Answer:

Step-by-step explanation:
The vertex form of an equation of a parabola:

(h, k) - vertex
We have

We must use the formula: 

Use the distributive formula a(b + c) = ab + ac

Two fractions equivalent are 18/20 & 27/30
Answer:
See below
Step-by-step explanation:
It has something to do with the<em> </em><u><em>Weierstrass substitution</em></u>, where we have

First, consider the double angle formula for tangent:

Therefore,

Once the double angle identity for sine is

we know
, but sure, we can derive this formula considering the double angle identity

Recall

Thus,
Similarly for cosine, consider the double angle identity
Thus,

Hence, we showed 
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![5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ]](https://tex.z-dn.net/?f=5%5Ccos%28x%29%20%3D12%5Csin%28x%29%20%2B3%2C%20x%20%5Cin%20%5B0%2C%202%5Cpi%20%5D)
Solving





Just note that

and
is not defined for 
Is means equal to and of means multiply. So....your equation would be 12=60%x. Now since u can't multiply by a percent, u gotta do 12=.6x. Now divide by .6 so that it cancels out. Your answer is 20. Hope this helps!