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Masja [62]
3 years ago
11

A rhombus has coordinates A(-6, 3), B(-4, 4), C(-2, 3), D(-4, 2). What are the coordinates of rhombus A'B'C'D' AFTER A 90° COUNT

ERCLOCKWISE ROTATION ABOUT THE ORIGIN FOLLOWED BY A TRANSLATION 3 UNITS TO THE LEFT AND 2 UNITS DOWN ?
Mathematics
1 answer:
Effectus [21]3 years ago
5 0

Answer:

See below.

Step-by-step explanation:

Firstly the counterclockwise  rotation is (x, y)  ---> (-y, x), so A(-6, 3) moves to (-3, -6).

Then the translation (-3, -6) ---> ( -3-3), -6) = (-6, -6)   (3 units to the left).

Finally the second translation  moves (-6, -6) to (-6, -6-2) = (-6, -8)  = A'  ( 2 units down).

The coordinates of B', C' and D' are calculated in exactly the same way.

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If A+D=117kg , A+M=88kg and M+D=161kg whats there total weight<br><br>thx in advance
garri49 [273]
Think of A as a constant - with D it's 117 and with M it's 88

So the difference between D and M is 29   (117-88)

We also know that M + D = 161

Let's substitute in the problem  We will leave M equal to M and rewrite D as M+29 since it is 29 kg heavier.
M + M + 29 -161  or 2M +29 = 161 Subtracting 29 from both sides we get 2M=122
Now divide both sides two and get M=61
Since D is M+29 it is 61+29 or 100kg
Finally A + M = 88   A + 61 - 88   Subtract 61 from both sides and get A=17
So there you have it:
A=17kg
M=61kg
D=100kg and it works in all three equation in your problem (try it to check)


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