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Mumz [18]
3 years ago
14

The vertices of polygon ABCD are at A(1, 1), B(2, 3), C(3, 2), and D(2, 1). ABCD is reflected across the x-axis and translated 2

units up to form polygon A′B′C′D′. Match each vertex of polygon A′B′C′D′ to its coordinates.
Tiles
(2, 1)
A′
(1, 1)
B′
(3, 0)
C′
(2, 3)
D′
(-3, 4)
(2, -1)
(-2, 5)
Mathematics
2 answers:
rjkz [21]3 years ago
8 0
The vertices of polygon ABCD are at A(1, 1), B(2, 3), C(3, 2), and D(2, 1). ABCD is reflected across the x-axis and translated 2 units up to form polygon A′B′C′D′. Match each vertex of polygon A′B′C′D′ to its coordinates.Tiles
(2, 1)A′(2, -1)
(1, 1)B′<span>(-3, 4)</span>
(3, 0)C′(2, -1)
(2, 3)D′(-2, 5)

tresset_1 [31]3 years ago
8 0

Answer:  The required match is given by

A'         (1, 1)

B'         (2, -1)

C'         (3, 0)

D'         (2, 1).

Step-by-step explanation:  Given that the vertices of polygon ABCD are at A(1, 1), B(2, 3), C(3, 2), and D(2, 1).

ABCD is reflected across the x-axis and translated 2 units up to form polygon A′B′C′D′.

We are to match the vertices of polygon ABCD to its co-ordinates.

We know that

if a point (x, y) is reflected across X-axis, hen the sign before the y co-ordinate changes. Also, if there is an additional translation of 2 units up, then the required transformation will be

(x, y)  ⇒  (x, -y + 2).

So, after getting reflected across the X-axis, the co-ordinates of the vertices of ABCD will change as follows :

A(1, 1)  ⇒  A'(1, -1+2) = A'(1, 1)

B(2, 3)  ⇒ B'(2, -3+2) = B'(2, -1)

C(3, 2)  ⇒  C'(3, -2+2) = C'(3, 0)

and

D(2, 1)  ⇒  D'(2, -1+2) = D'(2, 1).

Thus, the required match is given by

A'         (1, 1)

B'         (2, -1)

C'         (3, 0)

D'         (2, 1).

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I need help ASAP!!!!!!!!!!
Lorico [155]

Answer:

00

Step-by-step explanation:

7 0
3 years ago
Which choice is equivalent to the expression below when y2 0?<br> √y^2 + √16y^3 – 4y√y
DanielleElmas [232]

Answer:

Option C.

Step-by-step explanation:

We start with the expression:

\sqrt{y^3}  + \sqrt{16*y^3} - 4*y\sqrt{y}

where y > 0. (this allow us to have y inside a square root, so we don't mess with complex numbers)

We want to find the equivalent expression to this one.

Here, we can do the next two simplifications:

\sqrt{16*y^3} = \sqrt{16} \sqrt{y^3} = 4*\sqrt{y^3}

And:

y*\sqrt{y} = \sqrt{y^2} *\sqrt{y} = \sqrt{y^2*y} = \sqrt{y^3}

If we apply these two to our initial expression, we can rewrite it as:

\sqrt{y^3}  + \sqrt{16*y^3} - 4*y\sqrt{y}

\sqrt{y^3}  + 4*\sqrt{y^3} - 4\sqrt{y^3} = \sqrt{y^3}

Here we can use the second simplification again, to rewrite:

\sqrt{y^3} = y*\sqrt{y}

So, concluding, we have:

\sqrt{y^3}  + \sqrt{16*y^3} - 4*y\sqrt{y} = y*\sqrt{y}

Then the correct option is C.

8 0
3 years ago
PLEASE HELP ASAP PLEASE
emmainna [20.7K]

Answer:

The equation that represents the line passing through the point (2, -4) with a slope of one half is

f(x) = \frac{1}{2}x - 5

Step-by-step explanation:

The equation of a line can be described by a first order equation in the following format:

f(x) = ax + b

In which a is the slope of the line.

Solution:

The line slope is \frac{1}{2}, so a = \frac{1}{2}.

The equation of the line now is:

f(x) = \frac{1}{2}x + b

The problem states that the line passes through the point(2,-4). This means that when x = 2, f(x) = -4

So:

f(x) = \frac{1}{2}x + b

-4 = \frac{1}{2}*(2) + b

-4 = 1 + b

b = -5

So, the equation that represents the line passing through the point (2, -4) with a slope of one half is

f(x) = \frac{1}{2}x - 5

3 0
3 years ago
An equation of a line with a slope of -2 and a y-intercept of 2.5 is
Masja [62]

Answer:

its D

Step-by-step explanation:

8 0
3 years ago
-10 + 20p + 6p = -16
pshichka [43]

Answer: p= -3/13

Step-by-step explanation:

-10+26p=-16

+10 +10

26p =-6

26 26

p= -3/13

6 0
4 years ago
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