A'(-6, -10), B'(-3,-13), and C'(-5,-1) are the vertices of the ΔA'B'C' under the translation rule (x,y)→(x,y-3). This can be obtained by putting the ΔABC's vertices' values in (x, y-3).
<h3>Calculate the vertices of ΔA'B'C':</h3>
Given that,
ΔABC : A(-6,-7), B(-3,-10), C(-5,2)
(x,y)→(x,y-3)
The vertices are:
- A(-6,-7 )⇒ (-6,-7-3) = A'(-6, -10)
- B(-3,-10) ⇒ (-3,-10-3) = B'(-3,-13)
- C(-5,2) ⇒ (-5,2-3) = C'(-5,-1)
Hence A'(-6, -10), B'(-3,-13), and C'(-5,-1) are the vertices of the ΔA'B'C' under the translation rule (x,y)→(x,y-3).
Learn more about translation rule:
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Answer:
<em>(D). x = 2 , y = - 5</em>
Step-by-step explanation:
A = - 8 + 48 = 40
= -76 + 156 = 80
= 104 - 304 = - 200
<em>x =</em> 
= <em>2</em>
<em>y =</em> 
= <em>- 5</em>
<u><em>( 2, - 5 )</em></u>
Answer:
1 fact per 3 seconds because you divide 300 by 100
Answer:
Step-by-step explanation:
6k^2
------- = 2k; and:
3k
-15k
------- = -5
3k
-5 cannot be evenly divided by 3k, so there's a remainder instead.
6k^2 - 15k - 5
-------------------- = 2k -5 with remainder -5.
3k
