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Art [367]
3 years ago
9

Rectangle ABCD is congruent to rectangle .

Mathematics
2 answers:
baherus [9]3 years ago
8 0
The correct answer for the question that is being presented above is this one: "A. Rectangle ABCD was translated 8 units left and then 7 units down." Rectangle ABCD is congruent to rectangle. The sequence of transformations that could have been used to transform rectangle ABCD to produce rectangle is that <span>Rectangle ABCD was translated 8 units left and then 7 units down.</span>
MAVERICK [17]3 years ago
4 0

Answer:- A.  Rectangle ABCD was translated 8 units left and then 7 units down.

Explanation:-

We can see vertex A(2,4) of rectangle ABCD moves to A"(-6,-3), in a way that A translated 8 units left (-8)and 7 units down(-7) to A".

i.e.(2-8,4-7)=(-6,-3)

Similarly the other vertices of rectangle ABCD moves to form rectangle A"B"C"D"

B (2,2) → B" (-6,-5)

C (6,2) → C" (-2,-5)

D(6,4) → D" (-2,-3)

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3.14*16.8*16.8*16.9
vitfil [10]

Answer: 3.14 *16.8*16.8*16.9=14977.3478

14977.3478= 14,977.35

6 0
3 years ago
The green triangle is a dilation of the red triangle with a scale factor of s=1/3 and the center of dilation is at the point (4,
klasskru [66]

Given:

The scale factor is s=\dfrac{1}{3} and the center of dilation is at the point (4,2).

Red is original figure and green is dilated figure.

To find:

The coordinates of point C' and point A.

Solution:

Rule of dilation: If a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

(x,y)\to (k(x-a)+a,k(y-b)+b)

According to the given information, the scale factor is \dfrac{1}{3} and the center of dilation is at (4,2).

(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)            ...(i)

Let us assume the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using (i), we get

C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)

C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)

C(-2,11)\to C'(-2+4,3+2)

C(-2,11)\to C'(2,5)

Therefore, the coordinates of Point C' are C'(2,5).

We assumed that point A is A(m,n).

Using (i), we get

A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)

From the given figure it is clear that the image of point A is (8,4).

A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)

On comparing both sides, we get

\dfrac{1}{3}(m-4)+4=8

\dfrac{1}{3}(m-4)=8-4

(m-4)=3(4)

m=12+4

m=16

And,

\dfrac{1}{3}(n-2)+2=4

\dfrac{1}{3}(n-2)=4-2

(n-2)=3(2)

n=6+2

n=8

Therefore, the coordinates of point A are (16,8).

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2 years ago
Is 23 prime or composite ?
dalvyx [7]
It is prime, it has no factors besides 1,23
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3 years ago
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Hey mate u wrote the answer in the same question

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2 years ago
Which of the following is the conjugate of a complex number with −1 as the real part and 3i as the imaginary part? (6 points) –1
dmitriy555 [2]

Complex numbers have the form a + ib.  Given a + ib, the complex conjugatet is a - ib.

Therefore, the complex conjugate of -1 + 3i is -1 - 3i.

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