The vertices of a parallelogram are shown below.

1 answer:

4 0

The vertices for the dilated image can be found by multiplying the coordinates of the original image by the scale factor. In this problem, the scale factor is 1/3.

(0,0) ⇒ 0 x 1/3 = 0 ; 0 x 1/3 = 0 ⇒ (0,0)
(27,0) ⇒ 27 x 1/3 = 9 ; 0 x 1/3 = 0 ⇒ (9,0)
(36,30) ⇒36 x 1/3 = 12 ; 30 x 1/3 = 10 ⇒ (12,10)
(9,30) ⇒ 9 x 1/3 = 3 ; 30 x 1/3 = 10 ⇒ (3,10)

Answer is choice D. (3,10)

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gregori [183]

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2 years ago

Let x1 = 12, y1 = 15, and y2 = 3. Let y vary inversely with x. Find x2.

Ostrovityanka [42]

We have that an inverse variation in a function is given by:
$y = k / x
$
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$15 = k / 12
$
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3 0

3 years ago

Which expression is equivalent to Cube root of 216 x cubed y Superscript 6 Baseline z Superscript 12?

yaroslaw [1]

Answer:

6xy²z^4

Step-by-step explanation:

$\sqrt[3]{216x^3y^6z^12}$ = ∛(6³)(x³)(y³)(y³)(z³)(z³)(z³)(z³) = 6x·y·y·z·z·z·z = 6xy²z^4