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3 years ago

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Mathematics

1 answer:

jeka57 [31]3 years ago

4 0

Answer:

Step-by-step explanation:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\(-5,1)\quad\implies\quad x_1=-5\,,\ \ y_1=1\\\\(-1,4)\quad\implies\quad x_2=-1\,,\ \ y_2=4\\\\d=\sqrt{(-1+5)^2+(4-1)^2}= \sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$

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Quadrilateral ABCD with vertices A(0, 6), B(-3, -6), C(-9, -6), and D(-12, -3): a) dilation with scale factor of 1/3 centered at

Oksanka [162]

a) The points of the new quadrillateral are $A'(x,y) = (0, 2)$ , $B'(x,y) = (-1, -2)$ , $C'(x,y) = \left(-3,-2\right)$ and $D'(x,y) = (-4, -1)$ , respectively.

b) The points of the new quadrillateral are $A'(x,y) = (-5, 5)$ , $B'(x,y) = (-8,-7)$ , $C'(x,y) = (-13, -7)$ and $D'(x,y) = (-17, -4)$ , respectively.

<h3>How to perform transformations with points</h3>

a) A dillation centered at the origin is defined by following operation:

$P'(x,y) = k\cdot P(x,y)$ (1)

Where:

$P(x,y)$ - Original point
$P'(x,y)$ - Dilated point.

If we know that $k = \frac{1}{3}$ , $A(x,y) = (0,6)$ , $B(x,y) = (-3,-6)$ , $C(x,y) = (-9, -6)$ and $D(x,y) = (-12, -3)$ , then the new points of the quadrilateral are:

$A'(x,y) = \frac{1}{3}\cdot (0,6)$

$A'(x,y) = (0, 2)$

$B'(x,y) = \frac{1}{3} \cdot (-3,-6)$

$B'(x,y) = (-1, -2)$

$C'(x,y) = \frac{1}{3}\cdot (-9,-6)$

$C'(x,y) = \left(-3,-2\right)$

$D'(x,y) = \frac{1}{3}\cdot (-12,-3)$

$D'(x,y) = (-4, -1)$

The points of the new quadrillateral are $A'(x,y) = (0, 2)$ , $B'(x,y) = (-1, -2)$ , $C'(x,y) = \left(-3,-2\right)$ and $D'(x,y) = (-4, -1)$ , respectively. $\blacksquare$

b) A translation along a vector is defined by following operation:

$P'(x,y) = P(x,y) +T(x,y)$ (2)

Where $T(x,y)$ is the transformation vector.

If we know that $T(x,y) = (-5,-1)$ , $A(x,y) = (0,6)$ , $B(x,y) = (-3,-6)$ , $C(x,y) = (-9, -6)$ and $D(x,y) = (-12, -3)$ ,

$A'(x,y) = (0,6) + (-5, -1)$

$A'(x,y) = (-5, 5)$

$B'(x,y) = (-3, -6) + (-5, -1)$

$B'(x,y) = (-8,-7)$

$C'(x,y) = (-9, -6) + (-5, -1)$

$C'(x,y) = (-13, -7)$

$D'(x,y) = (-12,-3)+(-5,-1)$

$D'(x,y) = (-17, -4)$

The points of the new quadrillateral are $A'(x,y) = (-5, 5)$ , $B'(x,y) = (-8,-7)$ , $C'(x,y) = (-13, -7)$ and $D'(x,y) = (-17, -4)$ , respectively. $\blacksquare$

To learn more on transformation rules, we kindly invite to check this verified question: brainly.com/question/4801277

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2 years ago

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Alchen [17]

Answer:

7. 4^2 = 6

8. 7^2 = 49

9. 6^4 =1296

10. 1^2 = 1

11. 5^3 =125

12. 1^3 =1

Step-by-step explanation:

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3 years ago

What is the grates common factor of 16 and 48

lisabon 2012 [21]

The gcf of 16 and 48 is the largest positive integer that divides the numbers 16 and 48 without a remainder<span>. Spelled out, it is the greatest common factor of 16 and 48 Is what google says.</span>

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2 years ago

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Helpp!! what is the domain and range for the graph

ivolga24 [154]

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One marble is randomly drawn and then replaced from a jar containing two white marbles and one

melisa1 [442]

Answer:

1/3.

Step-by-step explanation:

Prob(Drawing white marble) = 2/3

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So:

Prob(Drawing a white then a red marble) = 2/3 * 1/2

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2 years ago