Answer:
Solution Given:
To find the simplest form we need to open bracket
-2x²(3x²-2x-6)
-2x²*3x²-2x²*(-2x) -2x²*(-6)\
is a required answer.
Answer:
y = 1/6 x^2 + 8/3 x + 49/6
Step-by-step explanation:
This is a parabola which opens upwards.
The distance of a point (x, y) from the focus is
√[(x - -8)^2 + (y - -1)^2] and
the distance of the point from the line y = -4
= y - -4
These distances are equal for a parabola so:
√[(x - -8)^2 + (y - -1)^2] = y + 4
Squaring both sides:
(x + 8)^2 + (y + 1)^2 = (y + 4)^2
x^2 + 16x + 64 + y^2 + 2y + 1 = y^2 + 8y + 16
x^2 + 16x + 64 + 1 - 16 = 8y - 2y
6y = x^2 + 16x + 49
y = 1/6 x^2 + 8/3 x + 49/6 is the equation of the parabola.
To make it easier make 0.5 a 1 (0.5 x 2 = 1) and also double the 6 (6 x 2 = 12) so now you have and easier setup now divide 204/12 and should get 17 and multiply by 1 ( 17 x 1 = 17)
You answer is 17
Answer:
![\tan(x) = \frac{ -\sqrt{2} }{4}](https://tex.z-dn.net/?f=%20%5Ctan%28x%29%20%20%3D%20%20%5Cfrac%7B%20-%5Csqrt%7B2%7D%20%7D%7B4%7D%20)
Step-by-step explanation:
So an angle has two parts. Initial side and terminal side.
Inital side like on x axis. and terminal side shows how much it open up. Here the terminal angle terminates in second quadrant so we have the following
- A negative Cosine Value
- A positive Sine value
- A negative Tangent Value.
Now, using Pythagoras identity let solve for cos theta.
Here you on the right track but remeber that son theta=1/3 so sin theta squared would be 1/3 squared so we have
![(\frac{1}{3} ) {}^{2} + \cos {}^{2} (x) = 1](https://tex.z-dn.net/?f=%20%28%5Cfrac%7B1%7D%7B3%7D%20%29%20%7B%7D%5E%7B2%7D%20%20%2B%20%20%5Ccos%20%7B%7D%5E%7B2%7D%20%28x%29%20%20%3D%20%201)
![\frac{1}{9} + \cos {}^{2} (x) = 1](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B9%7D%20%20%2B%20%20%5Ccos%20%7B%7D%5E%7B2%7D%20%28x%29%20%20%3D%201)
![\cos {}^{2} (x) = \frac{8}{9}](https://tex.z-dn.net/?f=%20%5Ccos%20%7B%7D%5E%7B2%7D%20%28x%29%20%20%3D%20%20%5Cfrac%7B8%7D%7B9%7D%20)
![\cos(x) = \frac{2 \sqrt{2} }{3}](https://tex.z-dn.net/?f=%20%5Ccos%28x%29%20%20%3D%20%20%5Cfrac%7B2%20%5Csqrt%7B2%7D%20%7D%7B3%7D%20)
Note since cosine is negative in second quadrant, cos theta is
![- \frac{2 \sqrt{2} }{3}](https://tex.z-dn.net/?f=%20-%20%20%5Cfrac%7B2%20%5Csqrt%7B2%7D%20%7D%7B3%7D%20)
To find tan theta we do the following
![\tan(x) = \frac{ \sin(x) }{ \cos(x) }](https://tex.z-dn.net/?f=%20%5Ctan%28x%29%20%20%3D%20%20%5Cfrac%7B%20%5Csin%28x%29%20%7D%7B%20%5Ccos%28x%29%20%7D%20)
![\tan(x) = \frac{ \frac{1}{3} }{ \frac{2 \sqrt{2} }{3} }](https://tex.z-dn.net/?f=%20%5Ctan%28x%29%20%20%3D%20%20%20%5Cfrac%7B%20%5Cfrac%7B1%7D%7B3%7D%20%7D%7B%20%5Cfrac%7B2%20%5Csqrt%7B2%7D%20%7D%7B3%7D%20%7D%20)
![\tan(x) = \frac{1}{2 \sqrt{2} }](https://tex.z-dn.net/?f=%20%5Ctan%28x%29%20%3D%20%20%20%5Cfrac%7B1%7D%7B2%20%5Csqrt%7B2%7D%20%7D%20)
![\tan(x) = \frac{2 \sqrt{2} }{8}](https://tex.z-dn.net/?f=%20%5Ctan%28x%29%20%20%20%3D%20%5Cfrac%7B2%20%5Csqrt%7B2%7D%20%7D%7B8%7D%20)
![\frac{ \sqrt{2} }{4}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%5Csqrt%7B2%7D%20%7D%7B4%7D%20)
So
![\tan(x) = \frac{ -\sqrt{2} }{4}](https://tex.z-dn.net/?f=%20%5Ctan%28x%29%20%20%3D%20%20%5Cfrac%7B%20-%5Csqrt%7B2%7D%20%7D%7B4%7D%20)
Tan is negative in second quadrant