Answer:
Yes
Step-by-step explanation:
A triangle can have these lengths
Question 1:To change the new image back into the original image, just do the opposite.
To change the original image to the new image, we added x by 3 and subtracted 7 from y.
To change the new image back to the original image, subtract x by 3 and add 7 to y.
Writing this into an equation gives us the following:
Question 2:An isometry is a transformation of an image where the new image is congruent to the preimage.
The second and third choices are incorrect because the preimage is not congruent to the new image in these two choices.
The fourth choice is incorrect because having the same position is not an isometric transformation.
Thus, the answer is the first choice.
Answer:
DNE (does not exist)
Step-by-step explanation:
Our function is:
. We want to find the limit of this as x approaches 0. The first thing we would want to do is to substitute 0 in for x. But when we do that, we get 0/0, which is undefined:
![\frac{0^2+2*0}{0^4}=\frac{0}{0}](https://tex.z-dn.net/?f=%5Cfrac%7B0%5E2%2B2%2A0%7D%7B0%5E4%7D%3D%5Cfrac%7B0%7D%7B0%7D)
Let's divide the numerator and denominator both by x:
![\frac{x^2+2x}{x^4}=\frac{x+2}{x^3}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E2%2B2x%7D%7Bx%5E4%7D%3D%5Cfrac%7Bx%2B2%7D%7Bx%5E3%7D)
Now substitute 0 in again:
![\frac{0+2}{0^3}=\frac{2}{0}](https://tex.z-dn.net/?f=%5Cfrac%7B0%2B2%7D%7B0%5E3%7D%3D%5Cfrac%7B2%7D%7B0%7D)
Because we have a number divided by 0, this cannot exist. If we graph this function (see attachment), we'll also see that the graph diverges at x = 0, so the limit does not exist.
Answer:
-7.3
Step-by-step explanation: