Answer:
x = 3
< D = 26
< E = 54
Step-by-step explanation:
I did the math
Answer:
A
Step-by-step explanation:
Just add both matrices
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:
brainly.com/question/15161224
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OK. I did it, and I have the solution.
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The length of the deck is (5 + 2x) .
The width of the deck is (4 + 2x) .
If the deck didn't have that big hole in the middle where the pool is,
then its area would be
(5 + 2x) · (4 + 2x) .
When you multiply that all out, you get Area = 4x² + 18x + 20
and the question tells us that the area of the whole big rectangle is 90 yds² .
So we can write
4x² + 18x + 20 = 90 .
Subtract 90 from each side: 4x² + 18x - 70 = 0
Divide each side by 2 : 2x² + 9x - 35 = 0
You can use the quadratic equation to solve that and find out that
x = 2.5 yards, and that's what the question is asking you.
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That makes the deck 10 yds high and 9 yds wide.
Total area of the whole big rectangle, (deck + pool ), = 90 yds².
