Answer: No this is not a linear function as the y values have different first differences.
Step-by-step explanation:
First we look at the x values. We start off and look at the second x value and subtract the first x value from it (That would be 1-0) and that equals to 1. Then we look at the third x value and subtract the second x value from it (2-1) and that equals to 1. Then we look at the fourth x value and subtract the third x value from it (That would be 3-2) and that is 1. Next we look at the fifth x value and subtract the fourth x value from it (That would be 4-3) and that is 1.
Now we look at the y values and do the same thing as with the x values. We start off and look at the second y value and subtract the first y value from it (That would be 1-0) and that equals to 1. Then we look at the third y value and subtract the second y value from it (4-1) and that equals to 3. Then we look at the fourth y value and subtract the third y value from it (That would be 9-4) and that is 5. Next we look at the fifth y value and subtract the fourth y value from it (That would be 16-9) and that is 7.
Even though the x values all have the same first differences, the y values do not and so this cannot be a linear function.
I hope this helps. Tell me if something doesn't make sense.
Lines BC are equal to or congruent to AD, because the tick marks show they are equal to each other in size. Therefore, they are congruent.
Factor out the 4 in both equations
8a^2-20^2=(2^2 times a^2 times 2)-(2^2 times 5)
therefor it is also equal to
(2a)^2 times 2-(2^2 times 5)
we can force it into a difference of 2 perfect squares which is a^2-b^2=(a-b)(a+b)
(2a√2)^2-(2√5)^2=((2a√2)-(2√5))((2a√2)+(2√5))
