Question 148144: Hi,
Looking for help with these problems.
1: The equation of the line passing through the point (1,-2) and perpendicular to the line Y=(-2/3X)+(4/3)is?
2: Simplify the expression below (leave answer with positive exponents)
(7X^3Y/14XY^-2)^2 both numerator & denominator squared.
3: The trigonometric expression SinӨ+CotӨCosӨ
4: If SinӨ=(3/5) and Ө is in quadrant2 find Sin2Ө (Hint Sin2Ө=SinӨCosӨ)
Last one for trig! yes!
5: The point (-2,3) lies on the terminal side of an angle Ө in standard position. Find the exact value of Ө.
Thank YOU!!!!!!!!!!!!!!!!!!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1: The equation of the line passing through the point (1,-2) and perpendicular to the line Y=(-2/3X)+(4/3)is?
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slope = 3/2
Form: y = mx + b
You know m=3/2 and y = -2 when x = 1 ; solve for "b":
-2 = (3/2)*1 + b
b = -7/2
EQUATION: y = (3/2)x - (7/2)
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2: Simplify the expression below (leave answer with positive exponents)
(7X^3Y/14XY^-2)^2 both numerator & denominator squared.
Cancel the common factors: 7, X
= (X^2Y^5/2)^2
= (X^4 y^10 / 4)
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3: The trigonometric expression
SinӨ+CotӨCosӨ = sin(theta) + [cos(theta)/sin(theta)]*cos(theta)
= sin(theta) + cos^2(theta)/sin(theta)
= [sin^2(theta) + cos^2(theta)]/sin(theta)
= 1/sin(theta)
= csc(theta)
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4: If SinӨ=(3/5) and Ө is in quadrant2 find Sin2Ө (Hint Sin2Ө=SinӨCosӨ)
In 2nd Quadrant, if sin=43/5, cos=-4/5
Then sin(2theta)= (3/5)(-4/5) = -12/25
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5: The point (-2,3) lies on the terminal side of an angle Ө in standard position. Find the exact value of Ө.
theta = tan^-1(-3/2) = -56.31 degrees = 180 = 123.31 degrees
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Cheers,
Stan H.
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