

A reflection,rotation,translation,or dilation is called

Mathematics

2 answers:

azamat3 years ago

8 0

They are called transformations.

PSYCHO15rus [73]3 years ago

5 0

Answer:

it is called transformation

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Help~~~~~~~~~~~~~~~~~~~~~~~~~

igor_vitrenko [27]

Answer:

28.25 square units

Step-by-step explanation:

A circumference of the circle is

$C=2\pi r,$

where r is the radius of the circle.

So,

$18.84=2\pi r\\ \\r=\dfrac{18.84}{2\pi}=\dfrac{9.42}{\pi}\ cm$

The area of the circle is

$A=\pi r^2$

Substitute the value of the radius:

$A=\pi \cdot \left(\dfrac{9.42}{\pi}\right)^2=\dfrac{9.42^2}{\pi}\approx 28.25\ un^2$

8 0

3 years ago

Read 2 more answers

Rational expression. simplify<br> 36m-48m over 6m.

ladessa [460]

One way is to find common factors

example

8/6=4/3 because 8/6, 8 and 6 have common factor of 2 so divide that out to get 4/3
basically divide the LCM from each

so factor

we can combine 36m-48m into -12m
we have
-12m/6m

common factor is 6m
-12m/6m=(-2)/1 times (6m)/(6m)=-2 times 1=-2

answer is -2

3 0

3 years ago

I dont understand how to do this please help me i want to get a good grade

horsena [70]

Answer:

y =20

Step-by-step explanation:

you have to get the y by itself so what you do on one side you do to the other. subtract 12 on both sides and you get y=20

7 0

3 years ago

Read 2 more answers

Which equation represents 4x – 3y = 15 when solved for y

andrey2020 [161]

Answer:

y=4/3x-5

Step-by-step explanation:

4x-3y=15

-4x -4x

______________

-3y=-4x+15

---- --- ---- (divide all by -3)

-3 -3 -3

y=4/3x-5

5 0

3 years ago

Find the area of the region that lies inside the first curve and outside the second curve.

marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = $\dfrac{1}{2}$

$\theta = -\dfrac{\pi}{3}, \ \ \dfrac{\pi}{3}$

Now, the area of the region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \ \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ 5^2 d \theta$

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \ \theta d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ d \theta$

$A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} \dfrac{cos \ 2 \theta +1}{2} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}$

$A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} {cos \ 2 \theta +1} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}} \ \ - \dfrac{25}{2} \begin {bmatrix} \dfrac{2 \pi}{3} \end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3}) \end {bmatrix} - \dfrac{25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} + \dfrac{\pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}$

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx

7 0

3 years ago