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pav-90 [236]
3 years ago
9

Quadrilateral ABCD is reflected over the x-axis to create quadrilateral A'B'C'D!

Mathematics
1 answer:
Yuliya22 [10]3 years ago
3 0

The answer is option A the points are A^{1} (-6, -2) and B^{1} (-6, -5).

Step-by-step explanation:

Step 1:

To understand reflection across an axis we need to be able to know which quadrant the shape is already in and in which quadrant it will be in after reflection.

The given shape is currently in the second quadrant where the points in it will have a negative x coordinate and a positive y coordinate. So assume the values are in the format of (-x, y).

Step 2:

Due to reflection across the x-axis, the shape will be reflected on to the third quadrant. In this, the x and y coordinates both have a negative value.

So due to reflection, the x value remains constant and the y value remains the same but both have the opposite symbol i.e. negative.

Step 3:

So (-x, y) becomes (-x, -y).

A (-6, 2) becomes A^{1} (-6, -2),

B (-6, 5) becomes B^{1} (-6, -5),

C (-3, 4) becomes C^{1} (-3, -4),

D (-2.5, 2) becomes D^{1} (-2.5, 2).

So the points are A^{1} (-6, -2) and B^{1} (-6, -5). This is option A.

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dalvyx [7]

Answer:

x = -3

y = 6

Step-by-step explanation:

3x + y = -3

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3x + y = -3

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-3y + 9 + y = -3

-3y + y + 9 = -3

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y = -12/-2

<u>y = 6</u>

x = -y + 3

x = -6 + 3

<u>x = -3</u>

6 0
3 years ago
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Alex_Xolod [135]
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4 years ago
What is the probability of rolling an even number on a number cube and flipping a tails on a coin?
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Step-by-step explanation:

4 0
3 years ago
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Evaluate the function with the given the given value of x.<br><br> f(x)=x-7 when x= -2
aivan3 [116]
I hope this helps you


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8 0
3 years ago
2.) Fill in the blank.
Montano1993 [528]

Answer:

E.) 1

Step-by-step explanation:

Firstly we will solve for L.H.S.

L.H.S. =Csc^2\theta

Since we know that Csc^2\theta is the inverse of Sin^2\theta.

So we can say that;

csc^2\theta=\frac{1}{sin^2\theta}

Now For R.H.S.

Cot^2\theta+1

Since we  can rewrite cot^2\theta as \frac{cos^2\theta}{sin^2\theta}.

Now we can say that the R.H.S. is;

\frac{cos^2\theta}{sin^2\theta}+1

Now we add the fraction and get;

\frac{cos^2\theta+sin^2\theta}{sin^2\theta}

Now according to trigonometric identity;

cos^2\theta+sin^2\theta=1

So, \frac{cos^2\theta+sin^2\theta}{sin^2\theta}=\frac{1}{sin^2\theta}

Here,

csc^2\theta=\frac{1}{sin^2\theta}   and     Cot^2\theta+1 = \frac{1}{sin^2\theta}<em>  </em>

L.H.S. = R.H.S.

Hence csc^2\theta=cot^2\theta+1

4 0
3 years ago
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