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3 years ago

The graph shows the dilation of quadrilateral MNOP with the origin as its center. Which rule describes this transformation?

Mathematics

1 answer:

lana66690 [7]3 years ago

8 0

Answer:

D) (x, y) → (1/3x , 1/3 y)

Step-by-step explanation:

A dilation is a change of size, if the dilation factor is greater than 1, then the figure is enlarged. If the dilation factor is smaller than 1, the figure is shrinked. In both cases, the coordinates are MULTIPLIED by the dilation factor.

Among the 4 choices, only 2 are dilations. One is with a dilation factor of 3 (A), which means the shape was enlarged. And the other is with dilation factor of 1/3, meaning the shape was shrinked.

Since we went from MNOP (LARGE one) to M'N'O'P' (small one), the dilation factor was < 1... so 1/3 is the answer.

Answers B and C show a translation/movement of the shape, not a dilation.

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A number that, when multiplied with another number, yields a result of 1 is called its multiplicative ____.

Aliun [14]

Answer:

multiplicative inverse

Step-by-step explanation:

A number that is multiplied by it's multiplicative inverse equals one.

A multiplicative inverse is also called a reciprocal.

Example:

1/7 is 7's multiplicative inverse.

$\frac{7}{1} *\frac{1}{7} =\frac{7}{7} =1$

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3 years ago

Read 2 more answers

If a>b, c<b, and d>a, then which of the following is the correct relationship?

tatuchka [14]

x > y is the same like y < x

If a > b and b > c then a > b > c

If d > a then d > a > b > c

<h3>Answer: (D) c < b < a < d</h3>

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3 years ago

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

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