anonymous
  • anonymous
HELP PLEASE! Figure 1 and figure 2 are two congruent parallelograms drawn on a coordinate grid as shown below:
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
anonymous
  • anonymous
mathmate
  • mathmate
hints: 1. whenever there the image is the opposite hand of the preimage, there is ONE (or any other odd number) reflection involved. 2. To check simple cases like this, take one point of the preimage, and follow instructions to see if you arrive at the same point on the image. For example, if you chose A. reflection across the y-axis, followed by reflection across the x-axis. To check, try to transform the top vertex of fig. 1 (preimage) which has coordinages (-4, 7). Reflection across the y-axis brings it to (4,7), and subsequent reflection across the x-axis bring it to (4,-7) which is not part of the image, so we can rule out choice A above. NOTE: remember that if there is one point that satisfies the rule, it is not sufficient to claim that the choice is correct. ALL points of the preimage must map to the image according to the same rule.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Thanks!
mathmate
  • mathmate
You're welcome! :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.