Answer:
oh wow thx
Step-by-step explanation:
The product of trigonometric functions can be converted into sum ans vice versa. The given function is a product and we can convert it to a sum.One way is to expand the given expression and simplify it to find the result.
The second and easier option is to just remember the formulas of product to sum conversion identities. The 4 formulas for this case are listed below in the image.
Since we have the product of cos and sin, the sum will involve two sine functions connected by addition.
So the correct answer is option A
Okay to find the perpendicular bisector of a segment you first need to find the slope of the reference segment.
m=(y2-y1)/(x2-x1) in this case:
m=(-5-1)/(2-4)
m=-6/-2
m=3
Now for the the bisector line to be perpendicular its slope must be the negative reciprocal of the reference segment, mathematically:
m1*m2=-1 in this case:
3m=-1
m=-1/3
So now we know that the slope is -1/3 we need to find the midpoint of the line segment that we are bisecting. The midpoint is simply the average of the coordinates of the endpoints, mathematically:
mp=((x1+x2)/2, (y1+y2)/2), in this case:
mp=((4+2)/2, (1-5)/2)
mp=(6/2, -4/2)
mp=(3,-2)
So our bisector must pass through the midpoint, or (3,-2) and have a slope of -1/3 so we can say:
y=mx+b, where m=slope and b=y-intercept, and given what we know:
-2=(-1/3)3+b
-2=-3/3+b
-2=-1+b
-1=b
So now we have the complete equation of the perpendicular bisector...
y=-x/3-1 or more neatly in my opinion :P
y=(-x-3)/3
Answer
D. `(y+4) = (5)/(3)(x+ 1)`
Step by step explanation
The given points are (-1, -4) and (2, 1)
Slope = (y2 - y1)/(x2 - x1)
= (1 - (-4)) / (2 - (-1))
= (1 + 4) / (2 +1)
m = 5/3
When two points are given, the equation of the line is
(y - y1) = m (x - x1)
y -(-4) = 5/3 (x - (-1))
y + 4 = 5/3 (x + 1)
The answer is (y+4) = (5)/(3)(x+ 1)
Thank you.
In y=mx±b of slope-intercept form, the b is the y-intercept either positive or negative.
y=4x+12, in this case, 12 is your y-intercept!!
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I hope that helps you out!!
Any more questions, please feel free to ask me and I will gladly help you out!!
~Zoey
