There are 6, you'd think there would be 8 but the answer is actually 6. A diagonal is a line segment that connects to vertices, the two vertices chosen do not count when forming the triangle, two vertices that are adjacent to that chosen vertex are NOT included when forming the diagonal. It's confusing, but I hope this helps. If you have any follow up questions I'd love to answer them.
The range of the function is the set of all possible outputs, that is, the set of all values obtained by applying the function to elements of the domain. So the set of all values which can be obtained by applying h(x) to an element of its domain is {−4,0,5,60} , and thus that is the range of h(x) .
Answer:
Step-by-step explanation:
There is an error in the question. The table does not show two linear functions. y₁ is a linear function, but y₂ is not a straight line. It makes a bend at (-6,1).
Line 1 goes through (-12,-3) and (0,5).
slope = (5-(-3))/(0-(-12)) = 2/3
y-intercept = 5
y₁ = (2/3)x + 5
Line 2 goes through (-12,-2) and (-6,1).
slope = (1-(-2))/(-6-(-12)) = 1/2
y₂ = (1/2)x + 4
(2/3)x + 5 = (1/2)x + 4
x = -6
y = (2/3)x + 5 = 1
Solution: (-6,1)
Answer:
D
Step-by-step explanation:
-4 is listed twice with 2 different outputs
I'm assuming that the 1/3 is an exponent.
If so, then
![2^{1/3} = \sqrt[3]{2}](https://tex.z-dn.net/?f=2%5E%7B1%2F3%7D%20%3D%20%5Csqrt%5B3%5D%7B2%7D)
Which is the cube root of 2. Raising any value to the 1/3 power is the same as taking the cube root.