Lines of symmetry if the triangle is isosceles.
4, 2, 1
Answer: ![\bold{\sqrt[4]{2} }](https://tex.z-dn.net/?f=%5Cbold%7B%5Csqrt%5B4%5D%7B2%7D%20%7D)
<u>Step-by-step explanation:</u>
![\dfrac{1}{2}\sqrt[4]{32} =\dfrac{1}{2}\sqrt[4]{2\cdot 2\cdot 2\cdot 2\cdot 2}=\dfrac{1}{2}\cdot 2\sqrt[4]{2}=\boxed{\sqrt[4]{2} }](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%5Csqrt%5B4%5D%7B32%7D%20%3D%5Cdfrac%7B1%7D%7B2%7D%5Csqrt%5B4%5D%7B2%5Ccdot%202%5Ccdot%202%5Ccdot%202%5Ccdot%202%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%202%5Csqrt%5B4%5D%7B2%7D%3D%5Cboxed%7B%5Csqrt%5B4%5D%7B2%7D%20%7D)
Answer:
The focus of the parabola is at the point (0, 2)
Step-by-step explanation:
Recall that the focus of a parabola resides at the same distance from the parabola's vertex, as the distance from the parabola's vertex to the directrix, and on the side of the curve's concavity. In fact this is a nice geometrical property of the parabola and the way it can be constructed base of its definition: "All those points on the lane whose distance to the focus equal the distance to the directrix."
Then, the focus must be at a distance of two units from the vertex, (0,0), on in line with the parabola's axis of symmetry (x=0), and on the positive side of the y-axis (notice the directrix is on the negative side of the y-axis. So that puts the focus of this parabola at the point (0, 2)
Jamie because it’s showing you the same information but one in a table and on in a graph
