True. The image of a translation is always congruent to the pre-image.
Proof: every point of the pre-image is move exactly the same distance and the same direction. That means that the shape stay the same but it is just located in a different place.
Hope this helps :)
<u>Complete Question:</u>
Janeel has a 10 inch by 12 inch photograph. She wants to scan the photograph, then reduce the results by the same amount in each dimension to post on her Web site. Janeel wants the area of the image to be one eight of the original photograph. Write an equation to represent the area of the reduced image. Find the dimensions of the reduced image.
<u>Correct Answer:</u>
A) 
B) Dimensions are : Length = 10-x = 3 inch , Breadth = 12-x = 5 inch
<u>Step-by-step explanation:</u>
a. Write an equation to represent the area of the reduced image.
Let the reduced dimensions is by x , So the new dimensions are

According to question , Area of new image is :
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So the equation will be :
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b. Find the dimensions of the reduced image
Let's solve : 
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By Quadratic formula :
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x = 15 is rejected ! as 15 > 10 ! Side can't be negative
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Therefore, Dimensions are : Length = 10-x = 3 inch , Breadth = 12-x = 5 inch
Answer:(1,23)(2,21)(4,17)(6,13)
Step-by-step explanation:
Plug in the x values in the graph to the function. For example plug in 2 to it and you get 25-2(2) which transfers to 25-4. 25-4 equals 21
Answer:
x=7.5 y y=12.99
Step-by-step explanation:
i will try again if this is wrong