Linear decreasing
I haven't learned this is class yet, but I had a bf that taught me this
Answer:
72 yd^2
Step-by-step explanation:
Separate the shape into a square and a rectangle. Figure out the measurements of each shape then add the total.
I'm assuming that y2 means y^2 or y squared. In that case, your factors will be (6y+1) and (6y-1).
Answer: 
Step-by-step explanation:
For this exercise it is important to know the definition of "Dilation".
A Dilation is defined as a transformation in which the Image (The figure obtained after the transformation) and the Pre-Image (The original figure before the transformation) have the same shape but have different sizes.
If the Dilation is centered at the origin, and knowing the scale factor of
, you need to multiply each coordinate of the point T by the scale factor in order to find the coordinates of the Image T'.
Knowing that the point T has the following coordinates:

You get that the coordinates of the Image T' are the shown below:

Given expression is
![\sqrt[4]{\frac{16x^{11}y^8}{81x^7y^6}}](https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E%7B11%7Dy%5E8%7D%7B81x%5E7y%5E6%7D%7D)
Radical is fourth root
first we simplify the terms inside the radical


So the expression becomes
![\sqrt[4]{\frac{16x^4y^2}{81}}](https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E4y%5E2%7D%7B81%7D%7D)
Now we take fourth root
![\sqrt[4]{16} = 2](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B16%7D%20%3D%202)
![\sqrt[4]{81} = 3](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B81%7D%20%3D%203)
![\sqrt[4]{x^4} = x](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E4%7D%20%3D%20x)
We cannot simplify fourth root (y^2)
After simplification , expression becomes
![\frac{2x\sqrt[4]{y^2}}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B2x%5Csqrt%5B4%5D%7By%5E2%7D%7D%7B3%7D)
Answer is option B