Answer:
Well, the first thing you must do is find the slope
(0,2) (1,-3) I am going to make x1= 0, y1=2, x2=1 and y2= -3
Now, I am going to use the formula for the slope, which is
y2-y1/(x2-x1)
[-3-2]/(1-0)= -5/1 or just -1
Next, I am going to use the point slope form, which is
y-y1=m(x-x1). m=5, x1=0, y1=2
y-2= -5(x-0)
y-2=-5x
+2 +2
y= -5x+2 This is the slope intercept form.
we are done. I am glad to see that you are doing math on a Saturday night.. Good luck... Please rate my answer:)
Answer:
try 1 and 2 for supplementary
and 7 and 6 for vertical
Step-by-step explanation:
13pi/12 lies between pi and 2pi, which means sin(13pi/12) < 0
Recall the double angle identity,
sin^2(x) = (1 - cos(2x))/2
If we let x = 13pi/12, then
sin(13pi/12) = - sqrt[(1 - cos(13pi/6))/2]
where we took the negative square root because we expect a negative value.
Now, because cosine has a period of 2pi, we have
cos(13pi/6) = cos(2pi + pi/6) = cos(pi/6) = sqrt[3]/2
Then
sin(13pi/12) = - sqrt[(1 - sqrt[3]/2)/2]
sin(13pi/12) = - sqrt[2 - sqrt[3]]/2
Answer:
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