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ahrayia [7]
3 years ago
11

Quadrilateral PUMA has coordinates at P(-5,-2), U(-1,2), M(4,-3), A(0,-7) and it is transformed by (x+3,y-5). Determined the ima

ge of M
Mathematics
1 answer:
IgorLugansk [536]3 years ago
8 0

The image of M is (7 , -8)

Step-by-step explanation:

Let us revise some transformation

  • If the point (x , y) translated horizontally to the right by h units  then its image is (x + h , y)
  • If the point (x , y) translated horizontally to the left by h units  then its image is (x - h , y)
  • If the point (x , y) translated vertically up by k units  then its image is (x , y + k)
  • If the point (x , y) translated vertically down by k units  then its image is (x , y - k)

∵ The quadrilateral PUMA has coordinates at:

   P (-5 , -2) , U (-1 , 2) , M (4 , -3) , A (0 , -7)

∵ It is transformed by (x + 3 , y - 5)

- That means the quadrilateral translated 3 units right and 5 units

   down, then we add each x-coordinate by 3 and subtract 5 from

   each y-coordinate

∵ The coordinates of point M are (4 , -3)

∴ The image of point M = (4 + 3 , -3 - 5)

∴ The image of point M = (7 , -8)

The image of M is (7 , -8)

Learn more:

You can learn more about transformation in brainly.com/question/5563823

brainly.com/question/11203617

*LearnwithBrainly

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Answer:

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Step-by-step explanation:

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5 0
3 years ago
What is the difference? StartFraction x Over x squared minus 16 EndFraction minus StartFraction 3 Over x minus 4 EndFraction Sta
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Answer:

The option "StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction" is correct

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Therefore \frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}

Step-by-step explanation:

Given problem is StartFraction x Over x squared minus 16 EndFraction minus StartFraction 3 Over x minus 4 EndFraction

It can be written as below :

\frac{x}{x^2-16}-\frac{3}{x-4}

To solve the given expression

\frac{x}{x^2-16}-\frac{3}{x-4}

=\frac{x}{x^2-4^2}-\frac{3}{x-4}

=\frac{x}{(x+4)(x-4)}-\frac{3}{x-4}  ( using the property a^2-b^2=(a+b)(a-b) )

=\frac{x-3(x+4)}{(x+4)(x-4)}

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=\frac{-2(x+6)}{(x+4)(x-4)}

\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}

Therefore \frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}

Therefore the option "StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction" is correct

That is \frac{-2(x+6)}{(x+4)(x-4)}

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