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anonymous
 one year ago
∆ABC is reflected across the xaxis and translated 2 units up to give the image ∆A'B'C' with coordinates A'(1, 1), B'(2, 1), C'(2, 1). What are the coordinates of the preimage ∆ABC?
anonymous
 one year ago
∆ABC is reflected across the xaxis and translated 2 units up to give the image ∆A'B'C' with coordinates A'(1, 1), B'(2, 1), C'(2, 1). What are the coordinates of the preimage ∆ABC?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444064255953:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well for starters if something is reflected across the xaxis, then the ycoordinate of a point will become the opposite.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So in this simple figure note how reflecting it across the xaxis just corresponds to the YVALUE being negated

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Secondly, if something is translated up (or down) a certain amount then it corresponds to adding (or subtracting) that amount from the YVALUE of the point

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's just I have to use one of these answers, and I'm very confused and horrible at this. A(1, 1), B(2, 3), C(2, 1) A(1, 1), B(2, 3), C(2, 1) A(1, 3), B(2, 1), C(2, 3) A(1, 3), B(2, 5), C(2, 3)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh sorry. I didn't mean to interrupt.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so given the order in the problem, first reflect then translate, we must apply those two steps in the process IN THAT ORDER. Note if you reverse the order you may (most likely will in this case) get a different answer.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so please list the verticies of triangle A'B'C' after you NEGATE (find the opposite) of their ycoordinate please.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In other words, list them like this: A'(1, 1), B'(2, 1), C'(2, 1) But insert the opposite of the ycoordinate, which corresponds to reflecting about the xaxis.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ill give you one: A'(1, 1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm very bad at this.. I get confused easily. So I'm not completely sure how to do any of this. What am I supposed to do with A'(1, 1)? I apologize for making this complicated.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Its ok, but I cant just give you the answer or you will never get better, just like if you were lifting weights... if I lifted the weight for you you wouldn't gain the muscle.... that being said I want to take you step by step through the process: I want you to find the opposite (i.e. negate) the yvalues in the following points: A'(1, 1), B'(2, 1), C'(2, 1) To help you along, I will demonstrate what I am looking for by doing the first point for you: A'(1, 1) (note I negated the yvalue) Now find the other two points please.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you understand what I mean by negate?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not really... I'm just starting this class and it's not going well. And I understand. I am glad you're not just giving the answers. I really wanna understand this.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So when you negate something it is the same as just making it negative. For example, if I have 5, then the OPPOSITE of that number is 5. If I you are asked to NEGATE 5, then the answer is 5. If the quantity they give you is already negative, no matter :D. The OPPOSITE of 7 is (7)=7. If asked to NEGATE 3, then you have (3)=3. Understand?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@emma.farrell still there?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, sorry. Just had to help my mom with something. I think I get it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so please list the points so I can check to make sure

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The points from the problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find the opposite of the ycoordinate (the second value in the order pairs) for all three points A'B'C' (just copy paste what I posted above and just modify the corresponding yvalues)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A'(1, 1), B'(2, 1), C'(2, 1) A' (1, 1) B' (2, 1), C' (2, 1) Is that what you mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry try again... remember only the yvalue is negated because (looking at my picture above) the vertical axis is the yaxis and those were the values that changed. The xvalues (the first value in the ordered pair) say the same.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, thanks. A'(1, 1), B'(2, 1), C'(2, 1) A'(1, 1), B'(2, 1), C'(2, 1) Is that better? Or am I looking at/doing the wrong thing?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now we must translate these points UP by 2. Do you think you can do that? Give it a shot and post your answer and I will let you know.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait, what do you mean translate them up? Sorry, I don't mean to be difficult.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Its ok... when something is "translated (up/down/left/right)" that is the mathematical way of saying shifted or moved in that direction.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Imagine moving the whole object up by two units... or pointwise... imagine moving each of the the vertices up by two units

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Anyways about your message, I can empathize I am exactly the same. But feel good because you can go on to be amazing at whatever you choose. It is merely a question of getting interested in the boring stuff so you can master it so that when you get to the more interesting stuff you can truely enjoy it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so as for this problem please do be scared to take a guess. One of the worst things you can do in math is be scared to take a shot at a problem. If you get it wrong no worries Im here to help so what do you think should happen to the points if I shift the figure up by two?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@emma.farrell are you there?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm sorry! I don't mean to keep disappearing, but I've gotten a bit busy. Would it be a bother if I came back to this in a bit? I've got some personal things to take care of.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Emma if you focus for a minute or two you can finish this problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So all you need to do is answer the question what happens to these points when I translate UP by 2: A'(1, 1), B'(2, 1), C'(2, 1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have to go now... In order to translate up simply shift the yvalues of your verticies by 2... i.e. add two to the yvalues of your responses: A'(1, 1+2), B'(2, 1+2), C'(2, 1+2)= A'(1, 1), B'(2, 3), C'(2, 1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank fully before I left, I reread the question.... Unfortunately I misread it the first time. Everything we discussed was correct only I solved the problem backwards. We have the final position A'(1, 1), B'(2, 1), C'(2, 1) and our goal is to find the initial image (aka the preImage) ABC. This means we must do the opposite of everything, including the steps (i.e. we have to deal with the translation first, then the reflection). First off, A'(1, 1), B'(2, 1), C'(2, 1) has already been shifted up by two (since its the final image), thus we must undo this shift by shifting it DOWN by 2: A'(1, 12), B'(2, 12), C'(2, 12)=A'(1, 1), B'(2, 3), C'(2, 1) Then we reflect about the xaxis (again by negating the yvalues): Therefore, A(1, 1), B(2, 3), C(2, 1) As it turns out this is the same answer. Note once the process finished I dropped the primes on A, B, & C which I neglected to do on my last post above.
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