The point-slope form of a line is:
y-y1=m(x-x1), where m=slope and (x1,y1) is any point on the line
First we need to find the slope, which is (y2-y1)/(x2-x1)
m=(4--1)/(8-2)
m=5/6 and we can use either point, I'll use (8,4)
y-4=(5/6)(x-8)
That is your equation in point-slope form.
Now the standard equation of a line is ax+by=c
y-4=(5/6)(x-8) we can perform the indicated multiplication on the right side
y-4=(5x-40)/6 multiply both sides by 6
6y-24=5x-40 add 24 to both sides
6y=5x-16 subtract 5x from both sides
-5x+6y=-16 and by convention, the standard equation of a line should be expressed with a positive coefficient for x, so multiply both sides by -1
5x-6y=16
You divide 56.28 by 3 so 18.76 then multiply by 7 which is 131.32 hope this helps!
Answer: not a function; domain-2,-1,1 ; range-2,0,2,3
Step-by-step explanation:
Answer:
55.0 or 60 i don't know,i just got those answers
Step-by-step explanation:
Answer:
<u><em>y' = 1 / sinxcosx</em></u>
Step-by-step explanation:
<u>Given :</u>
y = sinx + cosx / sinx - cosx
<u>Applying quotient rule :</u>
y' = [(sinx - cosx)(cosx - sinx) - (sinx + cosx)(cosx + sinx)] / (sinx -
cosx)²
y' = [sinxcosx - cos²x - sin²x + sinxcosx - (sinxcosx + cos²x + sin²x + sinxcosx)] / (sinx - cosx)²
y' = 2sinxcosx - 2sinxcosx - 2cos²x - 2sin²x / (sinx -cosx)²
y' = -2(sin²x + cos²x) / (sinx - cosx)²
y' = -2 / (sinx - cosx)²
y' = -2 / sin²x - 2sinxcosx + cos²x
y' = -2 / -2sinxcosx
<u><em>y' = 1 / sinxcosx</em></u> [⇒ Final answer]
