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cricket20 [7]
3 years ago
12

part 3. Find the coordinates of the vertices of the triangle after a reflection across the y-axis and then across the line y= 2.

​

Mathematics
1 answer:
garik1379 [7]3 years ago
7 0

Answer:

  c.  S'(3, 1), T'(1, -1), U'(0, 1)

Step-by-step explanation:

Reflection across the y-axis negates the x-coordinate, so is equivalent to the transformation ...

  (x, y) ⇒ (-x, y)

Reflection across the horizontal line y=c is equivalent to the transformation ...

  (x, y) ⇒ (x, 2c-y)

So, the combined reflections are equivalent to the transformation ...

  (x, y) ⇒ (-x, 4 -y)

Then we have ...

  S(-3, 3) ⇒ S'(-(-3), 4-3) = S'(3, 1)

  T(-1, 5) ⇒ T'(-(-1), 4-5) = T'(1, -1)

  U(0, 3) ⇒ U'(-(0), 4-3) = U'(0, 1) . . . . matches choice C

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3 years ago
1)A System of equations is shown below . What is the solution to the system of equations? 5x+2y=-15 2x-2y=-6
meriva

Answer:

x= -3 and y= 0

Step-by-step explanation:

5x+2y=-15

<u>2x-2y=-6     </u>

<u>7x        =-21</u>

x= -3

Putting value of x in equation 1  

5(-3) +2y=-15

-15+2y= -15

2y= 0

y= 0

This can be solved with the help of matrices

In matrix form the above equations can be written in the form

\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]  \left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]

Let

\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right] = A  \left[\begin{array}{ccc}x\\y\\\end{array}\right]  = X  and  \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]= B

Then AX= B

or X= A⁻¹ B

where  A⁻¹= adj A/ ║A║   where mod A≠ 0

adj A=  \left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]

║A║= ( 5*-2- 2*2)= -10-4= -14≠0

X= A⁻¹ B

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]    =- 1/14  \left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]   \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]    =- 1/14     \left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]  =- 1/14 \left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]  =- 1/14 \left[\begin{array}{ccc}42\\0\\\end{array}\right]

\left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]

\left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-3\\0\\\end{array}\right]

From here x= -3 and y= 0

Solution Set = [(-3,0)]

3 0
3 years ago
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Answer:

c and d

Step-by-step explanation:

ok so i do not know how to explain

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3 years ago
What is the volume of the prism?
fgiga [73]

Answer:

67 ½ units³ or 135/2 units³

Step-by-step explanation:

Volume = Base area × height

In case:

width is 6 the height is 4 1/2 and the length is 2 1/2

2 ½ = 5/2

4 ½ = 9/2

Volume = (6 × 5/2) × 9/2

= 15 × 9/2

= 135/2

= 67 ½ units³

3 0
3 years ago
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