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Gre4nikov [31]
4 years ago
14

Polygons ABCD and A′B′C′D′ are shown on the following coordinate grid: What set of transformations is performed on ABCD to form

A′B′C′D′? A coordinate grid is shown from positive 6 to negative 6 on the x-axis and from positive 6 to negative 6 on the y-axis. A polygonABCD is shown with vertex A on ordered pair 2, negative 2, vertex B on ordered pair 4, negative 2, vertex C on ordered pair 1, negative 3 and vertex D on ordered pair 5, negative 3. A polygonA prime B prime C prime D prime is shown with vertex A prime on ordered pair 1, 2 , vertex B prime on ordered pair 1, 4, vertex C prime on ordered 2, 1 and vertex D prime on ordered pair 2, 5. A 90 degrees counterclockwise rotation about the origin followed by a translation 1 unit to the left A translation 1 unit to the left, followed by a 90 degrees counterclockwise rotation about the origin A 270 degrees counterclockwise rotation about the origin followed by a translation 1 unit to the left A translation 1 unit to the left followed by a 270 degrees counterclockwise rotation about the origin

Mathematics
1 answer:
mestny [16]4 years ago
4 0

Answer:

  A 90 degrees counterclockwise rotation about the origin followed by a translation 1 unit to the left

Step-by-step explanation:

The transformation of coordinates for figure ABCD ⇒ A'B'C'D' is ...

   (x, y) ⇒ (-y-1, x)

__

We know the transformation ...

  (x, y) ⇒ (-y, x)

represents a CCW rotation by 90°. And the transformation ...

  (x, y) ⇒ (x-1, y)

represents a translation 1 unit left.

Combined, the rotation and translation produce the transformation described above. This figure has undergone ...

   a 90° CCW rotation about the origin, followed by a translation 1 unit left.

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\sf{\qquad\qquad\huge\underline{{\sf Answer}}}

Here we go ~

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Volume of cone is :

\qquad \sf  \dashrightarrow \:v = 270 \pi

\qquad \sf  \dashrightarrow \: \dfrac{1}{3}   \cancel\pi {r}^{2} h = 270 \cancel\pi

\qquad \sf  \dashrightarrow \:r {}^{2}  \sdot10 = 270 \times 3

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\qquad \sf  \dashrightarrow \: { {r}^{2} }^{}  = 81

\qquad \sf  \dashrightarrow \:r =  \sqrt{81}

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Now, let's calculate volume of solid sphere with same radius is ~

\qquad \sf  \dashrightarrow \:vol =  \dfrac{4}{3}  \pi {r}^{3}

\qquad \sf  \dashrightarrow \:vol =  \dfrac{4} {3}   \sdot\pi \sdot  {9}^{3}

\qquad \sf  \dashrightarrow \:vol =  \dfrac{4} {3}   \sdot\pi \sdot  729

\qquad \sf  \dashrightarrow \:vol =  {4} {}    \sdot243 \sdot\pi

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So, volume of the solid sphere in terms of pi is :

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<u>note</u> : the solid figure attached below the cone is a hemisphere, so if the volume of hemisphere is asked then just dovide the result for sphere by 2. that is :

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