we know that
The area of the hexagon is equal to the sum of the areas of the six equilateral triangles
Let
x-------> area of one equilateral triangle
so
![6x=24a^{2} -18](https://tex.z-dn.net/?f=%206x%3D24a%5E%7B2%7D%20-18%20)
Divide by
both sides
-------> area of one equilateral triangle
To find an equivalent expression for the area of the hexagon based on the area of a triangle, multiply the area of one equilateral triangle by ![6](https://tex.z-dn.net/?f=%206%20)
therefore
the answer is
The equivalent expression is equal to
Answer:
data set C.
Step-by-step explanation:
This set has a bigger/wider range (1-5)
1. No.
2. No.
3. Yes.
4. No.
Answer:
![y = 3\left(x-\dfrac{3}{2}\right)^2+\dfrac{41}{4}](https://tex.z-dn.net/?f=y%20%3D%203%5Cleft%28x-%5Cdfrac%7B3%7D%7B2%7D%5Cright%29%5E2%2B%5Cdfrac%7B41%7D%7B4%7D)
Step-by-step explanation:
Given equation:
![y = 3x^2 - 9x + 17](https://tex.z-dn.net/?f=y%20%3D%203x%5E2%20-%209x%20%2B%2017)
Factor out 3 from the first 2 terms:
![y = 3(x^2 - 3x) + 17](https://tex.z-dn.net/?f=y%20%3D%203%28x%5E2%20-%203x%29%20%2B%2017)
Divide the coefficient of x by 2 and square it: (-3 ÷ 2)² = 9/4
Add this inside the parentheses and subtract the distributed value of it outside the parentheses:
![y = 3\left(x^2 - 3x+\dfrac{9}{4}\right) + 17-\dfrac{27}{4}](https://tex.z-dn.net/?f=y%20%3D%203%5Cleft%28x%5E2%20-%203x%2B%5Cdfrac%7B9%7D%7B4%7D%5Cright%29%20%2B%2017-%5Cdfrac%7B27%7D%7B4%7D)
Factor the parentheses and combine the constants:
![y = 3\left(x-\dfrac{3}{2}\right)^2+\dfrac{41}{4}](https://tex.z-dn.net/?f=y%20%3D%203%5Cleft%28x-%5Cdfrac%7B3%7D%7B2%7D%5Cright%29%5E2%2B%5Cdfrac%7B41%7D%7B4%7D)