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AlladinOne [14]
3 years ago
11

Quadrilateral ABCD is located at A (−2, 2), B (−2, 4), C (2, 4), and D (2, 2). The quadrilateral is then transformed using the r

ule (x−3, y+4) to form the image A'B'C'D'. What are the new coordinates of A', B', C', and D'? Describe what characteristics you would find if the corresponding vertices were connected with line segments.
Mathematics
1 answer:
Mademuasel [1]3 years ago
8 0
New cordinates are formed by adding 7 in x and subtracting 2 from y
A(−2, 2) =A ' (-2 +7 , 2 - 1 ) = A' (5,1)
B(−2, 4)  = B' (-2 + 7 , 4 -1 )= B' (5,3)
C(2, 4)   =  C' (2 + 7 , 4 -1 )= C' (9,3)
<span>D(2, 2)   D' (2 + 7 , 2 -1 ) = D' ( 9 , 1)</span><span>

</span>
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Answer:

30:1

Step-by-step explanation:

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7 0
3 years ago
Find the value of x
Paladinen [302]

\qquad \qquad  \bf \huge\star \:  \:  \large{  \underline{Answer} }  \huge \:  \: \star

  • x = 12°

\textsf{  \underline{\underline{Steps to solve the problem} }:}

\qquad❖ \:  \sf \:2x + 6 + 5x + 90 = 180

[ by linear pair ]

\qquad❖ \:  \sf \:7x + 6 = 180 - 90

\qquad❖ \:  \sf \:7x + 6 = 90

\qquad❖ \:  \sf \:7x = 90 - 6

\qquad❖ \:  \sf \:7x = 84

\qquad❖ \:  \sf \:x = 84 \div 7

\qquad❖ \:  \sf \:x = 12

\qquad \large \sf {Conclusion}  :

\qquad❖ \:  \sf \:x = 12 \degree

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2 years ago
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