Hello this is super easy i can help you
Here, we just use the following x values and put them into the equation.
y = - 0.05x + 16
y = -0.5(0) + 16
y = 16
y = - 0.05x + 16
y = -0.5(160) + 16
y = -80 + 16
y = -64
y = - 0.05x + 16
y = -0.5(320) + 16
y = - 160 + 16
y = -144
Now, to set up the table, you could list the x values and the y values.
x values :- 0,160, 320
y values:- 16, -64, -144
The first option is correct
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Solve the trigonometric equation:

Restriction for the solution:

Square both sides of
(i):

![\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bsin%5C%2Cx%7D%7Bcos%5E2%5C%2Cx%7D%5Ccdot%20%5Cleft%5B2%5Ccdot%20%281-sin%5E2%5C%2Cx%29-sin%5C%2Cx%20%5Cright%5D%3D0%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%5Cdfrac%7Bsin%5C%2Cx%7D%7Bcos%5E2%5C%2Cx%7D%5Ccdot%20%5Cleft%5B2-2%5C%2Csin%5E2%5C%2Cx-sin%5C%2Cx%20%5Cright%5D%3D0%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B-%5C%2C%5Cdfrac%7Bsin%5C%2Cx%7D%7Bcos%5E2%5C%2Cx%7D%5Ccdot%20%5Cleft%5B2%5C%2Csin%5E2%5C%2Cx%2Bsin%5C%2Cx-2%20%5Cright%5D%3D0%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7Bsin%5C%2Cx%5Ccdot%20%5Cleft%5B2%5C%2Csin%5E2%5C%2Cx%2Bsin%5C%2Cx-2%20%5Cright%5D%3D0%7D)
Let

So the equation becomes

Solving the quadratic equation:



You can discard the negative value for
t. So the solution for
(ii) is

Substitute back for
t = sin x. Remember the restriction for
x:

where
k is an integer.
I hope this helps. =)
1/2+ 1/4
= 2/4+ 1/4
= 3/4
1/12+ 1/6
= 1/12+ 2/12
= 3/12
= 1/4
Therefore, 1/4+ 1/6+ 1/12
= (1/12+ 1/6)+ (1/12+ 1/6)
= 2*(1/12+ 1/6)
= 2* (1/4)
= 2/4
= 1/2
Hope this helps~