Question

If T: (x,y)> (x+8, y+4)), then t1: (x,y)> ?
My options:
(X+8,y-4)
(2x,y)
(-x/2,-y)
(X-8, y-4)

Answer 1

you're looking for the inverse right?

Answer 2

I'm not sure I think it has something to do with reflection.

Answer 3

is there a negative 1 over the t?

Answer 4

No there is not

Answer 5

hmm just seems odd what they mean by t1

Answer 6

t^1 make more sense?

Answer 7

If it was \(\large T^{-1}\) then that's the inverse operation of T

Answer 8

Yeah! But its not negitive. I could be I could of wrote it down wrong..

Answer 9

double check please

Answer 10

I can't, the question is a online course and I can only excess it at school.

Answer 11

oh that's a shame

Answer 12

well let's assume that it's \(\large T^{-1}\) that's the only thing that makes sense in this context

Answer 13

I agree

Answer 14

Then what would be the answer them?

Answer 15

that would mean that \(\large T^{-1}\) undoes everything that operation T does

T: (x,y)> (x+8, y+4))

means

take any point and move it 8 units to the right and 4 units up

Answer 16

The exact opposite would be

take any point and move it 8 units left and 4 units down

Answer 17

which is exactly what (x,y) ----> (x-8, y-4) is saying

Answer 18

So if T: (x,y)> (x+8, y+4))

then \(\large T^{-1}\): (x,y) ----> (x-8, y-4)

Answer 19

So it is D?

Answer 20

yep

Answer 21

THANK YOU!:D

Answer 22

yw