We see that the differences are -9, -3, +3, and +9. Thus, we see that the function is symmetric about x=2 (I'm assuming the five values correspond to x=0, 1, 2, 3, 4) and increases at a rate similar to (x-2) squared. With that in mind, we classify this function as a parabola, as the standard form of a parabola (y=a(x-h)^2 + k) shows similar growth to this function.
Answer:
See below for the matrix
D = 100
Step-by-step explanation:
![\left[\begin{array}{cccc}3&2&1&-8\\-1&-2&4&7\\1&-6&-3&15\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D3%262%261%26-8%5C%5C-1%26-2%264%267%5C%5C1%26-6%26-3%2615%5Cend%7Barray%7D%5Cright%5D)
3 2 1
D = -1 -2 4 = 3(-2)(-3) + 2(4)(1) + 1(-1)(-6)
1 -6 -3
- 1(-2)(1) - 3(4)(-6) - 2(-1)(-3)
= 18 + 8 + 6 + 2 + 72 - 6
= 100
Sorry, I can't type the vertical bars.
I hope this is what you are looking for.
33.3% sugar
you do 5÷15 (part÷whole)
A relation is a function if you associate exactly one output for every input. This means that, when you choose a value for x, there must be only one correspondent value for y. This only happens in the top-right parabola.
