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slega [8]
3 years ago
12

You rotate triangle ABC, with vertices A(-3, 1), B(-2, 2), and C(-3, 4), 90° counterclockwise about the origin to form triangle

A′B′C′. Match each vertex of triangle A′B′C′ to its coordinates. Tiles (2, 2) (-1, -3) (1, 3) (4, 3) (-2, -2) (-4, -3)
Mathematics
2 answers:
katrin2010 [14]3 years ago
6 0
The rule for going 90 degree counter clockwise is:
(x, y) becomes (y, -x)

Therefore, the new points are:

A(-3, 1) to (1, 3)
B(-2, 2) to (2, 2)
C(-3, 4) to (4, 3) 
IRINA_888 [86]3 years ago
6 0

Answer:

The coordinates of <em>A' are (-1,-3) , B' are (-2,-2) , C' are (-4,-3).</em>

Step-by-step explanation:

You rotate triangle ABC, with vertices A(-3, 1), B(-2, 2), and C(-3, 4), 90° counterclockwise about the origin to form triangle A′B′C′.

<em>" When point M (h, k) is rotated about the origin O through 90° in anticlockwise direction. The new position of point M (h, k) will become M'(-k, h) ".</em>

As the coordinates of A are (-3,1)

⇒ h=-3 and k=1

Hence, the coordinates of A' will be (-1,-3)

similarly the coordinates of B are (-2,2)

⇒ h=2 and k=-2

Hence, coordinates of B' are: (-2,-2)

similarly the coordinates of C are (-3,4).

⇒ h=-3 and k=4

Hence, coordinates of C' are (-4,-3).


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When conducting a significance test in practice, how should you choose the alpha level?.
iVinArrow [24]

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1 year ago
On a map the distance from Fort Collins to Cheyenne is 3.2 inches. The angle formed at Laramie has measure 43.6º, and the angle
juin [17]
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3 years ago
Read 2 more answers
276x^2−136x+869+−5184x+6
iragen [17]

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3 years ago
A model rocket is launched with an initial velocity of 200 ft per second. The height h, in feet, of the rocket t seconds after t
Solnce55 [7]

Answer:

2.10 s, 10.40 s.

Step-by-step explanation:

We know that the height of the rocket is given by the function:

h=-16t^2+200t

We are asked to find the time for which the height of the rocket will be 350 ft. So, for that moment, we know the height but we don't know the time; however, we know that the equation can help us to find the time, doing h=350:

350=-16t^2+200t

The last is a quadratic equation, which can be put in the form at^2+bt+c=0 and solved applying the formula:

t=\frac{-b+-\sqrt{b^2-4ac} }{2a}

So, let's put the equation on the form at^2+bt+c=0 adding 16t^2 and subtracting 200t to each side of the equation; the result is:

16t^2-200t+350=0

So, we note that a=16, b=-200, and c=350.

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t_1=\frac{200-\sqrt{200^2-4*16*350} }{2*16}=2.10

t_2=\frac{200+\sqrt{200^2-4*16*350} }{2*16}=10.40

According to the equation, that are the times for which the height will be 350 ft; that is because the rocket is going to ascend and then to fail again to the ground.

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