Is it possible that the image of a point after a reflection could be the same point as the preimage?

1 answer:

3 0

Yes, it is possible. For example, let's say that point A is (0,4) and there's a reflection over the y-axis, then point A would still be (0,4).

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20POINTS<br> x-4y=1 and 2 x-8y=2

Rom4ik [11]

First you would take one and move it over to the left side and never leave for why and then you divide by one which would leave for why and it would also have X equals 4Y and then you divide ask my Y and I will just leave you which is XY equals four

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Cameras R Us has a sale for 25% off camera bags, which is a discount of $20. Explain how you can set up a diagram to find the or

Strike441 [17]

I don't know how to set up a diagram for this, but I could write an expression. 20*4= x
sorry, but I hope this helps. I not even sure my expression is correct.

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Komok [63]

Answer: ur right

Step-by-step explanation:

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3 years ago

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SVETLANKA909090 [29]

Answer:

x = 0

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rosijanka [135]

Answer:

They subtracted the values incorrectly

Step-by-step explanation:

The formula for finding the distance between two points is this:
$\sqrt{} (x_{2} -x_{1})^{2} + (y_{2} -y_{1} )^{2}$

The person subtracted the variables incorrectly and subtracted x1 from x2 and y1 from y2 instead of the other way around.

To correctly solve it, do this: $\sqrt{((-7-(-1))^2+(-6-2)^2} = \sqrt{(-6)^{2} + (-8)^{2} } = \sqrt{36+64} = \sqrt{100} = 10$

Hope this helped!

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