Step-by-step explanation:
Since
varies directly as
we can write the relation as
![x = k\sqrt{y}](https://tex.z-dn.net/?f=x%20%3D%20k%5Csqrt%7By%7D)
where k is the constant of proportionality.
a) To solve for the k, we substitute the given values:
![5 = k\sqrt{13} = k(13)](https://tex.z-dn.net/?f=5%20%3D%20k%5Csqrt%7B13%7D%20%3D%20k%2813%29)
![\Rightarrow k = \dfrac{5}{13}](https://tex.z-dn.net/?f=%5CRightarrow%20k%20%3D%20%5Cdfrac%7B5%7D%7B13%7D)
b) The equation relation x and y can be written as
![x = \dfrac{5}{13}\sqrt{y}](https://tex.z-dn.net/?f=x%20%3D%20%5Cdfrac%7B5%7D%7B13%7D%5Csqrt%7By%7D)
c) When y = 9,
![x = \dfrac{5}{13}\sqrt{9} = \dfrac{5}{13}(3) = \dfrac{15}{13}](https://tex.z-dn.net/?f=x%20%3D%20%5Cdfrac%7B5%7D%7B13%7D%5Csqrt%7B9%7D%20%3D%20%5Cdfrac%7B5%7D%7B13%7D%283%29%20%3D%20%5Cdfrac%7B15%7D%7B13%7D)
Let's begin with <span>f(x) = a(x-h)^2+ k. Note that we must use "^" to indicate exponentiation. Write (x-h)^2, not (x-h)2.
If (-3,4) is the vertex, then the above equation becomes f(x) = a(x-[-3])^2 + 4, or
f(x) = a(x+3)^2 + 4. We are told that the graph passes through (-1,0), so must now substitute those coordinates into the above equation:
f(-1) = a([-1]+3)^2 + 4 = 0 (0 is the value of f when x is -1)
Then we have a(2)^2 + 4 =0, or 4a + 4 = 0. Thus, a = -1.
The equation of this parabola is now f(x) = -(x+3)^2 + 4.
Write it in "standard form:" f(x) = -(x^2 + 6x + 9) + 4, or
f(x) = -x^2 - 6x - 9 + 4, or
answer => f(x) = -x^2 - 6x - 5 = ax^2 + bx + c
Thus, a=-1, b=-6 and c= -5.</span>
Answer:
A:4
Step-by-step explanation:
(12-4) + -8/2 = 8 + -4 = 4
hoped this helped!