The total was $29.
1. Write an equation. 24+(2.50*2)= x
2. Solve your equation using PEMDAS
*2.5*2=5
*5+24=
3. Simplify equation.
5+24=29
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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. =)
Answer:
<h2>1) y = 5</h2>
Step-by-step explanation:
x = 4 it's a vertical line.
Perpendicular line to a vertical line is a horizontal line.
A horizontal line has equation y = a.
A horizontal line passes through the point (3, 5) → x = 3 and y = 5.
Therefore the equation is y = 5
Slope: (y2-y1)/(x2-x1)
(-3+2)/(-1+1) = -1/0 = undefined
The slope is undefined
Answer:
I will attach the missing drawing with the answer.
9.b)
Plane JKM
Plane JLM
Plane KLM
Step-by-step explanation:
The drawing for this question is missing. I will attach it with the answer.
9.a) Plane JKL is not an appropriate name for the plane because all of three points lie in the same line.
Through a line pass infinite planes. The plane JKL doesn't define a unique plane. That's why plane JKL isn't an appropriate name for the plane.
9.b) We can name the plane using three points that don't lie in the same line.
Three possible names for the plane are :
Plane JKM
Plane JLM
Plane KLM