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anonymous
 one year ago
What set of reflections would carry hexagon ABCDEF onto itself? Hexagon ABCDEF on the coordinate plane with point A at negative 1, 1, point B at negative 3, 1, point C at negative 4, 2, point D at negative 3, 3, point E at negative 1, 3, and point F at 0, 2. xaxis, y=x, xaxis, y=x y=x, xaxis, y=x, yaxis yaxis, xaxis, yaxis xaxis, yaxis, yaxis
anonymous
 one year ago
What set of reflections would carry hexagon ABCDEF onto itself? Hexagon ABCDEF on the coordinate plane with point A at negative 1, 1, point B at negative 3, 1, point C at negative 4, 2, point D at negative 3, 3, point E at negative 1, 3, and point F at 0, 2. xaxis, y=x, xaxis, y=x y=x, xaxis, y=x, yaxis yaxis, xaxis, yaxis xaxis, yaxis, yaxis

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A ) xaxis, y=x, xaxis, y=x B ) y=x, xaxis, y=x, yaxis C ) yaxis, xaxis, yaxis D ) xaxis, yaxis, yaxis

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think its B, am I right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Nnesha @campbell_st @phi @ParthKohli would any of you please help me? i need an understanding for this aswell..please :(

phi
 one year ago
Best ResponseYou've already chosen the best response.1Yes , B sounds good. It is definitely not C or D Do you know how to do the various reflections?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why is it B? (im glad i guessed right lol but...its good to know) no, i do not :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A reflection is a type of transformation that creates a mirror image of the preimage across a line of reflection. You learned how images are created across four different lines of reflection: Across the xaxis (x, y) → (x, −y) Across the yaxis (x, y) → (−x, y) Across the line y = x (x, y) → (y, x) Across a horizontal or vertical line Count the distance between each point on the figure and the line of reflection. Then, for each point, count this same distance from the line of reflection to find the corresponding point.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it theses? i got this from the lesson

phi
 one year ago
Best ResponseYou've already chosen the best response.1To test a choice, I would pick one point on the figure, and do each of the transforms and see where the point lands after all of the transforms.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how would i test A= (1,1) ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1Let's test B ) y=x, xaxis, y=x, yaxis using point B (3,1) Across the line y = x (x, y) → (y, x) so we get (1,3) Across the xaxis (x, y) → (x, −y) we get (1,3) Across the line y = x (x, y) → (y, x) we get (3,1) Across the yaxis (x, y) → (−x, y) we get (3,1) that is back where we started. If we do that with all the points, we find the same thing. all of that hopping around landed us back in the original spot.

phi
 one year ago
Best ResponseYou've already chosen the best response.1can you try point A using choice B B ) y=x, xaxis, y=x, yaxis ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it keep reflecting...technically otating us back to where we started?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0want me to try point A? or answer A? lol

phi
 one year ago
Best ResponseYou've already chosen the best response.1point A (1,1) using the same transformations I used for point B

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry it took lon i wrote i out, im going to type it now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Point A (1,1) across the line y=x(x,y) > (y,x) (1,1) across the xaxis (x,y) > (x,y) (1,1) across the line y=x(x,y) > (y,x) (1,1) across the yaxis (x,y) > (x,y) (1,1) did I do this correctly? (it did bring me back to where I start but i wanted to make sure lol) also, these type of reflections are basically formulas, correct? just like shifting in translations?

phi
 one year ago
Best ResponseYou've already chosen the best response.1yes, looks good. if you plot the points, the flips across the xaxis or yaxis will be clear. the only one that is hard to visualize is the flip across y=x (a diagonal line)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0also, did we read the question right? it says "carry hexagon ABCDEF onto itself" does onto itself mean back to where it started? (i just wanna make sure)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh yay!~ so if I plant the points it will show me it diagnaly reflected? or it wouldnt?

phi
 one year ago
Best ResponseYou've already chosen the best response.1that means every point gets mapped back onto the hexagon. It might mean the hexagon comes back flipped around (for example F to C and vice versa) But in this case each point is mapped back onto itself (we checked 2 points, it is safe to assume it works for the other points)

phi
 one year ago
Best ResponseYou've already chosen the best response.1***so if I plot the points*** not sure which points you mean.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok~ and i mean, like if i plot points ABCDEF with the set of reflections of B , would it show me the diagnal reflection? or no?

phi
 one year ago
Best ResponseYou've already chosen the best response.1if you plot the final result of the reflections in choice B, each point will be where it started. In other words, the "new" hexagon will lie directly on top of the "old" hexagon. The hexagon was "carried onto itself"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0LOL right! right! ive been out in the sun for too long this morning XD thank you so so so much! i seriously appreciate your help! i jut couldnt get anywhere by just reading the video and some of the terms in the lesson confuse me, but you helped me alot, you were very clear and simple in instructing me! thank you so very very much!~ :)
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