Step-by-step explanation:
P varies directly as the cubic root of Q means that:
![p = x \times \sqrt[3]{q}](https://tex.z-dn.net/?f=p%20%3D%20x%20%5Ctimes%20%20%5Csqrt%5B3%5D%7Bq%7D%20)
, where x is a real constant number.
Which means that:
![x = \frac{p}{ \sqrt[3]{q} }](https://tex.z-dn.net/?f=x%20%3D%20%20%5Cfrac%7Bp%7D%7B%20%5Csqrt%5B3%5D%7Bq%7D%20%7D%20)
for q ≠0.
So we get that for q=8,p=4 which means:
![x = \frac{4}{ \sqrt[3]{8} } = \frac{4}{2} = 2](https://tex.z-dn.net/?f=x%20%3D%20%20%5Cfrac%7B4%7D%7B%20%5Csqrt%5B3%5D%7B8%7D%20%7D%20%20%3D%20%20%5Cfrac%7B4%7D%7B2%7D%20%20%3D%202)
As a result we get that for Q=64:
![p = 2 \times \sqrt[3]{64} = 2 \times 4 = 8 \\ since \: {4}^{3} = 4 \times 4 \times 4 = \\ 16 \times 4 = 64](https://tex.z-dn.net/?f=p%20%3D%202%20%5Ctimes%20%20%5Csqrt%5B3%5D%7B64%7D%20%20%3D%202%20%5Ctimes%204%20%3D%208%20%5C%5C%20since%20%5C%3A%20%20%7B4%7D%5E%7B3%7D%20%20%3D%204%20%5Ctimes%204%20%5Ctimes%204%20%20%3D%20%20%5C%5C%20%2016%20%5Ctimes%204%20%3D%2064)
So for q=64,p=8.
2x + 4x - 3x - 6x = -3x
When x = 3,
-3x = -3(3) = -9
Answer:
For a general quadratic equation like:
y = a*x^2 + b*x + c
a is the leading coefficient.
And we define the vertex of this as (h, k)
Where h is:
h = -b/(2*a)
and k is the y-value given when we use x = h.
Now, in this case we have:
f(x) = 5*x^2
We can see that the leading coefficient is 5, then a = 5.
To find the value of h we use:
h = -b/(2*a)
there is no lineal term, thus b = 0, then:
h = -0/(2*5) = -0/10 = 0
and k is given by
f(h) = f(0) = 5*0^2 = 0
then k = 0.
Concluding:
a = 5
h = 0
k = 0
