The correct answer is true because the number do not repeat
Answer:
okay thanks for letting me know
Step-by-step explanation:
Let s be the side length of the square. The dimensions of the rectangle are three times the side of the square (i.e. 3s), and two less than the side of the square (i.e. s-2).
So, the area of this rectangle is
![3s(s-2)=3s^2-6s](https://tex.z-dn.net/?f=3s%28s-2%29%3D3s%5E2-6s)
The area of the square is
, and we know that the two areas are the same, so we have
![3s^2-6s=s^2 \iff 2s^2-6s=0 \iff 2s(s-3)=0 \iff s=0 \lor s=3](https://tex.z-dn.net/?f=3s%5E2-6s%3Ds%5E2%20%5Ciff%202s%5E2-6s%3D0%20%5Ciff%202s%28s-3%29%3D0%20%5Ciff%20s%3D0%20%5Clor%20s%3D3)
The solution s=0 would lead to the extreme case where the rectangle and the square are actually a point, so we accept the solution s=3.
Answer:
<h2>A. 2Pi</h2>
Step-by-step explanation:
The given function is
![y=2sin(x)](https://tex.z-dn.net/?f=y%3D2sin%28x%29)
Notice that the indepedent variable doesn't have any transformation, that means the period doesn't change. In other words, this function has the same period than its parent function which is
.
Therefore, the answer is A.
The image attached shows the graph of this function, there you can observe the period of the function.
Also, notice that this function is verticall stretched by a scale of 2, which doesn't change its original period.